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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.

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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.

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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.

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