Galois Groups

# Galois Groups

Recall from the Ring Automorphisms that if $F$ is a field then a function $\phi : F \to F$ is said to be an automorphism if:

• $\phi (a + b) = \phi(a) + \phi(b)$.
• $\phi (a \cdot b) = \phi(a) \cdot \phi (b)$.
• $\phi (1) = 1$.
• $\phi$ is bijective.

Now let $K$ be a field and let $F$ be an extension field of $K$. Consider the set of all automorphisms $\phi : F \to F$ that fix $K$:

(1)
\begin{align} \quad \{ \phi \in \mathrm{Aut} (F) : \phi(k) = k, \: \forall k \in K \} \end{align}

The following proposition tells us that the set above forms a subgroup of the automorphism group $\mathrm{Aut} (F)$.

 Proposition 1: Let $K$ be a field and let $F$ be a field extension of $K$. Then $G = \{ \phi \in \mathrm{Aut} (F) : \phi(k) = k, \: \forall k \in K \}$ is a subgroup of $\mathrm{Aut} (F)$.
• Proof: Let $\phi, \psi \in G$. Then for each $k \in K$ we have that $\phi (k) = k$. So $\psi (\phi(k)) = \psi(k) = k$. So $G$ is closed under the operation $\cdot$.
• Now let $\phi \in G$. Since $\phi (k) = k$ for every $k \in K$ we see that $\phi^{-1}(k) = k$ for every $k \in K$, and so $\phi^{-1} \in G$.
• Hence $G$ is a subgroup of $\mathrm{Aut} (F)$. $\blacksquare$

The subset of $\mathrm{Aut}(F)$ described above is given a special name which we define below.

 Definition: Let $K$ be a field and let $F$ be a field extension of $K$. The Galois Group of $F$ over $K$ is the group of all automorphisms on $F$ that fix $K$ and is denoted $\mathrm{Gal}(F/K) = \{ \phi \in \mathrm{Aut} (F) : \phi(k) = k, \: \forall k \in K \}$.

Given a field $K$ and a polynomial $f(x) \in K[x]$ we can also define the Galois group of $f(x)$ over $K$:

 Definition: Let $K$ be a field and let $f(x) \in K[x]$. Let $F$ be the splitting field of $f(x)$ over $K$. Then the Galois Group of $f(x)$ over $K$ is $\mathrm{Gal} (F/K)$.

Under certain circumstances we can determine the order of the Galois group of $f(x)$ over $K$. This is summarized in the theorem below.

 Theorem 1: Let $K$ be a field, $f(x) \in K[x]$, and let $F$ be a splitting field of $f(x)$ over $K$. If every irreducible factor of $f(x)$ has no repeated roots then $|\mathrm{Gal} (F/K)| = [F:K]$.