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I am having trouble understanding the statement and the proof of Bertini's theorem in Griffiths & Harris book (p.$137$). Frankly, I do not understand a word even after I read several answers on stack. The theorem is

The generic element of a linear system is smooth away from the base locus of the system.

First question. Does the statement above refer to linear of general line bundles rather than just line bundles associated to divisors?

As far as I can say, it refers to a linear system of a line bundle associated to a divisor. Tell me if I am wrong.

Second question. What is the generic element? Or what is the generic pencil?

In the proof, the authors start with "If the generic element of a linear system is singular away from the base locus of the system, then the same will be true for a generic pencil contained in the system; thus it suffices to prove Bertini for a pencil."

Third question. What does the above sentence mean exactly?

Now suppose $\left \{D_{\lambda} \right \}_{\lambda \in \mathbb{P}^1}$ is a pencil

Fourth question. Why do the authors write $D_{\lambda} = (f+\lambda g = 0)$? What do $f,g$ mean here?

The last question relates to the degree of a variety (p.$171$).

Bertini applied to the smooth locus of $V$ the generic $(n-k)$-plane $\mathbb{P}^{n-k} \subset \mathbb{P}^n$ will intersect $V$ transversely and so will meet $V$ in exactly $\mathrm{deg}(V) = ^{\#}(\mathbb{P}^{n-k}.V)$ points.

Last question. What is generic $(n-k)$-plane? In this case, why does it intersect $V$ transversely?

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  1. In your setting (a complex manifold) all line bundles come from divisors and vice-versa.

  2. A generic element of a linear system means that in the $\mathbb P^r$ parametrizing members of that linear system, we consider some dense open subset of $\mathbb P^r$. Generic elements are those parametrized by a point in that dense open. A generic pencil similarly parametrized by a point in a dense open of the Grassmannian $G(2,r+1)$ of $2$-dimensional subspaces of $H^0(L)$ (where $L$ is the line bundle).

  3. The sentence is saying that any "bad" behavior will occur in a pencil, so we don't need to worry about higher-dimensional linear systems.

  4. They mean $f,g \in H^0(L)$, so taking linear combinations of $f$ and $g$ yields a pencil.

  5. A generic plane is parametrized by a dense open subset of the appropriate Grassmannian. The transversality is because transversality is an open condition.

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  • $\begingroup$ 3) I wonder about the possibility of this reduction, would you mind to formulate this and the 'open condition' in 5)? 4) As fas as I understand, f,g stand for two generators of global sections $H^0(L)$. $\endgroup$
    – Alexey Do
    Aug 16, 2020 at 12:08
  • $\begingroup$ There's not really anything more I can think to say. If most points of $\mathbb P^r$ parametrize singular divisors, then in particular most points of a randomly chosen line in $\mathbb P^r$ will parametrize singular divisors. As for 5), I don't think I can say anything more explicit; the subset of planes that don't meet $V$ transversely is a closed subset (I suppose there's some Jacobian determinant one could write down that cuts out the this subset). $\endgroup$ Aug 16, 2020 at 14:01

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