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Abstract Algebra: Theory and Applications

Thomas W. Judson

Chapter 21 Fields

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

Section 21.1 Extension Fields

A field \(E\) is an extension field of a field \(F\) if \(F\) is a subfield of \(E\text{.}\) The field \(F\) is called the base field. We write \(F \subset E\text{.}\)

Example 21.1.

For example, let
\begin{equation*} F = {\mathbb Q}( \sqrt{2}\,) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \} \end{equation*}
and let \(E = {\mathbb Q }( \sqrt{2} + \sqrt{3}\,)\) be the smallest field containing both \({\mathbb Q}\) and \(\sqrt{2} + \sqrt{3}\text{.}\) Both \(E\) and \(F\) are extension fields of the rational numbers. We claim that \(E\) is an extension field of \(F\text{.}\) To see this, we need only show that \(\sqrt{2}\) is in \(E\text{.}\) Since \(\sqrt{2} + \sqrt{3}\) is in \(E\text{,}\) \(1 / (\sqrt{2} + \sqrt{3}\,) = \sqrt{3} - \sqrt{2}\) must also be in \(E\text{.}\) Taking linear combinations of \(\sqrt{2} + \sqrt{3}\) and \(\sqrt{3} - \sqrt{2}\text{,}\) we find that \(\sqrt{2}\) and \(\sqrt{3}\) must both be in \(E\text{.}\)

Example 21.2.

Let \(p(x) = x^2 + x + 1 \in {\mathbb Z}_2[x]\text{.}\) Since neither 0 nor 1 is a root of this polynomial, we know that \(p(x)\) is irreducible over \({\mathbb Z}_2\text{.}\) We will construct a field extension of \({\mathbb Z}_2\) containing an element \(\alpha\) such that \(p(\alpha) = 0\text{.}\) By Theorem 17.22, the ideal \(\langle p(x) \rangle\) generated by \(p(x)\) is maximal; hence, \({\mathbb Z}_2[x] / \langle p(x) \rangle\) is a field. Let \(f(x) + \langle p(x) \rangle\) be an arbitrary element of \({\mathbb Z}_2[x] / \langle p(x) \rangle\text{.}\) By the division algorithm,
\begin{equation*} f(x) = (x^2 + x + 1) q(x) + r(x)\text{,} \end{equation*}
where the degree of \(r(x)\) is less than the degree of \(x^2 + x + 1\text{.}\) Therefore,
\begin{equation*} f(x) + \langle x^2 + x + 1 \rangle = r(x) + \langle x^2 + x + 1 \rangle\text{.} \end{equation*}
The only possibilities for \(r(x)\) are then \(0\text{,}\) \(1\text{,}\) \(x\text{,}\) and \(1 + x\text{.}\) Consequently, \(E = {\mathbb Z}_2[x] / \langle x^2 + x + 1 \rangle\) is a field with four elements and must be a field extension of \({\mathbb Z}_2\text{,}\) containing a zero \(\alpha\) of \(p(x)\text{.}\) The field \({\mathbb Z}_2( \alpha)\) consists of elements
\begin{align*} 0 + 0 \alpha & = 0\\ 1 + 0 \alpha & = 1\\ 0 + 1 \alpha & = \alpha\\ 1 + 1 \alpha & = 1 + \alpha\text{.} \end{align*}
Notice that \({\alpha}^2 + {\alpha} + 1 = 0\text{;}\) hence, if we compute \((1 + \alpha)^2\text{,}\)
\begin{equation*} (1 + \alpha)(1 + \alpha)= 1 + \alpha + \alpha + (\alpha)^2 = \alpha\text{.} \end{equation*}
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 \(E\text{.}\)
\begin{equation*} \begin{array}{c|cccc} + & 0 & 1 & \alpha & 1 + \alpha \\ \hline 0 & 0 & 1 & \alpha & 1 + \alpha \\ 1 & 1 & 0 & 1 + \alpha & \alpha \\ \alpha & \alpha & 1 + \alpha & 0 & 1 \\ 1 + \alpha & 1 + \alpha & \alpha & 1 & 0 \end{array} \end{equation*}
Figure 21.3. Addition Table for \({\mathbb Z}_2(\alpha)\)
\begin{equation*} \begin{array}{c|cccc} \cdot & 0 & 1 & \alpha & 1 + \alpha \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & 1 + \alpha \\ \alpha & 0 & \alpha & 1 + \alpha & 1 \\ 1 + \alpha & 0 & 1 + \alpha & 1 & \alpha \end{array} \end{equation*}
Figure 21.4. Multiplication Table for \({\mathbb Z}_2(\alpha)\)
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.

Theorem 21.5.

Proof.

To prove this theorem, we will employ the method that we used to construct Example 21.2. Clearly, we can assume that \(p(x)\) is an irreducible polynomial. We wish to find an extension field \(E\) of \(F\) containing an element \(\alpha\) such that \(p(\alpha) = 0\text{.}\) The ideal \(\langle p(x) \rangle\) generated by \(p(x)\) is a maximal ideal in \(F[x]\) by Theorem 17.22; hence, \(F[x]/\langle p(x) \rangle\) is a field. We claim that \(E = F[x]/\langle p(x) \rangle\) is the desired field.
We first show that \(E\) is a field extension of \(F\text{.}\) We can define a homomorphism of commutative rings by the map \(\psi:F \rightarrow F[x]/\langle p(x) \rangle\text{,}\) where \(\psi(a) = a + \langle p(x)\rangle\) for \(a \in F\text{.}\) It is easy to check that \(\psi\) is indeed a ring homomorphism. Observe that
\begin{equation*} \psi( a ) + \psi( b ) = (a + \langle p(x) \rangle) + (b + \langle p(x) \rangle) = (a + b) + \langle p(x) \rangle = \psi( a + b ) \end{equation*}
and
\begin{equation*} \psi( a ) \psi( b ) = (a + \langle p(x) \rangle) (b + \langle p(x) \rangle) = ab + \langle p(x) \rangle = \psi( ab )\text{.} \end{equation*}
To prove that \(\psi\) is one-to-one, assume that
\begin{equation*} a + \langle p(x) \rangle = \psi(a) = \psi(b) = b + \langle p(x) \rangle\text{.} \end{equation*}
Then \(a - b\) is a multiple of \(p(x)\text{,}\) since it lives in the ideal \(\langle p(x) \rangle\text{.}\) Since \(p(x)\) is a nonconstant polynomial, the only possibility is that \(a - b = 0\text{.}\) Consequently, \(a = b\) and \(\psi\) is injective. Since \(\psi\) is one-to-one, we can identify \(F\) with the subfield \(\{ a + \langle p(x) \rangle : a \in F \}\) of \(E\) and view \(E\) as an extension field of \(F\text{.}\)
It remains for us to prove that \(p(x)\) has a zero \(\alpha \in E\text{.}\) Set \(\alpha = x + \langle p(x) \rangle\text{.}\) Then \(\alpha\) is in \(E\text{.}\) If \(p(x) = a_0 + a_1 x + \cdots + a_n x^n\text{,}\) then
\begin{align*} p( \alpha ) & = a_0 + a_1( x + \langle p(x) \rangle) + \cdots + a_n ( x + \langle p(x) \rangle)^n\\ & = a_0 + ( a_1 x + \langle p(x) \rangle) + \cdots + (a_n x^n + \langle p(x) \rangle)\\ & = a_0 + a_1 x + \cdots + a_n x^n + \langle p(x) \rangle\\ & = 0 + \langle p(x) \rangle\text{.} \end{align*}
Therefore, we have found an element \(\alpha \in E = F[x]/\langle p(x) \rangle\) such that \(\alpha\) is a zero of \(p(x)\text{.}\)

Example 21.6.

