Galois theory

Revision as of 19:39, 4 August 2006 by ComplexZeta (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Galois theory is an important tool for the study of fields. The primary objects of study in Galois theory are automorphisms of fields.

Consider the field $K=\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$. Then the map $f:K\to K$ given by $f(a+b\sqrt{2})=a-b\sqrt{2}$ is a field automorphism; that is, $f(\alpha\beta)=f(\alpha)f(\beta)$ and $f(\alpha+\beta)=f(\alpha)+f(\beta)$, and $f$ is a bijection. Of course, the map $g:K\to K$ given by $g(\alpha)=\alpha$ is also a field automorphism. Both of these automorphisms are the identity automorphism on $\mathbb{Q}$, a subfield of $K$. It turns out that $f$ and $g$ are the only automorphisms of $K$ that fix $\mathbb{Q}$. Furthermore, the automorphisms $f$ and $g$ form a group, called the Galois group of $K$ over $\mathbb{Q}$.

We now define Galois groups more rigorously.

Let $L/K$ be a field extension. Then the set of field automorphisms of $L$ that fix $K$ form a group under composition. This group is called the Galois group of $L/K$ and is denoted $Gal(L/K)$.

One may wonder if the elements of $K$ are the only elements of $L$ fixed by every element of $Gal(L/K)$. It turns out that this is not always the case. For example, if $K=\mathbb{Q}$ and $L=\mathbb{Q}(\sqrt[3]{2})$, then $Gal(L/K)$ is the trivial group, so every element of $L$ is fixed by $Gal(L/K)$. If the elements of $K$ are the only elements of $L$ fixed by $Gal(L/K)$, then we say that $L/K$ is a Galois extension.

Many beautiful results can be obtained with a bit of Galois theory. For example, one can prove that it is impossible to trisect an angle using Galois theory.

This article is a stub. Help us out by expanding it.