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Suppose that $I$ is ideal sheaf of $\mathcal{O}_\Bbb{P^r}$. Let $X$ be a scheme defined by $I$.

Is it true that we can find hyperplane $H$ such that it doesn't contain any components of $X$?

Plurisubharmonic Friend
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david
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  • Over a finite field, there are only finitely many hyperplanes and so it is possible that all of them intersect $X$. I think it should work if you work over an infinite field or allow yourself to take a finite field extension. – Pol van Hoften Dec 09 '17 at 16:02
  • https://en.wikipedia.org/wiki/Theorem_of_Bertini – KReiser Dec 10 '17 at 00:12

1 Answers1

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You can do this when $k$ is infinite. Here you want to find a hyperplane that does not contain any of the generic points of the irreducible components of $X$. Equivalently we want to find a homogeneous polynomial of degree $1$ that is not contained in any of the primes defining the generic points of the irreducible components.

So let $\mathfrak{p}_i$ be the generic points for the irreducible components and suppose that $(S_+)_1 \subseteq \bigcup_{i=1}^n (\mathfrak{p}_i)_1$. Then in particular this means that $(S_+)_1$, as a $k-$vector space, is the union of $n$ proper subspaces. But this is not possible when $k$ is infinite.

When $k$ is finite the claim is not true. For example, take $\mathbb{P}^1_{\mathbb{F}_2}$ with coordinates $X,Y$ and consider the scheme $X$ defined by $V(XY(X+Y))$. Then it is easy to see that any hyperplane (which are $X$, $Y$ and $X+Y$) contain an irreducible component of $X$.

loch
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