If $L_1/K$ and $L_2/K$ are normal field extensions then $L_1L_2/K$ is a normal field extension

Question

Let $L/K$ be a field extension and let

$L_1, L_2$ be fields such that

$K \leq L_1 \leq L$ , $[L_1:K] \in \mathbb{N}$

$K \leq L_2 \leq L$, $[L_2:K]\in \mathbb{N}$

If $L_1/K$ and $L_2/K$ are normal field extensions then $L_1L_2/K$ is a normal field extension

My attepmt:

Definition of normal extension:
A field extension $L/K$ is called normal, if $L/K$ is algebraic and every irreducible polynomial $p\in K[X]$ which has a root in $L$ factors into linear factors

I know the following:
Let $L/K$ be a finite field extension. Then the following is equivalent:

(1) $L/K$ is a normal field extension

(2) $L$ is splitting field of $P \in K[X]$

(3) For every finite field extension $\tilde{L}/K$:
$\sigma_1, \sigma_2 \in Mor_K(L,\tilde{L})$ $\Rightarrow $$\sigma_1(L)=\sigma_2(L)$

(4) for all $\sigma_1, \sigma_2 \in Mor_K(L,\overline{K})$ is $\sigma_1(L)=\sigma_2(L)$

So let $L_1/K$ and $L_2/K$ be normal field extensions.
This means that every irreducible polynomial $p_1 \in K[x]$ such that $\exists$ $a \in L_1$ such that $p_1(a)=0$ can be written as $a_1,…,a_n \in L_1$ such that $p_1=(x-a_1)…(x-a_n)$

Further every irreducible polynomial $p_2 \in K[x]$ such that $\exists$ $b \in L_2$ such that $p_2(b)=0$ can be written as $b_1,…,b_n \in L_2$ such that $p_2=(x-b_1)…(x-b_n)$

Now $L_1L_2:=\{\sum_{k=0}^n l_1l_2 : l_1 \in L_1, l_2 \in L_2 \}$

Now lets assume that $p$ is irreducible polynomial in $K[X]$ and $c \in L_1L_2$ such that p(c)=0.
Then $\exists$ $l_{1} \in L_1$ and $l_{2} \in L_2$ such that $c=l_{1}l_{2}$

$p(c)=p(l_{1}l_{2})=\alpha_0 + \alpha_1 (l_{1}l_{2})+…+\alpha_{n} (l_{1}l_{2})^n=0$

Rewriting the polynomial to:
$p(l_{1}l_{2})=\alpha_0 + \alpha_1l_{1}(l_{2})+…+\alpha_{n} l_{1}^n(l_{2})^n=\beta_0+\beta_1 l_2 +…+\beta_n l_2^n=0$

where $\beta_k:=\alpha_k l_1^k$

The problem now is that $\beta_k$ don't need to be elements of $K$
My question is how to continue this proof or if there is a better/shorter proof.

Answer 1

Here is a short proof using the fourth condition for nomality of field extensions. It can be rewritten as follows, by taking $\sigma_2$ to be an inclusion map $L \rightarrow \overline{K}$.

Let $L/K$ be a finite field extension. Then $L/K$ is a normal extension if and only if $\sigma(L)=L$ holds for every $K$-embedding $\sigma : L \rightarrow \overline{K}$.

In fact, you don't need to assume $L/K$ to be finite. It suffices to assume $L/K$ is an algebraic extension.

To prove that the finite field extension $L_1L_2/K$ is normal, we shall prove that if $\sigma : L_1 L_2 \rightarrow \overline{K}$ is a $K$-embedding, then $\sigma(L_1 L_2 )= L_1 L_2$. It is quite immediate; since $L_i/K$ are normal extensions, $\sigma(L_1L_2) = \sigma(L_1)\sigma(L_2) =L_1 L_2$. Here $\sigma(L_i) = L_i$ because $\sigma(L_i) = \sigma \vert_{L_i} (L_i)$ and $\sigma \vert_{L_i} : L_i \rightarrow \overline{K}$ is a $K$-embedding.