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Let $A$ be a normal ring with quotient field $K$. Let $L/K$ be a finite separable extension. Let $E_1/K$ and $E_2/K$ be extensions of $K$ contained in $L$. Let $B_1$ (resp. $B_2$) be the normalization of $A$ in $E_1$ (resp. $E_2$) and let $C$ be the normalization of $A$ in $L$. Then $$K\subset E_i\subset L\ \mbox{and}\ A\subset B_i\subset C\ \mbox{for}\ i=1, 2.$$ Let $\mathfrak P$ be a prime ideal of $C$. Define $\mathfrak p_1:=\mathfrak P\cap B_1$, $\mathfrak p_2:=\mathfrak P\cap B_2$ and $\mathfrak p:=\mathfrak P\cap A$. Then, on the level of residue fields, we have $$k(\mathfrak p)\subset k(\mathfrak p_i)\subset k(\mathfrak P)\ \mbox{for}\ i=1, 2.$$

Assume that $B_1$ and $B_2$ are finite etale $A$-algebras.


a) Does $L=E_1E_2$ (composite field) imply $k(\mathfrak P)=k(\mathfrak p_1) k(\mathfrak p_2)$?

b) Does $K=E_1\cap E_2$ imply $k(\mathfrak p)=k(\mathfrak p_1)\cap k(\mathfrak p_2)$?


a) For my application it would be enough to know the answers in the special case where $E_1/K$ and $E_2/K$ are Galois extensions.

b) Somebody told me that the answer to a) would be ``no'', if we omitted the hypothesis that the algebras $B_i/A$ are etale, even in the case where $K$ is a number field.

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up vote 6 down vote accepted

a) Consider the canonical map $B_1\otimes_A B_2\to C$. It is finite ($C$ is finite over $A$) and is surjective on the generic fiber. The spectrum of the lefthand side is a disjoint union of normal schemes because of the étale hypothesis. This implies that the spectrum of $C$ is actually one of these connected components. So $\mathrm{Spec} C\to \mathrm{Spec}(B_1)\times_A \mathrm{Spec}(B_2)$ is a closed immersion. This implies easily a).

b) Counterexample: take two quadratic extensions of $\mathbb Z$ defined by $\sqrt{a}$ and $\sqrt{a+p}$ with $a$ an integer prime to $p$. They have the same residue field at $p$.

[Add] In a), Galois hypothesis is not needed. The extensions $\mathbb Z[\sqrt{p}]$ and $\mathbb Z[\sqrt{ap}]$ with $a$ not a square mod $p$, give a non-étale counterexample to a).

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Thanks a lot for this fast answer! – Sebastian Petersen Jan 4 '11 at 15:23

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