Let $K$ be a non-algebraically closed (infinite) field of characteristic $0$ and $X$ a smooth, projective $K$-variety. Does there exist an ample invertible sheaf $\mathcal{L}$ on $X$ such that a general element of the linear system $|\mathcal{L}|$ is a smooth $K$-variety? If not true in general, is there any condition on $X$ under which this holds true?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ This is true for an infinite field of char 0 and uses only the fact that any non-empty Zariski open set of a variety contains a $K$-rational point. $\endgroup$– MohanCommented May 5, 2018 at 16:27
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
This is true both over finite and infinite fields.
- For infinite fields, see [Jou, Cor. I.6.11(2)]. It works for a general section of any very ample line bundle $\mathscr L$, using that over an infinite field a nonempty open subset of $|\mathscr L| \cong \mathbb P^N$ contains a rational point.
- For finite fields, see [Poo, Thm. 1.1]. It requires taking a high power of your given very ample line bundle $\mathscr L$.
References.
[Jou] Jouanolou, Jean-Pierre, Théorèmes de Bertini et applications. Progress in Mathematics, 42. Birkhäuser, Boston-Basel-Stuttgart (1983). ZBL0519.14002.
[Poo] Poonen, Bjorn, Bertini theorems over finite fields. Ann. Math. (2) 160.3 (2005), p. 1099-1127. ZBL1084.14026.
answered May 5, 2018 at 17:03
-
$\begingroup$ What happens when you replace "smooth" by "regular"? The reason I ask is that [Flenner–O'Carroll–Vogel, Cor. 3.4.14] seems to say that Bertini holds for regular varieties over infinite fields. $\endgroup$ Commented May 5, 2018 at 17:17
-
$\begingroup$ @TakumiMurayama: ah, I was probably thinking of the situation when you fix the line bundle (like in the classical Bertini theorem). $\endgroup$ Commented May 5, 2018 at 17:30