Art of Problem Solving

Galois theory

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.

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