Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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).

share|improve this answer
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

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.