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

Subsection Algebraic Elements

An element \(\alpha\) in an extension field \(E\) over \(F\) is algebraic over \(F\) if \(f(\alpha)=0\) for some nonzero polynomial \(f(x) \in F[x]\text{.}\) An element in \(E\) that is not algebraic over \(F\) is transcendental over \(F\text{.}\) An extension field \(E\) of a field \(F\) is an algebraic extension of \(F\) if every element in \(E\) is algebraic over \(F\text{.}\) If \(E\) is a field extension of \(F\) and \(\alpha_1, \ldots, \alpha_n\) are contained in \(E\text{,}\) we denote the smallest field containing \(F\) and \(\alpha_1, \ldots, \alpha_n\) by \(F( \alpha_1, \ldots, \alpha_n)\text{.}\) If \(E = F( \alpha )\) for some \(\alpha \in E\text{,}\) then \(E\) is a simple extension of \(F\text{.}\)
Example 21.7.
Both \(\sqrt{2}\) and \(i\) are algebraic over \({\mathbb Q}\) since they are zeros of the polynomials \(x^2 -2\) and \(x^2 + 1\text{,}\) respectively. Clearly \(\pi\) and \(e\) are algebraic over the real numbers; however, it is a nontrivial fact that they are transcendental over \({\mathbb Q}\text{.}\) Numbers in \({\mathbb R}\) that are algebraic over \({\mathbb Q}\) are in fact quite rare. Almost all real numbers are transcendental over \({\mathbb Q}\text{.}\)
 18 
(In many cases we do not know whether or not a particular number is transcendental; for example, it is still not known whether \(\pi + e\) is transcendental or algebraic.)
A complex number that is algebraic over \({\mathbb Q}\) is an algebraic number. A transcendental number is an element of \({\mathbb C}\) that is transcendental over \({\mathbb Q}\text{.}\)
Example 21.8.
We will show that \(\sqrt{2 + \sqrt{3} }\) is algebraic over \({\mathbb Q}\text{.}\) If \(\alpha = \sqrt{2 + \sqrt{3} }\text{,}\) then \(\alpha^2 = 2 + \sqrt{3}\text{.}\) Hence, \(\alpha^2 - 2 = \sqrt{3}\) and \(( \alpha^2 - 2)^2 = 3\text{.}\) Since \(\alpha^4 - 4 \alpha^2 + 1 = 0\text{,}\) it must be true that \(\alpha\) is a zero of the polynomial \(x^4 - 4 x^2 + 1 \in {\mathbb Q}[x]\text{.}\)
It is very easy to give an example of an extension field \(E\) over a field \(F\text{,}\) where \(E\) contains an element transcendental over \(F\text{.}\) The following theorem characterizes transcendental extensions.
Theorem 21.9.
Proof.
Let \(\phi_{\alpha} : F[x] \rightarrow E\) be the evaluation homomorphism for \(\alpha\text{.}\) Then \(\alpha\) is transcendental over \(F\) if and only if \(\phi_{\alpha} (p(x)) = p(\alpha) \neq 0\) for all nonconstant polynomials \(p(x) \in F[x]\text{.}\) This is true if and only if \(\ker \phi_{\alpha} = \{ 0 \}\text{;}\) that is, it is true exactly when \(\phi_{\alpha}\) is one-to-one. Hence, \(E\) must contain a copy of \(F[x]\text{.}\) The smallest field containing \(F[x]\) is the field of fractions \(F(x)\text{.}\) By Theorem 18.4, \(E\) must contain a copy of this field.
We have a more interesting situation in the case of algebraic extensions.
Theorem 21.10.
Proof.
Let \(\phi_{\alpha} : F[x] \rightarrow E\) be the evaluation homomorphism. The kernel of \(\phi_{\alpha}\) is a principal ideal generated by some \(p(x) \in F[x]\) with \(\deg p(x) \geq 1\text{.}\) We know that such a polynomial exists, since \(F[x]\) is a principal ideal domain and \(\alpha\) is algebraic. The ideal \(\langle p(x) \rangle\) consists exactly of those elements of \(F[x]\) having \(\alpha\) as a zero. If \(f( \alpha ) = 0\) and \(f(x)\) is not the zero polynomial, then \(f(x) \in \langle p(x) \rangle\) and \(p(x)\) divides \(f(x)\text{.}\) So \(p(x)\) is a polynomial of minimal degree having \(\alpha\) as a zero. Any other polynomial of the same degree having \(\alpha\) as a zero must have the form \(\beta p( x)\) for some \(\beta \in F\text{.}\)
Suppose now that \(p(x) = r(x) s(x)\) is a factorization of \(p(x)\) into polynomials of lower degree. Since \(p( \alpha ) = 0\text{,}\) \(r( \alpha ) s( \alpha ) = 0\text{;}\) consequently, either \(r( \alpha )=0\) or \(s( \alpha ) = 0\text{,}\) which contradicts the fact that \(p\) is of minimal degree. Therefore, \(p(x)\) must be irreducible.
Let \(E\) be an extension field of \(F\) and \(\alpha \in E\) be algebraic over \(F\text{.}\) The unique monic polynomial \(p(x)\) of the last theorem is called the minimal polynomial for \(\alpha\) over \(F\text{.}\) The degree of \(p(x)\) is the degree of \(\alpha\) over \(F\).
Example 21.11.
Let \(f(x) = x^2 - 2\) and \(g(x) = x^4 - 4 x^2 + 1\text{.}\) These polynomials are the minimal polynomials of \(\sqrt{2}\) and \(\sqrt{2 + \sqrt{3} }\text{,}\) respectively.
Proposition 21.12.
Proof.
Let \(\phi_{\alpha} : F[x] \rightarrow E\) be the evaluation homomorphism. The kernel of this map is \(\langle p(x) \rangle\text{,}\) where \(p(x)\) is the minimal polynomial of \(\alpha\text{.}\) By the First Isomorphism Theorem for rings, the image of \(\phi_{\alpha}\) in \(E\) is isomorphic to \(F( \alpha )\) since it contains both \(F\) and \(\alpha\text{.}\)
Theorem 21.13.
Proof.
Since \(\phi_{\alpha} ( F[x] ) \cong F( \alpha )\text{,}\) every element in \(E = F( \alpha )\) must be of the form \(\phi_{\alpha} ( f(x) ) = f( \alpha )\text{,}\) where \(f(\alpha)\) is a polynomial in \(\alpha\) with coefficients in \(F\text{.}\) Let
\begin{equation*} p(x) = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0 \end{equation*}
be the minimal polynomial of \(\alpha\text{.}\) Then \(p( \alpha ) = 0\text{;}\) hence,
\begin{equation*} {\alpha}^n = - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0\text{.} \end{equation*}
Similarly,
\begin{align*} {\alpha}^{n + 1} & = {\alpha} {\alpha}^n\\ & = - a_{n - 1} {\alpha}^n - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}\\ & = - a_{n - 1}( - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0) - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}\text{.} \end{align*}
Continuing in this manner, we can express every monomial \({\alpha}^m\text{,}\) \(m \geq n\text{,}\) as a linear combination of powers of \({\alpha}\) that are less than \(n\text{.}\) Hence, any \(\beta \in F( \alpha )\) can be written as
\begin{equation*} \beta = b_0 + b_1 \alpha + \cdots + b_{n - 1} \alpha^{n - 1}\text{.} \end{equation*}
To show uniqueness, suppose that
\begin{equation*} \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} \end{equation*}
for \(b_i\) and \(c_i\) in \(F\text{.}\) Then
\begin{equation*} g(x) = (b_0 - c_0) + (b_1 - c_1) x + \cdots + (b_{n - 1} - c_{n - 1})x^{n - 1} \end{equation*}
is in \(F[x]\) and \(g( \alpha ) = 0\text{.}\) Since the degree of \(g(x)\) is less than the degree of \(p( x )\text{,}\) the irreducible polynomial of \(\alpha\text{,}\) \(g(x)\) must be the zero polynomial. Consequently,
\begin{equation*} b_0 - c_0 = b_1 - c_1 = \cdots = b_{n - 1} - c_{n - 1} = 0\text{,} \end{equation*}
or \(b_i = c_i\) for \(i = 0, 1, \ldots, n-1\text{.}\) Therefore, we have shown uniqueness.
Example 21.14.
Since \(x^2 + 1\) is irreducible over \({\mathbb R}\text{,}\) \(\langle x^2 + 1 \rangle\) is a maximal ideal in \({\mathbb R}[x]\text{.}\) So \(E = {\mathbb R}[x]/\langle x^2 + 1 \rangle\) is a field extension of \({\mathbb R}\) that contains a root of \(x^2 + 1\text{.}\) Let \(\alpha = x + \langle x^2 + 1 \rangle\text{.}\) We can identify \(E\) with the complex numbers. By Proposition 21.12, \(E\) is isomorphic to \({\mathbb R}( \alpha ) = \{ a + b \alpha : a, b \in {\mathbb R} \}\text{.}\) We know that \(\alpha^2 = -1\) in \(E\text{,}\) since
\begin{align*} \alpha^2 + 1 & = (x + \langle x^2 + 1 \rangle)^2 + (1 + \langle x^2 + 1 \rangle)\\ & = (x^2 + 1) + \langle x^2 + 1 \rangle\\ & = 0\text{.} \end{align*}
Hence, we have an isomorphism of \({\mathbb R}( \alpha )\) with \({\mathbb C}\) defined by the map that takes \(a + b \alpha\) to \(a + bi\text{.}\)
