TITLE: Does $\operatorname{Spec}(B/\mathfrak{p}^e)=\operatorname{Spec}((A/\mathfrak{p}) \otimes_A  B)$?
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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.