Bertini's theorem: reduction to pencils I'm studying the proof of Bertini's theorem on "Principles of Algebraic Geometry" by Griffiths and Harris (page 137). The statement is as follows:
The generic element of a linear system is smooth away from the base locus of the system.
The authors claim that it suffices to prove Bertini for a pencil (that is, a linear system of dimension one). Unfortunately, I am not able to understand how this reduction to pencils is possible. I am quite a newcomer in Algebraic Geometry, so I'm looking for a very detailed explanation. It is probably a triviality, nonetheless I couldn't manage to find a proof, either by myself, or by asking my colleagues for help.
Thank you in advance!
 A: I asked myself this question some time ago, and I asked for help from several people.
I will now try to reconstruct all the story, even though... 

I am sorry, but I do not know whether the reduction to pencils is
   actually possible; I am trying to outline why in my opinion it is
   not possible.
  In summary, here is the reason why I think it's not possible: The
  Bertini condition is not (visibly) open.

Here is the strategy: I want to give a (false!) proof of the fact that the reduction is possible, and I will mark with a $(\star)$ the point where, won by tempation, I will be deducing something that is likely to be false. 
So, let's get started: we want to show that
$$
\textrm{Bertini true for a generic pencil} \overset{(\bullet)}{\Longrightarrow}\textrm{Bertini true for the whole system.}
$$
Notation: $X$ is our variety, and we have a linear system $\psi:X\dashrightarrow \mathbb P^r$ of dimension $r>1$. We will call "good" a divisor with no singular points outside the base locus of the linear system it is regarded into; it will be called "bad" otherwise. Let us introduce the Grassmannian $G=\mathbb G(1,r)$, the space of lines in $\mathbb P^r$. We denote by $p:\mathbb P^r\times G\to \mathbb P^r$ and $q:\mathbb P^r\times G\to G$ the projections.
Proof of $(\bullet)$.
Let us assume that Bertini holds for a generic pencil $\ell\subset \mathbb P^r$ contained in the big system: this means that for $\lambda$ generic in $\ell$ the divisor
$D_\lambda$ is good with respect to $\ell$. We want to find a dense open subset of $\mathbb P^r$ parameterizing good divisors for $\mathbb P^r$. To do so, let us describe the "bad" points of $\mathbb P^r$. There are two possibilities:


*

*The point $\lambda\in\mathbb P^r$ is on a "bad" line, that is, a line for which Bertini does not hold: there is no dense open subset of this line
parameterizing good divisors for this pencil. Since by hypothesis we deal with a generic pencil, 
$$\textrm{the set of these lines is a proper closed subset } B\subset G.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\star)$$

*Let us call $U=G\setminus B$ the open set of "good" lines. If $[\ell]\in U$ is one of them, it might contain some bad points $P_1(\ell),\dots,P_{n_\ell}(\ell)$. However, these points are in a finite number, because of our assumption (Bertini): the generic $\lambda\in\ell$ is good. So we have a bad locus
$$
C=\{(P_i(\ell),[\ell])\,|\,1\leq i\leq n_\ell\,\textrm{ and }\,[\ell]\in U\}\subset\mathcal L,
$$
where $\mathcal L\subset \mathbb P^r\times G$ is the universal line. 
As the dependence of the points $P_i(\ell)$ on the line $\ell$ is algebraic (it is given by the equation of $\ell$), we conclude that $C$ is a closed
subset of $q^{-1}(U)$. 


Summary: $q^{-1}(B)$ consists exactly of the bad points on $\mathcal L$ which lie on some bad line (we get these through point 1); the only other possibility for being bad is to lie on $C$ (we get these through point 2). 
So $q^{-1}(B)\cup C$ is the bad locus inside $\mathcal L$; note that it is closed. The bad points of $\mathbb P^r$ which we are interested in are exactly the points in $p(q^{-1}(B)\cup C)$. But $p$ is a proper morphism, hence closed, so the bad points form a closed subset of $\mathbb P^r$. Hence the good points form an open, thus dense, subset of $\mathbb P^r$, which means that Bertini holds for the $r$-dimensional linear system. $\,\,\,\,\,\square$
Yes, but to assert $(\star)$ is to say that the Bertini condition is open, and this is by no means evident.
$$\ast\,\,\,\,\,\,\,\,\ast\,\,\,\,\,\,\,\,\ast$$
Another approach (in fact, just a translation): try to show $(\bullet)$ by contrast.
Let us give a name, say $V$, to the good locus, i.e. the locus of hyperplanes giving a smooth transversal intersection with our variety $X$:
$$
V:=\{\textrm{hyperplanes }H\,\textrm{such that}\,|\,\psi^{-1}(H)\,\textrm{is smooth of dimension}\, \dim X-1\}.
$$
Then, to say that Bertini is false for $\mathbb P^r$ is to say that $V$ has empty interior (i.e. there is no open subset of $\mathbb P^r$ parametrizing good divisors. I cannot see why this should imply that for a generic line $\ell\subset \mathbb P^{r}$
the corresponding $V_\ell$ has empty interior as well.