Let \(E\) be a field extension of a field \(F\text{.}\) If we regard \(E\) as a vector space over \(F\text{,}\) 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 \(E\) are vectors; the elements in the field \(F\) are scalars. We can think of addition in \(E\) as adding vectors. When we multiply an element in \(E\) by an element of \(F\text{,}\) we are multiplying a vector by a scalar. This view of field extensions is especially fruitful if a field extension \(E\) of \(F\) is a finite dimensional vector space over \(F\text{,}\) and Theorem 21.13 states that \(E = F(\alpha )\) is finite dimensional vector space over \(F\) with basis \(\{ 1, \alpha, {\alpha}^2, \ldots, {\alpha}^{n - 1} \}\text{.}\)
If an extension field \(E\) of a field \(F\) is a finite dimensional vector space over \(F\) of dimension \(n\text{,}\) then we say that \(E\) is a finite extension of degree \(n\) over \(F\). We write
\begin{equation*} [E:F]= n\text{.} \end{equation*}
to indicate the dimension of \(E\) over \(F\text{.}\)
Theorem 21.15.
Proof.
Let \(\alpha \in E\text{.}\) Since \([E:F] = n\text{,}\) the elements
\begin{equation*} 1, \alpha, \ldots, {\alpha}^n \end{equation*}
cannot be linearly independent. Hence, there exist \(a_i \in F\text{,}\) not all zero, such that
\begin{equation*} a_n {\alpha}^n + a_{n - 1} {\alpha}^{n - 1} + \cdots + a_1 \alpha + a_0 = 0\text{.} \end{equation*}
Therefore,
\begin{equation*} p(x) = a_n x^n + \cdots + a_0 \in F[x] \end{equation*}
is a nonzero polynomial with \(p( \alpha ) = 0\text{.}\)
Remark 21.16.
Theorem 21.15 says that every finite extension of a field \(F\) 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 \({\mathbb R}\) that are algebraic over \({\mathbb Q}\) forms an infinite field extension of \({\mathbb Q}\text{.}\)
The next theorem is a counting theorem, similar to Lagrange’s Theorem in group theory. Theorem 21.17 will prove to be an extremely useful tool in our investigation of finite field extensions.
Theorem 21.17.
Proof.
Let \(\{ \alpha_1, \ldots, \alpha_n \}\) be a basis for \(E\) as a vector space over \(F\) and \(\{ \beta_1, \ldots, \beta_m \}\) be a basis for \(K\) as a vector space over \(E\text{.}\) We claim that \(\{ \alpha_i \beta_j \}\) is a basis for \(K\) over \(F\text{.}\) We will first show that these vectors span \(K\text{.}\) Let \(u \in K\text{.}\) Then \(u = \sum_{j = 1}^{m} b_j \beta_j\) and \(b_j = \sum_{i = 1}^{n} a_{ij} \alpha_i\text{,}\) where \(b_j \in E\) and \(a_{ij} \in F\text{.}\) Then
\begin{equation*} 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 )\text{.} \end{equation*}
So the \(mn\) vectors \(\alpha_i \beta_j\) must span \(K\) over \(F\text{.}\)
We must show that \(\{ \alpha_i \beta_j \}\) are linearly independent. Recall that a set of vectors \(v_1, v_2, \ldots, v_n\) in a vector space \(V\) are linearly independent if
\begin{equation*} c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0 \end{equation*}
implies that
\begin{equation*} c_1 = c_2 = \cdots = c_n = 0\text{.} \end{equation*}
Let
\begin{equation*} u = \sum_{i,j} c_{ij} ( \alpha_i \beta_j ) = 0 \end{equation*}
for \(c_{ij} \in F\text{.}\) We need to prove that all of the \(c_{ij}\)’s are zero. We can rewrite \(u\) as
\begin{equation*} \sum_{j = 1}^{m} \left( \sum_{i = 1}^{n} c_{ij} \alpha_i \right) \beta_j = 0\text{,} \end{equation*}
where \(\sum_i c_{ij} \alpha_i \in E\text{.}\) Since the \(\beta_j\)’s are linearly independent over \(E\text{,}\) it must be the case that
\begin{equation*} \sum_{i = 1}^n c_{ij} \alpha_i = 0 \end{equation*}
for all \(j\text{.}\) However, the \(\alpha_j\) are also linearly independent over \(F\text{.}\) Therefore, \(c_{ij} = 0\) for all \(i\) and \(j\text{,}\) which completes the proof.
The following corollary is easily proved using mathematical induction.
Corollary 21.18.
Corollary 21.19.
Proof.
We know that \(\deg p(x) = [F( \alpha ) : F ]\) and \(\deg q(x) = [F( \beta ) : F ]\text{.}\) Since \(F \subset F( \beta ) \subset F( \alpha )\text{,}\)
\begin{equation*} [F( \alpha ) : F ]= [ F( \alpha ) : F( \beta ) ] [ F( \beta ) : F ]\text{.} \end{equation*}
Example 21.20.
Let us determine an extension field of \({\mathbb Q}\) containing \(\sqrt{3} + \sqrt{5}\text{.}\) It is easy to determine that the minimal polynomial of \(\sqrt{3} + \sqrt{5}\) is \(x^4 - 16 x^2 + 4\text{.}\) It follows that
\begin{equation*} [{\mathbb Q}( \sqrt{3} + \sqrt{5}\, ) : {\mathbb Q} ] = 4\text{.} \end{equation*}
We know that \(\{ 1, \sqrt{3}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}\, )\) over \({\mathbb Q}\text{.}\) Hence, \(\sqrt{3} + \sqrt{5}\) cannot be in \({\mathbb Q}( \sqrt{3}\, )\text{.}\) It follows that \(\sqrt{5}\) cannot be in \({\mathbb Q}( \sqrt{3}\, )\) either. Therefore, \(\{ 1, \sqrt{5}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = ( {\mathbb Q}(\sqrt{3}\, ))( \sqrt{5}\, )\) over \({\mathbb Q}( \sqrt{3}\, )\) and \(\{ 1, \sqrt{3}, \sqrt{5}, \sqrt{3} \sqrt{5} = \sqrt{15}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = {\mathbb Q}( \sqrt{3} + \sqrt{5}\, )\) over \({\mathbb Q}\text{.}\) This example shows that it is possible that some extension \(F( \alpha_1, \ldots, \alpha_n )\) is actually a simple extension of \(F\) even though \(n \gt 1\text{.}\)
Example 21.21.
Let us compute a basis for \({\mathbb Q}( \sqrt[3]{5}, \sqrt{5} \, i )\text{,}\) where \(\sqrt{5}\) is the positive square root of \(5\) and \(\sqrt[3]{5}\) is the real cube root of \(5\text{.}\) We know that \(\sqrt{5} \, i \notin {\mathbb Q}(\sqrt[3]{5}\, )\text{,}\) so
\begin{equation*} [ {\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i) : {\mathbb Q}(\sqrt[3]{5}\, )] = 2\text{.} \end{equation*}
It is easy to determine that \(\{ 1, \sqrt{5}i\, \}\) is a basis for \({\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )\) over \({\mathbb Q}( \sqrt[3]{5}\, )\text{.}\) We also know that \(\{ 1, \sqrt[3]{5}, (\sqrt[3]{5}\, )^2 \}\) is a basis for \({\mathbb Q}(\sqrt[3]{5}\, )\) over \({\mathbb Q}\text{.}\) Hence, a basis for \({\mathbb Q}(\sqrt[3]{5}, \sqrt{5}\, i )\) over \({\mathbb Q}\) is
\begin{equation*} \{ 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 \}\text{.} \end{equation*}
Notice that \(\sqrt[6]{5}\, i\) is a zero of \(x^6 + 5\text{.}\) We can show that this polynomial is irreducible over \({\mathbb Q}\) using Eisenstein’s Criterion, where we let \(p = 5\text{.}\) Consequently,
\begin{equation*} {\mathbb Q} \subset {\mathbb Q}( \sqrt[6]{5}\, i) \subset {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )\text{.} \end{equation*}
But it must be the case that \({\mathbb Q}( \sqrt[6]{5}\, i) = {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i )\text{,}\) since the degree of both of these extensions is \(6\text{.}\)
Theorem 21.22.
Proof.
(1) \(\Rightarrow\) (2). Let \(E\) be a finite algebraic extension of \(F\text{.}\) Then \(E\) is a finite dimensional vector space over \(F\) and there exists a basis consisting of elements \(\alpha_1, \ldots, \alpha_n\) in \(E\) such that \(E = F(\alpha_1, \ldots, \alpha_n)\text{.}\) Each \(\alpha_i\) is algebraic over \(F\) by Theorem 21.15.
(2) \(\Rightarrow\) (3). Suppose that \(E = F(\alpha_1, \ldots, \alpha_n)\text{,}\) where every \(\alpha_i\) is algebraic over \(F\text{.}\) Then
\begin{equation*} E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n - 1} ) \supset \cdots \supset F( \alpha_1 ) \supset F\text{,} \end{equation*}
where each field \(F(\alpha_1, \ldots, \alpha_i)\) is algebraic over \(F(\alpha_1, \ldots, \alpha_{i - 1})\text{.}\)
(3) \(\Rightarrow\) (1). Let
\begin{equation*} E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n - 1} ) \supset \cdots \supset F( \alpha_1 ) \supset F\text{,} \end{equation*}
where each field \(F(\alpha_1, \ldots, \alpha_i)\) is algebraic over \(F(\alpha_1, \ldots, \alpha_{i - 1})\text{.}\) Since
\begin{equation*} F(\alpha_1, \ldots, \alpha_i) = F(\alpha_1, \ldots, \alpha_{i - 1} )(\alpha_i) \end{equation*}
is simple extension and \(\alpha_i\) is algebraic over \(F(\alpha_1, \ldots, \alpha_{i - 1})\text{,}\) it follows that
\begin{equation*} [ F(\alpha_1, \ldots, \alpha_i) : F(\alpha_1, \ldots, \alpha_{i - 1} )] \end{equation*}
is finite for each \(i\text{.}\) Therefore, \([E : F]\) is finite.

