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A divisor $D$ is an element of the quotient sheaf $\mathcal{R^*}/\mathcal{O^*}$ on a compact complex manifold $X$. They can be also thought of via valuations as formal sums of hypersurfaces.

A complete linear system $\mathcal{L}(D)$ is defined to be all positive divisors equivalent to $D$. These are bijective with the subspace of meromorphic functions $f$ such that $(f)+D\geq0$. This is bijective with $\mathbb{P}(H^0(X,[D]))$ (Holomorphic sections of the line bundle corresponding to D). A linear system is a subset of $\mathcal{L}(D)$ corresponding to a subspace of either of the last two sets.

The base locus of such a system of positive divisors is the set of divisors which occur in all the divisors in the system.

Bertini's theorem states that the generic element of a linear system is smooth away from the base locus of the system. This means that on a dense open set in the vector spaces above, the corresponding divisors are non-singular away from the base locus.

How do I use this to prove the more familiar fact that given a submanifold $M$ of some $\mathbb{P}^N$, there is a hyperplane $H=\mathbb{P}^{N-1}$ with $H\cap M$ non singular.

I hope someone can make some suggestions on how to proceed.

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  • $\begingroup$ This is an easier case of Bertini. I suggest you look the proof in Hartshorne. The more difficult version is that a general hyperplane section of a such an $M$ possibly singular, but irreducible with $\dim M\geq 2$ is still irreducible. This proof is not in Hartshorne. $\endgroup$
    – Mohan
    Nov 16, 2016 at 1:09
  • $\begingroup$ My linear reading of Hartshorne has not got me that far yet. I will look at that proof in the future. But I want to avoid that since it's stated this way in Griffiths-Harris (page 162). $\endgroup$
    – Latimer
    Nov 16, 2016 at 16:02
  • $\begingroup$ If you are willing to use your version, just use the fact that the linear system of hyperplanes have base locus empty. $\endgroup$
    – Mohan
    Nov 16, 2016 at 16:10
  • $\begingroup$ So look at $\mathcal{L}(H_0)$ assuming $M$ is not in $H_0$, (To conclude that $H_0\cap M$ is a divisor in M, I would need Chow's thoerem to show M is algebraic and then use some dimension theorem?) then note that the divisors $(\frac{x_i}{x_0})+H_0$ are in here with base locus empty? Then use G-H Bertini to say that some divisor $(a_0\frac{x_0}{x_0}+a_1\frac{x_1}{x_0}+...+a_n\frac{x_n}{x_0})+H_0$ is smooth? This corresponds to the intersection of a hyperplane with $M$? Is this ok? $\endgroup$
    – Latimer
    Nov 16, 2016 at 18:15
  • $\begingroup$ @Mohan Would I really have to assume $M$ is algebraic? G-H don't. Another way to show this is codimension $1$ is to show the intersection is transversal. But If I had this, smoothness comes for free. $\endgroup$
    – Latimer
    Nov 16, 2016 at 18:18

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This is just paraphrasing Hartshorne's proof. Look at the hyperplanes tangential at a point p∈M. This is a codimension N−dimM−1 linear subspace in the dual space $\mathbb{P}^{N∗}$. So, as p varies, it sweeps out a subvariety of the dual space of dimension at most N−1. So, general hyperplane is transversal at every point of M.

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