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Field Automorphisms

Our first task is to establish a link between group theory and field theory by examining automorphisms of fields. Proposition The set of all automorphisms of a field F is a group under composition of functions.

Field Automorphisms

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

Proposition

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.

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 \(E\). Let \(\sigma\) and \(\tau\) be two automorphisms of \(E\) such that \(\sigma( \alpha ) = \alpha\) and \(\tau( \alpha ) = \alpha\) for all \(\alpha \in F\). Then \(\sigma \tau( \alpha ) = \sigma( \alpha) = \alpha\) and \(\sigma^{-1}( \alpha ) = \alpha\). Since the identity fixes every element of \(E\), the set of automorphisms of \(E\) that leave elements of \(F\) fixed is a subgroup of the entire group of automorphisms of \(E\).

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, \[\begin{aligned}\end{aligned}\]. 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 \(G(E/F)\). \(G(E/F)\) Galois group of \(E\) over \(F\)

Example

Complex conjugation, defined by \(\sigma : a + bi \mapsto a - bi\), is an automorphism of the complex numbers. Since \[\begin{aligned}\end{aligned}\], the automorphism defined by complex conjugation must be in \(G( {\mathbb C} / {\mathbb R} )\).

Example

Consider the fields \({\mathbb Q} \subset {\mathbb Q}(\sqrt{5}\, ) \subset {\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\). Then for \(a, b \in {\mathbb Q}( \sqrt{5}\, )\), \[\begin{aligned}\end{aligned}\] is an automorphism of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) leaving \({\mathbb Q}( \sqrt{5}\, )\) fixed. Similarly, \[\begin{aligned}\end{aligned}\] 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 \({\mathbb Z}_2 \times {\mathbb Z}_2\). \[\begin{aligned}\end{aligned}\] 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 .

Theorem

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 \[\begin{aligned}\end{aligned}\].

Proof

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

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

Corollary

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

Proof

Let \(p\) be the characteristic of \(E\) and \(F\) and assume that the orders of \(E\) and \(F\) are \(p^m\) and \(p^n\), respectively. Then \(nk = m\). We can also assume that \(E\) is the splitting field of \(x^{p^m} - x\) over a subfield of order \(p\). Therefore, \(E\) must also be the splitting field of \(x^{p^m} - x\) over \(F\). Applying , we find that \(|G(E/F)| = k\).

To prove that \(G(E/F)\) is cyclic, we must find a generator for \(G(E/F)\). Let \(\sigma : E \rightarrow E\) be defined by \(\sigma(\alpha) = \alpha^{p^n}\). We claim that \(\sigma\) is the element in \(G(E/F)\) that we are seeking. We first need to show that \(\sigma\) is in \(\aut(E)\). If \(\alpha\) and \(\beta\) are in \(E\), \[\begin{aligned}\end{aligned}\] by . Also, it is easy to show that \(\sigma(\alpha \beta) = \sigma( \alpha ) \sigma( \beta )\). 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 \(G(E/F)\), since \(F\) is the splitting field of \(x^{p^n} - x\) over the base field of order \(p\). This means that \(\sigma\) leaves every element in \(F\) fixed. Finally, we must show that the order of \(\sigma\) is \(k\). By , we know that \[\begin{aligned}\end{aligned}\] is the identity of \(G( E/F)\). However, \(\sigma^r\) cannot be the identity for \(1 \leq r \lt k\); otherwise, \(x^{p^{nr}} - x\) would have \(p^m\) roots, which is impossible.

Example

We can now confirm that the Galois group of \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\) over \({\mathbb Q}\) in 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 \[\begin{aligned}\end{aligned}\].

Example

Let us compute the Galois group of \[\begin{aligned}\end{aligned}\] over \({\mathbb Q}\). We know that \(f(x)\) is irreducible by . 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 \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\], 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.

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 \(F[x]\). Suppose that \(f(x)\) factors over \(E\) as \[\begin{aligned}\end{aligned}\]. 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 \(\gcd( f(x), f'(x)) = 1\) ().

Proposition

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.

Proof

First assume that \(\chr F = 0\). 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 \(\chr F = p\), then \(f'(x)\) can be the zero polynomial if every coefficient of \(f'(x)\) is a multiple of \(p\). This can happen only if we have a polynomial of the form \(f(x) = a_0 + a_1 x^p + a_2 x^{2p} + \cdots + a_n x^{np}\).

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

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Frågor folk frågar

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Delar av denna sida är anpassade från Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Kondenserad och återförklarad här; felen är våra.

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