fields.xml

<?xml version="1.0"?>
<!-- This file is part of the book                    -->
<!--                                                  -->
<!-- Abstract Algebra: Theory and Applications        -->
<!--                                                  -->
<!-- Text: Copyright (C) 1997-2025  Thomas W. Judson  -->
<!-- Sage: Copyright (C) 2010-2025  Robert A. Beezer  -->
<!-- See the file COPYING for copying conditions.     -->
<!-- This file is part of the book                    -->
<!--                                                  -->
<!-- See the file COPYING for copying conditions.     -->
<chapter xmlns:xi="http://www.w3.org/2001/XInclude" xml:id="fields">
  <title>Fields</title>
  <introduction>
    <p>
      It is natural to ask whether or not some field <m>F</m> is contained in a larger field.
      We think of the rational numbers,
      which reside inside the real numbers,
      while in turn, the real numbers live inside the complex numbers.
      We can also study the fields between <m>{\mathbb Q}</m> and
      <m>{\mathbb R}</m> and inquire as to the nature of these fields.
    </p>

    <p>
      More specifically if we are given a field <m>F</m> and a polynomial <m>p(x) \in F[x]</m>,
      we can ask whether or not we can find a field <m>E</m> containing <m>F</m> such that <m>p(x)</m> factors into linear factors over <m>E[x]</m>.
      For example, if we consider the polynomial
      <me>
        p(x) = x^4 -5 x^2 + 6
      </me>
      in <m>{\mathbb Q}[x]</m>,
      then <m>p(x)</m> factors as <m>(x^2 - 2)(x^2 - 3)</m>.
      However, both of these factors are irreducible in <m>{\mathbb Q}[x]</m>.
      If we wish to find a zero of <m>p(x)</m>,
      we must go to a larger field.
      Certainly the field of real numbers will work, since
      <me>
        p(x) = (x - \sqrt{2} ) (x + \sqrt{2} )( x - \sqrt{3})(x + \sqrt{3})
      </me>.
      It is possible to find a smaller field in which <m>p(x)</m> has a zero, namely
      <me>
        {\mathbb Q }( \sqrt{2} ) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \}
      </me>.
      We wish to be able to compute and study such fields for arbitrary polynomials over a field <m>F</m>.
    </p>
  </introduction>

  <section xml:id="fields-section-extension-fields">
    <title>Extension Fields</title>
    <introduction>
      <p>
        A field <m>E</m> is an <term>extension field</term><idx><h>Extension</h><h>field</h></idx><idx><h>Field</h><h>extension</h></idx>
        of a field <m>F</m> if <m>F</m> is a subfield of <m>E</m>.
        The field <m>F</m> is called the <idx><h>Field</h><h>base</h></idx><term>base field</term>.
        We write <m>F \subset E</m>.
      </p>

      <example xml:id="fields-example-q-sqrt2">
        <p>
          For example, let
          <me>
            F = {\mathbb Q}( \sqrt{2}\,) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \}
          </me>
          and let <m>E =  {\mathbb Q }( \sqrt{2} + \sqrt{3}\,)</m> be the smallest field containing both
          <m>{\mathbb Q}</m> and <m>\sqrt{2} + \sqrt{3}</m>.
          Both <m>E</m> and <m>F</m> are extension fields of the rational numbers.
          We claim that <m>E</m> is an extension field of <m>F</m>.
          To see this,
          we need only show that <m>\sqrt{2}</m> is in <m>E</m>.
          Since <m>\sqrt{2} + \sqrt{3}</m> is in <m>E</m>,
          <m>1 / (\sqrt{2} + \sqrt{3}\,) = \sqrt{3} - \sqrt{2}</m> must also be in <m>E</m>.
          Taking linear combinations of
          <m>\sqrt{2} + \sqrt{3}</m> and <m>\sqrt{3} - \sqrt{2}</m>,
          we find that <m>\sqrt{2}</m> and
          <m>\sqrt{3}</m> must both be in <m>E</m>.
        </p>
      </example>

      <example xml:id="fields-example-finite-f4">
        <p>
          Let <m>p(x) = x^2 + x + 1 \in {\mathbb Z}_2[x]</m>.
          Since neither 0 nor 1 is a root of this polynomial,
          we know that <m>p(x)</m> is irreducible over <m>{\mathbb Z}_2</m>.
          We will construct a field extension of
          <m>{\mathbb Z}_2</m> containing an element <m>\alpha</m> such that <m>p(\alpha) = 0</m>.
          By <xref ref="poly-theorem-max-ideal"/>,
          the ideal <m>\langle p(x) \rangle</m> generated by <m>p(x)</m> is maximal;
          hence, <m>{\mathbb Z}_2[x] / \langle p(x) \rangle</m> is a field.
          Let <m>f(x) + \langle p(x) \rangle</m> be an arbitrary element of <m>{\mathbb Z}_2[x] / \langle p(x) \rangle</m>.
          By the division algorithm,
          <me>
            f(x) = (x^2 + x + 1) q(x) + r(x)
          </me>,
          where the degree of <m>r(x)</m> is less than the degree of <m>x^2 + x + 1</m>.
          Therefore,
          <me>
            f(x) + \langle x^2 + x + 1 \rangle = r(x) + \langle x^2 + x + 1 \rangle
          </me>.
          The only possibilities for <m>r(x)</m> are then <m>0</m>,
          <m>1</m>, <m>x</m>, and <m>1 + x</m>.
          Consequently,
          <m>E = {\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle</m> is a field with four elements and must be a field extension of <m>{\mathbb Z}_2</m>,
          containing a zero <m>\alpha</m> of <m>p(x)</m>.
          The field <m>{\mathbb Z}_2(  \alpha)</m> consists of elements
          <md>
            <mrow>0 + 0 \alpha &amp; = 0</mrow>
            <mrow>1 + 0 \alpha &amp; = 1</mrow>
            <mrow>0 + 1 \alpha &amp; = \alpha</mrow>
            <mrow>1 + 1 \alpha &amp; = 1 + \alpha</mrow>
          </md>.
          Notice that <m>{\alpha}^2 + {\alpha} + 1 = 0</m>;
          hence, if we compute <m>(1 + \alpha)^2</m>,
          <me>
            (1 + \alpha)(1 + \alpha)= 1 + \alpha + \alpha + (\alpha)^2 = \alpha
          </me>.
          Other calculations are accomplished in a similar manner.
          We summarize these computations in the following tables,
          which tell us how to add and multiply elements in <m>E</m>.
        </p>
      </example>

      <figure xml:id="fields-table-addition-z2-alpha">
            <!-- math table -->
        <caption>Addition Table for <m>{\mathbb Z}_2(\alpha)</m></caption>
        <p>
          <me>
            \begin{array}{c|cccc}
            + &amp; 0 &amp; 1 &amp; \alpha &amp; 1 + \alpha \\
            \hline
            0 &amp; 0 &amp; 1 &amp; \alpha &amp; 1 + \alpha \\
            1 &amp; 1 &amp; 0 &amp; 1 + \alpha &amp; \alpha \\
            \alpha &amp; \alpha &amp; 1 + \alpha &amp; 0 &amp; 1  \\
            1 + \alpha &amp; 1 + \alpha &amp; \alpha &amp; 1 &amp; 0
            \end{array}
          </me>
        </p>
      </figure>

      <figure xml:id="fields-table-multiplication-z2-alpha">
            <!-- math table -->
        <caption>Multiplication Table for <m>{\mathbb Z}_2(\alpha)</m></caption>
        <p>
          <me>
            \begin{array}{c|cccc}
            \cdot &amp; 0 &amp; 1 &amp; \alpha &amp; 1 + \alpha \\
            \hline
            0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
            1 &amp; 0 &amp; 1 &amp; \alpha &amp; 1 + \alpha \\
            \alpha &amp; 0 &amp; \alpha &amp; 1 + \alpha &amp; 1  \\
            1 + \alpha &amp; 0 &amp; 1 + \alpha &amp; 1 &amp; \alpha
            \end{array}
          </me>
        </p>
      </figure>

      <p>
        The following theorem, due to Kronecker,
        is so important and so basic to our understanding of fields that it is often known as the Fundamental Theorem of Field Theory.
      </p>

      <theorem xml:id="fields-theorem-fundamental-of-field-theory">
        <statement>
          <p>
            Let <m>F</m> be a field and let <m>p(x)</m> be a nonconstant polynomial in <m>F[x]</m>.
            Then there exists an extension field <m>E</m> of <m>F</m> and an element
            <m>\alpha \in E</m> such that <m>p(\alpha) = 0</m>.
          </p>
        </statement>

        <proof>
          <p>
            To prove this theorem,
            we will employ the method that we used to construct <xref ref="fields-example-finite-f4"/>.
            Clearly, we can assume that <m>p(x)</m> is an irreducible polynomial.
            We wish to find an extension field <m>E</m> of <m>F</m> containing an element <m>\alpha</m> such that <m>p(\alpha) = 0</m>.
            The ideal <m>\langle p(x) \rangle</m> generated by <m>p(x)</m> is a maximal ideal in <m>F[x]</m> by <xref ref="poly-theorem-max-ideal"/>;
            hence, <m>F[x]/\langle p(x) \rangle</m> is a field.
            We claim that <m>E = F[x]/\langle p(x) \rangle</m> is the desired field.
          </p>

          <p>
            We first show that <m>E</m> is a field extension of <m>F</m>.
            We can define a homomorphism of commutative rings by the map <m>\psi:F \rightarrow F[x]/\langle p(x) \rangle</m>,
            where <m>\psi(a) = a + \langle p(x)\rangle</m> for <m>a \in F</m>.
            It is easy to check that <m>\psi</m> is indeed a ring homomorphism.
            Observe that
            <me>
              \psi( a ) + \psi( b ) = (a + \langle p(x) \rangle) + (b + \langle p(x) \rangle) = (a + b) + \langle p(x) \rangle = \psi( a + b )
            </me>
            and
            <me>
              \psi( a ) \psi( b ) = (a + \langle p(x) \rangle)  (b + \langle p(x) \rangle) = ab + \langle p(x) \rangle = \psi( ab )
            </me>.
            To prove that <m>\psi</m> is one-to-one, assume that
            <me>
              a + \langle p(x) \rangle = \psi(a) = \psi(b) = b + \langle p(x) \rangle
            </me>.
            Then <m>a - b</m> is a multiple of <m>p(x)</m>,
            since it lives in the ideal <m>\langle p(x) \rangle</m>.
            Since <m>p(x)</m> is a nonconstant polynomial,
            the only possibility is that <m>a - b = 0</m>.
            Consequently, <m>a = b</m> and <m>\psi</m> is injective.
            Since <m>\psi</m> is one-to-one,
            we can identify <m>F</m> with the subfield
            <m>\{ a + \langle p(x) \rangle : a \in F \}</m> of <m>E</m> and view <m>E</m> as an extension field of <m>F</m>.
          </p>
            <!-- \alpha is in E not F - TWJ 4/23/2011 -->
          <p>
            It remains for us to prove that <m>p(x)</m> has a zero <m>\alpha \in E</m>.
            Set <m>\alpha = x + \langle p(x) \rangle</m>.
            Then <m>\alpha</m> is in <m>E</m>.
            If <m>p(x) = a_0 + a_1 x + \cdots + a_n x^n</m>, then
            <md>
              <mrow>p( \alpha ) &amp; =  a_0 + a_1( x + \langle p(x) \rangle) + \cdots + a_n ( x + \langle p(x) \rangle)^n</mrow>
              <mrow>&amp; = a_0 + ( a_1 x + \langle p(x) \rangle) + \cdots + (a_n  x^n + \langle p(x) \rangle)</mrow>
              <mrow>&amp; = a_0 + a_1 x + \cdots + a_n x^n + \langle p(x) \rangle</mrow>
              <mrow>&amp; = 0 + \langle p(x) \rangle</mrow>
            </md>.
            Therefore, we have found an element
            <m>\alpha \in E = F[x]/\langle p(x) \rangle</m> such that <m>\alpha</m> is a zero of <m>p(x)</m>.
          </p>
        </proof>
      </theorem>

      <example xml:id="fields-example-finite-f8">
        <p>
          Let <m>p(x) = x^5 + x^4 + 1 \in {\mathbb Z}_2[x]</m>.
          Then <m>p(x)</m> has irreducible factors
          <m>x^2 + x + 1</m> and <m>x^3 + x + 1</m>.
          For a field extension <m>E</m> of
          <m>{\mathbb Z}_2</m> such that <m>p(x)</m> has a root in <m>E</m>,
          we can let <m>E</m> be either
          <m>{\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle</m> or <m>{\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle</m>.
          We will leave it as an exercise to show that
          <m>{\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle</m> is a field with <m>2^3 = 8</m> elements.
        </p>
      </example>
    </introduction>

    <subsection xml:id="fields-subsection-algebraic-elements">
      <title>Algebraic Elements</title>
      <p>
        An element <m>\alpha</m> in an extension field <m>E</m> over <m>F</m> is <term>algebraic</term>
        over <m>F</m> if <m>f(\alpha)=0</m> for some nonzero polynomial <m>f(x) \in F[x]</m>.
        An element in <m>E</m> that is not algebraic over <m>F</m> is
        <term>transcendental</term><idx><h>Element</h><h>transcendental</h></idx><idx><h>Transcendental element</h></idx>
        over <m>F</m>.
        An extension field <m>E</m> of a field <m>F</m> is an
        <term>algebraic extension</term><idx><h>Algebraic extension</h></idx><idx><h>Extension</h><h>algebraic</h></idx>
        of <m>F</m> if every element in <m>E</m> is algebraic over <m>F</m>.
        If <m>E</m> is a field extension of <m>F</m> and
        <m>\alpha_1, \ldots, \alpha_n</m> are contained in <m>E</m>,
        we denote the smallest field containing <m>F</m> and
        <m>\alpha_1, \ldots, \alpha_n</m> by <m>F( \alpha_1, \ldots, \alpha_n)</m>.

