## Tag Archives: intermediate subfield

### Intersection of subgroups and the one of intermediate subfields

$\mathbf{Theorem 1:}$ Show that in the fundamental theorem of Galois theory, the subfield corresponding to the intersection of two subgroups $H_1$ and $H_2$ is the subfield generated by the corresponding intermediate subfields $\text{Inv}H_1$ and $\text{Inv}H_2$.
$\mathbf{Proof:}$ Suppose $K$ is the subfield generated by $\text{Inv}H_1$ and $\text{Inv}H_2$. Then $K=\displaystyle \bigcap L$ here $L$ is intermediate subfield embarcing $\text{Inv}H_1$ and $\text{Inv}H_2$.
For any such $L$, $\text{Gal }E/L\leq H_1$ and $\text{Gal }E/L\leq H_2$, so $\text{Gal }E/L\leq H_1\cap H_2$.
Thus $K\subset \text{Inv}(H_1\cap H_2)$.
Conversely, $\text{Inv}(H_1\cap H_2)\supset \text{Inv}H_1$, $\text{Inv}(H_1\cap H_2)\supset \text{Inv}H_2$. So $\text{Inv}(H_1\cap H_2)\supset K$.
$\text{Q.E.D}\hfill \square$
$\mathbf{Theorem 2:}$ The intersection of two intermediate fields $K_1$ and $K_2$ corresponds to the subgroup generated by the corresponding subgroups $\text{Gal }E/K_1$ and $\text{Gal } E/K_2$.
The proof is very similar to the previous one.

$\mathbf{Corollary:}$ Suppose we also know $K_1$ is Galois over the groud field and $K_2$ is another intermediate field. Then $\text{Gal }K_1/K_2\cap K_2$ is isomorphic to $\text{Gal }K_1K_2/K_2$.

$\mathbf{Proof:}$ By assumption, $H$ is normal. From group theory,

$\displaystyle H_1H_2/H_1\cong H_2/H_1\cap H_2$

Theorem 2 says that the corresponding group to $K_1\cap K_2$ is $\langle H_1,H_2\rangle$. Since $H_1$ is normal then $\langle H_1,H_2\rangle=H_1H_2$. And $\text{Gal }K_1/K_1\cap K_2=H_1H_2/H_1$.

Theorem 1 says that the corresponding group to intermediate field $K_1K_2$ is $H_1\cap H_2$. And $\text{Gal }K_1K_2/K_2=H_2/H_1\cap H_2$.

So the conclusion holds.

$\text{Q.E.D}\hfill \square$

In fact this corollary is valid even $K_2$ is transcedental extension, namely

$\mathbf{Theorem 3:}$ Let $K$ be a Galois extension of $k$, let $F$ be an arbitrary extension and assume that $K,F$ are subfields of some other field. Then $KF$ is Galois over $F$ and $K$ is Galois over $K\cap F$. Let $H$ be the Galois group of $KF$ over $F$, and $G$ the Galois group of $K$ over $k$. If $\sigma\in H$ then the restriction of $\sigma$ to $K$ is in $G$ and the map

$\sigma\to \sigma|_K$

gives an isomorphism of $H$ on the Galois group of $K$ over $K\cap F$.

$\mathbf{Remark:}$ Jacobson Algebra I. p243

### Simple Extension

$\mathbf{Defination:}$ Let $E/F$ be a finte extension, if there is $u\in E$ such that $E=F(u)$, then we call $u$ is a primitive element of the extension $E/F$. And this extension is called simple exentsion.

It is very important to know when a extension is a simple extension.The following theorem Emil Artin

$\mathbf{Thm:(Primitive\, Element\, Thm)}$ $E/F$ is a simple extension if and only if there are finitely many intermediate field $\displaystyle K$ between $E$ and $F$.
We are not interested in the proof of this theorem. But in the consequence of this theorem.

$\mathbf{Corollary:}$ Let $E/F$ be a finite separable extension, then $E/F$ is a simple extension. Especially when $\displaystyle char(F)$ is 0, all finite extension is a simple one.