MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a question about support of modules under extension of scalars.

Let $f \colon A \to B$ be a homomorphism of commutative rings (with unity), and let $M$ be a finitely generated $A$-module. Recall that the support of $M$ is the set of prime ideals $\mathfrak{p}$ of $A$ such that the localization $M_{\mathfrak{p}}$ is nonzero. Then $\mathrm{Supp} _B(B \otimes_A M) = f^{*-1}(\mathrm{Supp}_A(M))$, the set of prime ideals of $B$ whose contractions are in the support of $M$.

The $\subseteq$ containment is true for any $M$. What's an obvious example of a non-finitely generated module where the other containment doesn't hold?

share|cite|improve this question
up vote 4 down vote accepted

Take $A = \mathbb Z$, $M = \mathbb Q$, $B = \mathbb Z/p\mathbb Z$, were $p$ is a prime.

share|cite|improve this answer
That... is indeed obvious. Embarrassingly so, for me. Thanks! – jdc Dec 26 '10 at 9:05

Okay, a further question. Despite its simplicity, it is not homework.

Atiyah and Macdonald (Ch. 2, Ex. 19.iv)) claim that if $M$ is a sum of submodules $M_i$, then $\text{Supp}(M) = \bigcup \text{Supp}(M_i)$, not specifying what size of set the indices $i$ are to run over. I assumed initially the index set was to be assumed arbitrary, and came up with an argument, which it now occurs to me may be wrong, as your example makes me wonder if the sum should really be finite. (Or I'm possibly just (continuing to be) stupid.)

Continue with your example, so $M = \mathbb Q$ is a $A =\mathbb Z$-module, and $B = \mathbb Z / p \mathbb Z$. Now $\mathbb Q$ is generated over $\mathbb Z$ by the elements $\frac 1 n \in \mathbb Z$, so it is a sum of the submodules $\frac 1 n \mathbb Z$, Thus the elements $z_n = 1 \otimes \frac 1 n$ generate $B \otimes_A M = \mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \mathbb Q$; in fact all are zero, since the generator $z_n = p \otimes \frac 1 {pn}$ is zero already in $\mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \frac{1}{pn}\mathbb Z$.

So $\mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \mathbb Q = 0$ has empty support, even though the $\mathbb Z / p \mathbb Z$-submodules $\mathbb Z / p \mathbb Z \ \otimes_{\mathbb Z} \frac 1{n} \mathbb Z \cong \frac 1{n} \mathbb Z / \frac p{n} \mathbb Z \cong \mathbb Z / p \mathbb Z$ all have $(0)$ in their support.

Is this right, or am I making an obvious mistake? Should the sum of modules be assumed finite for the proposition to hold?

share|cite|improve this answer
In linear algebra, if $V$ and $W$ are vector spaces, and $U$ is a subspace of $V$, then the canonical map $U\otimes W\to V\otimes W$ is injective, so $U\otimes W$ can be (and usually is) considered a subspace of $V\otimes W$. But over an arbitrary ring, this doesn't work anymore (tensoring is right-exact, not left-exact), so your submodules aren't submodules. -- Please open a new thread for any followup question you have, rather than posting it as an answer (which confuses people). – darij grinberg Dec 26 '10 at 23:39
Wow, that is a very obvious mistake. Thank you. I didn't open a new thread because I thought it would be bad to clog up MathOverflow with too many ridiculous questions, but I will try to follow better posting etiquette in the future. – jdc Dec 26 '10 at 23:49

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.