        <notation>
          <usage><m>F( \alpha_1, \ldots, \alpha_n)</m></usage>
          <description>smallest field containing <m>F</m> and <m>\alpha_1, \ldots, \alpha_n</m></description>
        </notation>

        If <m>E = F( \alpha )</m> for some <m>\alpha \in E</m>,
        then <m>E</m> is a <term>simple extension</term><idx><h>Simple extension</h></idx><idx><h>Extension</h><h>simple</h></idx>
        of <m>F</m>.
      </p>

      <example xml:id="fields-example-extension-pi">
        <p>
          Both <m>\sqrt{2}</m> and <m>i</m> are algebraic over
          <m>{\mathbb Q}</m> since they are zeros of the polynomials <m>x^2 -2</m> and <m>x^2 + 1</m>,
          respectively.
          Clearly <m>\pi</m> and <m>e</m> are algebraic over the real numbers;
          however, it is a nontrivial fact that they are transcendental over <m>{\mathbb Q}</m>.
          Numbers in <m>{\mathbb R}</m> that are algebraic over
          <m>{\mathbb Q}</m> are in fact quite rare.
          Almost all real numbers are transcendental over <m>{\mathbb Q}</m>.<fn>
          The probability that a real number chosen at random from the interval <m>[0, 1]</m> will be transcendental over the rational numbers is one.
          </fn>(In many cases we do not know whether or not a particular number is transcendental;
          for example,
          it is still not known whether <m>\pi + e</m> is transcendental or algebraic.)
        </p>
      </example>

      <p>
        A complex number that is algebraic over <m>{\mathbb Q}</m> is an
        <idx><h>Algebraic number</h></idx><term>algebraic number</term>.   
        A <term>transcendental number</term><idx><h>Transcendental number</h></idx>
        is an element of <m>{\mathbb C}</m> that is transcendental over <m>{\mathbb Q}</m>.
      </p>

      <example xml:id="fields-example-sqrt2-sqrt3">
        <p>
          We will show that <m>\sqrt{2 + \sqrt{3} }</m> is algebraic over <m>{\mathbb Q}</m>.
          If <m>\alpha = \sqrt{2 + \sqrt{3} }</m>,
          then <m>\alpha^2 = 2 + \sqrt{3}</m>.
          Hence, <m>\alpha^2 - 2 = \sqrt{3}</m> and <m>( \alpha^2 - 2)^2 = 3</m>.
          Since <m>\alpha^4 - 4 \alpha^2 + 1 = 0</m>,
          it must be true that <m>\alpha</m> is a zero of the polynomial <m>x^4 - 4 x^2 + 1 \in {\mathbb Q}[x]</m>.
        </p>
      </example>

      <p>
        It is very easy to give an example of an extension field <m>E</m> over a field <m>F</m>,
        where <m>E</m> contains an element transcendental over <m>F</m>.
        The following theorem characterizes transcendental extensions.
      </p>

      <theorem>
        <statement>
          <p>
            Let <m>E</m> be an extension field of <m>F</m> and <m>\alpha \in E</m>.
            Then <m>\alpha</m> is transcendental over <m>F</m> if and only if
            <m>F( \alpha )</m> is isomorphic to <m>F(x)</m>,
            the field of fractions of <m>F[x]</m>.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>\phi_{\alpha} : F[x] \rightarrow E</m> be the evaluation homomorphism for <m>\alpha</m>.
            Then <m>\alpha</m> is transcendental over <m>F</m> if and only if
            <m>\phi_{\alpha} (p(x)) = p(\alpha) \neq 0</m> for all nonconstant polynomials <m>p(x) \in F[x]</m>.
            This is true if and only if <m>\ker \phi_{\alpha} = \{ 0 \}</m>;
            that is, it is true exactly when <m>\phi_{\alpha}</m> is one-to-one.
            Hence, <m>E</m> must contain a copy of <m>F[x]</m>.
            The smallest field containing <m>F[x]</m> is the field of fractions <m>F(x)</m>.
            By <xref ref="domains-theorem-field-of-quotients"/>, <m>E</m> must contain a copy of this field.
          </p>
        </proof>
      </theorem>

      <p>
        We have a more interesting situation in the case of algebraic extensions.
      </p>

      <theorem>
        <statement>
          <p>
            Let <m>E</m> be an extension field of a field <m>F</m> and
            <m>\alpha \in E</m> with <m>\alpha</m> algebraic over <m>F</m>.
            Then there is a unique irreducible monic polynomial
            <m>p(x) \in F[x]</m> of smallest degree such that <m>p( \alpha ) = 0</m>.
            If <m>f(x)</m> is another polynomial in <m>F[x]</m> such that <m>f(\alpha) = 0</m>,
            then <m>p(x)</m> divides <m>f(x)</m>.
          </p>
        </statement>
            <!--  Removed the hypothesis that f(x) be monic.  Suggested by Davis Schubert. - TWJ 4/24/2013 -->
        <proof>
          <p>
            Let <m>\phi_{\alpha} : F[x] \rightarrow E</m> be the evaluation homomorphism.
            The kernel of <m>\phi_{\alpha}</m> is a principal ideal generated by some
            <m>p(x) \in F[x]</m> with <m>\deg p(x) \geq 1</m>.
            We know that such a polynomial exists,
            since <m>F[x]</m> is a principal ideal domain and <m>\alpha</m> is algebraic.
            The ideal <m>\langle p(x) \rangle</m> consists exactly of those elements of <m>F[x]</m> having <m>\alpha</m> as a zero.
            If <m>f( \alpha ) = 0</m> and <m>f(x)</m> is not the zero polynomial,
            then <m>f(x) \in \langle p(x) \rangle</m> and <m>p(x)</m> divides <m>f(x)</m>.
            So <m>p(x)</m> is a polynomial of minimal degree having <m>\alpha</m> as a zero.
            Any other polynomial of the same degree having <m>\alpha</m> as a zero must have the form
            <m>\beta p( x)</m> for some <m>\beta \in F</m>.
          </p>

          <p>
            Suppose now that <m>p(x) = r(x) s(x)</m> is a factorization of <m>p(x)</m> into polynomials of lower degree.
            Since <m>p( \alpha ) = 0</m>,
            <m>r( \alpha ) s( \alpha ) = 0</m>;
            consequently,
            either <m>r( \alpha )=0</m> or <m>s( \alpha ) = 0</m>,
            which contradicts the fact that <m>p</m> is of minimal degree.
            Therefore, <m>p(x)</m> must be irreducible.
          </p>
        </proof>
      </theorem>

      <p>
        Let <m>E</m> be an extension field of <m>F</m> and
        <m>\alpha \in E</m> be algebraic over <m>F</m>.
        The unique monic polynomial <m>p(x)</m> of the last theorem is called the
        <term>minimal polynomial</term><idx><h>Polynomial</h><h>minimal</h></idx><idx><h>Minimal polynomial</h></idx>
        for <m>\alpha</m> over <m>F</m>.
        The degree of <m>p(x)</m> is the
        <term>degree of <m>\alpha</m> over <m>F</m></term>.
      </p>

      <example xml:id="fields-example-minimal-poly">
        <p>
          Let <m>f(x) = x^2 - 2</m> and <m>g(x) = x^4 - 4 x^2 + 1</m>.
          These polynomials are the minimal polynomials of
          <m>\sqrt{2}</m> and <m>\sqrt{2 + \sqrt{3} }</m>, respectively.
        </p>
      </example>

      <proposition xml:id="fields-proposition-minimal-poly">
        <statement>
          <p>
            Let <m>E</m> be a field extension of <m>F</m> and
            <m>\alpha \in E</m> be algebraic over <m>F</m>.
            Then <m>F( \alpha ) \cong F[x] / \langle p(x) \rangle</m>,
            where <m>p(x)</m> is the minimal polynomial of <m>\alpha</m> over <m>F</m>.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>\phi_{\alpha} : F[x] \rightarrow E</m> be the evaluation homomorphism.
            The kernel of this map is <m>\langle p(x) \rangle</m>,
            where <m>p(x)</m> is the minimal polynomial of <m>\alpha</m>.
            By the First Isomorphism Theorem for rings,
            the image of <m>\phi_{\alpha}</m> in <m>E</m> is isomorphic to
            <m>F( \alpha )</m> since it contains both <m>F</m> and <m>\alpha</m>.
          </p>
        </proof>
      </proposition>
            <!--  Corrected the kernel of <m>\phi_{\alpha}</m>.  Suggested by Aleks Vlasev. - TWJ 8/10/2011 -->
      <theorem xml:id="fields-theorem-simple_extension">
        <statement>
          <p>
            Let <m>E = F( \alpha )</m> be a simple extension of <m>F</m>,
            where <m>\alpha \in E</m> is algebraic over <m>F</m>.
            Suppose that the degree of <m>\alpha</m> over <m>F</m> is <m>n</m>.
            Then every element <m>\beta \in E</m> can be expressed uniquely in the form
            <me>
              \beta = b_0 + b_1 \alpha + \cdots + b_{n-1} \alpha^{n - 1}
            </me>
            for <m>b_i \in F</m>.
          </p>
        </statement>

        <proof>
          <p>
            Since <m>\phi_{\alpha} ( F[x] ) \cong F( \alpha )</m>,
            every element in <m>E = F( \alpha )</m> must be of the form <m>\phi_{\alpha} ( f(x) ) = f( \alpha )</m>,
            where <m>f(\alpha)</m> is a polynomial in <m>\alpha</m> with coefficients in <m>F</m>.
            Let
            <me>
              p(x) = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0
            </me>
            be the minimal polynomial of <m>\alpha</m>.
            Then <m>p( \alpha ) = 0</m>; hence,
            <me>
              {\alpha}^n = - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0
            </me>.
            Similarly,
            <md>
              <mrow>{\alpha}^{n + 1} &amp; = {\alpha} {\alpha}^n</mrow>
              <mrow>&amp; = - a_{n - 1} {\alpha}^n - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}</mrow>
              <mrow>&amp; = - a_{n - 1}( - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0) - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}</mrow>
            </md>.
            Continuing in this manner,
            we can express every monomial <m>{\alpha}^m</m>, <m>m \geq n</m>,
            as a linear combination of powers of
            <m>{\alpha}</m> that are less than <m>n</m>.
            Hence, any <m>\beta \in F( \alpha )</m> can be written as
            <me>
              \beta = b_0 + b_1 \alpha + \cdots + b_{n - 1} \alpha^{n - 1}
            </me>.
          </p>

          <p>
            To show uniqueness, suppose that
            <me>
              \beta = b_0 + b_1 \alpha + \cdots + b_{n-1} \alpha^{n-1} = c_0 + c_1 \alpha + \cdots + c_{n - 1} \alpha^{n - 1}
            </me>
            for <m>b_i</m> and <m>c_i</m> in <m>F</m>.
            Then
            <me>
              g(x) = (b_0 - c_0) + (b_1 - c_1) x + \cdots + (b_{n - 1} - c_{n - 1})x^{n - 1}
            </me>
            is in <m>F[x]</m> and <m>g( \alpha ) = 0</m>.
            Since the degree of <m>g(x)</m> is less than the degree of <m>p( x )</m>,
            the irreducible polynomial of <m>\alpha</m>,
            <m>g(x)</m> must be the zero polynomial.
            Consequently,
            <me>
              b_0 - c_0 = b_1 - c_1 = \cdots = b_{n - 1} - c_{n - 1} = 0
            </me>,
            or <m>b_i = c_i</m> for <m>i = 0, 1, \ldots, n-1</m>.
            Therefore, we have shown uniqueness.
          </p>
        </proof>
      </theorem>
            <!--  Changed = to \cong.  Suggested by Aleks Vlasev. - TWJ 8/10/2011 -->
      <example xml:id="fields-example-isomorph">
        <p>
          Since <m>x^2 + 1</m> is irreducible over <m>{\mathbb R}</m>,
          <m>\langle x^2 + 1 \rangle</m> is a maximal ideal in <m>{\mathbb R}[x]</m>.
          So <m>E = {\mathbb R}[x]/\langle x^2 + 1 \rangle</m> is a field extension of
          <m>{\mathbb R}</m> that contains a root of <m>x^2 + 1</m>.
          Let <m>\alpha = x + \langle x^2 + 1 \rangle</m>.
          We can identify <m>E</m> with the complex numbers.
          By <xref ref="fields-proposition-minimal-poly"/>,
          <m>E</m> is isomorphic to <m>{\mathbb R}( \alpha ) = \{ a + b \alpha : a, b \in {\mathbb R} \}</m>.
          We know that <m>\alpha^2 = -1</m> in <m>E</m>, since
          <md>
            <mrow>\alpha^2 + 1 &amp; = (x + \langle x^2 + 1  \rangle)^2 + (1 + \langle x^2 + 1 \rangle)</mrow>
            <mrow>&amp; = (x^2 + 1) + \langle x^2 + 1  \rangle</mrow>
            <mrow>&amp; = 0</mrow>
          </md>.
          Hence, we have an isomorphism of <m>{\mathbb R}( \alpha )</m> with
          <m>{\mathbb C}</m> defined by the map that takes <m>a + b \alpha</m> to <m>a + bi</m>.
        </p>
      </example>

      <p>
        Let <m>E</m> be a field extension of a field <m>F</m>.
        If we regard <m>E</m> as a vector space over <m>F</m>,
        then we can bring the machinery of linear algebra to bear on the problems that we will encounter in our study of fields.
        The elements in the field <m>E</m> are vectors;
        the elements in the field <m>F</m> are scalars.
        We can think of addition in <m>E</m> as adding vectors.
        When we multiply an element in <m>E</m> by an element of <m>F</m>,
        we are multiplying a vector by a scalar.
        This view of field extensions is especially fruitful if a field extension <m>E</m> of <m>F</m> is a finite dimensional vector space over <m>F</m>,
        and <xref ref="fields-theorem-simple_extension"/> states that
        <m>E = F(\alpha )</m> is finite dimensional vector space over <m>F</m> with basis <m>\{ 1, \alpha, {\alpha}^2, \ldots, {\alpha}^{n - 1} \}</m>.
      </p>

      <p>
        If an extension field <m>E</m> of a field <m>F</m> is a finite dimensional vector space over <m>F</m> of dimension <m>n</m>,
        then we say that <m>E</m> is a 
        <idx><h>Extension</h><h>finite</h></idx><term>finite extension of degree <m>n</m> over <m>F</m></term>. 
        We write
        <me>
          [E:F]= n
        </me>.
        to indicate the dimension of <m>E</m> over <m>F</m>.

        <notation>
          <usage><m>[E:F]</m></usage>
          <description>dimension of a field extension of <m>E</m> over <m>F</m></description>
        </notation>

      </p>

      <theorem xml:id="fields-theorem-finite-extension">
        <statement>
          <p>
            Every finite extension field <m>E</m> of a field <m>F</m> is an algebraic extension.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>\alpha \in E</m>.
            Since <m>[E:F] = n</m>, the elements
            <me>
              1, \alpha, \ldots, {\alpha}^n
            </me>
            cannot be linearly independent.
            Hence, there exist <m>a_i \in F</m>, not all zero, such that
            <me>
              a_n {\alpha}^n + a_{n - 1} {\alpha}^{n - 1} + \cdots + a_1 \alpha + a_0 = 0
            </me>.
            Therefore,
            <me>
              p(x) = a_n x^n + \cdots + a_0 \in F[x]
            </me>
            is a nonzero polynomial with <m>p( \alpha ) = 0</m>.
          </p>
        </proof>
      </theorem>

      <remark>
        <p>
          <xref ref="fields-theorem-finite-extension"/> says that every finite extension of a field <m>F</m> is an algebraic extension.
          The converse is false, however.
          We will leave it as an exercise to show that the set of all elements in
          <m>{\mathbb R}</m> that are algebraic over
          <m>{\mathbb Q}</m> forms an infinite field extension of <m>{\mathbb Q}</m>.
        </p>
      </remark>

      <p>
        The next theorem is a counting theorem,
        similar to Lagrange's Theorem in group theory.
        <xref ref="fields-theorem-tower-indices"/> will prove to be an extremely useful tool in our investigation of finite field extensions.
      </p>
            <!--Changed the refrerence in the preceding paragraph to the next theorem instead of the previous theorem.  TWJ 13/6/2014-->
      <theorem xml:id="fields-theorem-tower-indices">
        <statement>
          <p>
            If <m>E</m> is a finite extension of <m>F</m> and <m>K</m> is a finite extension of <m>E</m>,
            then <m>K</m> is a finite extension of <m>F</m> and
            <me>
              [K:F]= [K:E] [E:F]
            </me>.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>\{ \alpha_1, \ldots, \alpha_n \}</m> be a basis for <m>E</m> as a vector space over <m>F</m> and
            <m>\{ \beta_1, \ldots, \beta_m \}</m> be a basis for <m>K</m> as a vector space over <m>E</m>.
            We claim that <m>\{ \alpha_i \beta_j \}</m> is a basis for <m>K</m> over <m>F</m>.
            We will first show that these vectors span <m>K</m>.
            Let <m>u \in K</m>.
            Then <m>u = \sum_{j = 1}^{m} b_j \beta_j</m> and <m>b_j = \sum_{i = 1}^{n} a_{ij} \alpha_i</m>,
            where <m>b_j \in E</m> and <m>a_{ij} \in F</m>.
            Then
            <me>
              u = \sum_{j = 1}^{m} \left( \sum_{i = 1}^{n} a_{ij} \alpha_i  \right) \beta_j = \sum_{i,j} a_{ij} ( \alpha_i \beta_j )
            </me>.
            So the <m>mn</m> vectors <m>\alpha_i \beta_j</m> must span <m>K</m> over <m>F</m>.
          </p>

          <p>
            We must show that <m>\{ \alpha_i \beta_j \}</m> are linearly independent.
            Recall that a set of vectors
            <m>v_1, v_2, \ldots,
            v_n</m> in a vector space <m>V</m> are linearly independent if
            <me>
              c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0
            </me>
            implies that
            <me>
              c_1 = c_2 = \cdots = c_n = 0
            </me>.
            Let
            <me>
              u = \sum_{i,j} c_{ij} ( \alpha_i \beta_j ) = 0
            </me>
            for <m>c_{ij} \in F</m>.
            We need to prove that all of the <m>c_{ij}</m>'s are zero.
            We can rewrite <m>u</m> as
            <me>
              \sum_{j = 1}^{m} \left( \sum_{i = 1}^{n} c_{ij} \alpha_i \right) \beta_j = 0
            </me>,
            where <m>\sum_i c_{ij} \alpha_i \in E</m>.
            Since the <m>\beta_j</m>'s are linearly independent over <m>E</m>,
            it must be the case that
            <me>
              \sum_{i = 1}^n c_{ij} \alpha_i = 0
            </me>
            for all <m>j</m>.
            However, the <m>\alpha_j</m> are also linearly independent  over <m>F</m>.
            Therefore, <m>c_{ij} = 0</m> for all <m>i</m> and <m>j</m>,
            which completes the proof.
          </p>
        </proof>
      </theorem>

      <p>
        The following corollary is easily proved using mathematical induction.
      </p>

      <corollary>
        <statement>
          <p>
            If <m>F_i</m> is a field for
            <m>i = 1, \dots,
            k</m> and <m>F_{i+1}</m> is a finite extension of <m>F_i</m>,
            then <m>F_k</m> is a finite extension of <m>F_1</m> and
            <me>
              [F_k : F_1] = [F_k : F_{k-1} ] \cdots [F_2 : F_1 ]
            </me>.
          </p>
        </statement>
      </corollary>

      <corollary>
        <statement>
          <p>
            Let <m>E</m> be an extension field of <m>F</m>.
            If <m>\alpha \in E</m> is algebraic over <m>F</m> with minimal polynomial <m>p(x)</m> and
            <m>\beta \in F( \alpha )</m> with minimal polynomial <m>q(x)</m>,
            then <m>\deg q(x)</m> divides <m>\deg p(x)</m>.
          </p>
        </statement>

        <proof>
          <p>
            We know that <m>\deg p(x) = [F( \alpha ) : F ]</m> and <m>\deg q(x) = [F( \beta ) : F ]</m>.
            Since <m>F \subset F( \beta ) \subset F( \alpha )</m>,
            <me>
              [F( \alpha ) : F ]= [ F( \alpha ) : F( \beta ) ] [ F( \beta ) : F ]
            </me>.
          </p>
        </proof>
      </corollary>

      <example xml:id="fields-example-sqrt3-sqrt5">
        <p>
          Let us determine an extension field of
          <m>{\mathbb Q}</m> containing <m>\sqrt{3} + \sqrt{5}</m>.
          It is easy to determine that the minimal  polynomial of
          <m>\sqrt{3} + \sqrt{5}</m> is <m>x^4 - 16 x^2 + 4</m>.
          It follows that
          <me>
            [{\mathbb Q}( \sqrt{3} + \sqrt{5}\, ) : {\mathbb Q} ] = 4
          </me>.
          We know that <m>\{ 1, \sqrt{3}\, \}</m> is a basis for 
          <m>{\mathbb Q}( \sqrt{3}\, )</m> over <m>{\mathbb Q}</m>.
          Hence, <m>\sqrt{3} + \sqrt{5}</m> cannot be in <m>{\mathbb Q}( \sqrt{3}\, )</m>.
          It follows that <m>\sqrt{5}</m> cannot be in <m>{\mathbb Q}( \sqrt{3}\, )</m> either.
          Therefore, <m>\{ 1, \sqrt{5}\, \}</m> is a basis for <m>{\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = ( {\mathbb Q}(\sqrt{3}\, ))( \sqrt{5}\, )</m> over
          <m>{\mathbb Q}( \sqrt{3}\, )</m> and <m>\{ 1, \sqrt{3}, \sqrt{5}, \sqrt{3} \sqrt{5} = \sqrt{15}\, \}</m> is a basis for
          <m>{\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = {\mathbb Q}( \sqrt{3} + \sqrt{5}\, )</m> over <m>{\mathbb Q}</m>.
          This example shows that it is possible that some extension
          <m>F( \alpha_1, \ldots, \alpha_n )</m> is actually a simple extension of <m>F</m> even though <m>n \gt 1</m>.
        </p>
      </example>
            <!-- Changed minimal polynomial from  <m>x^4 - 16 x + 4</m> to <m>x^4 - 16 x^2 + 4</m>.  Discovered by Bradley Noyes - TWJ 3/25/2011 -->
      <example xml:id="fields-example-cubert5-sqrt5-i">
        <p>
          Let us compute a basis for <m>{\mathbb Q}( \sqrt[3]{5}, \sqrt{5} \, i )</m>,
          where <m>\sqrt{5}</m> is the positive square root of <m>5</m> and
          <m>\sqrt[3]{5}</m> is the real cube root of <m>5</m>.
          We know that <m>\sqrt{5} \, i \notin {\mathbb Q}(\sqrt[3]{5}\, )</m>, so
          <me>
            [ {\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i) : {\mathbb Q}(\sqrt[3]{5}\, )] = 2
          </me>.
          It is easy to determine that <m>\{ 1, \sqrt{5}i\, \}</m> is a basis for
          <m>{\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )</m> over <m>{\mathbb Q}( \sqrt[3]{5}\, )</m>.
          We also know that <m>\{ 1, \sqrt[3]{5}, (\sqrt[3]{5}\, )^2  \}</m> is a basis for
          <m>{\mathbb Q}(\sqrt[3]{5}\, )</m> over <m>{\mathbb Q}</m>.
          Hence, a basis for <m>{\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i )</m> over <m>{\mathbb Q}</m> is
          <me>
            \{ 1, \sqrt{5}\, i, \sqrt[3]{5}, (\sqrt[3]{5}\, )^2, (\sqrt[6]{5}\, )^5 i, (\sqrt[6]{5}\, )^7 i = 5 \sqrt[6]{5}\, i \text{ or } \sqrt[6]{5}\, i \}
          </me>.
          Notice that <m>\sqrt[6]{5}\, i</m> is a zero of <m>x^6 + 5</m>.
          We can show that this polynomial is irreducible over
          <m>{\mathbb Q}</m> using Eisenstein's Criterion,
          where we let <m>p = 5</m>.
          Consequently,
          <me>
            {\mathbb Q} \subset {\mathbb Q}( \sqrt[6]{5}\, i) \subset {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )
          </me>.
          But it must be the case that <m>{\mathbb Q}( \sqrt[6]{5}\, i) = {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )</m>,
          since the degree of both of these extensions is <m>6</m>.
        </p>
      </example>
            <!-- Corrected typo in the last set inclusion.  Suggest2/9/2013 -->
      <theorem>
        <statement>
          <p>
            Let <m>E</m> be a field extension of <m>F</m>.
            Then  the following statements are equivalent.

            <ol>
              <li>
                <p>
                  <m>E</m> is a finite extension of <m>F</m>.
                </p>
              </li>

              <li>
                <p>
                  There exists a finite number of algebraic elements
                  <m>\alpha_1, \ldots, \alpha_n \in E</m> such that <m>E = F(\alpha_1, \ldots, \alpha_n)</m>.
                </p>
              </li>

              <li>
                <p>
                  There exists a sequence of fields
                  <me>
                    E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n-1} ) \supset \cdots \supset F( \alpha_1 ) \supset F
                  </me>,
                  where each field <m>F(\alpha_1, \ldots, \alpha_i)</m> is algebraic over <m>F(\alpha_1, \ldots, \alpha_{i-1})</m>.
                </p>
              </li>
            </ol>
          </p>
        </statement>

        <proof>
          <p>
            (1) <m>\Rightarrow</m> (2).
            Let <m>E</m> be a finite algebraic extension of <m>F</m>.
            Then <m>E</m> is a finite dimensional vector space over <m>F</m> and there exists a basis consisting of elements
            <m>\alpha_1, \ldots, \alpha_n</m> in <m>E</m> such that <m>E = F(\alpha_1, \ldots, \alpha_n)</m>.
            Each <m>\alpha_i</m> is algebraic over <m>F</m> by <xref ref="fields-theorem-finite-extension"/>.
          </p>

          <p>
            (2) <m>\Rightarrow</m> (3).
            Suppose that <m>E = F(\alpha_1, \ldots, \alpha_n)</m>,
            where every <m>\alpha_i</m> is algebraic over <m>F</m>.
            Then
            <me>
              E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n - 1} ) \supset \cdots \supset F( \alpha_1 ) \supset F
            </me>,
            where each field <m>F(\alpha_1, \ldots, \alpha_i)</m> is algebraic over <m>F(\alpha_1, \ldots, \alpha_{i - 1})</m>.
          </p>

          <p>
            (3) <m>\Rightarrow</m> (1).
            Let
            <me>
              E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n - 1} ) \supset \cdots \supset F( \alpha_1 ) \supset F
            </me>,
            where each field <m>F(\alpha_1, \ldots, \alpha_i)</m> is algebraic over <m>F(\alpha_1, \ldots, \alpha_{i - 1})</m>.
            Since
            <me>
              F(\alpha_1, \ldots, \alpha_i) =  F(\alpha_1, \ldots, \alpha_{i - 1} )(\alpha_i)
            </me>
            is simple extension and <m>\alpha_i</m> is algebraic over <m>F(\alpha_1, \ldots, \alpha_{i - 1})</m>,
            it follows that
            <me>
              [ F(\alpha_1, \ldots, \alpha_i) : F(\alpha_1, \ldots, \alpha_{i - 1} )]
            </me>
            is finite for each <m>i</m>.
            Therefore, <m>[E : F]</m> is finite.
          </p>
        </proof>
      </theorem>
    </subsection>

    <subsection xml:id="fields-subsection-algebraic-closure">
      <title>Algebraic Closure</title>
      <p>
        Given a field <m>F</m>,
        the question arises as to whether or not we can find a field <m>E</m> such that every polynomial <m>p(x)</m> has a root in <m>E</m>.
        This leads us to the following theorem.
      </p>

      <theorem>
        <statement>
          <p>
            Let <m>E</m> be an extension field of <m>F</m>.
            The set of elements in <m>E</m> that are algebraic over <m>F</m> form a field.
          </p>
        </statement>
            <!--  Changed <m>\alpha \pm \beta</m>, $\alpha / -->
            <!-- \beta<m>, and </m>\alpha / \beta<m> to </m>\alpha \pm \beta<m>, </m>\alpha  -->
            <!-- \beta<m>, and </m>\alpha / \beta$  Suggested by Aleks Vlasev. - TWJ 8/10/2011 -->
        <proof>
          <p>
            Let <m>\alpha, \beta \in E</m> be algebraic over <m>F</m>.
            Then <m>F( \alpha, \beta )</m> is a finite extension of <m>F</m>.
            Since every element of <m>F( \alpha, \beta )</m> is algebraic over <m>F</m>,
            <m>\alpha \pm \beta</m>, <m>\alpha \beta</m>,
            and <m>\alpha / \beta</m> (<m>\beta \neq 0</m>) are all algebraic over <m>F</m>.
            Consequently,
            the set of elements in <m>E</m> that are algebraic over <m>F</m> form a field.
          </p>
        </proof>
      </theorem>

      <corollary xml:id="fields-corollary-algebraic-numbers-field">
        <statement>
          <p>
            The set of all algebraic numbers forms a field;
            that is, the set of all complex numbers that are algebraic over <m>{\mathbb Q}</m> makes up a field.
          </p>
        </statement>
      </corollary>

      <p>
        Let <m>E</m> be a field extension of a field <m>F</m>.
        We define the <term>algebraic closure</term><idx><h>Algebraic closure</h></idx>
        of a field <m>F</m> in <m>E</m> to be the field consisting of all elements in <m>E</m> that are algebraic over <m>F</m>.
        A field <m>F</m> is <term>algebraically closed</term><idx><h>Field</h><h>algebraically closed</h></idx> if every nonconstant polynomial in <m>F[x]</m> has a root in <m>F</m>.
      </p>

      <theorem>
        <statement>
          <p>
            A field <m>F</m> is algebraically closed if and only if every nonconstant polynomial in <m>F[x]</m> factors into linear factors over <m>F[x]</m>.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>F</m> be an algebraically closed field.
            If <m>p(x) \in F[x]</m> is a nonconstant polynomial,
            then <m>p(x)</m> has a zero in <m>F</m>, say <m>\alpha</m>.
            Therefore, <m>x-\alpha</m> must be a factor of <m>p(x)</m> and so <m>p(x) = (x - \alpha) q_1(x)</m>,
            where <m>\deg q_1(x) = \deg p(x) - 1</m>.
            Continue this process with <m>q_1(x)</m> to find a factorization
            <me>
              p(x) = (x - \alpha)(x - \beta)q_2(x)
            </me>,
            where <m>\deg q_2(x) = \deg p(x) -2</m>.
            The process must eventually stop since the degree of <m>p(x)</m> is finite.
          </p>

          <p>
            Conversely, suppose that every nonconstant polynomial <m>p(x)</m> in <m>F[x]</m> factors into linear factors.
            Let <m>ax - b</m> be such a factor.
            Then <m>p( b/a ) = 0</m>.
            Consequently, <m>F</m> is algebraically closed.
          </p>
        </proof>
      </theorem>

      <corollary>
        <statement>
          <p>
            An algebraically closed field <m>F</m> has no proper algebraic extension <m>E</m>.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>E</m> be an algebraic extension of <m>F</m>;
            then <m>F \subset E</m>.
            For <m>\alpha \in E</m>,
            the minimal polynomial of <m>\alpha</m> is <m>x - \alpha</m>.
            Therefore, <m>\alpha \in F</m> and <m>F = E</m>.
          </p>
        </proof>
      </corollary>

