Let Y be an affine noetherian scheme, $Z = V_+(F_1, \ldots, F_r)$ a closed subscheme of $\mathbb{P}^d_Y$ that is flat over Y. Let $y_0 \in Y$ be a point such that $Z_{y_0}$ is a complete intersection in $\mathbb{P}^d_{k(y_0)}$. Set $r = dim Z_{y_0}$. I am trying to show that this implies that there is an open neighborhood V of Y such that for all $y \in V$we have that $Z_y$ is a complete intersection in $\mathbb{P}^d_{k(y)}$ of dimension $r$ for every $ y \in V$. I am having no luck here, however. I have tried the following:

  1. Look at the case $r=2$ and show it there. Here still no luck, and the methods I thought of was quite ugly. One was to try to consider $(F_2)/rad(F_1)$ as some sort of quasicoherent sheaf and show something regarding the support. It didn't work however.
  2. Trying to just invert coefficients in polynomials in $F_i$. However, I couldn't show that inverting certain coefficients (those coefficients of $F_i$ not vanishing on $y_0$) gave a local complete intersection.

Any hints or solutions are welcome!

  • $\begingroup$ Do you really mean "local complete intersection" or do you mean that the $Y$-scheme $Z$ is a complete intersection Zariski locally on $Y$? $\endgroup$ – Ariyan Javanpeykar Mar 9 '14 at 21:16
  • $\begingroup$ @Ari I mean that it is a complete intersection zariski locally on Y. $\endgroup$ – user101036 Mar 12 '14 at 17:30

This is only a comment, not a full answer. I hope it will help you a bit on your way.

Let $Y$ be an affine noetherian scheme, and $y_0$ a regular point of $Y$ such that codim $y_0 =1$. Let $Z\to Y$ be a flat morphism of schemes such that $Z_{y_0}\to $ Spec $k(y_0)$ is a complete intersection. Then, we can use Proposition 2.1.12 in Olivier Benoist's thesis; see http://www-irma.u-strasbg.fr/~benoist/articles/Thesefinale.pdf to see that $Z\to Y$ is a complete intersection over the local ring $\mathcal O_{Y,y_0}$ of $y_0$.

In fact, Benoist proves that, since $T= $ Spec $\mathcal O_{Y,y_0}$ is a discrete valuation ring (by the regularity and codimension assumptions on $y_0$), the scheme $Z_T:= Z\times_Y T$ is a complete intersection over $T$, and moreover that the equations for $Z_{y_0}\to $ Spec $k(y_0)$ lift to equations for $Z_T\to T$. (Here we use that there is a canonical morphism $T\to Y$.) Actually, we probably don't need the second part of the result of Benoist to answer your question, but it's good to note it.


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.