0
$\begingroup$

What is the correct definition of a regular point of a map in algebraic geometry?

More specifically, let $f:X\to Y$ be a map of varieties with $f(p) = q$, and let $Z=f^{-1}(q)$. Let $\hat{X}$ be the formal completion of $X$ at $p$, $\hat{Y}$ the formal completion of $Y$ at $q$, and $\hat{Z}$ the formal completion of $Z$ at $p$. I would like a nice condition on $f$ that implies that $$\hat{X}\cong \hat{Y}\times\hat{Z}.$$ Intuitively, this should be an algebraic version of the statement that $p$ is a regular point of $f$.

I do not want to make any smoothness assumptions about $X$ or $Z$. In particular, if $X = Y \times Z$ and $f$ is the projection onto $Y$, then every point of $Y\times Z$ should be a regular point of $f$, even if $Z$ is singular. (The conclusion about formal neighborhoods is obviously true in this trivial case!)

$\endgroup$
5
  • $\begingroup$ A necessary condition should be that for each formal isomorphism class of point, the locus in $X$ of points $p$ such that isomorphism class of $f^{-1}(f(p))$ at $p$ is that isomorphism class should be smooth over $Y$. I'm not sure if that's sufficient. $\endgroup$
    – Will Sawin
    Nov 20, 2014 at 18:51
  • 1
    $\begingroup$ Are you sure this condition is what you want? Take $Y= \operatorname{Spec} k[t]$ and $X= \operatorname{Spec} k[x,y,t]/(xy(x-y)(x-ty))$ and $f: X \to Y$ the obvious map. Then none of the fibers are formally isomorphic to their neighbors near $0$, so $f$ is not regular at $0$ for any $t$, even though for most practical purposes this map is regular away from $t=0,1,\infty$. You could consider weakening the definition. $\endgroup$
    – Will Sawin
    Nov 20, 2014 at 18:55
  • $\begingroup$ @WillSawin: I'm not sure that I understand your example. We have $Y = \mathbb{A}^1$, and for all $q\in\mathbb{A}^1\smallsetminus\{0,1\}$, $Z$ is a union of four lines, glued at the origin. I agree that $(0,0,q)$ feels like it should be a regular point of $f$, and it seems to me that the formal completion of $X$ at $(0,0,q)$ does look like the product of the formal completion of $Z$ at 0 and the formal completion of $\mathbb{A}^1$ at $q$. Is this not the case? $\endgroup$ Nov 20, 2014 at 19:46
  • $\begingroup$ It is not the case. Take any two elements of the formal completion that generate the maximal ideal $(x,y)$. They will satisfy a polynomial relation of degree $4$. The leading term will be, up to a linear change of coordinates, equal to $xy(x-y)(x-ty)$. The action of $GL_2$ on binary quartic polynomials has a nontrivial invariant, the $j$ invariant, which is a nonconstant function of $t$, and so no change of variables can make this a constant family. $\endgroup$
    – Will Sawin
    Nov 20, 2014 at 21:48
  • $\begingroup$ Okay, that's an interesting example. I guess you've convinced me that the isomorphism I requested will be less common than I had thought. For what it's worth, the conclusion that I really care about is that the cohomology of the stalk of $IC_X$ at $p$ agrees with the cohomology of the stalk of $IC_Y$ at $q$ times the cohomology of the stalk of $IC_Z$ at $p$. Perhaps my hope for an actual formal trivialization of $\,f$ near $p$ is unrealistic. $\endgroup$ Nov 20, 2014 at 22:01

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.