Let $X$ be a scheme of finite type (say over the complex numbers). The set of points for which the local ring is reduced is then an open subset $U\subseteq X$.

Is it true that there is a closed subscheme $Z\hookrightarrow X$ such that

  1. $Z$ is supported on $X\smallsetminus U$ and
  2. $X$ is the coproduct $X\simeq X_{\mathrm{red}}\cup_{Z_{\mathrm{red}}}Z$ in the category of schemes?

An affine example:

$$X=\mathrm{Spec}\frac{k[x,y]}{(xy,y^2)}, \quad Z=\mathrm{Spec}\frac{k[y]}{(y^2)},\quad X_{\mathrm{red}}=\mathrm{Spec}(k[x])=\mathbb{A}^1_k, \quad Z_{\mathrm{red}}=\mathrm{Spec}(k),$$

$Z\to X$ given by $y\mapsto y$, $Z_{\mathrm{red}}\to Z$ given by $y\mapsto 0$, $Z_{\mathrm{red}}\to X$ given by $x\mapsto 0, y\mapsto 0$.

so $X$ is a line with an embedded point sticking out from the origin, $U=\mathbb{A}^1\smallsetminus\{0\}$, and $Z$ is the infinitesimal first-order "segment". Now, if I'm not mistaken,

$$\frac{k[x,y]}{(xy,y^2)}\simeq k[x]\times_k \frac{k[y]}{(y^2)}:=\{ (a(x),b(y))\in k[x]\times \frac{k[y]}{(y^2)} \mid a(0)=b(0)\}$$

by $x\leftrightarrow(x,0), y\leftrightarrow (0,y)$, so the algebra of $X$ is the fibered product of those of $X_{\mathrm{red}}$ and $Z$ over the evaluations to $k$, hence $X$ is the corresponding coproduct of schemes.

  • $\begingroup$ bubscheme ? typo $\endgroup$ – meh May 19 '19 at 21:14
  • $\begingroup$ yeah, of course - Fixed, thank you :) $\endgroup$ – Qfwfq May 19 '19 at 21:48
  • 1
    $\begingroup$ How is the scheme structure on $Z$ defined? $\endgroup$ – Piotr Achinger May 19 '19 at 22:00
  • $\begingroup$ @PiotrAchinger: you're right, I failed to define it; in fact, I have now edited the OP to include the question that you're asking in the above comment. $\endgroup$ – Qfwfq May 19 '19 at 23:02

If this happens then the cotangent bundle $\mathcal I_{Z_{red}} / \mathcal I_{Z_{red}}^2$ of $Z_{red}$ in $X$ is equal to the sum of the contangent bundle of $Z_{red}$ in $Z$ and the cotangent bundle of $Z_{red}$ in $X$. This is because locally a coproduct of schemes gives a fibered product of rings which therefore gives a product of ideals.

So if we chose $X$, letting $Y$ be the induced reduced subscheme structure on $X \setminus U$, where the natural map from the cotangent bundle of $Y$ in $X$ to the cotangent bundle of $Y$ in $X^{red}$ does not split, such $Z$ will not exist. For instance we can choose $X = \operatorname{Spec} k[x,y]/(x^2y)$. Or we can pick a non-split exact sequence of vector bundles $0 \to V_1 \to W \to V_2 \to 0$ on some other scheme and take the relative spectrum of the symmetric algebra of $W$ modulo the ideal generated by $W V_1$, so that the cotangent space is $W$ and the cotangent space of the reduced subscheme of the non-reduced locus is $V_2$.

  • $\begingroup$ Thank you for the answer: it's what I was looking for. (By "cotangent bundle" I assume you meant the conormal sheaf $N^*$). A question: does also the reverse implication hold? That is, if $N^*_{Y/X}=N^*_{Y/X_{red}}\oplus N^*_{Y/Z}$ then does the coproduct condition hold? I imagine it doesn't. $\endgroup$ – Qfwfq May 20 '19 at 20:55
  • $\begingroup$ @Qfwfq Re: conormal, yes, of course. For your second question, I would imagine one can find an obstruction in the next level of the filtration or in the extension class. $\endgroup$ – Will Sawin May 21 '19 at 4:25

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