Can isomorphism classes of varieties be detected on field-valued points?

Question

Let $X$ and $Y$ be schemes over an algebraically closed field $k$, and let $\phi: X \to Y$ be a morphism of schemes such that for all field extensions $K/k$, the induced map $\phi(K): X(K) \to Y(K)$ is a bijection. Are there situations in which we can deduce that $X$ and $Y$ are isomorphic? (For example, if $X$ and $Y$ are reduced smooth varieties).

Answer 1

(All rings below are commutative.)

Edit, 7/3: Okay, thanks to this discussion I've learned some more things. It is actually very easy to construct counterexamples where $A, B$ are reduced, so my guess below was wildly wrong.

Here is the issue. In a ring $A$ say that a weak inverse of an element $a \in R$ is an element $b$ satisfying $aba = a$ and $bab = b$. If $a$ has weak inverse $b$, these conditions imply that if $\varphi : A \to K$ is any homomorphism from $A$ to a field then exactly one of the following happens:

• If $\varphi(a) \neq 0$ then $\varphi(b) = \varphi(a)^{-1}$.
• If $\varphi(a) = 0$, then $\varphi(b) = 0$.

Either way, the image $\varphi(b)$ is uniquely determined by $\varphi(a)$. This just says that weak inverses are unique in fields and must be given by either the ordinary inverse on nonzero elements or $0$ for $0$. It follows that:

Corollary: A map $f : A \to B$ where $B$ is obtained from $A$ by adjoining weak inverses is always field-local.

Moreover, adjoining weak inverses to a reduced ring produces another reduced ring, so this produces a lot of reduced counterexamples. To rule them out we need to require at the very least that $B$ is a domain. So I think "$A, B$ are domains and $A$ is integrally closed" is likely the best we can hope for (and I'm still not sure if that's enough to make the proof work).

I find this weak inverse business very unfamiliar already in the case that $A = \mathbb{Z}$ (previously I was convinced that the result would hold in this special case but couldn't prove it). We can consider, for example, the ring obtained from $A$ by adjoining a weak inverse to a prime $p$, which explicitly is

$$B = \mathbb{Z}[a_p]/(p^2 a_p = p, p a_p^2 = a_p).$$

The unit inclusion $f : \mathbb{Z} \to \mathbb{Z}[a_p]$ is field-local, a bijection on prime ideals, and an isomorphism on each residue field $\text{Frac}(A/P)$ (I believe this is true of all field-local maps). The image of $a_p$ in any field is either $\frac{1}{p}$ if the characteristic is $\neq p$ or $0$ if the characteristic is $p$, and in particular is uniquely determined, as above. The induced map on local rings is an isomorphism at every prime except $(p)$.

We can see a little more clearly what's going on geometrically by noticing that the definition of weak inverse implies that if $a$ has weak inverse $b$ then $e = ab$ is idempotent. Any idempotent in a commutative ring factors it as the product of two rings which can be described as the quotients by $e$ and $1 - e$; applied to the ring $B = A[b]/(aba = a, bab = b)$ given by adjoining a weak inverse to $a \in A$ this gives

$$B \cong B/e \times B/(1-e)$$

where $B/e \cong A/a$ and $B/(1-e) \cong A[b]/(ab = 1)$. So the first ring is just the quotient by $a$ while the second ring is just the localization at $a$. It's clear that the inclusion of $A$ is field-local since morphisms from $B$ into any field just split into two cases depending on whether the image of $a$ is zero or nonzero. Geometrically this has the effect of "disconnecting" the zero set of $a$ off from the rest of $\text{Spec } A$, while the rest of $\text{Spec } A$ only remembers the localization away from the zero set. In particular our mysterious "weak localization" of $\mathbb{Z}$ above is just

$$\mathbb{Z}[a_p]/(p^2 a_p = p, p a_p^2 = a_p) \cong \mathbb{Z}/p \times \mathbb{Z}[p^{-1}].$$

This construction, when applied to a polynomial ring $K[t]$ and adjoining a weak inverse of $t$, also produces Dori Bejleri's counterexample of $\bullet \sqcup \mathbb{G}_m \to \mathbb{A}^1$.

