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?

  • 5
    $\begingroup$ I guess, the nodal singular cubic in ${\mathbb P}^2$ is seminormal. $\endgroup$ Feb 11, 2014 at 16:05

2 Answers 2


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.

  • 1
    $\begingroup$ Thanks a lot for your answer. How does one S2-ify a ring? Is there a canonical procedure? $\endgroup$
    – Adam
    Feb 11, 2014 at 17:30
  • $\begingroup$ Sure, let $X = \text{Spec} R$, $Z$ be the singular locus (note it is of codimension 2), $U = X \setminus Z$ and compute $\Gamma(U, O_X)$. $\endgroup$ Feb 11, 2014 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.