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If $A\subseteq B$ where $A,B$ are commutative domains and $B$ is a finitely generated $A$ module, is $\operatorname{Frac}(A)\subseteq \operatorname{Frac}(B)$ a finite field extension?

I know this extension is algebraic and every element of Frac$(B)$ satisfies a monic polynomial with coefficients in $A$. I am not sure how to prove this extension is finite and not aware of any counterexamples.

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    @QiaochuYuan: Thanks, that's precisely what I was trying, but it helps to know that I am on the right track.2011-12-09

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Yes. Here is why.

Let $S=A\setminus \lbrace 0 \rbrace$. Your finiteness hypothesis implies that also $S^{-1}B$ is finitely generated as a module over $S^{-1}A$.
Hence $S^{-1}B$ is a field since it is a domain of finite dimension over a field.
But then you have $S^{-1}A=Frac(A) \subset S^{-1}B\subset Frac (B)$ with $ S^{-1}B$ a field: necessarily then $ Frac ( B )= S^{-1}B$ and since we already know that $S^{-1}B$ is finite dimensional over $S^{-1}A=Frac(A)$, the same holds for $Frac(B)$.

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    @Georges: I had a follow up question. Your argument shows that $[Frac(B):Frac(A)]\leq n$ where $n$ is the number of generators of $B$ as an $A$ module. Under what conditions does equality hold? Thanks.2012-01-27