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This is related to the answer in this question: Showing that a power of an ample sheaf is equivalent to an effective Cartier divisor

Let $X$ be a quasiprojective scheme over a Noetherian ring A and suppose we have a very ample sheaf $\mathcal{L} \cong i^\ast O(1)$ for $i$ an immersion into projective space. Then, given a finite set of points $F$ (say the associated points of $X$) I want to show that there is a hyperplane $H \in \mathcal{O}(1)$ such that it does not meet any of these points, i.e that $Supp H \cap F = \emptyset.$

I was told that this is really tautologous, and I believe it is, but I am afraid I don't see it. I have seen arguments of the form previously, but never felt completely comfortable with them and thus I would be interested to see a careful proof (or as careful as you have the energy to give) of doing this.

In the comments, it seems as if the statement I am making here is not true. Basically, I am interested in this just to get a detailed answer for the previous question, so feel free to reinterpret the question as long as the previous question gets a detailed answer.

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  • This fails if $A$ is a finite field. You need to take a hypersurface. – Cantlog May 06 '14 at 21:53
  • @Cantlog OK - I posted this in reference to http://math.stackexchange.com/questions/781382/showing-that-a-power-of-an-ample-sheaf-is-equivalent-to-an-effective-cartier-div where I read it as one could find such a hyperplane. Maybe I interpreted the answer incorrectly. But this question is mostly to clear up the details on that question, so feel free to modify it so that the previous question gets a more detailed answer. – user101036 May 06 '14 at 22:25
  • In the post you link to, I read the answer too quickly, sorry. For any $A$, you can use Avoidance Lemma for homogeneous ideals to find such a hypersurface. – Cantlog May 07 '14 at 06:58
  • @Cantlog I am not sure I understood correctly: If we consider $\mathcal{O}(1)$ we want to find an element $H$ that avoids all these associated points. Sections of $\mathcal{O}(1)$ corresponds to elements of degree 1. How does the prime avoidance lemma make sure that we find an element of degree 1? (and what is the ideal I am using it for? you are welcome to post this as an answer if you want!9 – user101036 May 07 '14 at 11:30
  • @Tedar: Dear Tedar, I have to confess that when thinking about your linked question, I was thinking about the case when the base is a field, rather than an arbitrary ring $A$, which is slightly more technical. Regards, – Matt E May 09 '14 at 02:24

2 Answers2

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Embed $X$ into a projective scheme $\mathrm{Proj}(B)$ where $B$ is a homogeneous $A$-algebra. Let $\mathfrak p_1, \dots, \mathfrak p_n$ be the homogeneous prime ideals corresponding to the fintely many points you want to avoid.

As $B_+$ (the irrelevant ideal) is not contained in any $\mathfrak p_i$, the homogeneous prime avoidance lemma says that there exists a homogeneous element $f\in B$, of some degree $d$, such that $f\notin \cup_i \mathfrak p_i$. Then the hypersurface of degree $d$ defined by $f$ avoids the given points.

As said in the comment, in general it is not possible to find such an $f$ of degree $1$. Indeed, if $A$ is a finite field, and $X$ is a projective space over $A$, there is no hyperplane avoinding the (finite) set of the rational points of $X$. If $A$ contains an infinite field $k$, then we can view $B_1$ (homogeneous elements of degree $1$) and the $\mathfrak p_i$ as vector spaces over $k$, and of course there will then exists an $f\in B_1$ not in any $\mathfrak p_i$, thus a hyperplane avoinding the given points.

  • @Cantlog: Ah, so if I I understood correctly, the suggestion given in the other question was not correct? I understand the answer you give fine, thanks! – user101036 May 08 '14 at 12:58
  • @Tedar: Dear Tedar, cant_log is correct that over a finite field, or other particularly finite situations, you may not be able to find a hyperplane. It is standard in these situations to do as cant_log does and look for a hypersurface instead. As a general remark, I wouldn't focus on the case of finite fields (where things like Bertini and related issues can be a little delicate) if I was trying to build up geometric intuition. Regards, – Matt E May 09 '14 at 02:10
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As noted in cant_log's comment and answer, the statement is not true for all $A$, although a modified version is. (One has to replace $\mathcal O(1)$ by $\mathcal O(d)$ for an appropriate choice of $d$.)

Here is a proof when $k$ is an infinite field.

Let $X$ be a closed subset of $\mathbb P^n$, and let $X_i$ ($i = 1,\ldots,n$) denote the finite collection of irreducible and embedded components of $X$.

Let $(\mathbb P^n)^*$ denote the dual projective space, parameterizing hyperplanes in $\mathbb P^n$. Let $Z_i$ denote the collection of hyperplanes which contain $X_i$. Since $X_i$ is non-empty, this is a proper closed subset of $(\mathbb P^n)^*$, hence its complement $U_i$ is a non-empty open subset. The intersection $U$ of all the $U_i$ is thus also a non-empty open subset.

If $k$ is infinite, we may then find a $k$-point of $U$, and hence find a hyperplane which does not contain any of the $X_i$, as required.

If $k$ is finite, the problem is that $U$ may not contain any $k$-point. (E.g. $X$ might be the union of all the rational hyper lanes!) It will contain a $k'$-point for some finite extension $k'$ of $k$, and if $\ell$ is the equation (over $k'$) of the corresponding hyperplane, then the product of all the conjugates (by elements of Gal$(k'/k)$) of $\ell$ will be a higher degree hypersurface which does not contain any of the components of $X$.

For general $A$ you have to a little more careful with this geometric approach, since after all if Spec $A$ is not irreducible, then neither $\mathbb P^n$ nor $(\mathbb P^n)^*$ will be irreducible either.

One could cut down to the irreducible components of Spec $A$ (after all, the image of each component of $X$ lies (at least) one of the components of Spec $A$), and then, assuming $A$ is a domain and working first over the generic point of $A$ (where the case of a field applies) one could proceed by Noetherian induction on $A$ (essentially, induction on dimension) to prove the result.

I like this geometric way of thinking, but in terms of writing out the details, its probably simpler at this point to just use the algebraic argument suggested by cant_log.

Matt E
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