I have a question which should be pretty basic commutative algebra, but I can't find a reference and I'm stuck on proving a result myself, so here it goes:

Let $\varphi: S \to R$ be a morphism of commutative rings and let $M$ be a finitely generated $S$-module. Then, we can "extend the scalars of $M$" by considering $M \otimes_S R$, which is naturally an $R$-module. I would like to get a description of $\text{Supp}(M \otimes_S R)$ (as an $R$-module) involving $\text{Supp}(M)$ and $\varphi$.

Maybe this helps: by Eisenbud's Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a description of $\text{Ann}(M \otimes_S R)$ and it is not difficult to see that $\varphi(\text{Ann}(M))R \subset \text{Ann}(M \otimes_S R)$, but I fail to prove the other inclusion.

  • $\begingroup$ Probably there is no description unless $\phi$ is flat. $\endgroup$ – Martin Brandenburg Sep 8 '10 at 10:10
  • $\begingroup$ $\text{Ann}(M\otimes_S R)=\text{Ann}(M)\otimes_S R$. $\endgroup$ – user641 Sep 8 '10 at 12:37
  • $\begingroup$ @Martin: I think I can live with the assumption that $\varphi$ is flat. @Steve: How do you identify $\text{Ann}(M)\otimes_S R$ with an ideal of $R$? I can only see that if $\varphi$ is flat, when the multiplication map $\text{Ann}(M)\otimes_S R \to R$ is injective. Even then I think I'm still stuck with the same inclusion as pointed out in my question... $\endgroup$ – Sebastian Sep 8 '10 at 12:57
  • $\begingroup$ @Каднон Мстакуи, please don't edit dozens of questions in a short space of time. Maybe three or four a day. $\endgroup$ – Gerry Myerson Dec 7 '12 at 5:01

The support of $M$ is the set $\{ \mathfrak q \in \text{ Spec } S \, | \, \kappa(q)\otimes_S M \neq 0\}$, where $\kappa(\mathfrak q)$ denotes the fraction field of the domain $S/\mathfrak q$. (Since $M$ is finitely generated, one sees by Nakayama's lemma that the stalk $M_{\mathfrak q}$ is non-zero if and only if $\kappa(\mathfrak q)\otimes_S M$ is non-zero).

The support of $R\otimes_S M$ is defined analogously.

Now if $\mathfrak p$ is an element of Spec $R$, then $$\kappa(\mathfrak p)\otimes_R (R\otimes _S M) = \kappa(\mathfrak p)\otimes_S M = \kappa(\mathfrak p)\otimes_{\kappa(\mathfrak q)} (\kappa(\mathfrak q)\otimes_S M),$$ where $\mathfrak q$ is the preimage of $\mathfrak p$ in $S$.

Since $\kappa(\mathfrak q) \to \kappa(\mathfrak p)$ is just an inclusion of fields, we see that $\kappa(\mathfrak p)\otimes_R (R\otimes_S M) \neq 0$ if and only if $\kappa(q)\otimes_S M \neq 0$.

In conclusion, the support of $R\otimes_S M$ is the preimage in Spec $R$ of the support of $M$ in Spec $S$.

All of this has a geometric interpretation:

The first description of the support says that, when $M$ is finitely generated, we can check that the stalks of $M$ are non-zero by instead checking if the fibres are non-zero. Then the computation relating the fibre of $R\otimes_S M$ at $\mathfrak p$ to the fibre of $M$ at $\mathfrak q$ just says that the fibre of the pull-back is the pull-back of the fibre.

Finally, note that if you think geometrically, the result makes intuitive sense: we are pulling back a sheaf (i.e. applying $\varphi^*$), and we can think of the sheaf as being a subset of Spec $R$ (namely its support) with extra decoration (i.e. each point has a certain fibre of the sheaf lying over it). When we pull back the sheaf, we pull back the subset (i.e. we pull back the support), and then at each point of the pull back we also pull back the extra decoration. Thinking about supports just involves forgetting the extra decoration.

  • $\begingroup$ Thank you, this is a very nice answer! $\endgroup$ – Sebastian Sep 8 '10 at 13:34
  • $\begingroup$ Would we obtain the same answer without the assumption that $M$ is finetely generated but knowing that $Tor_S(M,R) = 0$? $\endgroup$ – Damien L Jul 25 '14 at 17:14
  • $\begingroup$ I'm pretty curious whether we can obtain some relation between $\operatorname{Ann}_A(M)$ and $\operatorname{Ann}_B(M\otimes_AB)$? $\endgroup$ – Yai0Phah Feb 3 '16 at 8:51
  • $\begingroup$ Except the obvious one indicating something about the radical ideal. $\endgroup$ – Yai0Phah Feb 3 '16 at 8:59

Since you say that you can't find a reference, look at Atiyah Macdonald Chapter 3, Exercise 19(viii).


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