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If $X$ is a (reduced) scheme and $P$ is a point of $X$ (not necessarily closed) such that the local ring $\mathcal{O}_{X,P}$ is a regular domain, then must there exist an open affine neighborhood $U = \text{Spec }A$ of $P$ such that $A$ is an integral domain?

I'm almost certain this is true, since the local ring being regular means that it doesn't sit in the intersection of irreducible components, and hence it must be "locally" irreducible...but I can't think how to prove it.

If needed, we can assume $X$ is also Noetherian.

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Over an algebraically closed, perfect field, regular is equivalent to nonsingular, which is an open condition. –  Andrew Jul 30 '12 at 21:50
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up vote 4 down vote accepted

The theory of irreducible components can be tricky when $X$ is not Noetherian, so I will suppose that $X$ is Noetherian, and so is the union of finitely many irreducible components.

Now the components on which $P$ lies are in bijection with the minimal primes of $\mathcal O_{X,P}$, and hence $P$ lies on a single component. The complement of all the other components is an integral open subscheme of $X$ containing $P$, and (like an scheme) it contains an open affine n.h. containing $P$, which will also be integral.

If we don't assume that $X$ is Noetherian, there can be topological complications in the theory of irreducible compoments. E.g. if $A$ is any Boolean ring (i.e. a ring of characteristic $2$ in which every element is idempontent), then each local ring of Spec $A$ is a domain, in fact a copy of $\mathbb F_2$, but Spec $A$ can contain points that are not contained in any integral open subscheme; one concrete example is given by $A = \prod_{n = 1}^{\infty} \mathbb F_2$ (and there are many others).

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That makes sense, thanks. Just to clarify, the reason I ask, is that to define the Weil divisor of a rational function, you need to be able to take valuations of the function, which can be tricky if you can't talk about function fields. –  oxeimon Jul 31 '12 at 1:21
    
Ie, given a rational function $f$ defined on an open set $U\subseteq X$ which meets every irred. component of $X$, and given a codimension 1 irreducible closed subset $Y\subset X$, you can find an integral neighborhood $V$ of the generic point of $Y$ (call it $\eta$). Since $U$ meets every irreducible component of $X$, $U\cap V\ne\emptyset$, so we can consider $f|_{U\cap V}$, which is now a function in the function field $k(V)$ of $V$. Since $V$ contains the generic point $\eta$ of $Y$, $\mathcal{O}_{X,\eta}\subseteq k(V)$, so you can now talk about the valuation of $f$ at $Y$. –  oxeimon Jul 31 '12 at 1:24
    
(in the above, $X$ is assumed to be regular in codimension 1, or that at least the $\mathcal{O}_{X,\eta}$ is regular) –  oxeimon Jul 31 '12 at 1:25
    
@oxeimon: Dear oxeimon, Why are you bringing up Weil divisors? That concept doesn't come up in any of the arguments or facts I discussed? Regards, –  Matt E Jul 31 '12 at 1:44
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