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?
 A: 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.  
