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](https://math216.wordpress.com/2011/06/10/favorite-open-or-closed-conditions-2/) 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?




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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.



Are schemes obtained from reduced schemes by gluing infinitesimal stuff along a reduced closed subscheme?
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 $V\_2$.


