Non-isomorphic locally ringed spaces which represent isomorphic functors $\mathsf{CommRing} \to \mathsf{Set}$.

Question

It's well known that the restricted Yoneda functor $よ : \mathsf{Schemes} \to \operatorname{Fun}(\mathsf{CommRing},\mathsf{Set})$ is an embedding, so that (in particular) if $X$ and $Y$ are schemes such that $よ(X) \cong よ(Y)$, then $X \cong Y$. This is the key idea of the "functor of points approach" to algebraic geometry (which then proceeds to classify precisely which co-presheaves on commutative rings are in the essential image of $よ$).

It's my understanding that the restricted Yoneda $よ : \mathsf{LRS} \to \operatorname{Fun}(\mathsf{CommRing},\mathsf{Set})$ is no longer an embedding, where $\mathsf{LRS}$ is the category of locally ringed spaces. Moreover, I expect there to exist non-isomorphic locally ringed spaces $X$ and $Y$ such that $よ(X) \cong よ(Y)$.

Does anyone know a simple example of this failure of essential injectivity?

Edit: Just realized I messed up the variance at first, hopefully I got it right now.

Answer 1

If $A$ is a local ring, let $X_A$ denote the locally ringed space with one point with $\mathcal{O}_{X_A}(X_A)=A$. If $B$ is a ring, then a morphism of ringed spaces $\operatorname{Spec} B\to X$ is just given by a homomorphism $f:A\to B$. In order for this to be a morphism of locally ringed spaces, $f(m)$ must be contained in every prime ideal of $B$, where $m$ is the maximal ideal of $A$. That is, $f(m)$ must be contained in the nilradical of $B$. If $m$ is finitely generated, this means $f$ must factor through the quotient $A/m^n$ for sufficiently large $n$.

In particular, this means that morphisms of locally ringed spaces $\operatorname{Spec} B\to X$ actually depend only on the completion $\hat{A}$ of $A$ with respect to $m$, since $\hat{A}/(m\hat{A})^n\cong A/m^n$ for all $n$. So, the morphism $X_{\hat{A}}\to X_A$ becomes an isomorphism after applying the functor $よ : \mathsf{LRS} \to \operatorname{Fun}(\mathsf{CommRing},\mathsf{Set})$.

Or, let $A$ be a valuation ring whose value group $\Gamma$ is strongly nonarchimedean in the sense that for each positive $x\in\Gamma$, there is a positive $y\in\Gamma$ such that $ny<x$ for all $n\in\mathbb{N}$. (For instance, $\Gamma$ could be the ordered ring $\mathbb{Z}[x]$ where $x$ is infinitesimal.) Note that then if $I\subset A$ is any ideal whose radical is the maximal ideal $m$, then $I$ must be equal to $m$, since $I$ must contain elements of arbitrarily small positive valuation. This means that if $f:A\to B$ is any homomorphism such that $f(m)$ is nilpotent, $f(m)$ must actually be $0$. By the first paragraph above, this means that morphisms $\operatorname{Spec} B\to X$ are just in bijection with homomorphisms $A/m\to B$. In other words, the natural morphism $\operatorname{Spec} A/m\to X_A$ becomes an isomorphism after applying $よ$.

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Here is a rather different sort of example, inspired by a now-deleted answer by Aragogh. Let $X$ be a $T_1$ locally ringed space with the following property: for each $x\in X$, there is a neighborhood $U$ of $x$ and a section $a\in \mathcal{O}_X(U)$ that vanishes at $x$ but at no other point of $U$ (where by "vanishing at a point" I mean its image in the stalk at that point is in the maximal ideal). For instance, $X$ could be a nice (say, metrizable) topological space equipped with the sheaf of continuous real-valued functions. I claim that if $Y$ is an affine scheme and $f:Y\to X$ is a morphism of locally ringed spaces, then the underling map of topological spaces is locally constant.

To prove this, suppose $y\in Y$ and $f$ is not locally constant at $y$. We know that $f$ induces a local homomorphism $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$. Moreover, each prime ideal $p$ of $\mathcal{O}_{Y,y}$ corresponds to a generization of $y$ which $f$ must map to the same point (since $X$ is $T_1$), and the corresponding homomorphism $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,p}$ is just obtained by composing with the localization map $\mathcal{O}_{Y,y}\to \mathcal{O}_{Y,p}$. Since all of these compositions with localizations are local homomorphisms, our homomorphism $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$ must actually map the maximal ideal of $\mathcal{O}_{X,f(y)}$ into the nilradical of $\mathcal{O}_{Y,y}$ (since its image must be contained in every prime ideal).

