##### Proposition23.1

The set of all automorphisms of a field \(F\) is a group under composition of functions.

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

The set of all automorphisms of a field \(F\) is a group under composition of functions.

Let \(E\) be a field extension of \(F\). Then the set of all automorphisms of \(E\) that fix \(F\) elementwise is a group; that is, the set of all automorphisms \(\sigma : E \rightarrow E\) such that \(\sigma( \alpha ) = \alpha\) for all \(\alpha \in F\) is a group.

Let \(E\) be a field extension of \(F\). We will denote the full group of automorphisms of \(E\) by \(\aut(E)\). 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 \(F\), then we define the Galois group of \(f(x)\) to be \(G(E/F)\).

Complex conjugation, defined by \(\sigma : a + bi \mapsto a - bi\), 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 \(G( {\mathbb C} / {\mathbb R} )\).

Consider the fields \({\mathbb Q} \subset {\mathbb Q}(\sqrt{5}\, ) \subset {\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\). Then for \(a, b \in {\mathbb Q}( \sqrt{5}\, )\), \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 \(\sqrt{5}\). It will soon be clear that \(\{ \identity, \sigma, \tau, \mu \}\) is the Galois group of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) over \({\mathbb Q}\). The following table shows that this group is isomorphic to \({\mathbb Z}_2 \times {\mathbb Z}_2\). \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 \(\{ 1, \sqrt{3}, \sqrt{5}, \sqrt{15}\, \}\). It is no coincidence that \(|G( {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) /{\mathbb Q})| = [{\mathbb Q}(\sqrt{3}, \sqrt{5}\, ):{\mathbb Q})] = 4\).

Let \(E\) be a field extension of \(F\) and \(f(x)\) be a polynomial in \(F[x]\). Then any automorphism in \(G(E/F)\) defines a permutation of the roots of \(f(x)\) that lie in \(E\).

Let \(E\) be an algebraic extension of a field \(F\). 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 \(x^2 - 2\).

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

If \(\alpha\) and \(\beta\) are conjugate over \(F\), there exists an isomorphism \(\sigma : F( \alpha ) \rightarrow F( \beta )\) such that \(\sigma\) is the identity when restricted to \(F\).

Let \(f(x)\) be a polynomial in \(F[x]\) and suppose that \(E\) is the splitting field for \(f(x)\) over \(F\). If \(f(x)\) has no repeated roots, then \begin{equation*}|G(E/F)| = [E:F].\end{equation*}

Let \(F\) be a finite field with a finite extension \(E\) such that \([E:F]=k\). Then \(G(E/F)\) is cyclic of order \(k\).

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 \({\mathbb Z}_2 \times {\mathbb Z}_2\). Certainly the group \(H = \{ \identity, \sigma, \tau, \mu \}\) is a subgroup of \(G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})\); however, \(H\) must be all of \(G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})\), since \begin{equation*}|H| = [{\mathbb Q}( \sqrt{3}, \sqrt{5}\, ):{\mathbb Q}] = |G({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )/{\mathbb Q})| = 4.\end{equation*}

Let us compute the Galois group of \begin{equation*}f(x) = x^4 + x^3 + x^2 + x + 1\end{equation*} over \({\mathbb Q}\). We know that \(f(x)\) is irreducible by Exercise 17.4.20 in Chapter 17. Furthermore, since \((x -1)f(x) = x^5 -1\), we can use DeMoivre's Theorem to determine that the roots of \(f(x)\) are \(\omega^i\), 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 \({\mathbb Q}(\omega)\). We can define automorphisms \(\sigma_i\) of \({\mathbb Q}(\omega )\) by \(\sigma_i( \omega ) = \omega^i\) for \(i = 1, \ldots, 4\). It is easy to check that these are indeed distinct automorphisms in \(G( {\mathbb Q}( \omega) / {\mathbb Q} )\). 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 \(G( {\mathbb Q}( \omega) / {\mathbb Q} )\). Therefore, \(G({\mathbb Q}( \omega) / {\mathbb Q})\cong {\mathbb Z}_4\) since \(\omega\) is a generator for the Galois group.

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 \(F[x]\). 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 \(n_i\). 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 \(E\). Equivalently, \(f(x)\) is separable if it factors into distinct linear factors over \(E[x]\). An extension \(E\) of \(F\) is a *separable extension* of \(F\) if every element in \(E\) is the root of a separable polynomial in \(F[x]\). Also recall that \(f(x)\) is separable if and only if \(\gcd( f(x), f'(x)) = 1\) (Lemma 22.5).

Let \(f(x)\) be an irreducible polynomial over \(F\). If the characteristic of \(F\) is \(0\), then \(f(x)\) is separable. If the characteristic of \(F\) is \(p\) and \(f(x) \neq g(x^p)\) for some \(g(x)\) in \(F[x]\), then \(f(x)\) is also separable.

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 \(F\), the obvious question to ask is when it is possible to find an element \(\alpha \in E\) such that \(E = F( \alpha )\). 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.

Let \(E\) be a finite separable extension of a field \(F\). Then there exists an \(\alpha \in E\) such that \(E=F( \alpha )\).