If $K\subset F_i\subset L$, $i=1,2$ are two normal extension of $K$, show that $F_1\cap F_2$ is normal on $K$.

Question

If $K\subset F_i\subset L$, $i=1,2$ are two normal extensions of $K$, show that $F_1\cap F_2$ is normal on $K$.

I have absolutely now idea how to proceed. First, is a normal extension an algebraic extension ?

Answer 1

Let $p$ be a polynomial with coefficients in $K$, irreducible over $K$, with a zero in $F_1\cap F_2$. Then that zero is in $F_1$, but $F_1/K$ is normal, so $p$ splits into linear factors over $F_1$; ditto over $F_2$. But $F_1$ and $F_2$ are contained in the same field $L$, so the factorization over $F_1$ is the same as over $F_2$, so it's a factorization over the intersection, so the intersection is normal over $K$.

Answer 2

Hint : $$\Phi: \text{Hom}_K(F_1,K^a)\longrightarrow \text{Hom}_K(F_1\cap F_2, K^a)$$ is surjective. Since $F_1/K$ is normale $$\text{Hom}_K(F_1,K^a)=\text{Aut}_K(F_1).$$

Using the same argument with $F_2$, you can easily conclude that $$\text{Hom}_K(F_1\cap F_2, K^a)=\text{Aut}_K(F_1\cap F_2)$$ what prove the claim.

Added

To show that $\Phi$ is surjective, you need Zorn's lemma. The fact that $$\text{Aut}_K(F_1\cap F_2)\subset \text{Hom}_K(F_1\cap F_2, K^a)$$ is obvious. For the other inclusion, let $$\psi\in\text{Hom}_K(F_1\cap F_2, K^a).$$ Then, there is a $\varphi_1\in\text{Aut}_K(F_1)$ and $\varphi_2\in\text{Aut}_K(F_2)$ such that $$\varphi_1=\psi|_{F_1}\quad\text{and}\quad\varphi_2=\psi|_{F_2}.$$ In particular, $$\psi=\psi|_{F_1\cap F_2}=\underbrace{\varphi_1|_{F_2}}_{\in F_1}=\underbrace{\varphi_2|_{F_1}}_{\in F_2}\in F_1\cap F_2$$

and thus $\Psi\in \text{Hom}_K(F_1\cap F_2, F_1\cap F_2)$. Since $\Psi$ is a field homomorphism, it's necessarily injective and thus bijective. Finally $\psi\in\text{Aut}_K(F_1\cap F_2)$ what prove the claim.