3
$\begingroup$

Let $X$ be an affine, irreducible (complex), generically reduced, scheme containing an embedded point, say at $x \in X$. Suppose further dimension of $X$ is strictly positive (can assume to be one dimensional). Assume that the associated reduced scheme $X_{\mbox{red}}$ is non-singular. Note that, there is a natural closed immersion $X_{\mbox{red}} \to X$. Then,

1) Are there examples of $X$ as above such that there exists a morphism $X \to X_{\mbox{red}}$ such that the composition $$ X_{\mbox{red}} \to X \to X_{\mbox{red}}$$ is the identity? I know of such examples in zero dimension. I am interested to see if such a phenomenon can happen in the setup of embedded points in schemes of positive dimension.

2) If such a map $X \to X_{\mbox{red}}$ exists, then does the structure sheaf of $X$ decompose (as an $\mathcal{O}_{X_{\mbox{red}}}$-module) as $\mathcal{O}_{X_{\mbox{red}}} \oplus I\mathcal{O}_{X_{\mbox{red}}}$ for some nilpotent ideal $I$?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ There's plenty of examples. Let $k = \mathbf{C}$. Take $R = k[x_1,...,x_n]$, an artinian local $k$-algebra $A$ with residue field $k$, and consider the fibre product $S = R \times_k A$, where the map $A \to k$ is the obvious one and $R \to k$ sets all $x_i = 0$. Geometrically, you've tacked on a copy of $\mathrm{Spec}(A)$ at the origin in $\mathbf{A}^n_k$. So the projection $S \to R$ is a nilpotent thickening with kernel $\mathfrak{m}_A$ (and there's a section). For (2), in what category do you want the splitting? There's such a splitting after pushing down to $X_{red}$ for trivial reasons... $\endgroup$
    – Anonymous
    Commented May 26, 2020 at 17:00
  • $\begingroup$ @Anonymous Is the scheme in your example really generically reduced? It seems it is generically non-reduced. $\endgroup$
    – Jana
    Commented May 26, 2020 at 17:44
  • 1
    $\begingroup$ @Jana fibre product, not tensor product. If $A = k[\varepsilon]/(\varepsilon^2)$, then $S \cong R[y]/(x_1y, \ldots, x_ny, y^2)$, which has an embedded point at the origin and the reduction is $S[y]/(y) = R$. Anonymous's construction is just a generalisation of that example. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 26, 2020 at 19:44
  • $\begingroup$ @R.vanDobbendeBruyn Thanks. I was thinking of tensor product. $\endgroup$
    – Jana
    Commented May 26, 2020 at 19:50

0

Reset to default

You must log in to answer this question.