Let $R$ be a Noetherian commutative ring and $M$ a finitely generated $R$-module.

What is known about the following subset of $ \mathrm {Spec}(R)$:

$$\mathrm{supp}_{fl}(M)=\{P\in \mathrm {Spec}(R):\ \mathrm{Tor}^R_1(M, R/P)= 0\}?$$

In particular,

Is $\mathrm{supp}_{fl}(M)$ open in $\mathrm {Spec}(R)$?


I suspect that Graham Denham's argument only shows that the weaker condition $\mathrm{Tor}_1(M,\kappa(P))=0$ is open, where $\kappa(P)$ is the residue field of $P$. With the problem as stated, here is a counterexample: for a field $k$, take $R=k[x,y]$ and $M=R/(x,y)$. Then it is easy to check that $\mathrm{supp}_{fl}(M)$ consists of:
$\bullet$ the generic point of $\mathrm{Spec}(R)$,
$\bullet$ all closed points except the origin $(x,y)$, and
$\bullet$ all principal primes $(f)$ where $f$ is irreducible and not in $(x,y)$, i.e. all generic points of curves not containing the origin.

This is not open in $\mathrm{Spec}(R)$.

| cite | improve this answer | |
  • $\begingroup$ Yes, you're right. I'll edit my answer accordingly. $\endgroup$ – Graham Denham Oct 21 '16 at 11:40

At least $\mathrm{supp}_{fl}(M)$ is open in $\mathrm{mSpec}(R)$. To see this, let $(C_\cdot,d)$ be a resolution of $M$ by finitely-generated free modules. By looking at matrices over $R$, one sees $\mathrm{Tor}^R_1(M,R/P)\neq0$ if and only if a maximal prime $P$ contains the (maximal) Fitting ideal of $d_1\oplus d_0$.

These closed sets are the homology jump loci of the complex $C_\cdot$, and the argument I sketch appears in

Nero Budur, MR 2875860 Complements and higher resonance varieties of hyperplane arrangements, Math. Res. Lett. 18 (2011), no. 5, 859--873.

though it goes back further.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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