21
$\begingroup$

Let $R$ be an excellent Noetherian ring. A property $P$ is said to be open if the set $\{q \in \operatorname{Spec}(R) \ | \ R_q \ \text{satisfies} \ (P)\}$ is Zariski open. Examples of open properties include:

  1. Regular (well-known), complete intersections, Gorenstein, Cohen-Macaulay, Serre's condition $(S_n)$ (Matsumura's book). These imply openness of other properties, for example normality, which means $(R_1)$ and $(S_2)$.

  2. Factorial (for $R$ of characteristic $0$, since the proof uses resolution of singularities). UPDATE: in a recent very interesting preprint, the factorial and $\mathbb Q$-factorial property are proved to be open for varieties over any algebraically closed field.

  3. $\mathbb Q$-Gorenstein, i.e. the canonical module is torsion in the class group (I don't know a convenient reference, please provide if you happen to know one).

  4. Being a rational singularity.

My questions are:

Question 1: Do you know other interesting class of open properties?

Question 2: Are there good heuristic reasons for why a certain property should be open? Phrased a bit more ambitiously, are there common techniques for proving openness for certain class of properties?

Some comments: The excellent condition is quite mild but crucial. Question 2 was motivated by another question of mine.

$\endgroup$
4
  • $\begingroup$ Hi Long, I looked at the question of yours you linked in the last sentence and it has an accepted answer. What is it that you still don't know? $\endgroup$ Commented Feb 6, 2011 at 20:17
  • 1
    $\begingroup$ Sandor, I think that answer was posted after Long posted this question. $\endgroup$ Commented Feb 7, 2011 at 1:24
  • $\begingroup$ Karl, I see. I hadn't realized this was an old question. I see now that it just became "new" since you answered it. :) $\endgroup$ Commented Feb 7, 2011 at 11:15
  • $\begingroup$ The link to Gorenstein's paper at springerlink.com is broken. Perhaps you could take a look, whenever possible... $\endgroup$ Commented Jun 5, 2022 at 12:25

2 Answers 2

11
$\begingroup$

to give a shamelessly trivial answer to your question 2, I would say that the point is usually that the failure of these properties is closed, often "obviously". I would also add one more auxiliary property, that acts as a meta property for some of these:

A coherent sheaf being locally free is an open property. This follows from Nakayama's lemma.

So, here is your list. The properties on the left fail along the loci on the right. [Caveat: I did not include conditions that are sometimes necessary, but I figured that this is a philosophical question and so the answer does not have to be stated in the most precise way.]

regular/smooth -------------- zero set of the Jacobian ideal, also the locus where $\Omega_X$ is not locally free

Cohen-Macaulay, $S_n$ ----- support of appropriate Ext sheaves

Gorenstein ------------------ {not CM} $\bigcup$ {CM but $\omega_X$ is not a line bundle}

$\mathbb Q$-Gorenstein --------------- $\bigcap_m$ {$\omega_X^{[m]}$ is not a line bundle}

rational singularity --------- (in char $0$) $\bigcup_{i=1}^{\dim X}{\rm supp\,}R^i\phi_*\mathcal O_{\widetilde X}$ where $\phi:\widetilde X\to X$ is a resolution.

klt singularity ---------------- zero set of the multiplier ideal.

Du Bois singularity --------- $\bigcup_{i\neq 0} {\rm supp\,} h^i(\underline\Omega_X^0)\bigcup \,{\rm supp\,}{\rm coker\,}[\mathcal O_X\to h^0(\underline\Omega_X^0)]$

(semi-)normality ------------ ${\rm supp\,}{\rm coker\,}[\mathcal O_X\to \pi_*\mathcal O_{Y}]$, where $\pi:Y\to X$ is the (semi-)normalization.

etcetera...

$\endgroup$
3
  • $\begingroup$ Nice explanations. But one should replace "regular" with "smooth" in the first item. $\endgroup$
    – Qing Liu
    Commented Feb 7, 2011 at 10:38
  • $\begingroup$ You're right. Although I call your attention to the "caveat" included for exactly this reason.... $\endgroup$ Commented Feb 7, 2011 at 11:13
  • $\begingroup$ Dear Sandor: thank you for the lovely answer, it is very well worth the wait! Do you have some insight on example 2)? $\endgroup$ Commented Feb 8, 2011 at 1:55
8
$\begingroup$

Long,

For the $\mathbb{Q}$-Gorenstein question, I don't know of a reference either but it should be easy, if $\omega_R^{(n)}$ is locally free at a point, then it's locally free in a neighborhood of that point (of course, I'm probably assuming normal or G1 + S2 to make sense of $\omega_R^{(n)}$).

A couple other that jump to mind are the following.

I. Seminormality / weak normality.

II. Being $F$-split in characteristic $p$ (at least in the $F$-finite case, $F$-finite is another decent one on its own). Of course, all the sings of the MMP (canonical, terminal, lc, klt, slc, rational, Du Bois, etc.) Most of the singularities of tight closure theory (strong F-regularity, F-purity, F-injectivity, F-rationality, some of these requiring again the $F$-finite case).

With reguards to your second question: The general common thing that virtually all the singulraity classes mentioned above possess, which makes them open is the following:

Almost all of these singularities are checked by either showing that a particular module $M$ is isomorphic to $R$ or $\omega_R$. Alternately, that a particular module is zero. For example, klt and log canonical singularities can both be checked by this (is the multiplier ideal / non-lc ideal equal to $R$). This also holds in the characteristic $p$ world, although its not the usual way things are phrased.

For example, strong $F$-regularity can be checked by looking at the ideal $$ J = \sum_{e \geq 0} \sum_{\phi} \phi(F^e_* cR) $$ where $\phi$ runs over all elements of $Hom_R(F^e_* R, R)$ and $c$ is chosen such that $R_c$ is regular. If this ideal equals $R$, then $R$ is strongly $F$-regular.

$\endgroup$
1
  • $\begingroup$ Karl, thanks a lot for resurrecting this, it was very helpful, probably not only for me. $\endgroup$ Commented Feb 8, 2011 at 1:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .