The goal of this section is to prove the Fundamental Theorem of Galois Theory. This theorem explains the connection between the subgroups of $G(E/F)$ and the intermediate fields between $E$ and $F$.

The subfield $F_{ \{\sigma_i \} }$ of $F$ is called the fixed field of $\{ \sigma_i \}$. The field fixed for a subgroup $G$ of $\aut(F)$ will be denoted by $F_G$.

##### Example23.15

Let $\sigma : {\mathbb Q}(\sqrt{3}, \sqrt{5}\, ) \rightarrow {\mathbb Q}(\sqrt{3}, \sqrt{5}\, )$ be the automorphism that maps $\sqrt{3}$ to $-\sqrt{3}$. Then ${\mathbb Q}( \sqrt{5}\, )$ is the subfield of ${\mathbb Q}(\sqrt{3}, \sqrt{5}\, )$ left fixed by $\sigma$.

A large number of mathematicians first learned Galois theory from Emil Artin's monograph on the subject [1]. The very clever proof of the following lemma is due to Artin.

Let $E$ be an algebraic extension of $F$. If every irreducible polynomial in $F[x]$ with a root in $E$ has all of its roots in $E$, then $E$ is called a normal extension of $F$; that is, every irreducible polynomial in $F[x]$ containing a root in $E$ is the product of linear factors in $E[x]$.

##### Proof

Before we determine the exact correspondence between field extensionsand automorphisms of fields, let us return to a familiar example.

##### Example23.20

In Example 23.4 we examined the automorphisms of ${\mathbb Q}( \sqrt{3}, \sqrt{5}\, )$ fixing ${\mathbb Q}$. Figure 23.21 compares the lattice of field extensions of ${\mathbb Q}$ with the lattice of subgroups of $G( {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) /{\mathbb Q})$. The Fundamental Theorem of Galois Theory tells us what the relationship is between the two lattices.

We are now ready to state and prove the Fundamental Theorem of Galois Theory.

##### Example23.24

In this example we will illustrate the Fundamental Theorem of Galois Theory by determining the lattice of subgroups of the Galois group of $f(x) = x^4 - 2$. We will compare this lattice to the lattice of field extensions of ${\mathbb Q}$ that are contained in the splitting field of $x^4-2$. The splitting field of $f(x)$ is ${\mathbb Q}( \sqrt[4]{2}, i )$. To see this, notice that $f(x)$ factors as $(x^2 + \sqrt{2}\, )(x^2 - \sqrt{2}\, )$; hence, the roots of $f(x)$ are $\pm \sqrt[4]{2}$ and $\pm \sqrt[4]{2}\, i$. We first adjoin the root $\sqrt[4]{2}$ to ${\mathbb Q}$ and then adjoin the root $i$ of $x^2 + 1$ to ${\mathbb Q}(\sqrt[4]{2}\, )$. The splitting field of $f(x)$ is then ${\mathbb Q}(\sqrt[4]{2}\, )(i) = {\mathbb Q}( \sqrt[4]{2}, i )$.

Since $[ {\mathbb Q}( \sqrt[4]{2}\, ) : {\mathbb Q}] = 4$ and $i$ is not in ${\mathbb Q}( \sqrt[4]{2}\, )$, it must be the case that $[ {\mathbb Q}( \sqrt[4]{2}, i ): {\mathbb Q}(\sqrt[4]{2}\, )] = 2$. Hence, $[ {\mathbb Q}( \sqrt[4]{2}, i ):{\mathbb Q}] = 8$. The set \begin{equation*}\{ 1, \sqrt[4]{2}, (\sqrt[4]{2}\, )^2, (\sqrt[4]{2}\, )^3, i, i \sqrt[4]{2}, i (\sqrt[4]{2}\, )^2, i(\sqrt[4]{2}\, )^3 \}\end{equation*} is a basis of ${\mathbb Q}( \sqrt[4]{2}, i )$ over ${\mathbb Q}$. The lattice of field extensions of ${\mathbb Q}$ contained in ${\mathbb Q}( \sqrt[4]{2}, i)$ is illustrated in Figure 23.25(a).

The Galois group $G$ of $f(x)$ must be of order 8. Let $\sigma$ be the automorphism defined by $\sigma( \sqrt[4]{2}\, ) = i \sqrt[4]{2}$ and $\sigma( i ) = i$, and $\tau$ be the automorphism defined by complex conjugation; that is, $\tau(i ) = -i$. Then $G$ has an element of order 4 and an element of order 2. It is easy to verify by direct computation that the elements of $G$ are $\{ \identity, \sigma, \sigma^2, \sigma^3, \tau, \sigma \tau, \sigma^2 \tau, \sigma^3 \tau \}$ and that the relations $\tau^2 = \identity$, $\sigma^4 = \identity$, and $\tau \sigma \tau = \sigma^{-1}$ are satisfied; hence, $G$ must be isomorphic to $D_4$. The lattice of subgroups of $G$ is illustrated in Figure 23.25(b).