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, $$F( \alpha, \beta ) = F( \gamma )\text{.}$$