# Are schemes obtained from reduced schemes by gluing infinitesimal stuff along a reduced closed subscheme?

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.

• bubscheme ? typo – meh May 19 '19 at 21:14
• yeah, of course - Fixed, thank you :) – Qfwfq May 19 '19 at 21:48
• How is the scheme structure on $Z$ defined? – Piotr Achinger May 19 '19 at 22:00
• @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. – 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 $$V_2$$.
• 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. – Qfwfq May 20 '19 at 20:55