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

Let $X$ be a variety in $\mathbb{C}^{10}$ defined by the ideal $I=\left<xz'-x'z, y'(u+z)-y(u'+z'), t'(u-z)-t(u'-z')+xy'-x'y\right>$ of $\mathbb{C}[x,y,z,u,t,x',y',z',u',t']$. Note that $I+\left<u-z, u'-z'\right>$ gives a determinantal variety $D$ defined over the matrix

$ \left( \begin{array}{ccc} x & y & z \\ x' & y' & z' \end{array} \right).$

It is well-known that $D$ has rational singularities. Is there any machinery theorem to deduce the property of having rational singularities for $X$ from that of $D$?

Any reference is greatly appreciated. Thanks in advance.

share|cite|improve this question
If $u-z, u'-z'$ happen to be a regular sequence on $\mathbb{C}[x,..., t']/I$, then this follows by Elkik's result on deformation of rational singularities (at least then $X$ is rational near $D$). Otherwise I think you are out of luck in terms of general tools. – Karl Schwede Sep 22 '13 at 22:35
However, you might be able to check this using a computer... (let me know if I could help with this). – Karl Schwede Sep 22 '13 at 22:35
I checked this with a computer, it appears to be rational. – Karl Schwede Sep 23 '13 at 2:41
Dear Karl, thanks so much for your answers. They are not a regular sequence because dim(D)=dim(X)-1. Anyway, it's very helpful that you know it has rational singularities. Could you please show me how to verify it? Thanks again. – NN guest Sep 23 '13 at 9:47
up vote 5 down vote accepted

Ok, $X$ is a complete intersection (dimension 7, cut out by three equations in $\mathbb{C}^{10}$).

You can then use the hasRationalSing function described HERE in the D-modules package of Macaulay2.

You can also verify it by the following method. Use the F-singularities PosChar.m2 package which can be downloaded HERE, choose a characteristic (I choose 3), define the same singularity, and then run the isFRegularQGor(R/I, 1) command. This proves that the equation is F-regular and hence F-rational and thus pseudo-rational in characteristic 3. It should follow from some general result (that I'll have to track down a precise reference for that later, it should be contained in a dissertation from the University of Michigan last year) that this implies that R/I has rational singularities in characteristic 0.

In this case, I checked both commands and they both agreed that it had rational singularities.

share|cite|improve this answer
This is cool! Whose thesis was that? – Hailong Dao Sep 23 '13 at 15:09
I think it was the thesis of Zhixian Zhu, in particular it should follow from a variant of Corollary 4.2 of I thought she was graduating last year, but maybe not (she's still listed as a student). – Karl Schwede Sep 23 '13 at 16:32
Many thanks, Karl! This helps a ton. – NN guest Sep 24 '13 at 9:32

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.