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Section23.1Field Automorphisms

Our first task is to establish a link between group theory and field theory by examining automorphisms of fields.

If \(\sigma\) and \(\tau\) are automorphisms of \(F\text{,}\) then so are \(\sigma \tau\) and \(\sigma^{-1}\text{.}\) The identity is certainly an automorphism; hence, the set of all automorphisms of a field \(F\) is indeed a group.

We need only show that the set of automorphisms of \(E\) that fix \(F\) elementwise is a subgroup of the group of all automorphisms of \(E\text{.}\) Let \(\sigma\) and \(\tau\) be two automorphisms of \(E\) such that \(\sigma( \alpha ) = \alpha\) and \(\tau( \alpha ) = \alpha\) for all \(\alpha \in F\text{.}\) Then \(\sigma \tau( \alpha ) = \sigma( \alpha) = \alpha\) and \(\sigma^{-1}( \alpha ) = \alpha\text{.}\) Since the identity fixes every element of \(E\text{,}\) the set of automorphisms of \(E\) that leave elements of \(F\) fixed is a subgroup of the entire group of automorphisms of \(E\text{.}\)

Let \(E\) be a field extension of \(F\text{.}\) We will denote the full group of automorphisms of \(E\) by \(\aut(E)\text{.}\) We define the Galois group of \(E\) over \(F\) to be the group of automorphisms of \(E\) that fix \(F\) elementwise; that is,

\begin{equation*} G(E/F) = \{ \sigma \in \aut(E) : \sigma(\alpha) = \alpha \text{ for all } \alpha \in F \}. \end{equation*}

If \(f(x)\) is a polynomial in \(F[x]\) and \(E\) is the splitting field of \(f(x)\) over \(F\text{,}\) then we define the Galois group of \(f(x)\) to be \(G(E/F)\text{.}\)

Example23.3

Complex conjugation, defined by \(\sigma : a + bi \mapsto a - bi\text{,}\) is an automorphism of the complex numbers. Since

\begin{equation*} \sigma(a) = \sigma(a + 0i) = a - 0i = a, \end{equation*}

the automorphism defined by complex conjugation must be in \(G( {\mathbb C} / {\mathbb R} )\text{.}\)

Example23.4

Consider the fields \({\mathbb Q} \subset {\mathbb Q}(\sqrt{5}\, ) \subset {\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\text{.}\) Then for \(a, b \in {\mathbb Q}( \sqrt{5}\, )\text{,}\)

\begin{equation*} \sigma( a + b \sqrt{3}\, ) = a - b \sqrt{3} \end{equation*}

is an automorphism of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) leaving \({\mathbb Q}( \sqrt{5}\, )\) fixed. Similarly,

\begin{equation*} \tau( a + b \sqrt{5}\, ) = a - b \sqrt{5} \end{equation*}

is an automorphism of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) leaving \({\mathbb Q}( \sqrt{3}\, )\) fixed. The automorphism \(\mu = \sigma \tau\) moves both \(\sqrt{3}\) and \(\sqrt{5}\text{.}\) It will soon be clear that \(\{ \identity, \sigma, \tau, \mu \}\) is the Galois group of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) over \({\mathbb Q}\text{.}\) The following table shows that this group is isomorphic to \({\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\)

\begin{equation*} \begin{array}{c|cccc} & \identity & \sigma & \tau & \mu \\ \hline \identity & \identity & \sigma & \tau & \mu \\ \sigma & \sigma & \identity & \mu & \tau \\ \tau & \tau & \mu & \identity & \sigma \\ \mu & \mu & \tau & \sigma & \identity \end{array} \end{equation*}

We may also regard the field \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\) as a vector space over \({\mathbb Q}\) that has basis \(\{ 1, \sqrt{3}, \sqrt{5}, \sqrt{15}\, \}\text{.}\) It is no coincidence that \(|G( {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) /{\mathbb Q})| = [{\mathbb Q}(\sqrt{3}, \sqrt{5}\, ):{\mathbb Q})] = 4\text{.}\)

