I always believed the following statement: if $X$ is a smooth variety over an algebraically closed field of positive characteristic, assuming we know that the general member of a base point free linear system $|L|$ is reduced, then indeed a general member is smooth.
However, I realize this is not obvious, though all the examples I know which fail this Bertini theorem has non-reduced fibers.
So I was wondering whether indeed this statement is true. Or counterexamples are known.
[REEDIT:] After Laurent Moret-Bailly's nice counterexample. I was then wondering whether a counterexample exists when the linear system induces a birational morphism to its image. The assumption now is closer to the one in Bertini's theorem which says the statement is true when we assume the morphism is an embedding.
This sort of counterexamples will always help birational geometers to think what's going on in characteristic $p$.
Best Answer
In characteristic 3, consider the surface $V\subset \mathbb{P}^2\times\mathbb{A}^1$ with equation $y^2z=x^3-tz^3$ (where $t$ is the coordinate on $\mathbb{A}^1$). It is easily seen to be smooth. The general fiber of the projection on $\mathbb{A}^1$ is a plane cubic which is reduced but not smooth. Taking a suitable projective completion $V\hookrightarrow X$ you get a morphism $X\to\mathbb{P}^1$ which is a counterexample.