1
$\begingroup$

Let $X$ be a normal, projective variety (can take $X$ to be a hypersurface in a projective space) of dimension at least $3$. Let $L$ be a very ample line bundle on $X$, hence base-point free. What can we say about the singularity of a general element of the linear system $|L|$? For example, can a general element be smooth or has singularities of codimension at least $2$?

$\endgroup$
  • 2
    $\begingroup$ The usual Bertini theorem would imply that the general element in the linear system $|L|$ will have singularities of codimension at least two. Unless $X$ has only isolated singularities, you will not get smoothness. $\endgroup$ – Mohan May 28 at 13:58
  • 2
    $\begingroup$ To emphasize the last point @Mohan made: if any (i.e., not just general) Cartier divisor on $X$ is non-singular, then $X$ is non-singular along that Cartier divisor and hence if that Cartier divisor is ample, then $X$ can only have isolated singularities. $\endgroup$ – Sándor Kovács May 28 at 19:47
0
$\begingroup$

1. General hypersurface of a normal variety is normal;

see ``The hyperplane sections of normal varieties'', Seidenberg, A. (1950), Transactions of the American Mathematical Society, 69(2), 357-386. (the main result is in the 2ed line page 358). It is true in any characteristic.

In the case of positive characteristic, the following is known:

2. The general hypersurface of a quasi-projective 3-fold with terminal (resp.~canonical, klt) singularities has the same singularities;

see Main Theorem of "General hyperplane sections of threefolds in positive characteristic", Sato, Kenta, and Shunsuke Takagi, Journal of the Institute of Mathematics of Jussieu (2018): 1-15.

Remark: Terminal singularity implies the singular locus is at least codimension 3. In particular, surface with terminal singularities is smooth. The above is true in char $=0$ and higher dimensional case; see Lemma 5.17(1), Kollar and Mori's book: Birational geometry of algebraic varieties.

Caution: Bertini theorem for base point free linear systems fails in positive characteristics. In higher dimensional, we don't have the tools of resolution of singularities and also the minimal model program. So the above is still open I think.

$\endgroup$

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.