Let

\begin{equation*} f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \end{equation*}

and suppose that \(\alpha \in E\) is a zero of \(f(x)\text{.}\) Then for \(\sigma \in G(E/F)\text{,}\)

\begin{align*} 0 & = \sigma( 0 )\\ & = \sigma( f( \alpha ))\\ & = \sigma(a_0 + a_1\alpha + a_2 \alpha^2 + \cdots + a_n \alpha^n)\\ & = a_0 + a_1 \sigma(\alpha) + a_2 [\sigma(\alpha)]^2 + \cdots + a_n [\sigma(\alpha)]^n; \end{align*}

therefore, \(\sigma( \alpha )\) is also a zero of \(f(x)\text{.}\)

Let \(E\) be an algebraic extension of a field \(F\text{.}\) Two elements \(\alpha, \beta \in E\) are conjugate over \(F\) if they have the same minimal polynomial. For example, in the field \({\mathbb Q}( \sqrt{2}\, )\) the elements \(\sqrt{2}\) and \(-\sqrt{2}\) are conjugate over \({\mathbb Q}\) since they are both roots of the irreducible polynomial \(x^2 - 2\text{.}\)

A converse of the last proposition exists. The proof follows directly from Lemma 21.32.

We will use mathematical induction on the degree of \(f(x)\text{.}\) If the degree of \(f(x)\) is 0 or 1, then \(E = F\) and there is nothing to show. Assume that the result holds for all polynomials of degree \(k\) with \(0 \leq k \lt n\text{.}\) Suppose that the degree of \(f(x)\) is \(n\text{.}\) Let \(p(x)\) be an irreducible factor of \(f(x)\) of degree \(r\text{.}\) Since all of the roots of \(p(x)\) are in \(E\text{,}\) we can choose one of these roots, say \(\alpha\text{,}\) so that \(F \subset F( \alpha ) \subset E\text{.}\) Then

\begin{equation*} [E: F(\alpha)] = n/r \quad \text{and} \quad [F(\alpha): F] = r. \end{equation*}

If \(\beta\) is any other root of \(p(x)\text{,}\) then \(F \subset F( \beta ) \subset E\text{.}\) By Lemma 21.32, there exists a unique isomorphism \(\sigma: F( \alpha ) \rightarrow F( \beta )\) for each such \(\beta\) that fixes \(F\) elementwise. Since \(E\) is a splitting field of \(p(x)\text{,}\) there are exactly \(r\) such isomorphisms. For each of these automorphisms, we can use our induction hypothesis on \([E: F(\alpha)] = n/r \lt n\) to conclude that

\begin{equation*} |G(E/F(\alpha))| = [E:F(\alpha)]. \end{equation*}

Consequently, there are

\begin{equation*} [E:F] = [E:F(\alpha)] [F( \alpha):F] = n \end{equation*}

possible automorphisms of \(E\) that fix \(F\text{,}\) or \(|G(E/F)| = [E:F]\text{.}\)

Let \(p\) be the characteristic of \(E\) and \(F\) and assume that the orders of \(E\) and \(F\) are \(p^m\) and \(p^n\text{,}\) respectively. Then \(nk = m\text{.}\) We can also assume that \(E\) is the splitting field of \(x^{p^m} - x\) over a subfield of order \(p\text{.}\) Therefore, \(E\) must also be the splitting field of \(x^{p^m} - x\) over \(F\text{.}\) Applying Theorem 23.7, we find that \(|G(E/F)| = k\text{.}\)

To prove that \(G(E/F)\) is cyclic, we must find a generator for \(G(E/F)\text{.}\) Let \(\sigma : E \rightarrow E\) be defined by \(\sigma(\alpha) = \alpha^{p^n}\text{.}\) We claim that \(\sigma\) is the element in \(G(E/F)\) that we are seeking. We first need to show that \(\sigma\) is in \(\aut(E)\text{.}\) If \(\alpha\) and \(\beta\) are in \(E\text{,}\)

\begin{equation*} \sigma(\alpha + \beta) = (\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n} = \sigma(\alpha) + \sigma(\beta) \end{equation*}

