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 $\endgroup$ – aginensky May 19 at 21:14
  • $\begingroup$ yeah, of course - Fixed, thank you :) $\endgroup$ – Qfwfq May 19 at 21:48
  • 1
    $\begingroup$ How is the scheme structure on $Z$ defined? $\endgroup$ – Piotr Achinger May 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 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 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 at 4:25

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.