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

Non-singularity of an algebraic variety can be characterised in intrinsic terms by the fact that all local rings are regular local rings.

By a theorem of Serre, any localization of a regular local ring at a prime ideal is again a regular local ring.

If ones proves that the local ring at any non-closed point is a localization of a local ring at a closed point, by the previous theorem it suffices to check non-singularity at closed points.

I am confused as to how to prove the former statement.

share|cite|improve this question
Which statement is the "former"? – Steven Landsburg Mar 13 '12 at 19:57
If you mean the statement about local rings being localizations of local rings at closed points, it amounts to the following statement: if $A$ is a ring, $\mathfrak{p}$ a prime of $A$, and $\mathfrak{m}$ a maximal ideal containing $\mathfrak{p}$, then $(A_\mathfrak{m})_{\mathfrak{p}A_{\mathfrak{m}}}$ is canonically isomorphic to $A_\mathfrak{p}$. – Keenan Kidwell Mar 13 '12 at 20:15
up vote 2 down vote accepted

Since problem is local assume scheme is $Spec(R)$.Nonclosed point is prime ideal $P$ and can find maximal ideal $M \supset P$. Then $R_P=(R_M)_{PR_M}$ and use Serre theorem and hypothese that $R_M$ is regular.

share|cite|improve this answer

I guess that what you call a variety is (at least) a scheme of finite over a field. If $x$ is a point in such a scheme $X$, its closure $Y$ is an irreducible scheme of finite type and hence contains a closed point $y$. Then any open affine neighbourhood $U=Spec(A)$ of $y$ in $X$ contains $x$ (the generic point of $Y$). Let $p,q$ be the primes of $A$ corresponding to $x,y$. The fact that $x$ specializes to $y$ tells you that $p\subset q$ and hence the localized rings satisfy $O_{X,x}=A_p=(A_q)_p=(O_{X,y})_p$, a localization of $O_{X,y}$ as desired.

share|cite|improve this answer

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.