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Let $R$ be a local ring with residue field $k$. Let $A$ be an $R$-algebra which is finitely generated as $R$-module. I want to show that the maximal ideals of $A$ are in one-to-one correspondence with the maximal ideals of $A \otimes_R k$.

I have a gut feeling that one should use the Nakayama Lemma, but I fail to find the right way in which to apply it. I would be thankful if someone could help me out here.

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Let $m$ be the maximal ideal of $R$. We have $A\otimes_Rk\simeq A/mA$. You want to show that every maximal ideal of $A$ contains $mA$, or equivalently that every maximal ideal of $A$ contracts to $m$. Let $M$ be a maximal ideal of $A$. Then $M\cap R$ is a prime ideal of $R$, and the extension $R/M\cap R\subset A/M$ is finite. Since $A/M$ is a field we get that $R/M\cap R$ is also a field, so $M\cap R=m$.

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  • $\begingroup$ Why is $R/(M \cap R) \to A/M$ injective? The kernel of $R \to A/M$ might be bigger than $M \cap R$, or am I mistaken? $\endgroup$ – sf1 Apr 24 '15 at 19:04

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