23.1: Field Automorphisms
Our first task is to establish a link between group theory and field theory by examining automorphisms of fields.
Proposition \(23.1\).
The set of all automorphisms of a field \(F\) is a group under composition of functions.
- Proof
-
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.
Proposition \(23.2\).
Let \(E\) be a field extension of \(F\text{.}\) 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.
- Proof
-
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,
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 \(G(E/F)\text{.}\)
Example \(23.3\).
Complex conjugation, defined by \(\sigma : a + bi \mapsto a - bi\text{,}\) is an automorphism of the complex numbers. Since
Solution
the automorphism defined by complex conjugation must be in \(G( {\mathbb C} / {\mathbb R} )\text{.}\)
Example \(23.4\).
Consider the fields \({\mathbb Q} \subset {\mathbb Q}(\sqrt{5}\, ) \subset {\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\text{.}\)
Solution
Then for \(a, b \in {\mathbb Q}( \sqrt{5}\, )\text{,}\)
is an automorphism of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) leaving \({\mathbb Q}( \sqrt{5}\, )\) fixed. Similarly,
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 \({\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\)
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 \(|G( {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) /{\mathbb Q})| = [{\mathbb Q}(\sqrt{3}, \sqrt{5}\, ):{\mathbb Q})] = 4\text{.}\)
Proposition \(23.5\).
Let \(E\) be a field extension of \(F\) and \(f(x)\) be a polynomial in \(F[x]\text{.}\) Then any automorphism in \(G(E/F)\) defines a permutation of the roots of \(f(x)\) that lie in \(E\text{.}\)
- Proof
-
Let
\[ f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \nonumber \]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 \(21.32\).
Proposition \(23.6\).
If \(\alpha\) and \(\beta\) are conjugate over \(F\text{,}\) there exists an isomorphism \(\sigma : F( \alpha ) \rightarrow F( \beta )\) such that \(\sigma\) is the identity when restricted to \(F\text{.}\)
Theorem \(23.7\).
Let \(f(x)\) be a polynomial in \(F[x]\) and suppose that \(E\) is the splitting field for \(f(x)\) over \(F\text{.}\) If \(f(x)\) has no repeated roots, then
- Proof
-
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 \(23.8\)).
\(Figure \text { } 23.8.\)
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 \(\phi: F(\alpha) \to F(\beta)\text{.}\)
Corollary \(23.9\).
Let \(F\) be a finite field with a finite extension \(E\) such that \([E:F]=k\text{.}\) Then \(G(E/F)\) is cyclic of order \(k\text{.}\)
- Proof
-
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 \(E\text{,}\)
\[ \sigma(\alpha + \beta) = (\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n} = \sigma(\alpha) + \sigma(\beta) \nonumber \]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
\[ \sigma^k( \alpha ) = \alpha^{p^{nk}} = \alpha^{p^m} = \alpha \nonumber \]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.
Example \(23.10\).
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 \({\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\)
Solution
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
Example \(23.11\).
Let us compute the Galois group of
over \({\mathbb Q}\text{.}\)
Solution
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
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
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.
Separable 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
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 \(22.5\)).
Proposition \(23.12\).
Let \(f(x)\) be an irreducible polynomial over \(F\text{.}\) If the characteristic of \(F\) is \(0\text{,}\) then \(f(x)\) is separable. If the characteristic of \(F\) is \(p\) and \(f(x) \neq g(x^p)\) for some \(g(x)\) in \(F[x]\text{,}\) then \(f(x)\) is also separable.
- Proof
-
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,
and
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.
Theorem \(23.13\). Primitive Element Theorem.
Let \(E\) be a finite separable extension of a field \(F\text{.}\) Then there exists an \(\alpha \in E\) such that \(E=F( \alpha )\text{.}\)
- Proof
-
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
\[ a \neq \frac{\alpha_i - \alpha}{\beta - \beta_j} \nonumber \]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
\[ \gamma = \alpha + a \beta \neq \alpha_i + a \beta_j; \nonumber \]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, \(F( \alpha, \beta ) = F( \gamma )\text{.}\).