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Let $X\subset\mathbb{P}^n$ be a smooth projective variety of dimension $\geq 2$ and assume that it is not contained in any hyperplane. Now, take some hyperplane $H\subset\mathbb{P}^n$ and consider the set $X\cap H$. Is it possible that there exists another hyperplane $H'$ containing the set $X\cap H$ if we assume furthermore that $X\cap H$ does not contain ruled components?

EDIT1: As @Lev Borisov pointed out in his answer, such examples exist if we don't put any restrictions on the geometry of the intersection set.

EDIT2: The example can be generalized in such a way that the set $X\cap H$ is almost arbitrary. So the assumption about ruled components does not make any difference.

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  • $\begingroup$ While an answer has been accepted, what happens if the intersection is interpreted scheme-theoretically? $\endgroup$
    – Jack Huizenga
    Commented Oct 7, 2013 at 9:15
  • $\begingroup$ Then, I guess, it is not possible to find such an example. Note, that on a projective variety any two global sections, of an invertible sheaf, have the same divisor of zeros if and only if, they differ by a nonzero scalar. $\endgroup$
    – Tomasz Lenarcik
    Commented Oct 7, 2013 at 12:49

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Take Veronese embedding of ${\mathbb P}^2$ into ${\mathbb P}^5$. Take the hyperplane given by $x_0^2$, where $(x_0:x_1:x_2)$ are coordinates on ${\mathbb P}^2$. Then $x_0^2=0$ will also be (as a set) contained in $x_0x_1=0$.

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  • $\begingroup$ You're right! It wasn't that hard after all :) It also made me realize that I forgot about some assumptions about the geometry of $X\cap H$. Please check out the updated question. $\endgroup$
    – Tomasz Lenarcik
    Commented Oct 6, 2013 at 22:48
  • $\begingroup$ If you consider any $X$ and any ample divisor $D$ on it, then you can take an embedding by $kD$. One of the hyperplanes will be just $kD$, and so on... $\endgroup$
    – Lev Borisov
    Commented Oct 6, 2013 at 23:41
  • $\begingroup$ Ok, now I see that this example is in fact pretty general and can be modified so that the intersection set is almost arbitrary. If $\gamma(x)=0$, $\deg\gamma=d$ is any curve, then consider Veronese embedding of degree $2d$. The hypersurfaces $\gamma(x)^2$ and $\gamma(x)\delta(x)$ (sopposing that $\deg\delta=d$) both contain the curve $\gamma(x)=0$. $\endgroup$
    – Tomasz Lenarcik
    Commented Oct 7, 2013 at 8:46

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