by Lemma 22.3 Also, it is easy to show that \(\sigma(\alpha \beta) = \sigma( \alpha ) \sigma( \beta )\text{.}\) Since \(\sigma\) is a nonzero homomorphism of fields, it must be injective. It must also be onto, since \(E\) is a finite field. We know that \(\sigma\) must be in \(G(E/F)\text{,}\) since \(F\) is the splitting field of \(x^{p^n} - x\) over the base field of order \(p\text{.}\) This means that \(\sigma\) leaves every element in \(F\) fixed. Finally, we must show that the order of \(\sigma\) is \(k\text{.}\) By Theorem 23.7, we know that

\begin{equation*} \sigma^k( \alpha ) = \alpha^{p^{nk}} = \alpha^{p^m} = \alpha \end{equation*}

is the identity of \(G( E/F)\text{.}\) However, \(\sigma^r\) cannot be the identity for \(1 \leq r \lt k\text{;}\) otherwise, \(x^{p^{nr}} - x\) would have \(p^m\) roots, which is impossible.

Example23.9

We can now confirm that the Galois group of \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\) over \({\mathbb Q}\) in Example 23.4 is indeed isomorphic to \({\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\) Certainly the group \(H = \{ \identity, \sigma, \tau, \mu \}\) is a subgroup of \(G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})\text{;}\) however, \(H\) must be all of \(G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})\text{,}\) since

\begin{equation*} |H| = [{\mathbb Q}( \sqrt{3}, \sqrt{5}\, ):{\mathbb Q}] = |G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})| = 4. \end{equation*}
Example23.10

Let us compute the Galois group of

\begin{equation*} f(x) = x^4 + x^3 + x^2 + x + 1 \end{equation*}

over \({\mathbb Q}\text{.}\) We know that \(f(x)\) is irreducible by Exercise 17.4.20 in Chapter 17. Furthermore, since \((x -1)f(x) = x^5 -1\text{,}\) we can use DeMoivre's Theorem to determine that the roots of \(f(x)\) are \(\omega^i\text{,}\) where \(i = 1, \ldots, 4\) and

\begin{equation*} \omega = \cos(2 \pi / 5 ) + i \sin(2 \pi / 5 ). \end{equation*}

Hence, the splitting field of \(f(x)\) must be \({\mathbb Q}(\omega)\text{.}\) We can define automorphisms \(\sigma_i\) of \({\mathbb Q}(\omega )\) by \(\sigma_i( \omega ) = \omega^i\) for \(i = 1, \ldots, 4\text{.}\) It is easy to check that these are indeed distinct automorphisms in \(G( {\mathbb Q}( \omega) / {\mathbb Q} )\text{.}\) Since

\begin{equation*} [{\mathbb Q}( \omega) : {\mathbb Q}] = | G( {\mathbb Q}( \omega) / {\mathbb Q})| = 4, \end{equation*}

the \(\sigma_i\)'s must be all of \(G( {\mathbb Q}( \omega) / {\mathbb Q} )\text{.}\) Therefore, \(G({\mathbb Q}( \omega) / {\mathbb Q})\cong {\mathbb Z}_4\) since \(\omega\) is a generator for the Galois group.

SubsectionSeparable Extensions

Many of the results that we have just proven depend on the fact that a polynomial \(f(x)\) in \(F[x]\) has no repeated roots in its splitting field. It is evident that we need to know exactly when a polynomial factors into distinct linear factors in its splitting field. Let \(E\) be the splitting field of a polynomial \(f(x)\) in \(F[x]\text{.}\) Suppose that \(f(x)\) factors over \(E\) as

\begin{equation*} f(x) = (x - \alpha_1)^{n_1} (x - \alpha_2)^{n_2} \cdots (x - \alpha_r)^{n_r} = \prod_{i = 1}^{r} (x - \alpha_i)^{n_i}. \end{equation*}

We define the multiplicity of a root \(\alpha_i\) of \(f(x)\) to be \(n_i\text{.}\) A root with multiplicity 1 is called a simple root. Recall that a polynomial \(f(x) \in F[x]\) of degree \(n\) is separable if it has \(n\) distinct roots in its splitting field \(E\text{.}\) Equivalently, \(f(x)\) is separable if it factors into distinct linear factors over \(E[x]\text{.}\) An extension \(E\) of \(F\) is a separable extension of \(F\) if every element in \(E\) is the root of a separable polynomial in \(F[x]\text{.}\) Also recall that \(f(x)\) is separable if and only if \(\gcd( f(x), f'(x)) = 1\) (Lemma 22.5).