      <theorem>
        <statement>
          <p>
            Every field <m>F</m> has a unique algebraic closure.
          </p>
        </statement>
      </theorem>

      <p>
        It is a nontrivial fact that every field has a unique algebraic closure.
        The proof is not extremely difficult,
        but requires some rather sophisticated set theory.
        We refer the reader to [3], [4], or [8] for a proof of this result.
      </p>
            <!-- not sure how to number these references within the chapter - 19 August 2010 - TWJ -->
      <p>
        We now state the Fundamental Theorem of Algebra,
        first proven by Gauss at the age of 22 in his doctoral thesis.
        This theorem states that every polynomial with coefficients in the complex numbers has a root in the complex numbers.
        The proof of this theorem will be given in <xref ref="galois"/>.
      </p>

      <theorem>
        <title>Fundamental Theorem of Algebra</title>
        <idx>
        <h>Fundamental Theorem</h>
        <h>of Algebra</h>
        </idx>
        <statement>
          <p>
            The field of complex numbers is algebraically closed.
          </p>
        </statement>
      </theorem>
    </subsection>
  </section>

  <section xml:id="fields-section-splitting-fields">
    <title>Splitting Fields</title>
    <p>
      Let <m>F</m> be a field and <m>p(x)</m> be a nonconstant polynomial in <m>F[x]</m>.
      We already know that we can find a field extension of <m>F</m> that contains a root of <m>p(x)</m>.
      However, we would like to know whether an extension <m>E</m> of <m>F</m> containing all of the roots of <m>p(x)</m> exists.
      In other words,
      can we find a field extension of <m>F</m> such that <m>p(x)</m> factors into a product of linear polynomials?
      What is the <q>smallest</q> extension containing all the roots of <m>p(x)</m>?
    </p>

    <p>
      Let <m>F</m> be a field and
      <m>p(x) = a_0 + a_1 x + \cdots + a_n x^n</m> be a nonconstant polynomial in <m>F[x]</m>.
      An extension field <m>E</m> of <m>F</m> is a
      <term>splitting field</term><idx><h>Splitting field</h></idx><idx><h>Field</h><h>splitting</h></idx>
      of <m>p(x)</m> if there exist elements
      <m>\alpha_1, \ldots, \alpha_n</m> in <m>E</m> such that <m>E = F( \alpha_1, \ldots, \alpha_n )</m> and
      <me>
        p(x) = ( x - \alpha_1 )(x - \alpha_2) \cdots (x - \alpha_n)
      </me>.
      A polynomial <m>p(x) \in F[x]</m> <term>splits</term>
      in <m>E</m> if it is the product of linear factors in <m>E[x]</m>.
    </p>

    <example xml:id="fields-example-splitting">
      <p>
        Let <m>p(x) = x^4 + 2x^2 - 8</m> be in <m>{\mathbb Q}[x]</m>.
        Then <m>p(x)</m> has irreducible factors <m>x^2 -2</m> and <m>x^2 + 4</m>.
        Therefore, the field <m>{\mathbb Q}( \sqrt{2}, i )</m> is a splitting field for <m>p(x)</m>.
      </p>
    </example>

    <example xml:id="fields-example-cube-root-not-splitting">
      <p>
        Let <m>p(x) = x^3 - 3</m> be in <m>{\mathbb Q}[x]</m>.
        Then <m>p(x)</m> has a root in the field <m>{\mathbb Q}( \sqrt[3]{3}\, )</m>.
        However, this field is not a splitting field for <m>p(x)</m> since the complex cube roots of 3,
        <me>
          \frac{ -\sqrt[3]{3} \pm (\sqrt[6]{3}\, )^5 i }{2}
        </me>,
        are not in <m>{\mathbb Q}( \sqrt[3]{3}\, )</m>.
      </p>
    </example>

    <theorem>
      <statement>
        <p>
          Let <m>p(x) \in F[x]</m> be a nonconstant polynomial.
          Then there exists a splitting field <m>E</m> for <m>p(x)</m>.
        </p>
      </statement>

      <proof>
        <p>
          We will use mathematical induction on the degree of <m>p(x)</m>.
          If <m>\deg p(x) = 1</m>,
          then <m>p(x)</m> is a linear polynomial and <m>E = F</m>.
          Assume that the theorem is true for all polynomials of degree <m>k</m> with
          <m>1 \leq k \lt n</m> and let <m>\deg p(x) = n</m>.
          We can assume that <m>p(x)</m> is irreducible;
          otherwise, by our induction hypothesis, we are done.
          By <xref ref="fields-theorem-fundamental-of-field-theory"/>,
          there exists a field <m>K</m> such that <m>p(x)</m> has a zero <m>\alpha_1</m> in <m>K</m>.
          Hence, <m>p(x) = (x - \alpha_1)q(x)</m>,
          where <m>q(x) \in K[x]</m>.
          Since <m>\deg q(x) = n -1</m>,
          there exists a splitting field
          <m>E \supset K</m> of <m>q(x)</m> that contains the zeros
          <m>\alpha_2, \ldots, \alpha_n</m> of <m>p(x)</m> by our induction hypothesis.
          Consequently,
          <me>
            E = K(\alpha_2, \ldots, \alpha_n) = F(\alpha_1, \ldots, \alpha_n)
          </me>
          is a splitting field of <m>p(x)</m>.
        </p>
      </proof>
    </theorem>

    <p>
      The question of uniqueness now arises for splitting fields.
      This question is answered in the affirmative.
      Given two splitting fields <m>K</m> and <m>L</m> of a polynomial <m>p(x) \in F[x]</m>,
      there exists a field isomorphism
      <m>\phi : K \rightarrow L</m> that preserves <m>F</m>.
      In order to prove this result,
      we must first prove a lemma.
    </p>

    <lemma xml:id="fields-lemma-field-isomorphism">
      <statement>
        <p>
          Let <m>\phi : E \rightarrow F</m> be an isomorphism of fields.
          Let <m>K</m> be an extension field of <m>E</m> and
          <m>\alpha \in K</m> be algebraic over <m>E</m> with minimal polynomial <m>p(x)</m>.
          Suppose that <m>L</m> is an extension field of <m>F</m> such that <m>\beta</m> is root of the polynomial in <m>F[x]</m> obtained from <m>p(x)</m> under the image of <m>\phi</m>.
          Then <m>\phi</m> extends to a unique isomorphism
          <m>\overline{\phi} : E( \alpha ) \rightarrow F( \beta )</m> such that <m>\overline{\phi}( \alpha ) = \beta</m> and
          <m>\overline{\phi}</m> agrees with <m>\phi</m> on <m>E</m>.
        </p>
      </statement>

      <proof>
        <p>
          If <m>p(x)</m> has degree <m>n</m>,
          then by <xref ref="fields-theorem-simple_extension"/> we can write any element in
          <m>E( \alpha )</m> as a linear combination of <m>1, \alpha, \ldots, \alpha^{n - 1}</m>.
          Therefore, the isomorphism that we are seeking must be
          <me>
            \overline{\phi}( a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1}) = \phi(a_0) + \phi(a_1) \beta + \cdots + \phi(a_{n - 1}) \beta^{n - 1}
          </me>,
          where
          <me>
            a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1}
          </me>
          is an element in <m>E(\alpha)</m>.
          The fact that <m>\overline{\phi}</m> is an isomorphism could be checked by direct computation;
          however, it is easier to observe that
          <m>\overline{\phi}</m> is a composition of maps that we already know to be isomorphisms.
        </p>

        <p>
          We can extend <m>\phi</m> to be an isomorphism from <m>E[x]</m> to <m>F[x]</m>,
          which we will also denote by <m>\phi</m>, by letting
          <me>
            \phi( a_0 + a_1 x + \cdots + a_n x^n ) =  \phi( a_0 ) + \phi(a_1) x + \cdots + \phi(a_n) x^n
          </me>.
          This extension agrees with the original isomorphism <m>\phi : E \rightarrow F</m>,
          since constant polynomials get mapped to constant polynomials.
          By assumption, <m>\phi(p(x)) = q(x)</m>;
          hence, <m>\phi</m> maps <m>\langle p(x) \rangle</m> onto <m>\langle q(x) \rangle</m>.
          Consequently,
          we have an isomorphism <m>\psi : E[x] / \langle p(x) \rangle \rightarrow F[x]/\langle q(x) \rangle</m>.
          By <xref ref="fields-proposition-minimal-poly"/>,
          we have isomorphisms <m>\sigma: E[x]/\langle p(x) \rangle \rightarrow E(\alpha)</m> and <m>\tau : F[x]/\langle q(x) \rangle \rightarrow F( \beta )</m>,
          defined by evaluation at <m>\alpha</m> and <m>\beta</m>,
          respectively.
          Therefore, <m>\overline{\phi} = \tau \psi \sigma^{-1}</m> is the required isomorphism (see <xref ref="fields-figure-splitting-fields-1"/>).
        </p>

        <figure xml:id="fields-figure-splitting-fields-1">
          <caption>Commutative diagram for <xref ref="fields-lemma-field-isomorphism"/></caption>
          <p><me>\begin{CD}
          E[x]/\langle p(x) \rangle    @>\psi>>  F[x]/\langle q(x) \rangle  \\
          @VV{\sigma}V        @VV{\tau}V\\
          E(\alpha)     @>\overline{\phi}>>  F(\beta) \\
          @VVV        @VVV\\
          E     @>\phi>>  F
          \end{CD}</me></p>
        </figure>

        <p>
          We leave the proof of uniqueness as a exercise.
        </p>
      </proof>
    </lemma>

    <theorem xml:id="fields-theorem-isomorphism-extension">
      <statement>
        <p>
          Let <m>\phi : E \rightarrow F</m> be an isomorphism of fields and let <m>p(x)</m> be a nonconstant polynomial in <m>E[x]</m> and <m>q(x)</m> the corresponding polynomial in <m>F[x]</m> under the isomorphism.
          If <m>K</m> is a splitting field of <m>p(x)</m> and <m>L</m> is a splitting field of <m>q(x)</m>,
          then <m>\phi</m> extends to an isomorphism <m>\psi : K \rightarrow L</m>.
        </p>
      </statement>

      <proof>
        <p>
          We will use mathematical induction on the degree of <m>p(x)</m>.
          We can assume that <m>p(x)</m> is irreducible over <m>E</m>.
          Therefore, <m>q(x)</m> is also irreducible over <m>F</m>.
          If <m>\deg p(x) = 1</m>, then by the definition of a splitting field,
          <m>K = E</m> and <m>L = F</m> and there is nothing to prove.
        </p>

        <p>
          Assume that the theorem holds for all polynomials of degree less than <m>n</m>.
          Since <m>K</m> is a splitting field of <m>p(x)</m>,
          all of the roots of <m>p(x)</m> are in <m>K</m>.
          Choose one of these roots,
          say <m>\alpha</m>, such that <m>E \subset E( \alpha ) \subset K</m>.
          Similarly, we can find a root <m>\beta</m> of <m>q(x)</m> in <m>L</m> such that <m>F \subset F( \beta) \subset L</m>.
          By <xref ref="fields-lemma-field-isomorphism"/>,
          there exists an isomorphism
          <m>\overline{\phi} : E(\alpha ) \rightarrow F( \beta)</m> such that <m>\overline{\phi}( \alpha ) = \beta</m> and
          <m>\overline{\phi}</m> agrees with <m>\phi</m> on <m>E</m>  (see <xref ref="fields-figure-splitting-fields-2"/>).
        </p>

        <figure xml:id="fields-figure-splitting-fields-2">
          <caption>Commutative diagram for <xref ref="fields-theorem-isomorphism-extension"/></caption>
          <p><me>\begin{CD}
          K  @>\psi>>  L  \\
          @VV{\sigma}V        @VV{\tau}V\\
          E(\alpha)     @>\overline{\phi}>>  F(\beta) \\
          @VVV        @VVV\\
          E     @>\phi>>  F
          \end{CD}</me></p>
        </figure>

        <p>
          Now write <m>p(x) = (x - \alpha ) f(x)</m> and <m>q(x) = ( x - \beta) g(x)</m>,
          where the degrees of <m>f(x)</m> and <m>g(x)</m> are less than the degrees of <m>p(x)</m> and <m>q(x)</m>,
          respectively.
          The field extension <m>K</m> is a splitting field for <m>f(x)</m> over <m>E( \alpha)</m>,
          and <m>L</m> is a splitting field for <m>g(x)</m> over <m>F( \beta )</m>.
          By our induction hypothesis there exists an isomorphism
          <m>\psi : K \rightarrow L</m> such that <m>\psi</m> agrees with
          <m>\overline{\phi}</m> on <m>E( \alpha)</m>.
          Hence, there exists an isomorphism
          <m>\psi : K \rightarrow L</m> such that <m>\psi</m> agrees with <m>\phi</m> on <m>E</m>.
        </p>
      </proof>
    </theorem>

    <corollary xml:id="fields-corollary-poly-splitting-field">
      <statement>
        <p>
          Let <m>p(x)</m> be a polynomial in <m>F[x]</m>.
          Then there exists a splitting field <m>K</m> of <m>p(x)</m> that is unique up to isomorphism.
        </p>
      </statement>
    </corollary>
  </section>

  <section xml:id="fields-section-constructions">
    <title>Geometric Constructions</title>
    <introduction>
      <p>
        In ancient Greece, three classic problems were posed.
        These problems are geometric in nature and involve straightedge-and-compass constructions from what is now high school geometry;
        that is, we are allowed to use only a straightedge and compass to solve them.
        The problems can be stated as follows.

