Compositum of normal extensions and embedding I want to show "If $E_1$ and $E_2$ are normal extensions of $F$, then the compositum $E_1 E_2$ and $E_1\cap E_2$ are normal extensions."
I already prove this, but I find another proof using an embedding $\sigma$ on $\overline{F}$(not F-embedding).
It says $\sigma (E_1 E_2)=\sigma (E_1) \sigma (E_2)$. I think it holds because $\sigma$ is an embedding, but it is hard to prove it. Do I need to use normalities of $E_i$?
Moreover, more generally, for embedding $\sigma$ on $\overline{F}$,  $ \sigma(F(S))=\sigma(F)(\sigma(S))$?
(I know it is holds for $\sigma(E_1\cap E_2)$..)
 A: TL;DR: 

Do I need to use normalities of $E_i$?

Yes.

for embedding $\sigma$ on $\overline{F}$,  $ \sigma(F(S))=\sigma(F)(\sigma(S))$?

I suppose here $S\subseteq \overline{F}$ is a set of generators for the extension $F\leq F(S)=:E$ and $\sigma:E\hookrightarrow\overline{F}$ is an embedding of $E$ into the algebraic closure $\overline{F}$ of $F$. However I believe the correct generalization should address how embeddings and compositums interact. See the proof below.

(I know it is holds for $\sigma(E_1\cap E_2)$...)

If you mean that whenever $\sigma$ is an embedding we have $\sigma(E_1\cap E_2)=\sigma(E_1)\cap\sigma(E_1)$, then this is an elementary set theoretical fact and you don't need any field structure to show this (e.g. see here). In the case of compositums though, you do need the field structure.

I believe you are citing Lang's Algebra, 3e (Theorem V.3.4 on p. 238), the relevant part of which is as follows:
Theorem: Let $k$ be a field, $k\leq\begin{smallmatrix} E_1\\E_2\end{smallmatrix}\leq K$, and $E_1,E_2$ be normal over $k$. Then $E_1E_2$ is normal over $k$.
Proof: Let $\sigma: E_1E_2\hookrightarrow \overline{k}:\sigma\vert_k\equiv \operatorname{id}_k$. Then we have $\sigma_1:=\sigma\vert_{E_1}: E_1\hookrightarrow \overline{k}$ and $\sigma_2:=\sigma\vert_{E_2}: E_2\hookrightarrow \overline{k}$ (both over $k$). Since $E_1$ and $E_2$ are normal over $k$, $\sigma_1\in\operatorname{Aut}(K_1)$ and $\sigma_2\in\operatorname{Aut}(K_2)$, so that $\sigma_1(E_1)=E_1,\sigma_2(E_2)=E_2$.
$E_1,E_2\leq E_1E_2 \implies \sigma_1(E_1),\sigma_2(E_2)\leq \sigma(E_1E_2)$. Since $\sigma_1(E_1)\sigma_2(E_2)$ is by definition the smallest field (contained in some larger field that contains both $\sigma_1(E_1)$ and $\sigma_2(E_2)$) that contains both $\sigma_1(E_1)$ and $\sigma_2(E_2)$, $E_1E_2=\sigma_1(E_1)\sigma_2(E_2)\leq\sigma(E_1E_2)$. We have
\begin{align}
[E_1E_2:k]&=[\sigma(E_1E_2):\sigma(k)]\\
&=[\sigma(E_1E_2):\sigma_1(E_1)\sigma_2(E_2)][\sigma_1(E_1)\sigma_2(E_2):\sigma(k)]\\
&=[\sigma(E_1E_2):\sigma_1(E_1)\sigma_2(E_2)][E_1E_2:k]\\
& \implies [\sigma(E_1E_2):\sigma_1(E_1)\sigma_2(E_2)]=1\\
& \implies E_1E_2=\sigma_1(E_1)\sigma_2(E_2)=\sigma(E_1E_2),
\end{align}
where the first equality holds since $\sigma$ in an embedding. Thus $\sigma\in\operatorname{Aut}(E_1E_2)$, so that $E_1E_2$ is normal over $k$.

To put the above theorem (in its complete form) in perspective here are some further details:
Let $\mathcal{F}$ be the class of all fields. $\mathcal{F}$ is partially ordered by inclusion (or extension) $\leq$:
$$k\leq F \iff k\subseteq F \mbox{ and } k \mbox{ is a subfield of } F.$$
A class of extensions $\mathcal{C}\subseteq\leq$ is distinguished if it is


*

*filterable: $k\leq F\leq E,k\leq_{\mathcal{C}}E \implies k\leq_{\mathcal{C}} F\leq_{\mathcal{C}} E$,

*transitive: $k\leq_{\mathcal{C}} F\leq_{\mathcal{C}} E\implies k\leq_{\mathcal{C}}E$,

*closed under liftings: $k\leq_{\mathcal{C}} E\leq K,k\leq F\leq K \implies F\leq_{\mathcal{C}} EF$.


Observe that a distinguished $\mathcal{C}$ is also closed under compositums:


*closed under compositums: $k\leq_{\mathcal{C}} E\leq K,k\leq_{\mathcal{C}} F\leq K \implies k\leq_{\mathcal{C}} EF$.


Examples of distinguished classes of field extensions include 


*

*finite extensions,

*finitely generated extensions,

*algebraic extensions.


The class $\mathcal{N}$ of normal extensions  on the other hand is not distinguished:
Example 1: If $\beta\in\Bbb{R}: X^3-2\in\ker(\operatorname{ev}_\beta),$ and $\omega$ is a primitive third root of unity (i.e. $\omega^3=1, \omega\neq1$), then $\Bbb{Q}\leq_\mathcal{N}\Bbb{Q}(\beta,\omega)$ and $\Bbb{Q}\leq\Bbb{Q}(\beta)\leq\Bbb{Q}(\beta,\omega)$ but $\Bbb{Q}\not\leq_{\mathcal{N}}\Bbb{Q}(\beta)$, so that filterability is not satisfied.
Example 2: If $\alpha\in\Bbb{R}:X^4-2\in\ker(\operatorname{ev}_\alpha)$, then $\Bbb{Q}\leq_{\mathcal{N}}\Bbb{Q}(\alpha^2)\leq_\mathcal{N}\Bbb{Q}(\alpha)$ but $\Bbb{Q}\not\leq_\mathcal{N}\Bbb{Q}(\alpha)$ so that transitivity is not satisfied.
Even though $\mathcal{N}$ is not filterable nor transitive, hence not distinguished, it is not completely undistinguished, so to speak:
Theorem V.3.4: $\mathcal{N}$ is
$ \quad$1'. half-filterable: $k\leq F\leq E,k\leq_{\mathcal{N}}E \implies F\leq_{\mathcal{N}} E$,


*closed under liftings,

*closed under compositums.
