For local ring $R$, does funcotor $\operatorname{Hom( Spec}R, X)$ characterize scheme $X$?

Let $\bf{Sch, Sets, Ring}$ be a category of schemes, sets, commutative rings.

By Yoneda's lemma, scheme $X$ is characterized by contravariant functor

$$\operatorname{Hom}(*, X): \bf{Sch}^{op}\to Sets$$

Now thinking glueing of schemes by affine schemes, $X$ can be characterized by covariant functor

$$\operatorname{Hom}({\operatorname{ Spec} }(*), X): \bf{Ring}\to Sets$$

I want to know that whether $X$ is characterized by only local rings?

i.e. for schemes $X, Y$, if for all local ring $R$, $\operatorname{Hom}({\operatorname{ Spec} }(R), X)\cong \operatorname{Hom}({\operatorname{ Spec} }(R), Y)$, then $X\cong Y$?

Solution 1:

After (a lot of) thinking about it, I think the answer to the specific question you ask is no – I think there are non-isomorphic schemes that yield isomorphic functors $\textbf{LocCRing} \to \textbf{Set}$ – but there is a sense in which the answer to your general question is yes.

First, let me sketch an argument that suggests why the answer to your specific question is no. Let $X$ be a scheme. For every point $x$ of $X$, let $\langle x \rangle$ denote $\operatorname{Spec} O_{X, x}$ considered as a scheme over $X$ in the canonical way. Let $P (X) = \coprod_{x \in X} \langle x \rangle$ and let $R (X) = P (X) \times_X P (X)$. We have a cofork: $$R (X) \rightrightarrows P (X) \rightarrow X$$ If this cofork were a coequaliser diagram for every scheme $X$, then it would be the case that $\textrm{Hom} (\operatorname{Spec} ({-}), {-})$ embeds the category of schemes fully faithfully into the category of functors $\textbf{LocCRing} \to \textbf{Set}$. I think the converse is also true but I haven't checked carefully. Anyway, this is all counterfactual, because the cofork is not always a coequaliser diagram. Consider $\mathbb{A}^1_k$ and $\mathbb{P}^1_k$ for an algebraically closed field $k$. It is well known that $P (\mathbb{A}^1_k) \cong P (\mathbb{P}^1_k)$. Furthermore $R (\mathbb{A}^1_k) \cong R (\mathbb{P}^1_k)$, and we can arrange for these isomorphisms to be compatible with the two projections $R \rightrightarrows P$. So if both coforks were coequaliser diagrams we would have $\mathbb{A}^1_k \cong \mathbb{P}^1_k$, which we know to be false. So at least one of these coforks is not a coequaliser.

But your specific question is not quite the same as asking whether $\textrm{Hom} (\operatorname{Spec} ({-}), {-})$ is a fully faithful embedding, and your general question can be interpreted in other ways. For instance, you might ask whether $\textrm{Hom} (\operatorname{Spec} ({-}), {-})$ is at least a faithful embedding, and the answer is yes:

Theorem. The class of local schemes is a separating class in the category of schemes, i.e. given morphisms $f_0, f_1 : X \to Y$ of schemes, if for every local ring $T$ and every morphism $t : \operatorname{Spec} T \to X$ we have $f_0 \circ t = f_1 \circ t$, then $f_0 = f_1$.

Proof. Let $\mathfrak{m}$ be the maximal ideal of $T$. So $\mathfrak{m}$ is also the unique closed point of $\operatorname{Spec} T$. Since the only open neighbourhood of $\mathfrak{m}$ is $\operatorname{Spec} T$ itself, the image of $t : \operatorname{Spec} T \to X$ is contained in every open neighbourhood of $t (\mathfrak{m})$. Furthermore, every point of $X$ is $t (\mathfrak{m})$ for some choice of $T$ and $t : \operatorname{Spec} T \to X$. Thus, it is enough to prove the theorem in the case where both $X$ and $Y$ are affine. But this is an easy consequence of the following lemma.

Lemma. Let $A$ be a commutative ring and let $a \in A$. The following are equivalent:

• $a = 0$ in $A$.
• For every prime ideal $\mathfrak{p} \subset A$, $a = 0$ in $A_\mathfrak{p}$.
• For every maximal ideal $\mathfrak{m} \subset A$, $a = 0$ in $A_\mathfrak{m}$.

Proof. The downward implications are trivial. We verify the upward implication to complete the cycle.

Suppose $a = 0$ in $A_\mathfrak{m}$. That means there is a $b_\mathfrak{m} \in A \setminus \mathfrak{m}$ such that $a b_\mathfrak{m} = 0$. Choose such a $b_\mathfrak{m}$ for every maximal ideal $\mathfrak{m} \subset A$. Consider the ideal generated by $\{ b_\mathfrak{m} : \mathfrak{m} \in \operatorname{MaxSpec} A \}$. It is not contained in any maximal ideal of $A$, so it must be the unit ideal. On the other hand, by construction, this ideal is annihilated by $a$. So $a = 0$.　◼