Suppose $F\subset K,F\subset E$ are Galois extensions. Show that EK is Galois.
Proof:
$K=F(a_1,..,a_r) $ where $a_i$ are separable and (are roots of $f(x)$) over $F$(Since K is a splitting field of some separable polynomial f(x)) .
Similarly, $E=F(b_1,..,b_n) $ where $b_i$ are separable and (are roots of $g(x)$) over $F$.
Hence, $EK=F(a_1,..,a_r,b_1,..,b_n)$ (Hence, EK is the splitting field of $f(x)g(x)$ and therefore normal over F )
This is because EK is the smallest field containing E and K and $F(a_1,..,a_r,b_1,..,b_n)\subset EK$
By Primitive element Theorem $F(a_1,..,a_r,b_1,..,b_n)=F(c)$ where $c$ is separable over F.
Let M be the splitting field of the separable minimal polynomial of $c$.
Hence, $F\subset EK=F(c ) \subset M$. Since M is Galois over F and EK is normal over F this implies that EK is Galois over F.
Is this proof valid? Thanks