2
$\begingroup$

Let $X\subset \mathbb{P}^n$ be a reduced closed subscheme. For a general hyperplane $H$, $X\cap H$ is again reduced (and of dimension one less). Is there an easy proof of this result?

Algebraically, we take an algebraically closed field $k$, an homogeneous radical ideal $I\subset k[X_0,\ldots,X_n]$ and ask that for $a_0,\ldots,a_n\in k$ general, the ideal generated by $I$ and $a_0X_0+\cdots +a_nX_n$ is again radical (and of dimension one less).

The result follows from Bertini theorem if $X$ is smooth, but I do not assume it here, and would like to have a simple proof of this.

$\endgroup$
  • $\begingroup$ Welcome new contributor. Variants of this question arise periodically on MO, cf. mathoverflow.net/questions/282781/… There are various approaches, as well as several "standard" references (EGA, the book by Jouanolou, etc.). Which of these is most useful to you depends on your background and what you consider to be "easy" and "simple". $\endgroup$ – Jason Starr Jan 27 at 23:05
  • $\begingroup$ Thanks for the help. As I am just asking reduced, which is much more weaker that the Bertini's theorem, I would be happy with a proof which uses essentially "nothing", so which could be in the beginning of a course in algebraic geometry, essentially after the definition of the Hilbert polynomial associated to ideals, I would be happy to say that the degree of a radical ideal is the number of points we get by intersecting a general linear space of the good dimension. But to prove this I need that the intersection is reduced... $\endgroup$ – user2718 Jan 28 at 9:02
  • $\begingroup$ "... which is much more weaker than the Bertini's theorem ..." That depends on your background. Every proof uses some algebra, e.g., Proposition 4.6.1, p. 68, of EGA IV_2. Most proofs also use some geometry, e.g., smoothness over $X$ of the incidence correspondence $I=\{(p,[H])\in X\times (\mathbb{P}^n)^\vee : p\in H \}$ and density of the open smooth locus $X_{\text{sm}}$ of $X$. Thus $I$ is reduced, and the inverse image $I_U$ of $U$ in $I$ is also dense. Taken together, these prove that the geometric generic fiber of the following projection is reduced: $I\to (\mathbb{P}^n)^\vee$. $\endgroup$ – Jason Starr Jan 28 at 11:53
  • $\begingroup$ Typo correction: The dense open $U$ at the end of the comment should be $X_{\text{sm}}$, the dense open in $X$ that is the smooth locus. Thus, $I_U$ is the inverse image in $I$ of $X_{\text{sm}}$. $\endgroup$ – Jason Starr Jan 28 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.