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I have been thinking on the following for a while and can't seem to crack it.

Let $F, F_1, F_2$ be function fields of one variable over a fixed perfect field, $F_i$ contain $F$, and $P_1, P_2$ are places of $F_1,F_2$ such that $P_1\cap F=P_2\cap F$. Also, assume $[F_1F_2:F_2]=[F_1:F]$. Show that there is a place of $F_1F_2$ above $P_1$ and $P_2$.

If $F_1F_2/F$ isn't separable we can first prove for the separable part, and then there is a unique lift to $F_1F_2$. So, assume everything is separable.

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Let's remove the function field business. It is a question of compatible discrete valuations. Let $F$ be a field with a discrete valuation $v$ and $(F_1,v_1)$, $(F_2,v_2)$ be extensions of $(F,v)$ such that $v_1$ and $v_2$ restrict to $v$ on $F$ (well, after normalizing $v_1$ on $F$ to make the value group $\mathbf Z$, etc.). Assume $[F_1F_2:F_2]=[F_1:F]$, which amounts to saying the natural $F$-algebra homomorphism $F_1 \otimes_F F_2 \rightarrow F_1F_2$ is an isomorphism. Then you want to show there's a discrete valuation on $F_1F_2$ extending both $v_1$ and $v_2$. Now use completions (cont.) –  KCd Apr 15 '12 at 3:18
    
Let $F_v$ be the completion of $F$ at $v$. Completing $F_1$ and $F_2$ at $v_1$ and $v_2$ provides us with $F$-embeddings $\varphi_1 \colon F_1 \rightarrow \overline{F_v}$ and $\varphi_2 \colon F_2 \rightarrow \overline{F_v}$. With these we obtain an $F$-algebra homomorphism $\varphi_1 \otimes \varphi_2 \colon F_1 \otimes_F F_2 \rightarrow \overline{F_v}$, which is an embedding since $F_1 \otimes_F F_2$ is a field: it's isomorphic to $F_1F_2$. Thus we obtain an $F$-embedding $F_1F_2 \rightarrow \overline{F_v}$ which restricts to $\varphi_i$ on $F_i$. The discrete valuation this induces (cont.) –  KCd Apr 15 '12 at 3:22
    
on $F_1F_2$ by pulling back the canonical valuation on $\overline{F_v}$ and renormalizing it to have value group $\mathbf Z$ will be your desired common lift of $v_1$ and $v_2$ to $F_1F_2$. –  KCd Apr 15 '12 at 3:23
    
Here's an example in the number field setting showing how things can break down without the condition $[F_1F_2:F_2] = [F_1:F]$. Let $F=\mathbf Q$, $F_1 = {\mathbf Q}(\sqrt[4]{2})$, and $F_2 = {\mathbf Q}(\zeta_8)$. Then $[F_1:F] = 4$ but $[F_1F_2:F_2] = 2$ because actually $F_1F_2 = {\mathbf Q}(\sqrt[4]{2},i)$ has degree 8 over $F$. (Note $\zeta_8 = (1+i)/\sqrt{2} = (1+i)/\sqrt[4]{2}^2$.) Consider the prime $p=41$. In $F_1$, $p = {\mathfrak p}_1{\mathfrak p}_2$. In $F_2$, $p = {\mathfrak q}_1{\mathfrak q}_2{\mathfrak q}_3{\mathfrak q}_4$. The field $F_1F_2$ is Galois over $F$, and (cont.) –  KCd Apr 15 '12 at 3:54
    
in $F_1F_2$ we have $p = {\mathfrak P}_1{\mathfrak P}_2{\mathfrak P}_3{\mathfrak P}_4$. Index these so ${\mathfrak q}_i$ extends to ${\mathfrak P}_i$ in $F_1F_2$. The primes ${\mathfrak p}_1$ and ${\mathfrak p}_2$ each extend to $F_1F_2$ as a product of two of the ${\mathfrak P}_i$'s. Therefore ${\mathfrak p}_1$ and two of the ${\mathfrak q}_i$'s do not lie under a common prime in $F_1F_2$. In terms of embeddings into extensions of ${\mathbf Q}_{41}$, there are four embeddings of $F_2$ (all with image in ${\mathbf Q}_{41}$) and there are two embeddings of $F_1$. These embeddings (cont.) –  KCd Apr 15 '12 at 4:00

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