Subsection Algebraic Closure

Given a field \(F\text{,}\) the question arises as to whether or not we can find a field \(E\) such that every polynomial \(p(x)\) has a root in \(E\text{.}\) This leads us to the following theorem.
Theorem 21.23.
Proof.
Let \(\alpha, \beta \in E\) be algebraic over \(F\text{.}\) Then \(F( \alpha, \beta )\) is a finite extension of \(F\text{.}\) Since every element of \(F( \alpha, \beta )\) is algebraic over \(F\text{,}\) \(\alpha \pm \beta\text{,}\) \(\alpha \beta\text{,}\) and \(\alpha / \beta\) (\(\beta \neq 0\)) are all algebraic over \(F\text{.}\) Consequently, the set of elements in \(E\) that are algebraic over \(F\) form a field.
Corollary 21.24.
Let \(E\) be a field extension of a field \(F\text{.}\) We define the algebraic closure of a field \(F\) in \(E\) to be the field consisting of all elements in \(E\) that are algebraic over \(F\text{.}\) A field \(F\) is algebraically closed if every nonconstant polynomial in \(F[x]\) has a root in \(F\text{.}\)
Theorem 21.25.
Proof.
Let \(F\) be an algebraically closed field. If \(p(x) \in F[x]\) is a nonconstant polynomial, then \(p(x)\) has a zero in \(F\text{,}\) say \(\alpha\text{.}\) Therefore, \(x-\alpha\) must be a factor of \(p(x)\) and so \(p(x) = (x - \alpha) q_1(x)\text{,}\) where \(\deg q_1(x) = \deg p(x) - 1\text{.}\) Continue this process with \(q_1(x)\) to find a factorization
\begin{equation*} p(x) = (x - \alpha)(x - \beta)q_2(x)\text{,} \end{equation*}
where \(\deg q_2(x) = \deg p(x) -2\text{.}\) The process must eventually stop since the degree of \(p(x)\) is finite.
Conversely, suppose that every nonconstant polynomial \(p(x)\) in \(F[x]\) factors into linear factors. Let \(ax - b\) be such a factor. Then \(p( b/a ) = 0\text{.}\) Consequently, \(F\) is algebraically closed.
Corollary 21.26.
Proof.
Let \(E\) be an algebraic extension of \(F\text{;}\) then \(F \subset E\text{.}\) For \(\alpha \in E\text{,}\) the minimal polynomial of \(\alpha\) is \(x - \alpha\text{.}\) Therefore, \(\alpha \in F\) and \(F = E\text{.}\)
Theorem 21.27.
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.
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 Chapter 23.
Theorem 21.28. Fundamental Theorem of Algebra.

Section 21.2 Splitting Fields

Let \(F\) be a field and \(p(x)\) be a nonconstant polynomial in \(F[x]\text{.}\) We already know that we can find a field extension of \(F\) that contains a root of \(p(x)\text{.}\) However, we would like to know whether an extension \(E\) of \(F\) containing all of the roots of \(p(x)\) exists. In other words, can we find a field extension of \(F\) such that \(p(x)\) factors into a product of linear polynomials? What is the “smallest” extension containing all the roots of \(p(x)\text{?}\)
Let \(F\) be a field and \(p(x) = a_0 + a_1 x + \cdots + a_n x^n\) be a nonconstant polynomial in \(F[x]\text{.}\) An extension field \(E\) of \(F\) is a splitting field of \(p(x)\) if there exist elements \(\alpha_1, \ldots, \alpha_n\) in \(E\) such that \(E = F( \alpha_1, \ldots, \alpha_n )\) and
\begin{equation*} p(x) = ( x - \alpha_1 )(x - \alpha_2) \cdots (x - \alpha_n)\text{.} \end{equation*}
A polynomial \(p(x) \in F[x]\) splits in \(E\) if it is the product of linear factors in \(E[x]\text{.}\)

Example 21.29.

Let \(p(x) = x^4 + 2x^2 - 8\) be in \({\mathbb Q}[x]\text{.}\) Then \(p(x)\) has irreducible factors \(x^2 -2\) and \(x^2 + 4\text{.}\) Therefore, the field \({\mathbb Q}( \sqrt{2}, i )\) is a splitting field for \(p(x)\text{.}\)

Example 21.30.

Let \(p(x) = x^3 - 3\) be in \({\mathbb Q}[x]\text{.}\) Then \(p(x)\) has a root in the field \({\mathbb Q}( \sqrt[3]{3}\, )\text{.}\) However, this field is not a splitting field for \(p(x)\) since the complex cube roots of 3,
\begin{equation*} \frac{ -\sqrt[3]{3} \pm (\sqrt[6]{3}\, )^5 i }{2}\text{,} \end{equation*}
are not in \({\mathbb Q}( \sqrt[3]{3}\, )\text{.}\)

Theorem 21.31.

