0
$\begingroup$

I am studying first order deformations and a natural question arises.

Situation: Let $X_1$ be a scheme. $\pi: X_1 \to {\rm Spec}~ k[t]/(t^2)$ is a flat morphism of finite type, where $k$ is an algebraically closed field. $X_0$ is the central fiber, i.e., $X_1 \times_{{\rm Spec}~ k} {\rm Spec~} k[t]/(t^2)$. The subscript $0$ will always denote the central fiber of the corresponding object. Assume $X_0$ is regular. $U_1\subset X_1$ is an open subset and $Z_1\subset U_1$ is an effective Cartier divisor.

Question: Is the scheme-theoretic closure of $Z_1$ in $X_1$ an effective Cartier divisor?

My Observations:

  1. An effective Cartier divisor $W_1 \subset X_1$ is flat over ${\rm Spec~}k[t]/(t^2)$.
  2. As in 1, $W_0$ is an effective Cartier divisor in $X_0$.
  3. If $W_1$ is a subscheme of $X_1$, flat over ${\rm Spec~}k[t]/(t^2)$, assume $W_0$ is an effective Cartier divisor, then $W_1$ is an effective Cartier divisor. [Lemma 10.93.1, Stacks Project, Tag 00MD]
  4. $\bar {Z_0}$ need not equal $\bar {Z_1}\cap X_0$. Example: $X_1=\mathbb P^2_{{\rm Spec~}k[t]/(t^2)}$, $U_1={\rm Spec~} k[x,y,t]/(t^2)$ and $Z_1={\rm Spec~} k[x,y,t]/(t^2,x+ty^2)$.
  5. [This is wrong. See the answer.]If for every $x\in X_0$, there is a neighborhood $U$ such that the class group $Cl(U)=0$, then the answer to the above question is positive. This is because locally the deformation is trivial, so we may assume that $X_1={\rm Spec~}\left( A\oplus tA\right)$ where $A$ is some $k-$alegebra which is an UFD. Then we can compute the scheme theoretic closure easily.

[This is wrong. See the answer.]By observation 5, if there would be a counterexample, $X_0$ cannot be as simple as affine planes.

$\endgroup$

1 Answer 1

0
$\begingroup$

Sorry, I do not understand what you say in 5. What if A = k[x, y] and X_0 = Spec(A) and U_0 = D(x) and the effective Cartier divisor is given by the zero locus of y + t/x ? Then (y + t/x)(y - t/x) = y^2 and the scheme theoretic closure of Z_1 is given by an ideal containing both y^2 and xy + t which cannot be principal.

There are results in the literature concerning this question when one takes out something of sufficiently high codimension and X_0 is sufficiently nice. Look in Grothendieck's Cohomologie locale des faisceaux coh´erents et th´eoremes de Lefschetz locaux et globaux (SGA 2) for example.

$\endgroup$
1
  • $\begingroup$ Yes you are definitely right! I made a mistake here. Thank you! $\endgroup$
    – Yang Zhou
    Commented Jan 2, 2014 at 0:21

You must log in to answer this question.

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