Now take a neighborhood $U$ of $f(y)$ and a section $a\in \mathcal{O}_X(U)$ that vanishes at $f(y)$ and nowhere else. Since the image of $a$ in $\mathcal{O}_{Y,y}$ is nilpotent, we may replace $a$ with a power of itself and assume there is some neighborhood $V\subseteq f^{-1}(U)$ of $y$ such that the image of $a$ in $\mathcal{O}_Y(V)$ is $0$. Since $f$ is not locally constant at $y$, there exists some $z\in V$ such that $f(z)\neq f(y)$. But then $a$ is invertible in a neighborhood of $f(z)$, so the image of $a$ in $\mathcal{O}_{Y,z}$ is invertible. This is a contradiction, since the map $\mathcal{O}_X(U)\to\mathcal{O}_{Y,z}$ factors through the map $\mathcal{O}_X(U)\to\mathcal{O}_Y(V)$ that maps $a$ to $0$.

So, every morphism from an affine scheme to $X$ is locally constant. Now let $D(X)$ be $X$ with the discrete topology, equipped with the sheaf of local rings that has the same stalks as $X$. The identity function $i:D(X)\to X$ has an obvious structure of a morphism of locally ringed spaces, and a locally constant morphism to $X$ is just one that factors (uniquely) through $i$. So, $よ(i)$ is an isomorphism.

This gives rise to a bunch of other related examples. For instance, if $M$ and $N$ are two nonempty manifolds of the same dimension $d>0$, equipped with the sheaves of continuous (or smooth) real-valued functions, note that $D(M)$ and $D(N)$ are isomorphic: they are each just a disjoint union of $2^{\aleph_0}$ singletons whose local ring is the ring of germs of real-valued functions at a point of $\mathbb{R}^d$. It follows that $よ(M)$ and $よ(N)$ are isomorphic (though this isomorphism does not come from a map between $M$ and $N$).

In fact, we can go further. If $X$ is a topological space, $x\in X$, and $(A,m)$ is the local ring of germs of real-valued continuous functions at $x$, then $A$ has the following property. For each nonzero $a\in m$, there exists a nonzero $b\in m$ such that $a$ is divisible by $b^n$ for all $n\in\mathbb{N}$. Indeed, you can take $b=g(a)$ where $g:\mathbb{R}\to\mathbb{R}$ is a continuous function such that $g(0)=0$ and as $t\to 0$, $g(t)$ approaches $0$ more slowly than $|t|^{1/n}$ for each $n$. This property means that the only ideal of $A$ whose radical is $m$ is $m$ itself. As in the third paragraph above, this means the canonical morphism $\operatorname{Spec} \mathbb{R}\to X_A$ becomes an isomorphism after applying $よ$. It follows that if $X$ is a nice space as above equipped with its sheaf of continuous real-valued functions, then actually the canonical morphism $\coprod_{x\in X}\operatorname{Spec}\mathbb{R}\to X$ becomes an isomorphism after applying $よ$, since $よ$ can't see the difference between the points of $D(X)$ and copies of $\operatorname{Spec}\mathbb{R}$. In particular, this means that if $X$ and $Y$ are two metrizable spaces of the spaces of the same cardinality equipped with their sheaves of continuous real-valued functions, then $よ(X)\cong よ(Y)$.

(In fact, you can generalize this a bit further: all of this applies to spaces on which real-valued continuous functions separate points, not just metrizable spaces. To sketch the proof, the argument above shows that if $Y$ is an affine scheme and $f:Y\to X$ is a morphism of locally ringed spaces, then every continuous real-valued function on $X$ actually pulls back to a locally constant real-valued function on $Y$. If $f$ were not locally constant, then its image would be infinite, and then since functions separate points on $X$, you can conclude that the functions on $X$ pull back to an infinite-dimensional algebra of locally constant functions on $Y$, which must come from an infinite subalgebra of the Boolean algebra of clopen subsets of $Y$. But then you can construct a uniformly convergent infinite sum of such functions which is not locally constant, and this sum will still come from a continuous function on $X$, giving a contradiction.)