        <ol>
          <li>
            <p>
              Given an arbitrary angle,
              can one trisect the angle into three equal subangles using only a straightedge and compass?
            </p>
          </li>

          <li>
            <p>
              Given an arbitrary circle,
              can one construct a square with the same area using only a straightedge and compass?
            </p>
          </li>

          <li>
            <p>
              Given a cube,
              can one construct the edge of another cube having twice the volume of the original?
              Again, we are only allowed to use a straightedge and compass to do the construction.
            </p>
          </li>
        </ol>

        After puzzling mathematicians for over two thousand years,
        each of these constructions was finally shown to be impossible.
        We will use the theory of fields to provide a proof that the solutions do not exist.
        It is quite remarkable that the long-sought solution to each of these three geometric problems came from abstract algebra.
      </p>

      <p>
        First, let us determine more specifically what we mean by a straightedge and compass,
        and also examine the nature of these problems in a bit more depth.
        To begin with, <em>a straightedge is not a ruler</em>.
        We cannot measure arbitrary lengths with a straightedge.
        It is merely a tool for drawing a line through two points.
        The statement that the trisection of an arbitrary angle is impossible means that there is at least one angle that is impossible to trisect with a straightedge-and-compass construction.
        Certainly it is possible to trisect an angle in special cases.
        We can construct a <m>30^\circ</m> angle;
        hence, it is possible to trisect a <m>90^\circ</m> angle.
        However, we will show that it is impossible to construct a <m>20^\circ</m> angle.
        Therefore, we cannot trisect a <m>60^\circ</m> angle.
      </p>
    </introduction>

    <subsection xml:id="fields-subsection-constructible-numbers">
      <title>Constructible Numbers</title>
      <p>
        A real number <m>\alpha</m> is <term>constructible</term><idx><h>Constructible number</h></idx>
        if we can construct a line segment of length
        <m>| \alpha |</m> in a finite number of steps from a segment of unit length by using a straightedge and compass.
      </p>

      <theorem xml:id="fields-theorem-constructible-numbers">
        <statement>
          <p>
            The set of all constructible real numbers forms a subfield <m>F</m> of the field of real numbers.
          </p>
        </statement>

        <proof>
          <p>
            Let <m>\alpha</m> and <m>\beta</m> be constructible numbers.
            We must show that <m>\alpha + \beta</m>,
            <m>\alpha - \beta</m>, <m>\alpha \beta</m>,
            and <m>\alpha / \beta</m> (<m>\beta \neq 0</m>) are also constructible numbers.
            We can assume that both <m>\alpha</m> and <m>\beta</m> are positive with <m>\alpha \gt \beta</m>.
            It is quite obvious how to construct
            <m>\alpha + \beta</m> and <m>\alpha - \beta</m>.
            To find a line segment with length <m>\alpha \beta</m>,
            we assume that <m>\beta \gt 1</m> and construct the triangle in <xref ref="fields-figure-multiplication-construction"/> such that triangles
            <m>\triangle ABC</m> and <m>\triangle ADE</m> are similar.
            Since <m>\alpha / 1 = x / \beta</m>,
            the line segment <m>x</m> has length <m>\alpha \beta</m>.
            A similar construction can be made if <m>\beta \lt 1</m>.
            We will leave it as an exercise to show that the same triangle can be used to construct
            <m>\alpha / \beta</m> for <m>\beta \neq 0</m>.
          </p>
        </proof>
      </theorem>

      <figure xml:id="fields-figure-multiplication-construction">
        <caption>Construction of products</caption>
            <!-- Replaced figure with tikz figure and corrected figure - TWJ 8/19/2010 -->
        <image width="80%" xml:id="fields-multiply">
        	<description>A triangle ADE with B on the line segment AD and C on the line segment AE such that BC is parallel to DE.  The length of AB is 1, the length of AD is beta, the length of AC is alpha and the length of AE is x.</description>
            <latex-image>
                <xi:include href="tikz/fields-multiply.tex" parse="text"/>
            </latex-image>
        </image>

      </figure>

      <lemma>
        <statement>
          <p>
            If <m>\alpha</m> is a constructible number,
            then <m>\sqrt{\alpha}</m> is a constructible number.
          </p>
        </statement>

        <proof>
          <p>
            In <xref ref="fields-figure-root-construction"/> the triangles <m>\triangle ABD</m>,
            <m>\triangle BCD</m>, and <m>\triangle ABC</m> are similar;
            hence, <m>1 /x = x / \alpha</m>, or <m>x^2 = \alpha</m>.
          </p>
        </proof>
      </lemma>

      <figure xml:id="fields-figure-root-construction">
        <caption>Construction of roots</caption>
            <!-- Replaced figure with tikz figure and corrected figure - TWJ 8/19/2010 -->
        <image width="80%" xml:id="fields-root">
        	<description>A semi-circle with AC on the diameter and B on the arc of the circle.  D is on eht line segment AC such that BD is perpendicular to AC.  The length of AD is 1, the length of DC is alpha, and the length of BD is x. Triangles ABC, ADB, and BDC arre similar right triangle.</description>
            <latex-image>
                <xi:include href="tikz/fields-root.tex" parse="text"/>
            </latex-image>
        </image>

      </figure>

      <p>
        By <xref ref="fields-theorem-constructible-numbers"/>,
        we can locate in the plane any point
        <m>P =( p, q)</m> that has rational coordinates <m>p</m> and <m>q</m>.
        We need to know what other points can be constructed with a compass and straightedge from points with rational coordinates.
      </p>

      <lemma>
        <statement>
          <p>
            Let <m>F</m> be a subfield of <m>{\mathbb R}</m>.

            <ol>
              <li>
                <p>
                  If a line contains two points in <m>F</m>,
                  then it has the equation <m>a x + by + c = 0</m>,
                  where <m>a</m>, <m>b</m>, and <m>c</m> are in <m>F</m>.
                </p>
              </li>

              <li>
                <p>
                  If a circle has a center at a point with coordinates in <m>F</m> and a radius that is also in <m>F</m>,
                  then it has the equation <m>x^2 + y^2 + d x + e y + f = 0</m>,
                  where <m>d</m>, <m>e</m>, and <m>f</m> are in <m>F</m>.
                </p>
              </li>
            </ol>
          </p>
        </statement>

        <proof>
          <p>
            Let <m>(x_1, y_1)</m> and <m>(x_2, y_2)</m> be points on a line whose coordinates are in <m>F</m>.
            If <m>x_1 = x_2</m>,
            then the equation of the line through the two points is <m>x - x_1 = 0</m>,
            which has the form <m>a x + by + c = 0</m>.
            If <m>x_1 \neq x_2</m>,
            then the equation of the line through the two points is given by
            <me>
              y - y_1 = \left( \frac{y_2 - y_1}{x_2 - x_1} \right) (x - x_1)
            </me>,
            which can also be put into the proper form.
          </p>

          <p>
            To prove the second part of the lemma,
            suppose that <m>(x_1, y_1)</m> is the center of a circle of radius <m>r</m>.
            Then the circle has the equation
            <me>
              (x - x_1)^2 + (y - y_1)^2 - r^2 = 0
            </me>.
            This equation can easily be put into the appropriate form.
          </p>
        </proof>
      </lemma>

      <p>
        Starting with a field of constructible numbers <m>F</m>,
        we have three possible ways of constructing additional points in
        <m>{\mathbb R}</m> with a compass and straightedge.

        <ol>
          <li>
            <p>
              To find possible new points in <m>{\mathbb R}</m>,
              we can take the intersection of two lines,
              each of which passes through two known points with coordinates in <m>F</m>.
            </p>
          </li>

          <li>
            <p>
              The intersection of a line that passes through two points that have coordinates in <m>F</m> and a circle whose center has coordinates in <m>F</m> with radius of a length in <m>F</m> will give new points in <m>{\mathbb R}</m>.
            </p>
          </li>

          <li>
            <p>
              We can obtain new points in
              <m>{\mathbb R}</m> by intersecting two circles whose centers have coordinates in <m>F</m> and whose radii are of lengths in <m>F</m>.
            </p>
          </li>
        </ol>

        The first case gives no new points in <m>{\mathbb R}</m>,
        since the solution of two equations of the form
        <m>a x + by + c = 0</m> having coefficients in <m>F</m> will always be in <m>F</m>.
        The third case can be reduced to the second case.
        Let
        <md>
          <mrow>x^2 + y^2 + d_1 x + e_1 y + f_1 = 0</mrow>
          <mrow>x^2 + y^2 + d_2 x + e_2 y + f_2 = 0</mrow>
        </md>
 <!-- corrected equation - 28 April 2012 - TWJ -->
        be the equations of two circles,
        where <m>d_i</m>, <m>e_i</m>,
        and <m>f_i</m> are in <m>F</m> for <m>i = 1, 2</m>.
        These circles have the same intersection as the circle
        <me>
          x^2 + y^2 + d_1 x +e_1 x + f_1 = 0
        </me>
        and the line
        <me>
          (d_1 - d_2) x + b(e_2 - e_1)y + (f_2 - f_1) = 0
        </me>.
        The last equation is that of the chord passing through the intersection points of the two circles.
        Hence, the intersection of two circles can be reduced to the case of an intersection of a line with a circle.
      </p>

      <p>
        Considering the case of the intersection of a line and a circle,
        we must determine the nature of the solutions of the equations
        <md>
          <mrow>a x + by + c &amp; = 0</mrow>
          <mrow>x^2 + y^2 + d x + e y + f &amp; = 0</mrow>
        </md>.
        If we eliminate <m>y</m> from these equations,
        we obtain an equation of the form <m>Ax^2 + B x + C = 0</m>,
        where <m>A</m>, <m>B</m>, and <m>C</m> are in <m>F</m>.
        The <m>x</m> coordinate of the intersection points is given by
        <me>
          x = \frac{- B \pm \sqrt{B^2 - 4 A C} }{2 A}
        </me>
        and is in <m>F( \sqrt{\alpha}\, )</m>,
        where <m>\alpha = B^2 - 4 A C \gt 0</m>.
        We have proven the following lemma.
      </p>

      <lemma xml:id="fields-lemma-lines-and-circles">
        <statement>
          <p>
            Let <m>F</m> be a field of constructible numbers.
            Then the points determined by the intersections of lines and circles in <m>F</m> lie in the field
            <m>F( \sqrt{\alpha}\, )</m> for some <m>\alpha</m> in <m>F</m>.
          </p>
        </statement>
      </lemma>

      <theorem>
        <statement>
          <p>
            A real number <m>\alpha</m> is a constructible number if and only if there exists a  sequence of fields
            <me>
              {\mathbb Q} = F_0 \subset F_1 \subset \cdots \subset F_k
            </me>
            such that <m>F_i = F_{i-1}( \sqrt{ \alpha_i}\, )</m> with
            <m>\alpha_i \in F_{i-1}</m> and <m>\alpha \in F_k</m>.
            In particular,
            there exists an integer <m>k \gt 0</m> such that <m>[{\mathbb Q}(\alpha) : {\mathbb Q} ] = 2^k</m>.
          </p>
        </statement>

        <proof>
          <p>
            The existence of the <m>F_i</m>'s and the
            <m>\alpha_i</m>'s is a direct consequence of <xref ref="fields-lemma-lines-and-circles"/> and of the fact that
            <me>
              [F_k: {\mathbb Q}] = [F_k : F_{k - 1}][F_{k - 1} : F_{k - 2}] \cdots [F_1: {\mathbb Q} ] = 2^k
            </me>.
          </p>
        </proof>
      </theorem>
            <!-- Clarified where <m>\alpha_i</m> lives - TWJ 4/23/2011 -->
      <corollary>
        <statement>
          <p>
            The field of all constructible numbers is an algebraic extension of <m>{\mathbb Q}</m>.
          </p>
        </statement>
      </corollary>

      <p>
        As we can see by the field of constructible numbers,
        not every algebraic extension of a field is a finite extension.
      </p>
    </subsection>

    <subsection xml:id="fields-subsection-doubling-the-cube">
      <title>Doubling the Cube and Squaring the Circle</title>
      <p>
        We are now ready to investigate the classical problems of doubling the cube and squaring the circle.
        We can use the field of constructible numbers to show exactly when a particular geometric construction can be accomplished.
      </p>

      <paragraphs>
        <title>Doubling the cube is impossible</title>
        <idx>
        <h>Doubling the cube</h>
        </idx>

        <p>
          Given the edge of the cube,
          it is impossible to construct with a straightedge and compass the edge of the cube that has twice the volume of the original cube.
          Let the original cube have an edge of length <m>1</m> and,
          therefore, a volume of <m>1</m>.
          If we could construct a cube having a volume of <m>2</m>,
          then this new cube would have an edge of length <m>\sqrt[3]{2}</m>.
          However, <m>\sqrt[3]{2}</m> is a zero of the irreducible polynomial <m>x^3 -2</m> over <m>{\mathbb Q}</m>; hence,
          <me>
            [{\mathbb Q}(\sqrt[3]{2}\, ) : {\mathbb Q}] = 3
          </me>
          This is impossible, since <m>3</m> is not a power of <m>2</m>.
        </p>
      </paragraphs>

      <paragraphs>
        <title>Squaring the circle</title>
        <idx>
        <h>Squaring the circle is impossible</h>
        </idx>

        <p>
          Suppose that we have a circle of radius <m>1</m>.
          The area of the circle is <m>\pi</m>;
          therefore, we must be able to construct a square with side <m>\sqrt{\pi}</m>.
          This is impossible since <m>\pi</m> and consequently
          <m>\sqrt{\pi}</m> are both transcendental.
          Therefore, using a straightedge and compass,
          it is not possible to construct a square with the same area as the circle.
        </p>
      </paragraphs>
    </subsection>

    <subsection xml:id="fields-subsection-trisect-angle">
      <title>Trisecting an Angle</title>
      <idx>
      <h>Trisection of an angle</h>
      </idx>

      <p>
        <em>Trisecting an arbitrary angle is impossible</em>.
        We will show that it is impossible to construct a <m>20^\circ</m> angle.
        Consequently, a <m>60^{\circ}</m> angle cannot be trisected.
        We first need to calculate the triple-angle formula for the cosine:
        <md>
          <mrow>\cos 3 \theta &amp; = \cos( 2 \theta + \theta )</mrow>
          <mrow>&amp; = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta</mrow>
          <mrow>&amp; = ( 2 \cos^2 \theta - 1) \cos \theta - 2 \sin^2 \theta \cos \theta</mrow>
          <mrow>&amp; = ( 2 \cos^2 \theta - 1) \cos \theta - 2 (1- \cos^2 \theta) \cos \theta</mrow>
          <mrow>&amp; = 4 \cos^3 \theta - 3 \cos \theta</mrow>
        </md>.
        The angle <m>\theta</m> can be constructed if and only if <m>\alpha = \cos \theta</m> is constructible.
        Let <m>\theta = 20^{\circ}</m>.
        Then <m>\cos 3 \theta = \cos 60^\circ = 1/2</m>.
        By the triple-angle formula for the cosine,
        <me>
          4 \alpha^3 - 3 \alpha = \frac{1}{2}
        </me>.
        Therefore, <m>\alpha</m> is a zero of <m>8 x^3 - 6 x -1</m>.
        This polynomial has no factors in <m>{\mathbb Z}[x]</m>,
        and hence is irreducible over <m>{\mathbb Q}[x]</m>.
        Thus, <m>[{\mathbb Q}( \alpha ) : {\mathbb Q }] = 3</m>.
        Consequently, <m>\alpha</m> cannot be a constructible number.
      </p>

      <paragraphs component="sage-blurb">
        <title>Sage</title>
        <p>
          Extensions of the field of rational numbers are a central object of study in number theory,
          so with Sage's roots in this discipline,
          it is no surprise that there is extensive support for fields and for extensions of the rationals.
          Sage also contains an implementation of the entire field of algebraic numbers,
          with exact representations.
        </p>
      </paragraphs>
    </subsection>

    <subsection xml:id="fields-subsection-historical-note">
      <title>Historical Note</title>
      <p>
        Algebraic number theory uses the tools of algebra to solve problems in number theory.
        Modern algebraic number theory began with Pierre de Fermat<idx><h>Fermat, Pierre de</h></idx>
        (1601<ndash/>1665).
        Certainly we can find many positive integers that satisfy the equation
        <m>x^2 + y^2 = z^2</m>; Fermat conjectured that the equation
        <m>x^n + y^n = z^n</m> has no positive integer solutions for <m>n \geq 3</m>.
        He stated in the margin of his copy of the Latin translation of Diophantus' <pubtitle>Arithmetica</pubtitle> that he had found a marvelous proof of this theorem,
        but that the margin of the book was too narrow to contain it.
        Building on work of other mathematicians,
        it was Andrew Wiles who finally succeeded in proving Fermat's Last Theorem in the 1990s.
        Wiles's achievement was reported on the front page of the <pubtitle>New York Times</pubtitle>.
      </p>

