Let $K$ be a nonalgebraically 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?
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1$\begingroup$ This is true for an infinite field of char 0 and uses only the fact that any nonempty Zariski open set of a variety contains a $K$rational point. $\endgroup$ – Mohan May 5 '18 at 16:27
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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, JeanPierre, Théorèmes de Bertini et applications. Progress in Mathematics, 42. Birkhäuser, BostonBaselStuttgart (1983). ZBL0519.14002.
[Poo] Poonen, Bjorn, Bertini theorems over finite fields. Ann. Math. (2) 160.3 (2005), p. 10991127. ZBL1084.14026.

$\begingroup$ 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. $\endgroup$ – Takumi Murayama May 5 '18 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). $\endgroup$ – R. van Dobben de Bruyn May 5 '18 at 17:30