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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 \}\text{.} \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\text{,} \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 $[E:F]\text{.}$ If $[E:F] = 1\text{,}$ then $E = F$ and there is nothing to show. If $[E:F] \gt 1\text{,}$ let $f(x) = p(x)q(x)\text{,}$ where $p(x)$ is irreducible of degree $d\text{.}$ We may assume that $d \gt 1\text{;}$ otherwise, $f(x)$ splits over $F$ and $[E:F] = 1\text{.}$ Let $\alpha$ be a root of $p(x)\text{.}$ If $\phi: F(\alpha) \to E$ is any injective homomorphism, then $\phi( \alpha) = \beta$ is a root of $p(x)\text{,}$ and $\phi: F(\alpha) \to F(\beta)$ is a field automorphism. Since $f(x)$ has no repeated roots, $p(x)$ has exactly $d$ roots $\beta \in E\text{.}$ By Proposition 23.5, there are exactly $d$ isomorphisms $\phi: F(\alpha) \to F(\beta_i)$ that fix $F\text{,}$ one for each root $\beta_1, \ldots, \beta_d$ of $p(x)$ (see Figure 23.8).

Since $E$ is a splitting field of $f(x)$ over $F\text{,}$ it is also a splitting field over $F(\alpha)\text{.}$ Similarly, $E$ is a splitting field of $f(x)$ over $F(\beta)\text{.}$ Since $[E: F(\alpha)] = [E:F]/d\text{,}$ induction shows that each of the $d$ isomorphisms $\phi$ has exactly $[E:F]/d$ extensions, $\psi : E \to E\text{,}$ and we have constructed $[E:F]$ isomorphisms that fix $F\text{.}$ Finally, suppose that $\sigma$ is any automorphism fixing $F\text{.}$ Then $\sigma$ restricted to $F(\alpha)$ is $\phi$ for some $\phi: F(\alpha) \to F(\beta)\text{.}$

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.

###### Example23.10.

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\text{.} \end{equation*}
###### Example23.11.

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 17.5.20 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 )\text{.} \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\text{,} \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}\text{.} \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{5}, \sqrt{5}\, i ) = {\mathbb Q}( \sqrt{5}\, i )\text{.} \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 $1\text{,}$ 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, $F( \alpha, \beta ) = F( \gamma )\text{.}$