      <p>
        Attempts to prove Fermat's Last Theorem have led to important contributions to algebraic number theory by such notable mathematicians as Leonhard Euler<idx><h>Euler, Leonhard</h></idx>
        (1707<ndash/>1783).
        Significant advances in the understanding of Fermat's Last Theorem were made by Ernst Kummer<idx><h>Kummer, Ernst</h></idx>
        (1810<ndash/>1893).
        Kummer's student, Leopold Kronecker<idx><h>Kronecker, Leopold</h></idx>
        (1823<ndash/>1891), became one of the leading algebraists of the nineteenth century.
        Kronecker's theory of ideals and his study of algebraic number theory added much to the understanding of fields.
      </p>

      <p>
        David Hilbert<idx><h>Hilbert, David</h></idx>
        (1862<ndash/>1943) and Hermann Minkowski<idx><h>Minkowski, Hermann</h></idx>
        (1864<ndash/>1909) were among the mathematicians who led the way in this subject at the beginning of the twentieth century.
        Hilbert and Minkowski were both mathematicians at Göttingen University in Germany.
        Göttingen was truly one the most important centers of mathematical research during the last two centuries.
        The large number of exceptional mathematicians who studied there included Gauss, Dirichlet, Riemann, Dedekind, Noether,
        and Weyl.
      </p>

      <p>
        André Weil<idx><h>Weil, André</h></idx>
        answered questions in number theory using algebraic geometry,
        a field of mathematics that studies geometry by studying commutative rings.
        From about 1955 to 1970, Alexander Grothendieck<idx><h>Grothendieck, Alexander</h></idx>
        dominated the field of algebraic geometry.
        Pierre Deligne,<idx><h>Deligne, Pierre</h></idx>
        a student of Grothendieck,
        solved several of Weil's number-theoretic conjectures.
        One of the most recent contributions to algebra and number theory is Gerd Faltings'<idx><h>Faltings, Gerd</h></idx>
        proof of the Mordell conjecture.<idx><h>Mordell conjecture</h></idx>
        This conjecture of Mordell, now known as Faltings' theorem, essentially says that certain polynomials <m>p(x, y)</m> in
        <m>{\mathbb Z}[x,y]</m> have only a finite number of integral solutions.
      </p>
    </subsection>
  </section>

    <reading-questions>

    <exercise label="rq-fields-simple-extension">
      <statement>
        <p>
          What does it mean for an extension field <m>E</m> of a field <m>F</m> to be a simple extension of <m>F</m>?
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-fields-min-poly">
      <statement>
        <p>
          What is the definition of a minimal polynomial of an element <m>\alpha\in E</m>,
          where <m>E</m> is an extension of <m>F</m>,
          and <m>\alpha</m> is algebraic over <m>F</m>?
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-fields-linear-algebra">
      <statement>
        <p>
          Describe how linear algebra enters into this chapter.
          What critical result relies on a proof that is almost entirely linear algebra?
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-fields-algebraically-closed">
      <statement>
        <p>
          What is the definition of an algebraically closed field?
        </p>
      </statement>
      <response/>
    </exercise>

    <exercise label="rq-fields-splitting-field">
      <statement>
        <p>
          What is a splitting field of a polynomial <m>p(x)\in F[x]</m>?
        </p>
      </statement>
      <response/>
    </exercise>

  </reading-questions>
    <!-- Exercises with Solutions  -->
    <!-- File: fields.xml  -->
    <!-- Title: Fields -->
  <exercises xml:id="fields-exercises" filenamebase="fields">
    <title>Exercises</title>

    <exercise number="1" xml:id="fields-exercise-algebraic-numbers">
      <statement>
        <p>
          Show that each of the following numbers is algebraic over
          <m>{\mathbb Q}</m> by finding the minimal polynomial of the number over <m>{\mathbb Q}</m>.

          <ol>
            <li>
              <p>
                <m>\sqrt{ 1/3 + \sqrt{7} }</m>
              </p>
            </li>

            <li>
              <p>
                <m>\sqrt{ 3} + \sqrt[3]{5}</m>
              </p>
            </li>

            <li>
              <p>
                <m>\sqrt{3} + \sqrt{2}\, i</m>
              </p>
            </li>

            <li>
              <p>
                <m>\cos \theta + i \sin \theta</m> for
                <m>\theta = 2 \pi /n</m> with <m>n \in {\mathbb N}</m>
              </p>
            </li>

            <li>
              <p>
                <m>\sqrt{ \sqrt[3]{2} - i }</m>
              </p>
            </li>
          </ol>
        </p>
      </statement>
      <hint>
        <p>
          (a) <m>x^4 - (2/3) x^2 - 62/9</m>; (c) <m>x^4 - 2 x^2 + 25</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="2" xml:id="fields-exercise-basis-field-extension">
      <statement>
        <p>
          Find a basis for each of the following field extensions.
          What is the degree of each extension?

          <ol>
            <li>
              <p>
                <m>{\mathbb Q}( \sqrt{3}, \sqrt{6}\, )</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}( \sqrt[3]{2}, \sqrt[3]{3}\, )</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}( \sqrt{2}, i)</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}( \sqrt{3}, \sqrt{5}, \sqrt{7}\, )</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}( \sqrt{2}, \root 3 \of{2}\, )</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}( \sqrt{8}\, )</m> over <m>{\mathbb Q}(\sqrt{2}\, )</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}(i, \sqrt{2} +i, \sqrt{3} + i )</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}( \sqrt{2} + \sqrt{5}\, )</m> over <m>{\mathbb Q} ( \sqrt{5}\, )</m>
              </p>
            </li>

            <li>
              <p>
                <m>{\mathbb Q}( \sqrt{2}, \sqrt{6} + \sqrt{10}\, )</m> over <m>{\mathbb Q} ( \sqrt{3} + \sqrt{5}\, )</m>
              </p>
            </li>
          </ol>
        </p>
      </statement>
      <hint>
        <p>
          (a) <m>\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6}\, \}</m>; (c)
          <m>\{ 1, i, \sqrt{2}, \sqrt{2}\, i \}</m>; (e) <m>\{1, 2^{1/6}, 2^{1/3}, 2^{1/2}, 2^{2/3}, 2^{5/6} \}</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="3" xml:id="fields-exercise-splitting-field">
      <statement>
        <p>
          Find the splitting field for each of the following polynomials.

          <ol cols="2">
            <li>
              <p>
                <m>x^4 - 10 x^2 + 21</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>x^4 + 1</m> over <m>{\mathbb Q}</m>
              </p>
            </li>

            <li>
              <p>
                <m>x^3 + 2x + 2</m> over <m>{\mathbb Z}_3</m>
              </p>
            </li>

            <li>
              <p>
                <m>x^3 - 3</m> over <m>{\mathbb Q}</m>
              </p>
            </li>
          </ol>
        </p>
      </statement>
      <hint>
        <p>
          (a) <m>{\mathbb Q}(\sqrt{3}, \sqrt{7}\, )</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="4" xml:id="fields-exercise-root-4-3">
      <statement>
        <p>
          Consider the field extension
          <m>{\mathbb Q}( \sqrt[4]{3}, i )</m> over <m>\mathbb Q</m>.

          <ol>
            <li>
              <p>
                Find a basis for the field extension
                <m>{\mathbb Q}( \sqrt[4]{3}, i )</m> over <m>\mathbb Q</m>.
                Conclude that <m>[{\mathbb Q}( \sqrt[4]{3}, i ): \mathbb Q] = 8</m>.
              </p>
            </li>

            <li>
              <p>
                Find all subfields <m>F</m> of
                <m>{\mathbb Q}( \sqrt[4]{3}, i )</m> such that <m>[F:\mathbb Q] = 2</m>.
              </p>
            </li>

            <li>
              <p>
                Find all subfields <m>F</m> of
                <m>{\mathbb Q}( \sqrt[4]{3}, i )</m> such that <m>[F:\mathbb Q] = 4</m>.
              </p>
            </li>
          </ol>
        </p>
      </statement>
    </exercise>

    <exercise number="5" xml:id="fields-exercise-gf-8">
      <statement>
        <p>
          Show that <m>{\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle</m> is a field with eight elements.
          Construct a multiplication table for the multiplicative group of the field.
        </p>
      </statement>
      <hint>
        <p>
          Use the fact that the elements of
          <m>{\mathbb Z}_2[x]/ \langle x^3 + x + 1 \rangle</m> are 0, 1, <m>\alpha</m>,
          <m>1 + \alpha</m>,
          <m>\alpha^2</m>, <m>1 + \alpha^2</m>, <m>\alpha + \alpha^2</m>,
          <m>1 + \alpha + \alpha^2</m> and the fact that <m>\alpha^3 + \alpha + 1 = 0</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="6" xml:id="fields-exercise-9-gon">
      <statement>
        <p>
          Show that the regular <m>9</m>-gon is not constructible with a straightedge and compass,
          but that the regular <m>20</m>-gon is constructible.
        </p>
      </statement>
    </exercise>

    <exercise number="7" xml:id="fields-exercise-one-degree">
      <statement>
        <p>
          Prove that the cosine of one degree (<m>\cos 1^\circ</m>) is algebraic over
          <m>{\mathbb Q}</m> but not constructible.
        </p>
      </statement>
    </exercise>

    <exercise number="8" xml:id="fields-exercise-3-times-cube">
      <statement>
        <p>
          Can a cube be constructed with three times the volume of a given cube?
        </p>
      </statement>
      <hint>
        <p>
          False.
        </p>
      </hint>
    </exercise>

    <exercise number="9" xml:id="fields-exercise-algebraic-not-finite">
      <statement>
        <p>
          Prove that <m>{\mathbb Q}(\sqrt{3}, \sqrt[4]{3}, \sqrt[8]{3}, \ldots )</m> is an algebraic extension of
          <m>{\mathbb Q}</m> but not a finite extension.
        </p>
      </statement>
    </exercise>

    <exercise number="10" xml:id="fields-exercise-Q-pi-3">
      <statement>
        <p>
          Prove or disprove:
          <m>\pi</m> is algebraic over <m>{\mathbb Q}(\pi^3)</m>.
        </p>
      </statement>
    </exercise>

    <exercise number="11" xml:id="fields-exercise-splitting-field-p-x">
      <statement>
        <p>
          Let <m>p(x)</m> be a nonconstant polynomial of degree <m>n</m> in <m>F[x]</m>.
          Prove that there exists a splitting field <m>E</m> for <m>p(x)</m> such that <m>[E : F] \leq n!</m>.
        </p>
      </statement>
    </exercise>

    <exercise number="12" xml:id="fields-exercise-Q-root-2-Q-root-3">
      <statement>
        <p>
          Prove or disprove:
          <m>{\mathbb Q}( \sqrt{2}\, ) \cong {\mathbb Q}( \sqrt{3}\, )</m>.
        </p>
      </statement>
    </exercise>

    <exercise number="13" xml:id="fields-exercise-isomorphic-fields">
      <statement>
        <p>
          Prove that the fields <m>{\mathbb Q}(\sqrt[4]{3}\, )</m> and
          <m>{\mathbb Q}(\sqrt[4]{3}\, i)</m> are isomorphic but not equal.
        </p>
      </statement>
    </exercise>

    <!--Todo Finish the proof.-->

    <exercise number="14" xml:id="fields-exercise-K-E-F-algebraic">
      <statement>
        <p>
          Let <m>K</m> be an algebraic extension of <m>E</m>,
          and <m>E</m> an algebraic extension of <m>F</m>.
          Prove that <m>K</m> is algebraic over <m>F</m>. [<em> Caution</em>: Do not assume that the extensions are finite.]
        </p>
      </statement>
      <hint>
        <p>
          Suppose that <m>E</m> is algebraic over <m>F</m> and <m>K</m> is algebraic over <m>E</m>.
          Let <m>\alpha \in K</m>.
          It suffices to show that <m>\alpha</m> is algebraic over some finite extension of <m>F</m>.
          Since <m>\alpha</m> is algebraic over <m>E</m>,
          it must be the zero of some polynomial <m>p(x) = \beta_0 + \beta_1 x + \cdots + \beta_n x^n</m> in <m>E[x]</m>.
          Hence <m>\alpha</m> is algebraic over <m>F(\beta_0, \ldots, \beta_n)</m>.
        </p>
      </hint>
    </exercise>

    <exercise number="15" xml:id="fields-exercise-Q-mod-poly-field">
      <statement>
        <p>
          Prove or disprove:
          <m>{\mathbb Z}[x] / \langle x^3 -2 \rangle</m> is a field.
        </p>
      </statement>
    </exercise>

    <!--Todo Finish the proof.-->

    <exercise number="16" xml:id="fields-exercise-char-p">
      <statement>
        <p>
          Let <m>F</m> be a field of characteristic <m>p</m>.
          Prove that <m>p(x) = x^p - a</m> either is irreducible over <m>F</m> or splits in <m>F</m>.
        </p>
      </statement>
    </exercise>

    <!--Todo Finish the proof.-->

    <exercise number="17" xml:id="fields-exercise-p-x-splits-alg-closure">
      <statement>
        <p>
          Let <m>E</m> be the algebraic closure of a field <m>F</m>.
          Prove that every polynomial <m>p(x)</m> in <m>F[x]</m> splits in <m>E</m>.
        </p>
      </statement>
    </exercise>

    <exercise number="18" xml:id="fields-exercise-linear-poly-irred">
      <statement>
        <p>
          If every irreducible polynomial <m>p(x)</m> in <m>F[x]</m> is linear,
          show that <m>F</m> is an algebraically closed field.
        </p>
      </statement>
    </exercise>

    <!--Todo Fix this proof.-->

    <exercise number="19" xml:id="fields-exercise-alpha-beta-construct">
      <statement>
        <p>
          Prove that if <m>\alpha</m> and <m>\beta</m> are constructible numbers such that <m>\beta \neq 0</m>,
          then so is <m>\alpha / \beta</m>.
        </p>
      </statement>
    </exercise>

    <!--Todo Fix this proof.-->

    <exercise number="20" xml:id="fields-exercise-algebraic-elements-over-Q">
      <statement>
        <p>
          Show that the set of all elements in
          <m>{\mathbb R}</m> that are algebraic over
          <m>{\mathbb Q}</m> form a field extension of <m>{\mathbb Q}</m> that is not finite.
        </p>
      </statement>
    </exercise>

    <exercise number="21" xml:id="fields-exercise-auto-permute-zeros">
      <statement>
        <p>
          Let <m>E</m> be an algebraic extension of a field <m>F</m>,
          and let <m>\sigma</m> be an automorphism of <m>E</m> leaving <m>F</m> fixed.
          Let <m>\alpha \in E</m>.
          Show that <m>\sigma</m> induces a permutation of the set of all zeros of the minimal polynomial of <m>\alpha</m> that are in <m>E</m>.
        </p>
      </statement>
    </exercise>

