# Unramification and compositum

The background is: a field $K$ complete with respect to a discrete valuation $|\ |$. We write $A$ and $k$ for his discret valuation ring and the residue field of $A$. We assume that $K$ and $k$ are perfect. We have $L$ an algebraic extension of $K$ with $B$ and $\mathfrak{p}$ for the valuation ring and the residue field.

So my question is: let $K'$ and $K''$ unramified (finite) extensions of $K$ in $L$ with the same residue field $k'$. Why do we have that the compositum $K'\cdot K''$ is an unramified extension of $K$ and that $k'$ is his residue field?

## 1 Answer

Let $A', A''$ be the respective valuation rings of $K', K''$. Then $A'\otimes_A A''$ is unramified over $A$ (used discriminant for example), so it is integrally closed (but not integral in general) because something étale over normal ring is normal. In particular it is a finite product of discrete valuation rings $\prod_i A_i$. Its total ring of fractions is $K'\otimes_K K''=\prod_i \mathrm{Frac}(A_i)$. Consider the canonical map $$A'\otimes_A A''\to B, \quad a'\otimes a''\mapsto a'a''.$$ Its image is an integral quotient of $A'\otimes A'''$ hence is one of the factors, say, $A_1$. We then have $K'K''=\mathrm{Frac}(A_1)$. The residue field of $K'K''$ is the residue field of $A_1$, equal to the image of $k'\otimes_k k'$ in the residue field of $B$, and this is just $k'$.

P.S. I don't have enough reputations to comment. This proof is maybe too complicate. You should unaccept this answer, so other people could provide something simpler. I am not convinced by the first proof because we can lift two different roots of $f(T)\in k[T]$ defining $k'$ to two differents roots $t_1, t_2\in B$ of a lifting $F(T)\in A[T]$ of $f(T)$, and consider $K'=K[t_1]$, $K''=K[t_2]$. If $k'/k$ is not Galois, $K'$ will be different from $K''$ in general.

• The proof is incorrect, but the results is true. – Cantlog Mar 14 '13 at 15:56
• @ Cantlog: In fact I don't see why your first proof wasn't correct. My problem take place in the demonstration of the fact (that you used in your first proof) that K' (or K'') is determined by his residue field. So I'm happy that you propose another proof. – Macadam Mar 14 '13 at 19:34
• @ Cantlog: I don't understand many things on your proof: why $A'\otimes_A A''$ is unramified over $A$? Why can we deduce that $A'\otimes_A A''$ is integrally close? Why can we decude that it is a finite product of DVR? Why the residue filed of $A_1$ is $k'\otimes_k k'$? So maybe could you tell me were I can find all these details... thanks – Macadam Mar 14 '13 at 19:39