Proof.

We will use mathematical induction on the degree of \(p(x)\text{.}\) If \(\deg p(x) = 1\text{,}\) then \(p(x)\) is a linear polynomial and \(E = F\text{.}\) Assume that the theorem is true for all polynomials of degree \(k\) with \(1 \leq k \lt n\) and let \(\deg p(x) = n\text{.}\) We can assume that \(p(x)\) is irreducible; otherwise, by our induction hypothesis, we are done. By Theorem 21.5, there exists a field \(K\) such that \(p(x)\) has a zero \(\alpha_1\) in \(K\text{.}\) Hence, \(p(x) = (x - \alpha_1)q(x)\text{,}\) where \(q(x) \in K[x]\text{.}\) Since \(\deg q(x) = n -1\text{,}\) there exists a splitting field \(E \supset K\) of \(q(x)\) that contains the zeros \(\alpha_2, \ldots, \alpha_n\) of \(p(x)\) by our induction hypothesis. Consequently,
\begin{equation*} E = K(\alpha_2, \ldots, \alpha_n) = F(\alpha_1, \ldots, \alpha_n) \end{equation*}
is a splitting field of \(p(x)\text{.}\)
The question of uniqueness now arises for splitting fields. This question is answered in the affirmative. Given two splitting fields \(K\) and \(L\) of a polynomial \(p(x) \in F[x]\text{,}\) there exists a field isomorphism \(\phi : K \rightarrow L\) that preserves \(F\text{.}\) In order to prove this result, we must first prove a lemma.

Lemma 21.32.

Proof.

If \(p(x)\) has degree \(n\text{,}\) then by Theorem 21.13 we can write any element in \(E( \alpha )\) as a linear combination of \(1, \alpha, \ldots, \alpha^{n - 1}\text{.}\) Therefore, the isomorphism that we are seeking must be
\begin{equation*} \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}\text{,} \end{equation*}
where
\begin{equation*} a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1} \end{equation*}
is an element in \(E(\alpha)\text{.}\) The fact that \(\overline{\phi}\) is an isomorphism could be checked by direct computation; however, it is easier to observe that \(\overline{\phi}\) is a composition of maps that we already know to be isomorphisms.
We can extend \(\phi\) to be an isomorphism from \(E[x]\) to \(F[x]\text{,}\) which we will also denote by \(\phi\text{,}\) by letting
\begin{equation*} \phi( a_0 + a_1 x + \cdots + a_n x^n ) = \phi( a_0 ) + \phi(a_1) x + \cdots + \phi(a_n) x^n\text{.} \end{equation*}
This extension agrees with the original isomorphism \(\phi : E \rightarrow F\text{,}\) since constant polynomials get mapped to constant polynomials. By assumption, \(\phi(p(x)) = q(x)\text{;}\) hence, \(\phi\) maps \(\langle p(x) \rangle\) onto \(\langle q(x) \rangle\text{.}\) Consequently, we have an isomorphism \(\psi : E[x] / \langle p(x) \rangle \rightarrow F[x]/\langle q(x) \rangle\text{.}\) By Proposition 21.12, we have isomorphisms \(\sigma: E[x]/\langle p(x) \rangle \rightarrow E(\alpha)\) and \(\tau : F[x]/\langle q(x) \rangle \rightarrow F( \beta )\text{,}\) defined by evaluation at \(\alpha\) and \(\beta\text{,}\) respectively. Therefore, \(\overline{\phi} = \tau \psi \sigma^{-1}\) is the required isomorphism (see Figure 21.33).
\begin{equation*} \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} \end{equation*}
Figure 21.33. Commutative diagram for Lemma 21.32
We leave the proof of uniqueness as a exercise.

Theorem 21.34.

Proof.

We will use mathematical induction on the degree of \(p(x)\text{.}\) We can assume that \(p(x)\) is irreducible over \(E\text{.}\) Therefore, \(q(x)\) is also irreducible over \(F\text{.}\) If \(\deg p(x) = 1\text{,}\) then by the definition of a splitting field, \(K = E\) and \(L = F\) and there is nothing to prove.
Assume that the theorem holds for all polynomials of degree less than \(n\text{.}\) Since \(K\) is a splitting field of \(p(x)\text{,}\) all of the roots of \(p(x)\) are in \(K\text{.}\) Choose one of these roots, say \(\alpha\text{,}\) such that \(E \subset E( \alpha ) \subset K\text{.}\) Similarly, we can find a root \(\beta\) of \(q(x)\) in \(L\) such that \(F \subset F( \beta) \subset L\text{.}\) By Lemma 21.32, there exists an isomorphism \(\overline{\phi} : E(\alpha ) \rightarrow F( \beta)\) such that \(\overline{\phi}( \alpha ) = \beta\) and \(\overline{\phi}\) agrees with \(\phi\) on \(E\) (see Figure 21.35).
\begin{equation*} \begin{CD} K @>\psi>> L \\ @VV{\sigma}V @VV{\tau}V\\ E(\alpha) @>\overline{\phi}>> F(\beta) \\ @VVV @VVV\\ E @>\phi>> F \end{CD} \end{equation*}
Figure 21.35. Commutative diagram for Theorem 21.34
Now write \(p(x) = (x - \alpha ) f(x)\) and \(q(x) = ( x - \beta) g(x)\text{,}\) where the degrees of \(f(x)\) and \(g(x)\) are less than the degrees of \(p(x)\) and \(q(x)\text{,}\) respectively. The field extension \(K\) is a splitting field for \(f(x)\) over \(E( \alpha)\text{,}\) and \(L\) is a splitting field for \(g(x)\) over \(F( \beta )\text{.}\) By our induction hypothesis there exists an isomorphism \(\psi : K \rightarrow L\) such that \(\psi\) agrees with \(\overline{\phi}\) on \(E( \alpha)\text{.}\) Hence, there exists an isomorphism \(\psi : K \rightarrow L\) such that \(\psi\) agrees with \(\phi\) on \(E\text{.}\)

Corollary 21.36.

Section 21.3 Geometric Constructions

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.
  1. Given an arbitrary angle, can one trisect the angle into three equal subangles using only a straightedge and compass?
  2. Given an arbitrary circle, can one construct a square with the same area using only a straightedge and compass?
  3. 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.
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.
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, a straightedge is not a ruler. 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 \(30^\circ\) angle; hence, it is possible to trisect a \(90^\circ\) angle. However, we will show that it is impossible to construct a \(20^\circ\) angle. Therefore, we cannot trisect a \(60^\circ\) angle.