First assume that \(\chr F = 0\text{.}\) Since \(\deg f'(x) \lt \deg f(x)\) and \(f(x)\) is irreducible, the only way \(\gcd( f(x), f'(x)) \neq 1\) is if \(f'(x)\) is the zero polynomial; however, this is impossible in a field of characteristic zero. If \(\chr F = p\text{,}\) then \(f'(x)\) can be the zero polynomial if every coefficient of \(f'(x)\) is a multiple of \(p\text{.}\) This can happen only if we have a polynomial of the form \(f(x) = a_0 + a_1 x^p + a_2 x^{2p} + \cdots + a_n x^{np}\text{.}\)

Certainly extensions of a field \(F\) of the form \(F(\alpha)\) are some of the easiest to study and understand. Given a field extension \(E\) of \(F\text{,}\) the obvious question to ask is when it is possible to find an element \(\alpha \in E\) such that \(E = F( \alpha )\text{.}\) In this case, \(\alpha\) is called a primitive element. We already know that primitive elements exist for certain extensions. For example,

\begin{equation*} {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = {\mathbb Q}( \sqrt{3} + \sqrt{5}\, ) \end{equation*}

and

\begin{equation*} {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i ) = {\mathbb Q}( \sqrt[6]{5}\, i ). \end{equation*}

Corollary 22.12 tells us that there exists a primitive element for any finite extension of a finite field. The next theorem tells us that we can often find a primitive element.

We already know that there is no problem if \(F\) is a finite field. Suppose that \(E\) is a finite extension of an infinite field. We will prove the result for \(F(\alpha, \beta)\text{.}\) The general case easily follows when we use mathematical induction. Let \(f(x)\) and \(g(x)\) be the minimal polynomials of \(\alpha\) and \(\beta\text{,}\) respectively. Let \(K\) be the field in which both \(f(x)\) and \(g(x)\) split. Suppose that \(f(x)\) has zeros \(\alpha = \alpha_1, \ldots, \alpha_n\) in \(K\) and \(g(x)\) has zeros \(\beta = \beta_1, \ldots, \beta_m\) in \(K\text{.}\) All of these zeros have multiplicity 1, since \(E\) is separable over \(F\text{.}\) Since \(F\) is infinite, we can find an \(a\) in \(F\) such that

\begin{equation*} a \neq \frac{\alpha_i - \alpha}{\beta - \beta_j} \end{equation*}

for all \(i\) and \(j\) with \(j \neq 1\text{.}\) Therefore, \(a( \beta - \beta_j ) \neq \alpha_i - \alpha\text{.}\) Let \(\gamma = \alpha + a \beta\text{.}\) Then

\begin{equation*} \gamma = \alpha + a \beta \neq \alpha_i + a \beta_j; \end{equation*}

hence, \(\gamma - a \beta_j \neq \alpha_i\) for all \(i, j\) with \(j \neq 1\text{.}\) Define \(h(x) \in F( \gamma )[x]\) by \(h(x) = f( \gamma - ax)\text{.}\) Then \(h( \beta ) = f( \alpha ) = 0\text{.}\) However, \(h( \beta_j ) \neq 0\) for \(j \neq 1\text{.}\) Hence, \(h(x)\) and \(g(x)\) have a single common factor in \(F( \gamma )[x]\text{;}\) that is, the minimal polynomial of \(\beta\) over \(F( \gamma )\) must be linear, since \(\beta\) is the only zero common to both \(g(x)\) and \(h(x)\text{.}\) So \(\beta \in F( \gamma )\) and \(\alpha = \gamma - a \beta\) is in \(F( \gamma )\text{.}\) Hence, \(F( \alpha, \beta ) = F( \gamma )\text{.}\)