TITLE: Bertini theorems for base-point-free linear systems in positive characteristics
QUESTION [17 upvotes]: Suppose that $X$ is a smooth algebraic variety over an algebraically closed (uncountable if it helps) field of characteristic $p > 0$.  Suppose that $L$ is a line bundle, probably ample or at least positive, and that $\delta \subseteq |H^0(X, L)|$ is a linear system.  
It is well known that just because $\delta$ is base-point-free, it does not mean that a general member defines a smooth subscheme (it need not even be reduced, for example Frobenius pull-backs of linear systems, Remark 10.9.3 in Hartshorne's algebraic geometry).
However, let's suppose the following:
$f : X \to Y$ is a map, $M$ is a very ample line bundle on $Y$ and $L = f^* M$.  Suppose further that $\delta$ is the pull-back of the complete linear system $|H^0(Y, M)|$.  Are there any (separability?) conditions on the map $f$ which still guarantee that Bertini holds for $\delta$?  I imagine this must be well known, but I don't know the right references.
In particular, I am looking for conditions weaker than etale (/ etale outside a finite set of points)?  Say in the birational case, or the finite case?

REPLY [2 votes]: I thought it might be useful to post a statement of an answer.  This is contained within Cumino-Greco-Manaresi and also Spreafico's paper which Anton mentioned earlier, but can't really fit in a comment and I think might be useful to someone else.  I probably should have posted this earlier.
Indeed, in the context of my original question, if $f : X \to Y$ induces separable residue field extensions (or separably generated not necessarily algebraic residue field extensions), then everything is fine (say over an algebraically closed field).  I should note that even really nice maps may fail to have separably generated residue field extensions.
Indeed, consider the projection map $\mathbb{A}^2 \to \mathbb{A}^1$ (if you want, this can be an open chart from the projection $\mathbb{P}^1 \times  \mathbb{P}^1 \to \mathbb{P}^1$) over a field $k$ of characteristic $p > 0$.   This corresponds to the inclusion $k[x] \subseteq k[x,y]$.  This does not have separable residue field extensions.  
Consider the ideal $(x-y^p) \subseteq k[x,y]$.  This pulls back to the zero ideal in $k[x]$ and induces the residue field extension $k(x) \subseteq k(x^{1/p})$ which is clearly non-separable.
This indicates to me that even birational maps can fail this property horribly (for example, if I blow up a curve on a 3-fold similar things can be arranged).  Finite maps are much better behaved of course.