Subsection Constructible Numbers

A real number \(\alpha\) is constructible if we can construct a line segment of length \(| \alpha |\) in a finite number of steps from a segment of unit length by using a straightedge and compass.
Theorem 21.37.
Proof.
Let \(\alpha\) and \(\beta\) be constructible numbers. We must show that \(\alpha + \beta\text{,}\) \(\alpha - \beta\text{,}\) \(\alpha \beta\text{,}\) and \(\alpha / \beta\) (\(\beta \neq 0\)) are also constructible numbers. We can assume that both \(\alpha\) and \(\beta\) are positive with \(\alpha \gt \beta\text{.}\) It is quite obvious how to construct \(\alpha + \beta\) and \(\alpha - \beta\text{.}\) To find a line segment with length \(\alpha \beta\text{,}\) we assume that \(\beta \gt 1\) and construct the triangle in Figure 21.38 such that triangles \(\triangle ABC\) and \(\triangle ADE\) are similar. Since \(\alpha / 1 = x / \beta\text{,}\) the line segment \(x\) has length \(\alpha \beta\text{.}\) A similar construction can be made if \(\beta \lt 1\text{.}\) We will leave it as an exercise to show that the same triangle can be used to construct \(\alpha / \beta\) for \(\beta \neq 0\text{.}\)
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.
Figure 21.38. Construction of products
Lemma 21.39.
Proof.
In Figure 21.40 the triangles \(\triangle ABD\text{,}\) \(\triangle BCD\text{,}\) and \(\triangle ABC\) are similar; hence, \(1 /x = x / \alpha\text{,}\) or \(x^2 = \alpha\text{.}\)
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.
Figure 21.40. Construction of roots
By Theorem 21.37, we can locate in the plane any point \(P =( p, q)\) that has rational coordinates \(p\) and \(q\text{.}\) We need to know what other points can be constructed with a compass and straightedge from points with rational coordinates.
Lemma 21.41.
Proof.
Let \((x_1, y_1)\) and \((x_2, y_2)\) be points on a line whose coordinates are in \(F\text{.}\) If \(x_1 = x_2\text{,}\) then the equation of the line through the two points is \(x - x_1 = 0\text{,}\) which has the form \(a x + by + c = 0\text{.}\) If \(x_1 \neq x_2\text{,}\) then the equation of the line through the two points is given by
\begin{equation*} y - y_1 = \left( \frac{y_2 - y_1}{x_2 - x_1} \right) (x - x_1)\text{,} \end{equation*}
which can also be put into the proper form.
To prove the second part of the lemma, suppose that \((x_1, y_1)\) is the center of a circle of radius \(r\text{.}\) Then the circle has the equation
\begin{equation*} (x - x_1)^2 + (y - y_1)^2 - r^2 = 0\text{.} \end{equation*}
This equation can easily be put into the appropriate form.
Starting with a field of constructible numbers \(F\text{,}\) we have three possible ways of constructing additional points in \({\mathbb R}\) with a compass and straightedge.
  1. To find possible new points in \({\mathbb R}\text{,}\) we can take the intersection of two lines, each of which passes through two known points with coordinates in \(F\text{.}\)
  2. The intersection of a line that passes through two points that have coordinates in \(F\) and a circle whose center has coordinates in \(F\) with radius of a length in \(F\) will give new points in \({\mathbb R}\text{.}\)
  3. We can obtain new points in \({\mathbb R}\) by intersecting two circles whose centers have coordinates in \(F\) and whose radii are of lengths in \(F\text{.}\)
The first case gives no new points in \({\mathbb R}\text{,}\) since the solution of two equations of the form \(a x + by + c = 0\) having coefficients in \(F\) will always be in \(F\text{.}\) The third case can be reduced to the second case. Let
\begin{gather*} x^2 + y^2 + d_1 x + e_1 y + f_1 = 0\\ x^2 + y^2 + d_2 x + e_2 y + f_2 = 0 \end{gather*}
be the equations of two circles, where \(d_i\text{,}\) \(e_i\text{,}\) and \(f_i\) are in \(F\) for \(i = 1, 2\text{.}\) These circles have the same intersection as the circle
\begin{equation*} x^2 + y^2 + d_1 x +e_1 x + f_1 = 0 \end{equation*}
and the line
\begin{equation*} (d_1 - d_2) x + b(e_2 - e_1)y + (f_2 - f_1) = 0\text{.} \end{equation*}
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.
Considering the case of the intersection of a line and a circle, we must determine the nature of the solutions of the equations
\begin{align*} a x + by + c & = 0\\ x^2 + y^2 + d x + e y + f & = 0\text{.} \end{align*}
If we eliminate \(y\) from these equations, we obtain an equation of the form \(Ax^2 + B x + C = 0\text{,}\) where \(A\text{,}\) \(B\text{,}\) and \(C\) are in \(F\text{.}\) The \(x\) coordinate of the intersection points is given by
\begin{equation*} x = \frac{- B \pm \sqrt{B^2 - 4 A C} }{2 A} \end{equation*}
and is in \(F( \sqrt{\alpha}\, )\text{,}\) where \(\alpha = B^2 - 4 A C \gt 0\text{.}\) We have proven the following lemma.
Lemma 21.42.
Theorem 21.43.
Proof.
The existence of the \(F_i\)’s and the \(\alpha_i\)’s is a direct consequence of Lemma 21.42 and of the fact that
\begin{equation*} [F_k: {\mathbb Q}] = [F_k : F_{k - 1}][F_{k - 1} : F_{k - 2}] \cdots [F_1: {\mathbb Q} ] = 2^k\text{.} \end{equation*}
Corollary 21.44.
As we can see by the field of constructible numbers, not every algebraic extension of a field is a finite extension.

Subsection Doubling the Cube and Squaring the Circle

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.
Doubling the cube is impossible.
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 \(1\) and, therefore, a volume of \(1\text{.}\) If we could construct a cube having a volume of \(2\text{,}\) then this new cube would have an edge of length \(\sqrt[3]{2}\text{.}\) However, \(\sqrt[3]{2}\) is a zero of the irreducible polynomial \(x^3 -2\) over \({\mathbb Q}\text{;}\) hence,
\begin{equation*} [{\mathbb Q}(\sqrt[3]{2}\, ) : {\mathbb Q}] = 3 \end{equation*}
This is impossible, since \(3\) is not a power of \(2\text{.}\)
Squaring the circle.
Suppose that we have a circle of radius \(1\text{.}\) The area of the circle is \(\pi\text{;}\) therefore, we must be able to construct a square with side \(\sqrt{\pi}\text{.}\) This is impossible since \(\pi\) and consequently \(\sqrt{\pi}\) are both transcendental. Therefore, using a straightedge and compass, it is not possible to construct a square with the same area as the circle.

Subsection Trisecting an Angle

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

Subsection Historical Note

Algebraic number theory uses the tools of algebra to solve problems in number theory. Modern algebraic number theory began with Pierre de Fermat (1601–1665). Certainly we can find many positive integers that satisfy the equation \(x^2 + y^2 = z^2\text{;}\) Fermat conjectured that the equation \(x^n + y^n = z^n\) has no positive integer solutions for \(n \geq 3\text{.}\) He stated in the margin of his copy of the Latin translation of Diophantus’ Arithmetica 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 New York Times.
Attempts to prove Fermat’s Last Theorem have led to important contributions to algebraic number theory by such notable mathematicians as Leonhard Euler (1707–1783). Significant advances in the understanding of Fermat’s Last Theorem were made by Ernst Kummer (1810–1893). Kummer’s student, Leopold Kronecker (1823–1891), 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.
David Hilbert (1862–1943) and Hermann Minkowski (1864–1909) 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.
André Weil 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 dominated the field of algebraic geometry. Pierre Deligne, 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’ proof of the Mordell conjecture. This conjecture of Mordell, now known as Faltings’ theorem, essentially says that certain polynomials \(p(x, y)\) in \({\mathbb Z}[x,y]\) have only a finite number of integral solutions.

Reading Questions 21.4 Reading Questions

1.

What does it mean for an extension field \(E\) of a field \(F\) to be a simple extension of \(F\text{?}\)

2.

What is the definition of a minimal polynomial of an element \(\alpha\in E\text{,}\) where \(E\) is an extension of \(F\text{,}\) and \(\alpha\) is algebraic over \(F\text{?}\)

3.

Describe how linear algebra enters into this chapter. What critical result relies on a proof that is almost entirely linear algebra?

4.

What is the definition of an algebraically closed field?

5.

What is a splitting field of a polynomial \(p(x)\in F[x]\text{?}\)

Exercises 21.5 Exercises

1.

Show that each of the following numbers is algebraic over \({\mathbb Q}\) by finding the minimal polynomial of the number over \({\mathbb Q}\text{.}\)
  1. \(\displaystyle \sqrt{ 1/3 + \sqrt{7} }\)
  2. \(\displaystyle \sqrt{ 3} + \sqrt[3]{5}\)
  3. \(\displaystyle \sqrt{3} + \sqrt{2}\, i\)
  4. \(\cos \theta + i \sin \theta\) for \(\theta = 2 \pi /n\) with \(n \in {\mathbb N}\)
  5. \(\displaystyle \sqrt{ \sqrt[3]{2} - i }\)

2.

