Galois Theory

#Abstract Algebra, Galois Theory

2023/04/28

Table of Content

Theorem (the fundamental theorem of Galois theory)
- Corollary
- Theorem (more properties)

研究 field、subfield 時，往往牽涉到複雜的計算；相對於 field theory 來說，群的問題相對比較容易解決（像是找 subgroup，或是確認群的性質）。Galois theory 正可以把 field theory 的問題轉換為 group theory 的問題！

Theorem (the fundamental theorem of Galois theory)

Let \(E/F\) be a finite Galois extension with Galois group \(G\) and \(\text{char}F=0\). Let \(H\) be a subgroup of \(G\) and let \(K\) be an intermediate field between \(E\) and \(F\). The following holds:

\(G_{E^H} = H\) and \(\vert H\vert = [E : E^H]\).

\(E^{G_K} = K\) and \(\vert G_K\vert = [E:K]\).

\(K/F\) is Galois iff \(G_K\) is normal. In this case, \(\text{Gal}(K/F) = G/G_K\).

Consequently, there is a bijection between subgroups pf \(G\) and intermediate fields between \(E\) and \(F\).

Note that

\[\begin{align*} E^H &= \{a\in E\mid h(a) = a,\forall h\in H \}, \\ G_K &= \{g\in G\mid g(a) = a,\forall a\in K \}. \end{align*}\]

固定 field 中的元素！一個收集 field element，另一個收集 group element。想想 automorphism 的定義，正是固定 \(F\) 的 mapping！

Proof of 3.

To show \(G_K\) is normal, we can construct a map with \(G_K\) being the kernel. Consider the map:

\[\begin{gather} \phi: &\text{Aut}(E/F)& \to &\text{Aut}(K/F)& \\ &\rho& \mapsto &\rho\vert_K& \end{gather}\]

We can see that \(\rho\) is in the kernel if and only if \(\rho\vert_K\) is the identity map, i.e., \(\rho\) as well as \(\rho\vert_K\) fix all \(k\in K\). By definition,

\[G_K = \{g\in G\mid g(k)=k,\ \forall k\in K \},\]

which is exactly \(\text{ker}\phi\).

Then, by the first ring isomorphism theorem, we have \(\text{Gal}(K/F)\cong G/G_K\). ■

See a different proof: Northeastern University: p.35.

\(G_K = \text{Aut}(E/K)\), since now \(K\) is the one to be fixed.

Corollary

Let \(E/F\) be a finite Galois extension with Galois group \(G\), and let \(K\) be an intermediate field. Then for all \(\alpha\in E\), \(\alpha \in K\) iff \(\sigma(\alpha) = \alpha\) for all \(\sigma\in G_K\).

Theorem (more properties)

Let \(E/F\) be a finite Galois extension and \(K_1\) and \(K_2\) be intermediate fields. Let \(H_1=\text{Gal}(E/K_1)\) and \(H_2=\text{Gal}(E/K_2)\).

If \(K_1\cap K_2 = F\), then \(G=\langle H_1, H_2\rangle\).

If \(\langle K_1, K_2\rangle = E\), then \(H_1\cap H_2 = \{e\}\).

If the above two cases both hold and either \(K_1/F\) or \(K_2/F\) is Galois, then \(G=H_1H_2\), and \(G\) is isomorphic to a semidirect product of \(H_1\) and \(H_2\).