    <exercise number="22" xml:id="fields-exercise-Q-root-3-root-7">
      <statement>
        <p>
          Show that <m>{\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) = {\mathbb Q}( \sqrt{3} + \sqrt{7}\, )</m>.
          Extend your proof to show that <m>{\mathbb Q}( \sqrt{a}, \sqrt{b}\, ) = {\mathbb Q}( \sqrt{a} + \sqrt{b}\, )</m>,
          where <m>a \neq b</m> and neither <m>a</m> nor <m>b</m> is a perfect square.
        </p>
      </statement>
      <hint>
        <p>
          Since <m>\{ 1, \sqrt{3}, \sqrt{7}, \sqrt{21}\, \}</m> is a basis for
          <m>{\mathbb Q}( \sqrt{3}, \sqrt{7}\, )</m> over <m>{\mathbb Q}</m>,
          <m>{\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) \supset {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )</m>.
          Since <m>[{\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) : {\mathbb Q}] = 4</m>,
          <m>[{\mathbb Q}( \sqrt{3} + \sqrt{7}\, ) : {\mathbb Q}] = 2</m> or 4.
          Since the degree of the minimal polynomial of
          <m>\sqrt{3} +\sqrt{7}</m> is 4, <m>{\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) = {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )</m>.
        </p>
      </hint>
    </exercise>

    <!--Todo Finish the proof.-->

    <exercise number="23" xml:id="fields-exercise-extension-degree-2">
      <statement>
        <p>
          Let <m>E</m> be a finite extension of a field <m>F</m>.
          If <m>[E:F] = 2</m>,
          show that <m>E</m> is a splitting field of <m>F</m> for some polynomial <m>f(x) \in F[x]</m>.
        </p>
      </statement>
    </exercise>

    <exercise number="24" xml:id="fields-exercise-Z6">
      <statement>
        <p>
          Prove or disprove: Given a polynomial <m>p(x)</m> in <m>{\mathbb Z}_6[x]</m>,
          it is possible to construct a ring <m>R</m> such that <m>p(x)</m> has a root in <m>R</m>.
        </p>
      </statement>
    </exercise>

    <!--Todo Finish this proof.-->

    <exercise number="25" xml:id="fields-exercise-alpha-alpha-3">
      <statement>
        <p>
          Let <m>E</m> be a field extension of <m>F</m> and <m>\alpha \in E</m>.
          Determine <m>[F(\alpha): F(\alpha^3)]</m>.
        </p>
      </statement>
    </exercise>

    <!--Todo Finish the solution.-->

    <exercise number="26" xml:id="fields-exercise-transcendental">
      <statement>
        <p>
          Let <m>\alpha, \beta</m> be transcendental over <m>{\mathbb Q}</m>.
          Prove that either <m>\alpha \beta</m> or
          <m>\alpha + \beta</m> is also transcendental.
        </p>
      </statement>
    </exercise>

    <exercise number="27" xml:id="fields-exercise-not-transcendental">
      <statement>
        <p>
          Let <m>E</m> be an extension field of <m>F</m> and
          <m>\alpha \in E</m> be transcendental over <m>F</m>.
          Prove that every element in
          <m>F(\alpha)</m> that is not in <m>F</m> is also transcendental over <m>F</m>.
        </p>
      </statement>
      <hint>
        <p>
          Let <m>\beta \in F(\alpha)</m> not in <m>F</m>.
          Then <m>\beta = p(\alpha)/q(\alpha)</m>,
          where <m>p</m> and <m>q</m> are polynomials in <m>\alpha</m> with
          <m>q(\alpha) \neq 0</m> and coefficients in <m>F</m>.
          If <m>\beta</m> is algebraic over <m>F</m>,
          then there exists a polynomial
          <m>f(x) \in F[x]</m> such that <m>f(\beta) = 0</m>.
          Let <m>f(x) = a_0 + a_1 x + \cdots + a_n x^n</m>.
          Then
          <me>
            0 = f(\beta) = f\left( \frac{p(\alpha)}{q(\alpha)} \right) = a_0 + a_1 \left( \frac{p(\alpha)}{q(\alpha)} \right) + \cdots + a_n \left( \frac{p(\alpha)}{q(\alpha)} \right)^n
          </me>.
          Now multiply both sides by
          <m>q(\alpha)^n</m> to show that there is a polynomial in <m>F[x]</m> that has <m>\alpha</m> as a zero.
        </p>
      </hint>
    </exercise>

    <exercise number="28" xml:id="fields-exercise-degree-irred-poly">
      <statement>
        <p>
          Let <m>\alpha</m> be a root of an irreducible monic polynomial
          <m>p(x) \in F[x]</m>, with <m>\deg p = n</m>.
          Prove that <m>[F(\alpha) : F] = n</m>.
        </p>
      </statement>
      <hint>
        <p>
          See the comments following <xref ref="fields-theorem-simple_extension"/>.
        </p>
      </hint>
    </exercise>

  </exercises>

  <references xml:id="fields-references">
    <title>References and Suggested Readings</title>
    <biblio type="raw">
<!-- was [1] --><!-- Out of print  - 19 August 2010 - TWJ -->
      Dean, R. A.
      <title>Elements of Abstract Algebra</title>
      . Wiley, New York, 1966.
    </biblio>

    <biblio type="raw">
<!-- was [2] --><!-- Out of print  - 19 August 2010 - TWJ -->
      Dudley, U.
      <title>A Budget of Trisections</title>. Springer-Verlag, New York, 1987. An interesting and entertaining account of how not to  trisect an angle.
    </biblio>

    <biblio type="raw">
<!-- was [3] --><!-- Reference updated - 19 August 2010 - TWJ -->
      Fraleigh, J. B.
      <title>A First Course in Abstract Algebra</title>. 7th ed. Pearson, Upper Saddle River, NJ, 2003.
    </biblio>

    <biblio type="raw">
<!-- was [4] --><!-- Out of print  - 19 August 2010 - TWJ -->
      Kaplansky, I.
      <title>Fields and Rings</title>, 2nd ed. University of Chicago Press, Chicago, 1972.
    </biblio>

    <biblio type="raw">
<!-- was [5] --><!-- This book is available through Nabu Press, a print on demand publisher that sells through Amazon  - 19 August 2010 - TWJ -->
      Klein, F.
      <title>Famous Problems of Elementary Geometry</title>. Chelsea, New York, 1955.
    </biblio>

    <biblio type="raw">
<!-- was [6] --><!-- Reference added - 19 August 2010 - TWJ -->
      Martin, G.
      <title>Geometric Constructions</title>. Springer, New York, 1998.
    </biblio>

    <biblio type="raw">
<!-- was [7] --><!-- Reference updated - 19 August 2010 - TWJ -->
      H. Pollard and H. G. Diamond.
      <title>Theory of Algebraic Numbers</title>, Dover, Mineola, NY, 2010.
    </biblio>

    <biblio type="raw">
<!-- was [8] --><!-- Out of print  - 19 August 2010 - TWJ -->
      Walker, E. A.
      <title>Introduction to Abstract Algebra</title>. Random House,  New York, 1987. This work contains a proof showing that every field has an algebraic closure.
    </biblio>
  </references>



  <section xml:id="fields-sage" component="sage-info">
    <title>Sage</title>
    <introduction>
      <p>
        In Sage, and other places,
        an extension of the rationals is called a <q>number field.</q>
        They are one of Sage's most mature features.
      </p>
    </introduction>

    <subsection>
      <title>Number Fields</title>
      <p>
        There are several ways to create a number field.
        We are familiar with the syntax where we adjoin an irrational number that we can write with traditional combinations of arithmetic and roots.
      </p>

            <sage>
                <input>
        M.&lt;a&gt; = QQ[sqrt(2)+sqrt(3)]; M
        </input>
                <output>
        Number Field in a with defining polynomial x^4 - 10*x^2 + 1
        with a = 3.146264369941973?
        </output>
            </sage>

      <p>
        We can also specify the element we want to adjoin as the root of a monic irreducible polynomial.
        One approach is to construct the polynomial ring first so that the polynomial has the location of its coefficients specified properly.
      </p>

            <sage>
                <input>
        F.&lt;y&gt; = QQ[]
        p = y^3 - 1/4*y^2 - 1/16*y + 1/4
        p.is_irreducible()
        </input>
                <output>
        True
        </output>
            </sage>
            <sage>
                <input>
        N.&lt;b&gt; = NumberField(p, 'b'); N
        </input>
                <output>
        Number Field in b with
        defining polynomial y^3 - 1/4*y^2 - 1/16*y + 1/4
        </output>
            </sage>

      <p>
        Rather than building the whole polynomial ring,
        we can simply introduce a variable as the generator of a polynomial ring and then create polynomials from this variable.
        This spares us naming the polynomial ring.
        Notice in the example that both instances of <c>z</c> are necessary.
      </p>

            <sage>
                <input>
        z = polygen(QQ, 'z')
        q = z^3 - 1/4*z^2 - 1/16*z + 1/4
        q.parent()
        </input>
                <output>
        Univariate Polynomial Ring in z over Rational Field
        </output>
            </sage>
            <sage>
                <input>
        P.&lt;c&gt; = NumberField(q, 'c'); P
        </input>
                <output>
        Number Field in c with
        defining polynomial z^3 - 1/4*z^2 - 1/16*z + 1/4
        </output>
            </sage>

      <p>
        We can recover the polynomial used to create a number field,
        even if we constructed it by giving an expression for an irrational element.
        In this case,
        the polynomial is the minimal polynomial of the element.
      </p>

            <sage>
                <input>
        M.polynomial()
        </input>
                <output>
        x^4 - 10*x^2 + 1
        </output>
            </sage>
            <sage>
                <input>
        N.polynomial()
        </input>
                <output>
        y^3 - 1/4*y^2 - 1/16*y + 1/4
        </output>
            </sage>

      <p>
        For any element of a number field, Sage will obligingly compute its minimal polynomial.
      </p>

            <sage>
                <input>
        element = -b^2 + 1/3*b + 4
        element.parent()
        </input>
                <output>
        Number Field in b with
        defining polynomial y^3 - 1/4*y^2 - 1/16*y + 1/4
        </output>
            </sage>
            <sage>
                <input>
        r = element.minpoly('t'); r
        </input>
                <output>
        t^3 - 571/48*t^2 + 108389/2304*t - 13345/216
        </output>
            </sage>
            <sage>
                <input>
        r.parent()
        </input>
                <output>
        Univariate Polynomial Ring in t over Rational Field
        </output>
            </sage>
            <sage>
                <input>
        r.subs(t=element)
        </input>
                <output>
        0
        </output>
            </sage>

      <p>
        Substituting <c>element</c> back into the alleged minimal polynomial and getting back zero is not convincing evidence that it is the <em>minimal</em>
        polynomial, but it is heartening.
      </p>
    </subsection>

    <subsection>
      <title>Relative and Absolute Number Fields</title>
      <p>
        With Sage we can adjoin several elements at once and we can build nested towers of number fields.
        Sage uses the term <q>absolute</q>
        to refer to a number field viewed as an extension of the rationals themselves,
        and the term <q>relative</q> to refer to a number field constructed,
        or viewed,
        as an extension of another (nontrivial) number field.
      </p>

            <sage>
                <input>
        A.&lt;a,b&gt; = QQ[sqrt(2), sqrt(3)]
        A
        </input>
                <output>
        Number Field in sqrt2 with defining polynomial x^2 - 2 over
        its base field
        </output>
            </sage>
            <sage>
                <input>
        B = A.base_field(); B
        </input>
                <output>
        Number Field in sqrt3 with defining polynomial x^2 - 3
        </output>
            </sage>
            <sage>
                <input>
        A.is_relative()
        </input>
                <output>
        True
        </output>
            </sage>
            <sage>
                <input>
        B.is_relative()
        </input>
                <output>
        False
        </output>
            </sage>

      <p>
        The number field <c>A</c> has been constructed mathematically as what we would write as <m>{\mathbb Q}\subset{\mathbb Q}[\sqrt{3}]\subset{\mathbb Q}[\sqrt{3},\sqrt{2}]</m>.
        Notice the slight difference in ordering of the elements we are adjoining,
        and notice how the number fields use slightly fancier internal names (<c>sqrt2</c>,
        <c>sqrt3</c>) for the new elements.
      </p>

      <p>
        We can <q>flatten</q> a relative field to view it as an absolute field,
        which may have been our intention from the start.
        Here we create a new number field from <c>A</c> that makes it a pure absolute number field.
      </p>

            <sage>
                <input>
        C.&lt;c&gt; = A.absolute_field()
        C
        </input>
                <output>
        Number Field in c with defining polynomial x^4 - 10*x^2 + 1
        </output>
            </sage>

      <p>
        Once we construct an absolute number field this way,
        we can recover isomorphisms to and from the absolute field.
        Recall that our tower was built with generators <c>a</c> and <c>b</c>,
        while the flattened tower is generated by <c>c</c>.
        The <c>.structure()</c> method returns a pair of functions,
        with the absolute number field as the domain and codomain
        (in that order).
      </p>

            <sage>
                <input>
        fromC, toC = C.structure()
        fromC(c)
        </input>
                <output>
        sqrt2 - sqrt3
        </output>
            </sage>
            <sage>
                <input>
        toC(a)
        </input>
                <output>
        1/2*c^3 - 9/2*c
        </output>
            </sage>
            <sage>
                <input>
        toC(b)
        </input>
                <output>
        1/2*c^3 - 11/2*c
        </output>
            </sage>

      <p>
        This tells us that the single generator of the flattened tower,
        <c>c</c>, is equal to <m>\sqrt{2}-\sqrt{3}</m>, and further,
        each of <m>\sqrt{2}</m> and
        <m>\sqrt{3}</m> can be expressed as polynomial functions of <c>c</c>.
        With these connections,
        you might want to compute the final two expressions in <c>c</c> by hand,
        and appreciate the work Sage does to determine these for us.
        This computation is an example of the conclusion of the upcoming <xref ref="galois-theorem-primitive-element"/>.
      </p>

      <p>
        Many number field methods have both relative and absolute versions,
        and we will also find it more convenient to work in a tower or a flattened version,
        thus the isomorphisms between the two can be invaluable for translating both questions and answers.
      </p>

      <p>
        As a vector space over <m>{\mathbb Q}</m>,
        or over another number field,
        number fields that are finite extensions have a dimension,
        called the degree.
        These are easy to get from Sage,
        though for a relative field,
        we need to be more precise about which degree we desire.
      </p>

            <sage>
                <input>
        B.degree()
        </input>
                <output>
        2
        </output>
            </sage>
            <sage>
                <input>
        A.absolute_degree()
        </input>
                <output>
        4
        </output>
            </sage>
            <sage>
                <input>
        A.relative_degree()
        </input>
                <output>
        2
        </output>
            </sage>
    </subsection>

    <subsection>
      <title>Splitting Fields</title>
      <p>
        Here is a concrete example of how to use Sage to construct a splitting field of a polynomial.
        Consider <m>p(x)=x^4+x^2-1</m>.
        We first build a number field with a single root,
        and then factor the polynomial over this new, larger, field.
      </p>

            <sage>
                <input>
        x = polygen(QQ, 'x')
        p = x^4 + x^2 - 1
        p.parent()
        </input>
                <output>
        Univariate Polynomial Ring in x over Rational Field
        </output>
            </sage>
            <sage>
                <input>
        p.is_irreducible()
        </input>
                <output>
        True
        </output>
            </sage>
            <sage>
                <input>
        M.&lt;a&gt; = NumberField(p, 'a')
        y = polygen(M, 'y')
        p = p.subs(x = y)
        p
        </input>
                <output>
        y^4 + y^2 - 1
        </output>
            </sage>
            <sage>
                <input>
        p.parent()
        </input>
                <output>
        Univariate Polynomial Ring in y over Number Field in a with
        defining polynomial x^4 + x^2 - 1
        </output>
            </sage>
            <sage>
                <input>
        p.factor()
        </input>
                <output>
        (y - a) * (y + a) * (y^2 + a^2 + 1)
        </output>
            </sage>
            <sage>
                <input>
        a^2 + 1 in QQ
        </input>
                <output>
        False
        </output>
            </sage>