Find a basis for each of the following field extensions. What is the degree of each extension?
  1. \({\mathbb Q}( \sqrt{3}, \sqrt{6}\, )\) over \({\mathbb Q}\)
  2. \({\mathbb Q}( \sqrt[3]{2}, \sqrt[3]{3}\, )\) over \({\mathbb Q}\)
  3. \({\mathbb Q}( \sqrt{2}, i)\) over \({\mathbb Q}\)
  4. \({\mathbb Q}( \sqrt{3}, \sqrt{5}, \sqrt{7}\, )\) over \({\mathbb Q}\)
  5. \({\mathbb Q}( \sqrt{2}, \root 3 \of{2}\, )\) over \({\mathbb Q}\)
  6. \({\mathbb Q}( \sqrt{8}\, )\) over \({\mathbb Q}(\sqrt{2}\, )\)
  7. \({\mathbb Q}(i, \sqrt{2} +i, \sqrt{3} + i )\) over \({\mathbb Q}\)
  8. \({\mathbb Q}( \sqrt{2} + \sqrt{5}\, )\) over \({\mathbb Q} ( \sqrt{5}\, )\)
  9. \({\mathbb Q}( \sqrt{2}, \sqrt{6} + \sqrt{10}\, )\) over \({\mathbb Q} ( \sqrt{3} + \sqrt{5}\, )\)

3.

Find the splitting field for each of the following polynomials.
  1. \(x^4 - 10 x^2 + 21\) over \({\mathbb Q}\)
  2. \(x^4 + 1\) over \({\mathbb Q}\)
  3. \(x^3 + 2x + 2\) over \({\mathbb Z}_3\)
  4. \(x^3 - 3\) over \({\mathbb Q}\)

4.

Consider the field extension \({\mathbb Q}( \sqrt[4]{3}, i )\) over \(\mathbb Q\text{.}\)
  1. Find a basis for the field extension \({\mathbb Q}( \sqrt[4]{3}, i )\) over \(\mathbb Q\text{.}\) Conclude that \([{\mathbb Q}( \sqrt[4]{3}, i ): \mathbb Q] = 8\text{.}\)
  2. Find all subfields \(F\) of \({\mathbb Q}( \sqrt[4]{3}, i )\) such that \([F:\mathbb Q] = 2\text{.}\)
  3. Find all subfields \(F\) of \({\mathbb Q}( \sqrt[4]{3}, i )\) such that \([F:\mathbb Q] = 4\text{.}\)

5.

Show that \({\mathbb Z}_2[x] / \langle x^3 + x + 1 \rangle\) is a field with eight elements. Construct a multiplication table for the multiplicative group of the field.

6.

Show that the regular \(9\)-gon is not constructible with a straightedge and compass, but that the regular \(20\)-gon is constructible.

7.

Prove that the cosine of one degree (\(\cos 1^\circ\)) is algebraic over \({\mathbb Q}\) but not constructible.

8.

Can a cube be constructed with three times the volume of a given cube?

9.

Prove that \({\mathbb Q}(\sqrt{3}, \sqrt[4]{3}, \sqrt[8]{3}, \ldots )\) is an algebraic extension of \({\mathbb Q}\) but not a finite extension.

10.

Prove or disprove: \(\pi\) is algebraic over \({\mathbb Q}(\pi^3)\text{.}\)

11.

Let \(p(x)\) be a nonconstant polynomial of degree \(n\) in \(F[x]\text{.}\) Prove that there exists a splitting field \(E\) for \(p(x)\) such that \([E : F] \leq n!\text{.}\)

12.

Prove or disprove: \({\mathbb Q}( \sqrt{2}\, ) \cong {\mathbb Q}( \sqrt{3}\, )\text{.}\)

13.

Prove that the fields \({\mathbb Q}(\sqrt[4]{3}\, )\) and \({\mathbb Q}(\sqrt[4]{3}\, i)\) are isomorphic but not equal.

14.

Let \(K\) be an algebraic extension of \(E\text{,}\) and \(E\) an algebraic extension of \(F\text{.}\) Prove that \(K\) is algebraic over \(F\text{.}\) [Caution: Do not assume that the extensions are finite.]

15.

Prove or disprove: \({\mathbb Z}[x] / \langle x^3 -2 \rangle\) is a field.

16.

Let \(F\) be a field of characteristic \(p\text{.}\) Prove that \(p(x) = x^p - a\) either is irreducible over \(F\) or splits in \(F\text{.}\)

17.

Let \(E\) be the algebraic closure of a field \(F\text{.}\) Prove that every polynomial \(p(x)\) in \(F[x]\) splits in \(E\text{.}\)

18.

If every irreducible polynomial \(p(x)\) in \(F[x]\) is linear, show that \(F\) is an algebraically closed field.

19.

Prove that if \(\alpha\) and \(\beta\) are constructible numbers such that \(\beta \neq 0\text{,}\) then so is \(\alpha / \beta\text{.}\)

20.

Show that the set of all elements in \({\mathbb R}\) that are algebraic over \({\mathbb Q}\) form a field extension of \({\mathbb Q}\) that is not finite.

21.

Let \(E\) be an algebraic extension of a field \(F\text{,}\) and let \(\sigma\) be an automorphism of \(E\) leaving \(F\) fixed. Let \(\alpha \in E\text{.}\) Show that \(\sigma\) induces a permutation of the set of all zeros of the minimal polynomial of \(\alpha\) that are in \(E\text{.}\)

22.

Show that \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) = {\mathbb Q}( \sqrt{3} + \sqrt{7}\, )\text{.}\) Extend your proof to show that \({\mathbb Q}( \sqrt{a}, \sqrt{b}\, ) = {\mathbb Q}( \sqrt{a} + \sqrt{b}\, )\text{,}\) where \(a \neq b\) and neither \(a\) nor \(b\) is a perfect square.

23.

Let \(E\) be a finite extension of a field \(F\text{.}\) If \([E:F] = 2\text{,}\) show that \(E\) is a splitting field of \(F\) for some polynomial \(f(x) \in F[x]\text{.}\)

24.

Prove or disprove: Given a polynomial \(p(x)\) in \({\mathbb Z}_6[x]\text{,}\) it is possible to construct a ring \(R\) such that \(p(x)\) has a root in \(R\text{.}\)

25.

Let \(E\) be a field extension of \(F\) and \(\alpha \in E\text{.}\) Determine \([F(\alpha): F(\alpha^3)]\text{.}\)

26.

Let \(\alpha, \beta\) be transcendental over \({\mathbb Q}\text{.}\) Prove that either \(\alpha \beta\) or \(\alpha + \beta\) is also transcendental.

27.

Let \(E\) be an extension field of \(F\) and \(\alpha \in E\) be transcendental over \(F\text{.}\) Prove that every element in \(F(\alpha)\) that is not in \(F\) is also transcendental over \(F\text{.}\)

28.

Let \(\alpha\) be a root of an irreducible monic polynomial \(p(x) \in F[x]\text{,}\) with \(\deg p = n\text{.}\) Prove that \([F(\alpha) : F] = n\text{.}\)

References 21.6 References and Suggested Readings

[1]
  
Dean, R. A. Elements of Abstract Algebra . Wiley, New York, 1966.
[2]
  
Dudley, U. A Budget of Trisections. Springer-Verlag, New York, 1987. An interesting and entertaining account of how not to trisect an angle.
[3]
  
Fraleigh, J. B. A First Course in Abstract Algebra. 7th ed. Pearson, Upper Saddle River, NJ, 2003.
[4]
  
Kaplansky, I. Fields and Rings, 2nd ed. University of Chicago Press, Chicago, 1972.
[5]
  
Klein, F. Famous Problems of Elementary Geometry. Chelsea, New York, 1955.
[6]
  
Martin, G. Geometric Constructions. Springer, New York, 1998.
[7]
  
H. Pollard and H. G. Diamond. Theory of Algebraic Numbers, Dover, Mineola, NY, 2010.
[8]
  
Walker, E. A. Introduction to Abstract Algebra. Random House, New York, 1987. This work contains a proof showing that every field has an algebraic closure.

