Compositum of normal extensions is a normal extension I'm trying to prove that if $ F \subset K, F \subset M $ are normal extensions, $ K,M \subset E $, then $ KM$ is also a normal extension of $ F $.
I tried using the fact that $ F \subset KM $ is a normal extension iff for any $ f \in K[x] $ such that $ f(km) = 0 $ for some $ km \in KM $, $ f $ splits into linear factors in $ KM[x] $. So I would write 
$$ f(km) = a_0 + a_1km + \dots + a_n (km)^n = 0$$
$$ g(m) = a_0/k^n + a_1/k^{n-1} m + \dots +a_nm^n = 0 $$
And I thought it would lead me to using the fact that $ F\subset M $ is normal. But $ b_i = a_i/k^{n-i} $ are in $ K $, not necessarily in $ F $.
I'm not sure if that's the right way. I'd appreciate some suggestions

Let $\sigma$ be a homomorphism $KM \to \overline F$, that fixes $F$. We have $\sigma(KM)=\sigma(K)\sigma(M)$. By assumption we have $\sigma(K) \subset K$ and the same holds for $M$. So we have $\sigma(KM) \subset KM$, which is one characterisation of a normal extension.
