Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question/problem really comes from a fact in algebraic geometry, where it says that given an irreducible variety $V$ ($\dim V \geq 2$) then for any given pair of points $x,y\in V$ there is an irreducible curve $C$ connecting them.

This can proved by invoking Bertini's theorem [See Lazarsfeld R. - Positivity in Algebraic Geometry 1, example 3.3.5].

Translating this back to coordinate rings we get something like: Given a finitely generated $k$-algebra $R$ which is a domain, $k=\bar{k}$ (alg.closed) then for any pair of maximal ideals $m_1,m_2\subset R$ there is a prime ideal $p\subset m_1\cap m_2 $ with $\dim R/p =1$.

So my question is therefore: Is there a purely ring theoretical argument for this fact? In what generality does it hold?

The reason why i emphasize on purely is because, and i am no expert on this, but apparently the Bertini type argument can be (re-)formulated algebraically, however this, I have been informed, will not be pretty...

share|improve this question
    
Can you use Noether normalization and then "pull back" a hyperplane section in $\mathrm{Spec}k[x_1,\dots,x_n]$? –  David Hansen Jul 28 '11 at 23:14

2 Answers 2

This reminds me of what someone once said to me:

The geometers can always take a hyperplane section. We can't!

The purpose of this post is to analyze the question to see how close it is to Bertini's theorem (it is not obvious to me). The statement is immediately equivalent to: one can always find a nonzero prime ideal $P \subseteq m_1\cap m_2$ (since if we can, then induction on dimension proves the original).

Now, how can such $P$ exist? Let $U_1 = R-m_1$ and $U_2=R-m_2$. Let $U=\{xy \| x\in U_1, y\in U_2\}$. $U$ is multiplicative and our prime $P$ obviously just has to avoid $U$. So $P$ exists unless the localization $U^{-1}R$ has dimension $0$. But it is a domain, so we have to make sure $U^{-1}R$ is not a field. That statement is equivalent to the existence of some element $f\in R$ such that $f$ does not become an unit in $U^{-1}R$. In other words:

there are no $a,b \in U$ such that $af=b$.

Since $b$ is itself a product of elements in $U_1,U_2$, our condition is obvious if $fR$ is a prime ideal and $f\in m_1\cap m_2$. But it is technically weaker, although not clear to me by how much. Note that we do not require $f$ to be linear, which is common for Bertini's type statement.

This analysis would seem to rule out certain clever arguments. But may be one can find some!

For "algebraic" version of Bertini's theorem, I will look at the reference given here. Also, how to rescue Bertini over finite fields using hypersurface instead of hyperplane (which suits your purpose), looks here.

share|improve this answer

EDIT: The following is junk. I'm not deleting it because of all the discussion in the comments.

I think the following works. Choose any non-zero element $f \in m_1 \cap m_2$ (if $m_1 \cap m_2 = 0$, then you weren't a domain). As pointed out in the comments, this doesn't work. Set $p$ to be a minimal associated prime of $f$. Somewhere in Matsumura it proves that this is height 1 I think.

EDIT: I was being dumb. Sorry about that, I shouldn't try to answer mathoverflow early in the morning. Anyway, if you could do it, then you need to argue by induction. But you need to choose your $f$ carefully.

share|improve this answer
1  
This is the opposite of what I'm asking. I want $p$ to be such that if $p\subset q$ for a prime ideal $q$ then $q$ is maximal, i.e. depth $p$ =1. –  Tore Forbregd Jul 13 '11 at 12:12
    
This is usually called the dimension of $p$. I've edited the title of the post. –  Graham Leuschke Jul 13 '11 at 12:18
4  
@Karl, There is a small issue with your argument I think. Why should a minimal associated prime containing $f$ be in both the maximal ideals? For example, let the maximal ideals be $(x,y-1)$ and $(x-1,y)$ in $k[x,y]$. Let $f=xy$. –  Mohan Jul 13 '11 at 12:33
    
I think this issue can be fixed by localizing R at the multiplicative system given by the complement of the union of m1 and m2. Then one can reason as Karl does and it seems to me that one gets the solution. Or not? –  Tommaso Centeleghe Jul 13 '11 at 14:05
2  
Mohan's example still causes problems. The minimal primes of $f=xy$ are $(x)$ and $(y)$, each of which is contained in the union of $m_1$ and $m_2$, but neither of which is contained in the intersection. There is a choice of $f$ that works, namely $f=x+y-1$, but a random choice won't do it. –  Graham Leuschke Jul 13 '11 at 14:19

Your Answer

 
discard

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.