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Let $A$ be a seminormal ring. (Assume that $A$ is a finitely generated $k$-algebra, if it helps.) Is it true that $A$ is regular in codimension 1? I know this is true for normal rings. If the answer to the above question is no, then can we characterise seminormal rings that are regular in codimension 1?

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I guess, the nodal singular cubic in ${\mathbb P}^2$ is seminormal. –  Sasha Anan'in Feb 11 at 16:05

2 Answers 2

up vote 5 down vote accepted

Nope, in fact seminormal rings need not even be Gorenstein in codimension 1. For instance, $$k[x,y,z]/\langle x,y \rangle \cap \langle x, z \rangle \cap \langle y, z \rangle$$ is not Gorenstein.

In terms of a characterization of seminormal rings that are regular in codimension 1, I don't think such a thing is possible. They will all be obtained by some appropriate gluing of points, and that gluing will just happen along a codimension $\geq 2$ subscheme of a normal variety. You can see this discussion for some description of gluing.

Of course, if anything is regular in codimension 1 (and equidimensional) then you can normalize it by S2-ifying it.

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Thanks a lot for your answer. How does one S2-ify a ring? Is there a canonical procedure? –  Adam Feb 11 at 17:30
    
Sure, let $X = \text{Spec} R$, $Z$ be the singular locus (note it is of codimension 2), $U = X \setminus Z$ and compute $\Gamma(U, O_X)$. –  Karl Schwede Feb 11 at 19:20

The answer is no.

For instance, let $x \in X$ be a point of a plane curve defined over an algebraic closed field $k$. Then $\mathcal{O}_{X,x}$ is semi normal if and only if $x$ is an ordinary $n$-fold point, i.e. a point of multiplicity $n$ whose tangent cone is composed by $n$ distinct lines. For instance, a node is seminormal whereas a cusp is not.

The answer to your question remains no even replacing semi normality with the close notion of weak normality. For instance, the Withney umbrella, i.e. the affine surface $X \subset \mathbb{C}^3$ defined by $y^2=x^2z$, is weakly normal but it is not regular in codimension $1$, because its singular locus is the $z$-axis.

More generally, the following holds. Let $C={T_1, \ldots, T_r}$ be a non-empty collection of disjoint subsets of $\{1, \ldots, p\}$ and let $z_1, \ldots, z_p$ be indeterminates. Let $R_C$ be the complete local ring defined by $$R_C = k[[z_1, \ldots, z_p]]/ (z_{\alpha}z_{\beta} \, | \, \alpha \in T_i, \beta \in t_j, i \neq j).$$ We say that a point $x$ in a variety $X$ is a multicross if $\widehat{\mathcal{O}}_{X,x}$ is isomorphic to $R_C$ for some $C$ as above. Then we have the

Proposition. Assume $\textrm{char}(k)=0$. If $x \in X$ is a multicross, then the local ring $\mathcal{O}_{X,x}$ is weakly normal. Moreover, the set of singular points of $X$ which are not multicross form a closed subset of codimension at least $2$.

In other words, a weakly normal variety can be singular in codimension $1$, but in that case the general singular point is analitically a multicross.

For further detail you can look at the notes Weak Normality and Seminormality by M. Vitulli.

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