I've just read the definition of a closed subscheme in Hartshorne's recently and I collected here and there (notes that people put online) the following statement.
Claim. Suppose that $(X,\mathcal{O}_X)$ is a scheme, $Y \subset X$ is a closed subset. Then $Y$ can be equipped with a sheaf $\mathcal{O}_Y$ such that $(Y,\mathcal{O}_Y)$ is a reduced closed subscheme of $X$ with the following universal property: for any reduced scheme $Z$ together with a morphism $f: Z \rightarrow X$ such that $f(Z) \subset X$, $f$ factors uniquely through $i: Y \hookrightarrow X$ ($i^{\#}: \mathcal{O}_X \rightarrow i_{\ast}\mathcal{O}_Y$).
Here are my questions (I expect an answer to either (1) or (2) and an answer to (3). Thanks!
1) Can you provide me some good references for the statement of the result, for a proof or a construction of this result?
2) Since I haven't found a satisfactory description of the above fact in popular texts yet, I came up with my own try. I define an ideal sheaf $\mathcal{I}_Y(U) = \{s \in \mathcal{O}_X(U) \mid s_p \in \mathfrak{m}_{X,p} \text{ for all } p \in U \cap Y\}$ on $X$, and I define $\mathcal{O}_Y = i^{-1}(\mathcal{O}_X/\mathcal{I}_Y)$. I expect $\mathcal{O}_Y$ to be the desired sheaf on $Y$. So far, I have managed to show that $(Y,\mathcal{O}_Y)$ is a reduced subscheme of $X$, but I haven't had any idea how to prove the universal property yet. Do you know whether or not the above construction gives the right sheaf in the claim? If yes, could you provide me some idea how to prove the universal property?
3) Suppose that $\mathcal{F}$ is another sheaf on $Y$ that makes $(Y,\mathcal{F})$ a reduced subscheme of $X$, does this imply that $(Y,\mathcal{F})$ also has the universal property stated in the claim?
Best Answer
The best way to address your question might be to split the answer in two parts:
1) Given a closed subset $Y$ of a scheme $X$ (or more precisely of its underlying topological space $|X|$), there is a unique way to endow it with the structure of reduced scheme and with a closed embedding $i: Y \hookrightarrow X$ whose underlying set-theoretic map is the inclusion $|Y| \hookrightarrow |X|$ .
Note that unicity implies that necessarily $\mathcal F=\mathcal O_Y$ in your question 3), whose answer is thus "yes".
2) There is a functorial way to associate to a completely arbitrary morphism of schemes $f: Y\to X$ a morphism of the corresponding reduced schemes $f_{red}: Y_{red}\to X_{red}$ compatible with the two immersions $Y_{red}\hookrightarrow Y$ and $X_{red}\hookrightarrow X$.
It follows that any completely arbitrary morphism $f:Y\to X$ from a reduced scheme $Y$ (not related in any way to $X$) to $X$ can be factored uniquely as a composite $Y\to X_{red}\hookrightarrow X$.
This is a generalization of your claim : the fact that for you $Y$ is a subscheme of $X$ turns out to actually be irrelevant.
These results are proved in EGA I (surprise, surprise!), (5.2.1) and (5.1.5) respectively.
And to end on a pleasant note: your construction of $\mathcal O_Y$ is absolutely correct. Bravo!