MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
While an answer has been accepted, what happens if the intersection is interpreted scheme-theoretically? – Jack Huizenga Oct 7 '13 at 9:15
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. – Tomasz Lenarcik Oct 7 '13 at 12:49
up vote 3 down vote accepted

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 $x_0x_1=0$.

share|cite|improve this answer
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. – Tomasz Lenarcik Oct 6 '13 at 22:48
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... – Lev Borisov Oct 6 '13 at 23:41
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 $\gamma(x)=0$. – Tomasz Lenarcik Oct 7 '13 at 8:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.