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

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 $G(E/F)$.

##### Example23.3

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 $G( {\mathbb C} / {\mathbb R} )$.

##### Example23.4

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 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.

##### 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 ${\mathbb Z}_2 \times {\mathbb Z}_2$. 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*}

##### Example23.10

Let us compute the Galois group of \begin{equation*}f(x) = x^4 + x^3 + x^2 + x + 1\end{equation*} over ${\mathbb Q}$. 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).
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.