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This is from Atiyah-Macdonald Ex7.23. (It seems that the entire problem is not needed, but I will write it just in case: Let $A$ be a Notherian ring, $f:A \to B$ a ring homomorphism of finite type. $f^*:\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ be the mapping associated with $f$. Then the image under $f^*$ of a constructible subset $E$ of $Y$ is a constructible subset of $X$.)

In the hint of the book, or solution, the following is written without any exposition: $\operatorname{Spec}(B/\mathfrak{p}^e)=\operatorname{Spec}((A/\mathfrak{p}) \otimes_A B)$ where $\mathfrak{p}$ is a prime ideal of $A$. So it may be very basic and elementary, but I don't get it easily. (I think I'm not understanding tensor product well.) Why does it hold?

What I have found regarding this is that for $\alpha$ an ideal, $M$ an $A$-module, $M/\alpha M$ is isomorphic to $(A/ \alpha) \otimes_A M)$. (Ex2.2) But this does not say the ring isomorphism. Is this fact needed or is there a more easy way to prove it?

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up vote 6 down vote accepted

I'm assuming $\mathfrak{p}^e$ is what A and M call the extension of the ideal $\mathfrak{p}$ to $B$, so, using the notation from the third paragraph of your post, $\mathfrak{p}^e=\mathfrak{p}B$. The map $(A/\mathfrak{p})\otimes_AB\rightarrow B/\mathfrak{p}B$ given by $(a+\mathfrak{p})\otimes b\mapsto ab+\mathfrak{p}B$ is an isomorphism of $A$-modules (can you prove this?), so you just have to check that it respects the ring structure on both sides (in fact it's an isomorphism of $B$-algebras).

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No, I meant $A$-modules. I just wanted the OP to think about it in terms of modules first. –  Keenan Kidwell Jun 20 '11 at 13:09
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It should be easy to check that $B/\mathfrak{p}B$ and $(A/\mathfrak{p}) \otimes_A B$ are isomorphic as rings: at some point during the proof, you will probably get a map $B \to (A/\mathfrak{p}) \otimes_A B$ given by $b \mapsto 1 \otimes b$ with the correct kernel (as in the case where $B$ is a module), and this is very much a ring homomorphism.

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