Section 21.7 Sage

In Sage, and other places, an extension of the rationals is called a “number field.” They are one of Sage’s most mature features.

Subsection Number Fields

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.
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.
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 z are necessary.
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.
For any element of a number field, Sage will obligingly compute its minimal polynomial.
Substituting element back into the alleged minimal polynomial and getting back zero is not convincing evidence that it is the minimal polynomial, but it is heartening.

Subsection Relative and Absolute Number Fields

With Sage we can adjoin several elements at once and we can build nested towers of number fields. Sage uses the term “absolute” to refer to a number field viewed as an extension of the rationals themselves, and the term “relative” to refer to a number field constructed, or viewed, as an extension of another (nontrivial) number field.
The number field A has been constructed mathematically as what we would write as \({\mathbb Q}\subset{\mathbb Q}[\sqrt{3}]\subset{\mathbb Q}[\sqrt{3},\sqrt{2}]\text{.}\) Notice the slight difference in ordering of the elements we are adjoining, and notice how the number fields use slightly fancier internal names (sqrt2, sqrt3) for the new elements.
We can “flatten” 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 A that makes it a pure absolute number field.
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 a and b, while the flattened tower is generated by c. The .structure() method returns a pair of functions, with the absolute number field as the domain and codomain (in that order).
This tells us that the single generator of the flattened tower, c, is equal to \(\sqrt{2}-\sqrt{3}\text{,}\) and further, each of \(\sqrt{2}\) and \(\sqrt{3}\) can be expressed as polynomial functions of c. With these connections, you might want to compute the final two expressions in 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 Theorem 23.13.
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.
As a vector space over \({\mathbb Q}\text{,}\) 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.

Subsection Splitting Fields

Here is a concrete example of how to use Sage to construct a splitting field of a polynomial. Consider \(p(x)=x^4+x^2-1\text{.}\) We first build a number field with a single root, and then factor the polynomial over this new, larger, field.
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, \(a^2+1\text{,}\) so the quadratic factor properly belongs in the polynomial ring over M and not over QQ.
We build an extension containing a root of the quadratic factor, called q here. Then, rather than using the polygen() function, we build an entire polynomial ring R over N with the indeterminate z. The reason for doing this is we can illustrate how we “upgrade” the polynomial p with the syntax R(p) to go from having coefficients in M to having coefficients in N.
So we have a field, N, where our polynomial factors into linear factors with coefficients from the field. We can get another factorization by converting N to an absolute number field and factoring there. We need to recreate the polynomial over N, since a substitution will carry coefficients from the wrong ring.
This is an interesting alternative, in that the roots of the polynomial are expressions in terms of the single generator c. Since the roots involve a seventh power of c, we might suspect (but not be certain) that the minimal polynomial of c has degree \(8\) and that P is a degree \(8\) extension of the rationals. Indeed P (or N) is a splitting field for \(p(x)=x^4+x^2-1\text{.}\) The roots are not really as bad as they appear — lets convert them back to the relative number field.
First we want to rewrite a single factor (the first) in the form \((w-r)\) to identify the root with the correct signs.
With the conversion isomorphisms, we can recognize the roots for what they are.
So the rather complicated expression in 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).
This is a good opportunity to illustrate Theorem 21.17.

Subsection Algebraic Numbers

Corollary 21.24 says that the set of all algebraic numbers forms a field. This field is implemented in Sage as QQbar. This allows for finding roots of polynomials as exact quantities which display as inexact numbers.
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 multiplicities=False to turn them off.) Let us take a look under the hood and see how Sage manages the field of algebraic numbers.
Three items are associated with this initial root. First is a number field, with generator a and a defining polynomial similar to the polynomial we are finding the roots of, but not identical. Second is an expression in the generator a, which is the actual root. You might evaluate this expression with the numerical approximation of a, coming next, to verify this is a root. Finally, there is a ring homomorphism from the number field to the “Algebraic Real Field”, AA, the subfield of QQbar with just real elements, which associates the generator a with the number -1.272019649514069?. Let us verify, in two ways, that the root given is really a root.
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 (Corollary 21.24). So what is the resulting minimal polynomial? How is it computed in Sage? You could read the source code if you wanted the answer.

Subsection Geometric Constructions

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 \(60\) degree angle is impossible. We adjoin the cosine of a \(20\) degree angle (in radians) to the rationals, determine the degree of the extension, and check that it is not an integer power of \(2\text{.}\) In one line. Sweet.

Exercises 21.8 Sage Exercises

1.

Create the polynomial \(p(x)=x^5+2x^4+1\) over \({\mathbb Z}_3\text{.}\) Verify that it does not have any linear factors by evaluating \(p(x)\) with each element of \({\mathbb Z}_3\text{,}\) and then check that \(p(x)\) is irreducible.
Create a finite field of order \(3^5\) with the FiniteField() command, but include the modulus keyword set to the polynomial \(p(x)\) to override the default choice.
Recreate \(p(x)\) as a polynomial over this field. Check each of the \(3^5 = 243\) 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 \(p(x)\) over the field, and comment on the relationship between your list of roots and your factorization.

2.

This problem continues the previous one. Build the ring of polynomials over \({\mathbb Z}_3\) and within this ring use \(p(x)\) 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 Proposition 21.12 this quotient ring is isomorphic to the number field in the previous problem.
Borrowing from your results in the previous question, construct five roots of the polynomial \(p(x)\) 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.

3.

The subsection Algebraic Elements relies on techniques from linear algebra and contains Theorem 21.15: every finite extension is an algebraic extension. This exercise will help you understand this proof.
The polynomial \(r(x)=x^4+2x+2\) is irreducible over the rationals (Eisenstein’s criterion with prime \(p=2\)). Create a number field that contains a root of \(r(x)\text{.}\) By Theorem 21.15, 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 .minpoly() method). This exercise will show how we can use just linear algebra to determine this minimal polynomial.
Suppose that a is the generator of the number field you just created with \(r(x)\text{.}\) Then we will determine the minimal polynomial of t = 3a + 1 using just linear algebra. According to the proof, the first five powers of t (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 t 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.
Hints: The vector() and matrix() commands will create vectors and matrices, and the .solve_right() 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 a, the .vector() method of the element will provide the coefficients of this polynomial in a list.

4.

Construct the splitting field of \(s(x)=x^4+x^2+1\) and find a factorization of \(s(x)\) over this field into linear factors.

5.

Form the number field, \(K\text{,}\) which contains a root of the irreducible polynomial \(q(x)=x^3+3x^2+3x-2\text{.}\) Name your root a. Verify that \(q(x)\) factors, but does not split, over \(K\text{.}\) With \(K\) now as the base field, form an extension of \(K\) where the quadratic factor of \(q(x)\) has a root. Name this root b, and call this second extension of the tower \(L\text{.}\)
Use M.<c> = L.absolute_field() to form the flattened tower that is the absolute number field M. Find the defining polynomial of M with the .polynomial() method. From this polynomial, which must have the generator c as a root, you should be able to use elementary algebra to write the generator as a fairly simple expression.
\(M\) should be the splitting field of \(q(x)\text{.}\) To see this, start over, and build from scratch a new number field, \(P\text{,}\) using the simple expression for c that you just found. Use d as the name of the root used to construct P. Since d is a root of the simple minimal polynomial for c, you should be able to write an expression for d that a pre-calculus student would recognize.
Now factor the original polynomial \(q(x)\) (with rational coefficients) over \(P\text{,}\) to see the polynomial split (as expected). Using this factorization, and your simple expression for d write simplified expressions for the three roots of \(q(x)\text{.}\) See if you can convert between the two versions of the roots “by hand”, and without using the isomorphisms provided by the .structure() method on M.
The probability that a real number chosen at random from the interval \([0, 1]\) will be transcendental over the rational numbers is one.
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