Proving the Fundamental Theorem

Section 17.3 Proving the Fundamental Theorem

Subsection Artin’s Theorem

The following is an important theorem with many corollaries. In fact, the Fundamental Theorem of Galois Theory, which we will state shortly, will follow from this result.

Theorem 17.31. Artin’s Theorem.

Let \(L\) be any field. If \(G\) is a finite subgroup of \(\operatorname{Aut}(L)\text{,}\) then \(L^{G}\) is a subfield of \(L\text{,}\) the extension \(L / L^{G}\) is finite and Galois, and \(\operatorname{Gal}\left(L / L^{G}\right)=G\text{.}\)

Note that we really do mean equality here: both \(G\) and \(\operatorname{Gal}\left(L / L^{G}\right)\) are subgroups of \(\operatorname{Aut}(L)\text{,}\) and the theorem states that they coincide. The containment \(G \subseteq \operatorname{Gal}\left(L / L^{G}\right)\) is clear: if \(\sigma \in G\text{,}\) then by construction \(\sigma\) fixes every element of \(L^{G}\) and hence \(\sigma \in \operatorname{Gal}\left(L / L^{G}\right)\text{.}\) The point of the theorem is that the extension \(L^{G} \subseteq L\) is always Galois and that if \(\sigma \in \operatorname{Aut}(L)\) fixes every element of \(L^{G}\) then \(\sigma\) must belong to \(G\text{.}\)

We will not prove Artin’s Theorem right away. Instead, we will first deduce some of its consequences, including the Fundamental Theorem of Galois Theory. We will then illustrate the Fundamental Theorem with many examples and give some consequences of it too. Only then will we circle back to prove Artin’s Theorem.

Example 17.32.

The group \(G=\left\{\mathrm{id}_{\mathbb{C}}, \sigma\right\}\text{,}\) where \(\sigma\) is complex conjugation, is a finite subgroup of \(\Aut(\mathbb{C})\text{.}\) Artin’s Theorem tells us that \(\mathbb{C}^{G} \subseteq \mathbb{C}\) is finite and Galois with Galois group \(G\text{.}\) It follows that \(\left[\mathbb{C}: \mathbb{C}^{G}\right]=|G|=2\text{.}\) We already knew all this, since \(\mathbb{C}^{G}=\mathbb{R}\text{.}\)

As we head towards the Fundamental Theorem of Galois Theory, we start by stating a few helpful corollaries of Artin’s Theorem. These will also allow us to show that finite Galois extensions are precisely the splitting fields of separable polynomials.

Corollary 17.33.

Let \(L / F\) be any Galois extension. Then \(F=L^{\mathrm{Gal}(L / F)}\text{.}\)

Proof.

Note that \(F \subseteq L^{\mathrm{Gal}(L / F)}\) holds by definition, and so

\begin{equation*} [L: F]=\left[L: L^{\operatorname{Gal}(L / F)}\right]\left[L^{\operatorname{Gal}(L / F)}: F\right] \end{equation*}

by the Degree Formula. But Artin’s Theorem gives

\begin{equation*} \left[L: L^{\operatorname{Gal}(L / F)}\right]=|\operatorname{Gal}(L / F)| \end{equation*}

and we also know that \([L: F]=|\operatorname{Gal}(L / F)|\text{.}\) Therefore, \(\left[L^{\mathrm{Gal}(L / F)}: F\right]=1\) and thus \(F=L^{\operatorname{Gal}(L / F)}\text{.}\)

Example 17.34.

We know from Example 6.14 that \(L=\mathbb{Q}(\sqrt[4]{2}, i)\) is Galois over \(\mathbb{Q}\) with Galois group \(D_{8}\text{.}\) More precisely, this identification is given by writting

\begin{equation*} \alpha_{1}=\sqrt[4]{[2]}, \quad \alpha_{2}=i \sqrt[4]{2}, \quad \alpha_{3}=-\sqrt[4]{2}, \quad \alpha_{4}=-i \sqrt[4]{2} \end{equation*}

and labelling the four corners of a square with \(\alpha_{1}, \ldots, \alpha_{4}\text{,}\) counter-clockwise. Consider

\begin{equation*} \beta:=\alpha_{1}+\cdots \alpha_{4} \end{equation*}

and \(\gamma=\alpha_{1} \cdots \alpha_{4}\text{.}\) Then each of \(\beta\) and \(\gamma\) are fixed by every Galois automorphism and hence by Corollary \(6.24 \beta\) and \(\gamma\) must be rational. In fact, one can easily see that \(\beta=0\) and \(\gamma=2\text{,}\) but notice that the exact same reasoning would apply in general to the sum of roots and the product of roots in the splitting field of any separable polynomial.

Corollary 17.35.

Let \(F \subseteq L\) be a Galois extension. For every \(\alpha \in L, m_{\alpha, F}\) is separable and all of its roots belong to \(L\text{.}\) Moreover, \(\operatorname{Gal}(L / F)\) acts transitively on the set of roots of \(m_{\alpha, F}\text{.}\)

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