Given two normal (finite) extensions $E_1,E_2$ of $F$, such that $E_1,E_2 \subseteq \bar{F}$, prove that $E_1E_2 = \{\sum_{i=1}^d e_1e_2: e_1 \in E_1, e_2 \in E_2, d>0\}$ is normal.
I see the answer of Compositum of normal extensions is a normal extension. But why does $\sigma(KM) = \sigma(K)\sigma(M)$? An element in $KM$ is not a single product but finite sum of products.
But more importantly, can we prove this statement without using that characterization? Answer in Compositum of normal extensions is normal only prove infinite case assuming the finite case.
Best Answer
$E_1E_2 $ is the smallest subfield of $\overline {F} $ which contains elements of both $E_1$ and $E_2$. Thus any element of $E_1E_2 $ is made using elements of $E_1,E_2$ using field operations. So an arbitrary element of $E_1E_2 $ can be written as $$\left. \sum_{i=1}^m a_ib_i\middle/\sum_{i=1}^n c_id_i\right.\tag{1} $$ where $a_i, c_i\in E_1$ and $b_i,d_i\in E_2$. Now the denominator here contains $2n$ variables and each of them is a root of some irreducible polynomial in $F[x] $. Here we use the fact that the extensions $E_1/F,E_2/F$ are finite and hence algebraic.
Thus assuming these variables are roots of irreducible polynomials of degrees $k_1,k_2,\dots,k_{2n}$ we can observe that their conjugates also lie in same field. In other words cionjugates of $c_i$ lie in $E_1$ while those of $d_i$ lie in $E_2$. Thus we can replace each variable in denominator with its conjugate and multiply the denominator with such expressions. Finally the denominator will consist of a product of $$k_1k_2\dots k_{2n}$$ such expressions (the first of which is already the denominator of $(1)$). We need to use the same expressions except the first one to multiply the numerator of $(1)$ also so that the fraction does not change. The denominator is now a symmetric function of $c_i, d_i$ and their conjugates and hence belongs to $F$.
The numerator on the other hand is a product of linear combinations of elements of $E_1,E_2$ and hence the overall expression can be written in the form $$\sum_{i=1}^p e_if_i$$ where $e_i\in E_1,f_i\in E_2$. Moreover any expression of above form lies in $E_1E_2 $ as it is made up of elements of $E_1,E_2$ using sums and products.
Next we show that $E_1E_2 $ is normal over $F$. It is better to use the property that if $\sigma$ is an automorpshism of $\overline {F} $ which fixes $F$ then we have $$\sigma (E_1)\subseteq E_1,\sigma(E_2)\subseteq E_2\tag{1}$$ Next let us apply $\sigma$ on any arbitrary element $\sum_{i=1}^n a_ib_i$ (where $a_i\in E_1,b_i\in E_2$) of $E_1E_2 $ to get $$\sigma(\sum_{i=1}^{n}a_ib_i)=\sum_{i=1}^n\sigma(a_i)\sigma(b_i)$$ and the right hand side lies in $E_1E_2$ because $\sigma(a_i) \in E_1,\sigma(b_i)\in E_2$. Thus applying $\sigma$ on any arbitrary element of $E_1E_2 $ gives us another element of $E_1E_2 $. This means that $$\sigma(E_1E_2) \subseteq E_1E_2 $$ and this holds for any arbitrary automorphism $\sigma$ of $\overline {F} $ which fixes $F$. Thus $E_1E_2 $ is also normal over $F$.