      <p>
        So our polynomial factors partially into two linear factors and a quadratic factor.
        But notice that the quadratic factor has a coefficient that is irrational,
        <m>a^2+1</m>,
        so the quadratic factor properly belongs in the polynomial ring over <c>M</c> and not over <c>QQ</c>.
      </p>

      <p>
        We build an extension containing a root of the quadratic factor,
        called <c>q</c> here.
        Then, rather than using the <c>polygen()</c> function,
        we build an entire polynomial ring <c>R</c> over <c>N</c> with the indeterminate <c>z</c>.
        The reason for doing this is we can illustrate how we <q>upgrade</q>
        the polynomial <c>p</c> with the syntax <c>R(p)</c> to go from having coefficients in <c>M</c> to having coefficients in <c>N</c>.
      </p>

            <sage>
                <input>
        q = y^2 + a^2 + 1
        N.&lt;b&gt; = NumberField(q, 'b')
        R.&lt;z&gt; = N[]
        s = R(p)
        s
        </input>
                <output>
        z^4 + z^2 - 1
        </output>
            </sage>
            <sage>
                <input>
        s.parent()
        </input>
                <output>
        Univariate Polynomial Ring in z over Number Field in b with
        defining polynomial y^2 + a^2 + 1 over its base field
        </output>
            </sage>
            <sage>
                <input>
        s.factor()
        </input>
                <output>
        (z + b) * (z + a) * (z - a) * (z - b)
        </output>
            </sage>
            <sage>
                <input>
        a in N, b in N
        </input>
                <output>
        (True, True)
        </output>
            </sage>

      <p>
        So we have a field, <c>N</c>,
        where our polynomial factors into linear factors with coefficients from the field.
        We can get another factorization by converting <c>N</c> to an absolute number field and factoring there.
        We need to recreate the polynomial over <c>N</c>,
        since a substitution will carry coefficients from the wrong ring.
      </p>

            <sage>
                <input>
        P.&lt;c&gt; = N.absolute_field()
        w = polygen(P, 'w')
        p = w^4 + w^2- 1
        p.factor()
        </input>
                <output>
        (w - 7/18966*c^7 + 110/9483*c^5 +  923/9483*c^3 +  3001/6322*c) *
        (w - 7/37932*c^7 +  55/9483*c^5 + 923/18966*c^3 - 3321/12644*c) *
        (w + 7/37932*c^7 -  55/9483*c^5 - 923/18966*c^3 + 3321/12644*c) *
        (w + 7/18966*c^7 - 110/9483*c^5 -  923/9483*c^3 -  3001/6322*c)
        </output>
            </sage>

      <p>
        This is an interesting alternative,
        in that the roots of the polynomial are expressions in terms of the
        <em>single</em> generator <c>c</c>.
        Since the roots involve a seventh power of <c>c</c>, we might suspect
        (but not be certain)
        that the minimal polynomial of <c>c</c> has degree <m>8</m> and that <c>P</c> is a degree <m>8</m> extension of the rationals.
        Indeed <c>P</c>
        (or <c>N</c>)
        is a splitting field for <m>p(x)=x^4+x^2-1</m>.
        The roots are not really as bad as they appear <mdash/> lets convert them back to the relative number field.
      </p>

      <p>
        First we want to rewrite a single factor
        (the first)
        in the form <m>(w-r)</m> to identify the root with the correct signs.
      </p>

            <sage type="display">
                <input>
        (w -  7/18966*c^7 + 110/9483*c^5 + 923/9483*c^3 + 3001/6322*c) =
        (w - (7/18966*c^7 - 110/9483*c^5 - 923/9483*c^3 - 3001/6322*c))
        </input>
            </sage>

      <p>
        With the conversion isomorphisms,
        we can recognize the roots for what they are.
      </p>

            <sage>
                <input>
        fromP, toP = P.structure()
        fromP(7/18966*c^7 - 110/9483*c^5 - 923/9483*c^3 - 3001/6322*c)
        </input>
                <output>
        -b
        </output>
            </sage>

      <p>
        So the rather complicated expression in <c>c</c> is just the negative of the root we adjoined in the second step of constructing the tower of number fields.
        It would be a good exercise to see what happens to the other three roots
        (being careful to get the signs right on each root).
      </p>

      <p>
        This is a good opportunity to illustrate <xref ref="fields-theorem-tower-indices"/>.
      </p>

            <sage>
                <input>
        M.degree()
        </input>
                <output>
        4
        </output>
            </sage>
            <sage>
                <input>
        N.relative_degree()
        </input>
                <output>
        2
        </output>
            </sage>
            <sage>
                <input>
        P.degree()
        </input>
                <output>
        8
        </output>
            </sage>
            <sage>
                <input>
        M.degree()*N.relative_degree() == P.degree()
        </input>
                <output>
        True
        </output>
            </sage>
    </subsection>

    <subsection>
      <title>Algebraic Numbers</title>
      <p>
        <xref ref="fields-corollary-algebraic-numbers-field"/> says that the set of <em>all</em>
        algebraic numbers forms a field.
        This field is implemented in Sage as <c>QQbar</c>.
        This allows for finding roots of polynomials as exact quantities which display as inexact numbers.
      </p>

            <sage>
                <input>
        x = polygen(QQ, 'x')
        p = x^4 + x^2 - 1
        r = p.roots(ring=QQbar); r
        </input>
                <output>
        [(-0.7861513777574233?,  1), (0.7861513777574233?,  1),
         (-1.272019649514069?*I, 1), (1.272019649514069?*I, 1)]
        </output>
            </sage>

      <p>
        So we asked for the roots of a polynomial over the rationals,
        but requested any root that may lie outside the rationals and within the field of algebraic numbers.
        Since the field of algebraic numbers contains all such roots,
        we get a full four roots of the fourth-degree polynomial.
        These roots are computed to lie within an interval and the question mark indicates that the preceding digits are accurate.
        (The integers paired with each root are the multiplicities of that root.
        Use the keyword <c>multiplicities=False</c> to turn them off.)
        Let us take a look under the hood and see how Sage manages the field of algebraic numbers.
      </p>

            <sage>
                <input>
        r1 = r[0][0]; r1
        </input>
                <output>
        -0.7861513777574233?
        </output>
            </sage>
            <sage>
                <input>
        r1.as_number_field_element()
        </input>
                <output>
        (Number Field in a with defining polynomial y^4 - y^2 - 1,
         a^3 - a,
         Ring morphism:
           From: Number Field in a with defining polynomial y^4 - y^2 - 1
           To:   Algebraic Real Field
           Defn: a |--&gt; -1.272019649514069?)
        </output>
            </sage>

      <p>
        Three items are associated with this initial root.
        First is a number field,
        with generator <c>a</c> and a defining polynomial similar to the polynomial we are finding the roots of,
        but not identical.
        Second is an expression in the generator <c>a</c>, which is the actual root.
        You might evaluate this expression with the numerical approximation of <c>a</c>,
        coming next, to verify this is a root.
        Finally, there is a ring homomorphism from the number field to the
        <q>Algebraic Real Field</q>,
        <c>AA</c>, the subfield of <c>QQbar</c> with just real elements,
        which associates the generator <c>a</c> with the number <c>-1.272019649514069?</c>.
        Let us verify, in two ways, that the root given is really a root.
      </p>

            <sage>
                <input>
        r1^4 + r1^2 - 1
        </input>
                <output>
        0
        </output>
            </sage>
            <sage>
                <input>
        N, rexact, homomorphism = r1.as_number_field_element()
        (rexact)^4 + rexact^2 - 1
        </input>
                <output>
        0
        </output>
            </sage>

      <p>
        Now that we have enough theory to understand the field of algebraic numbers,
        and a natural way to represent them exactly,
        you might consider the operations in the field.
        If we take two algebraic numbers and add them together,
        we get another algebraic number (<xref ref="fields-corollary-algebraic-numbers-field"/>).
        So what is the resulting minimal polynomial?
        How is it computed in Sage?
        You could read the source code if you wanted the answer.
      </p>
    </subsection>

    <subsection>
      <title>Geometric Constructions</title>
      <p>
        Sage can do a lot of things,
        but it is not yet able to lay out lines with a straightedge and compass.
        However, we can very quickly determine that trisecting a <m>60</m> degree angle is impossible.
        We adjoin the cosine of a <m>20</m> degree angle
        (in radians)
        to the rationals, determine the degree of the extension,
        and check that it is not an integer power of <m>2</m>.
        In one line.
        Sweet.
      </p>

            <sage>
                <input>
        log(QQ[cos(pi/9)].degree(), 2) in ZZ
        </input>
                <output>
        False
        </output>
            </sage>
    </subsection>
  </section>

  <exercises xml:id="fields-sage-exercises" component="sage-exercises">
    <title>Sage Exercises</title>

    <exercise number="1" xml:id="fields-sage-exercise-Z3">
      <statement>
        <p>
          Create the polynomial <m>p(x)=x^5+2x^4+1</m> over <m>{\mathbb Z}_3</m>.
          Verify that it does not have any linear factors by evaluating <m>p(x)</m> with each element of <m>{\mathbb Z}_3</m>,
          and then check that <m>p(x)</m> is irreducible.
        </p>

        <p>
          Create a finite field of order <m>3^5</m> with the <c>FiniteField()</c> command,
          but include the <c>modulus</c> keyword set to the polynomial <m>p(x)</m> to override the default choice.
        </p>

        <p>
          Recreate <m>p(x)</m> as a polynomial over this field.
          Check each of the <m>3^5 = 243</m> elements of the field to see if they are roots of the polynomial and list all of the elements which are roots.
          Finally, request that Sage give a factorization of <m>p(x)</m> over the field,
          and comment on the relationship between your list of roots and your factorization.
        </p>
      </statement>
    </exercise>

    <exercise number="2" xml:id="fields-sage-exercise-Z3-PID">
      <statement>
        <p>
          This problem continues the previous one.
          Build the ring of polynomials over
          <m>{\mathbb Z}_3</m> and within this ring use <m>p(x)</m> to generate a principal ideal.
          Finally construct the quotient of the polynomial ring by the ideal.
          Since the polynomial is irreducible,
          this quotient ring is a field,
          and by <xref ref="fields-proposition-minimal-poly"/> this quotient ring is isomorphic to the number field in the previous problem.
        </p>

        <p>
          Borrowing from your results in the previous question,
          construct five roots of the polynomial <m>p(x)</m> within this quotient ring,
          but now as expressions in the generator of the quotient ring
          (which is technically a coset).
          Use Sage to verify that they are indeed roots.
          This demonstrates using a quotient ring to create a splitting field for an irreducible polynomial over a finite field.
        </p>
      </statement>
    </exercise>

    <exercise number="3" xml:id="fields-sage-exercise-finite-extension">
      <statement>
        <p>
          The subsection <xref ref="fields-subsection-algebraic-elements" text="title"/> relies on techniques from linear algebra and contains <xref ref="fields-theorem-finite-extension"/>:
          every finite extension is an algebraic extension.
          This exercise will help you understand this proof.
        </p>

        <p>
          The polynomial <m>r(x)=x^4+2x+2</m> is irreducible over the rationals (Eisenstein's criterion with prime <m>p=2</m>).
          Create a number field that contains a root of <m>r(x)</m>.
          By <xref ref="fields-theorem-finite-extension"/>,
          and the remark following,
          every element of this finite field extension is an algebraic number,
          and hence satisfies some polynomial over the base field (it is this polynomial that Sage will produce with the <c>.minpoly()</c> method).
          This exercise will show how we can use just linear algebra to determine this minimal polynomial.
        </p>

        <p>
          Suppose that <c>a</c> is the generator of the number field you just created with <m>r(x)</m>.
          Then we will determine the minimal polynomial of <c>t = 3a + 1</c> using just linear algebra.
          According to the proof, the first five powers of <c>t</c>
          (start counting from zero)
          will be linearly dependent. (Why?) So a nontrivial relation of linear dependence on these powers will provide the coefficients of a polynomial with <c>t</c> as a root.
          Compute these five powers,
          then construct the correct linear system to determine the coefficients of the minimal polynomial,
          solve the system, and suitably interpret its solutions.
        </p>

        <p>
          Hints: The <c>vector()</c> and <c>matrix()</c> commands will create vectors and matrices,
          and the <c>.solve_right()</c> method for matrices can be used to find solutions.
          Given an element of the number field,
          which will necessarily be a polynomial in the generator <c>a</c>,
          the <c>.vector()</c> method of the element will provide the coefficients of this polynomial in a list.
        </p>
      </statement>
    </exercise>

    <exercise number="4" xml:id="fields-sage-exercise-splitting-field">
      <statement>
        <p>
          Construct the splitting field of
          <m>s(x)=x^4+x^2+1</m> and find a factorization of <m>s(x)</m> over this field into linear factors.
        </p>
      </statement>
    </exercise>

    <exercise number="5" xml:id="fields-sage-exercise-irred-poly">
      <statement>
        <p>
          Form the number field, <m>K</m>,
          which contains a root of the irreducible polynomial <m>q(x)=x^3+3x^2+3x-2</m>.
          Name your root <c>a</c>.
          Verify that <m>q(x)</m> factors,
          but does not split, over <m>K</m>.
          With <m>K</m> now as the base field,
          form an extension of <m>K</m> where the quadratic factor of <m>q(x)</m> has a root.
          Name this root <c>b</c>,
          and call this second extension of the tower <m>L</m>.
        </p>

        <p>
          Use <c>M.&lt;c&gt; = L.absolute_field()</c> to form the flattened tower that is the absolute number field <c>M</c>.
          Find the defining polynomial of <c>M</c> with the <c>.polynomial()</c> method.
          From this polynomial, which must have the generator <c>c</c> as a root,
          you should be able to use elementary algebra to write the generator as a fairly simple expression.
        </p>

        <p>
          <m>M</m> should be the splitting field of <m>q(x)</m>.
          To see this, start over,
          and build from scratch a new number field, <m>P</m>,
          using the simple expression for <c>c</c> that you just found.
          Use <c>d</c> as the name of the root used to construct <c>P</c>.
          Since <c>d</c> is a root of the simple minimal polynomial for <c>c</c>,
          you should be able to write an expression for <c>d</c> that a pre-calculus student would recognize.
        </p>

        <p>
          Now factor the original polynomial <m>q(x)</m>
          (with rational coefficients)
          over <m>P</m>, to see the polynomial split
          (as expected).
          Using this factorization,
          and your simple expression for <c>d</c> write simplified expressions for the three roots of <m>q(x)</m>.
          See if you can convert between the two versions of the roots <q>by hand</q>,
          and without using the isomorphisms provided by the <c>.structure()</c> method on <c>M</c>.
        </p>
      </statement>
    </exercise>

  </exercises>
</chapter>