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Let $X$ be a nonsingular closed subvariety of $\mathbb{P}^n_k$, where $k$ algebraically closed field. Bertini Theorem say that there exists a hyperplane $H \subset \mathbb{P}^n_k$, not containing $X$ such that $H\cap X$ is nonsingular.

In proof, we for a closde point $x\in X$ define a map of $k$- vectorspaces $\Gamma(\mathbb{P}^n_k,\mathcal{O} _{\mathbb{P}^n_k}(1)) \rightarrow \mathcal{O}_{X, x}/\mathfrak{m}^2_x$. Using this map, we can know that $B_x :=\{H | X\subseteq H~ \mbox{or} ~X \nsubseteq H ~\mbox{but}~ x\in H\cap X,~ \mbox{and}~ x ~\mbox{is not a nonsigular point of}~ H\cap X \}$ is a linear system of hyperplanes with dimension $n-r-1$ where ${\rm dim}X=r$. Consider the complete linear system $|H|$ as a projective space and let $B$ be the subset of $X \times |H|$ consisting of all pair $(x,H)$ such that $x\in X$ is a closed point and $H\in B_x$. Then $B$ is the set of closed points of a closed subset of $X\times |H|$ say $B'$. we assume that $B'$ is reduced.

I have some question:

  1. what is $B'$? I think that $B'$ is the closure of $B$...
  2. The projection $P_1: B' \rightarrow X$ is surjective??? I know that the preimage of the closed point of X is nonempty....
  3. $B'$ is irreducible with dimension $n - 1$????
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First a remark. Bertini's Theorem says more than what you assert: it says that for $X\subset\mathbb P^n$ nonsingular projective over $k=\overline k$ there exists a dense open subset $$U\subset |H|=\mathbb P^{n\ast}=\{\textrm{Hyperplanes in }\mathbb P^n\}$$ such that $H\cap X$ is nonsingular of dimension $r-1$ for every $H\in U$.

The strategy of the proof is as follows:

  1. Show that $B$ has the structure of a projective variety.
  2. Show that the first projection $p_1:B\to X$ is surjective, and that we have $\dim\,B\leq n-1<n$.

If this is done, Bertini's Theorem is proved. Indeed $\dim\, B<n$ implies that the second projection $p_2:B\to |H|$ sends $B$ onto a proper subset of $|H|=\mathbb P^{n\ast}$. Moreover, $p_2(B)$ is closed because $p_2$ is a closed map. Then it is easy to see that we were looking for the "good" (open dense) locus $$U=|H|\setminus p_2(B).$$

Now, it is not restrictive to assume that $r=\dim\,X<n-1$. Indeed, if $X$ is a smooth hypersurface, we can just take $U=|H|\setminus X^*$, where $X^*$ is the dual variety (closed in $|H|$).

As for your first question, it goes as follows: you can consider the family $B'\subset X\times |H|$ of subvarieties of $|H|$ defined by $$B'=\bigcup_{P\in X}\{P\}\times B_P\to X.$$ The members ($=$ fibers) of this family are clearly the spaces $B_P\subset |H|$, which are projective spaces of dimension $n-r-1$: this answers affirmatively your third question. So $B'\to X$ is clearly surjective as we assumed $n-r-1>0$, and this answers your second question.

Note (if you want to) that to prove point 1. you just have to realize that $B=B'\cap\mathcal H$, where $\mathcal H\subset \mathbb P^{n}\times\mathbb P^{n\ast}$ is the universal hyperplane.

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  • $\begingroup$ My pleasure. Regards $\endgroup$ – Brenin Sep 17 '13 at 8:23

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