I consider the book "Algebra" by Serge Lang and on page 238 he has the theorem 3.4 saying that normal extensions remain normal under lifting. I don't see what he means by that, and therefore also his proof is not really clear to me.
Can maybe someone explain this a bit more to me, because my TA told me something about a diagram that commutes and then the proof should be clear but I can't think about a diagram in this case.
Thanks for your help.
By the lift of a field to another we mean that for $k \subset K$ and $k \subset F$ we look at the field extension $FK$ over $F$. We "lifted" the extension $K/k$ by building the compositum in both field to $KF/kF$ where $kF$ will just be $F$ because $k$ is a subset of $F$ and the compositum $kF$ is the smallest field containing both $k$ and $F$.
Now Lang claims (and proves) that if $K/k$ is normal then so is $KF/F$. He shows this with the help of the property of normal field that any embedding $\sigma: K \to k^a$ will fulfill $\sigma(K) =K$. What we have to show now is that any embedding $$\sigma: KF \to k^a$$ (now $F$ is the "base field" so with such an embedding we always mean a map that induces the identity on $F$ - not only on $k$ anymore) will fulfill $\sigma(KF) =KF$.
For that we look at the field $\sigma(KF)$. We claim that it is the same as $\sigma(K)\sigma(F)$ (the compositum). We know that $\sigma(K)=K$ and $\sigma(F)=F$ because $K$ is normal over $k$ and therefore any $\sigma$ which is the identity on $k$ will give $\sigma(K)=K$ - and here $\sigma$ is already the identity on $F$, so on an even bigger field. $\sigma(F)=F$ follows from just that.
To show $\sigma(KF)= \sigma(K)\sigma(F) = KF$ we need to show that $\sigma(K), \sigma(F) \subset \sigma(KF)$ - that is obvious because $KF \supset K, F$. This gives $\sigma(K)\sigma(F) \subset \sigma(KF)$ because the left side is the smallest field contatining both $\sigma(F)$ and $\sigma(K)$. On the other hand $KF$ is generated by the elements of $K$ over $F$ so $\sigma(KF)$ is generated by the elements $\sigma(K)$ over $F$ which gives the desired equality $\sigma(KF) = \sigma(K)\sigma(F) = KF$.
Another way to prove that would be by using the characterisation of normal extensions by being the splitting field of some family of polynomials. If $K/k$ is generated by the zeros $\{\alpha_i\}_{i \in I}$ of some polynomials over $k$ then $KF$ is generated by the same set over $F$ because $KF = k(F, \{\alpha_i\}_{i \in I}) = F(\{\alpha_i\}_{i \in I})$. Therefore $KF$ is the splitting field of the same family of polynomials that $K$ was the splitting field of - just now viewed as polynomials in $F$.