Generally this discussion implies that $f : A \to B$ is field-local iff it is field-local after adjoining all weak inverses to both $A$ and $B$. This is what one might call the "meadowization," and reduces the problem to understand which commutative rings have isomorphic meadowizations. Geometrically meadowization appears to have the effect of nearly completely disconnecting the spectrum into individual points.

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Outdated answer: Nice question! In my youth I never learned how to prove anything about non-affine schemes so please allow me to work with affine schemes only. In exchange I will work with arbitrary rings. So, the question for affine schemes is: if $f : A \to B$ is a morphism of rings inducing a bijection

$$\text{Hom}(f, K) : \text{Hom}(B, K) \cong \text{Hom}(A, K)$$

for all fields $K$, when can we conclude that $f$ is an isomorphism? Let's call $f$ satisfying this condition field-local.

As Kenta says in the comments, we cannot expect this to be true if either $A$ or $B$ are not reduced. I think the result should be true only assuming $A, B$ are reduced but I can't prove it. Here's what I can prove so far.

Lemma 1: A field-local map $f : A \to B$ is an isomorphism if $A = k$ is a field and $B$ is reduced.

Proof. $f : k \to B$ equips $B$ with the structure of a $k$-algebra, and in terms of this $k$-algebra structure the field-local condition means that if $K/k$ is any field extension of $k$ then there is a unique $k$-algebra map $B \to K$. In particular this is true if we take $K = k$, so there is a unique $k$-algebra map $\varepsilon : B \to k$. Composing this map with the embedding $k \hookrightarrow K$ of $k$ into any extension $K/k$ must therefore give the unique $k$-algebra map $B \to K$. So it follows that every morphism $B \to K$ from $B$ to a field factors uniquely through $\varepsilon$.

This means that every element of the kernel $\ker(\varepsilon)$ lies in the kernel of every morphism from $B$ to a field. These kernels give all the prime ideals of $B$, so since $B$ is reduced, their intersection is zero, so it follows that $\ker(\varepsilon) = 0$. Since $\varepsilon$ is a morphism of $k$-algebras it follows that $\varepsilon$ is an isomorphism with inverse $f$. $\Box$

Lemma 2: If $f : A \to B$ is field-local, then so is any localization $f[S^{-1}] : A[S^{-1}] \to B[f(S)^{-1}]$.

Proof. The field-local condition can be restated as the condition that every morphism $g : A \to K$ from $A$ to a field uniquely extends to a morphism $B \to K$. If $g : A[S^{-1}] \to K$ is a morphism from the localization to a field, then it still extends uniquely to a morphism $B[f(S)^{-1}] \to K$; the image of $B$ is uniquely determined by the unique extension of the restriction of $g$ to $A$, while the image of $f(S)^{-1}$ is uniquely determined by the uniqueness of inverses in $K$. $\Box$

Lemma 3: A field-local map $f : A \to B$ cannot invert any non-invertible elements of $A$. That is, if $a \in A$ is non-invertible, then so is $f(a)$.

Proof. If $a \in A$ is non-invertible, then $(a)$ is a proper ideal, so is contained in some maximal ideal $m$. Then the quotient map $A \to A/m$, whose kernel contains $a$, extends to a map $B \to A/m$, whose kernel contains $f(a)$. So $f(a) \in B$ is non-invertible. $\Box$

Now we can prove:

Theorem: A field-local map $f : A \to B$ is an isomorphism if $A, B$ are domains.

(Edit, 6/28: This is not right and the proof has a gap! As Dori Bejleri says in the comments we need $A$ to be integrally closed.)

Proof. By Lemma 2, $f$ remains field-local after localizing at every nonzero element of $A$, which produces a map

$$f' : \text{Frac}(A) \to B \otimes_A \text{Frac}(A).$$

Since the source of $f'$ is now a field, and since $B \otimes_A \text{Frac}(A)$ is a localization of a reduced ring and hence reduced, Lemma 1 applies to it, so we conclude that $f'$ is an isomorphism, giving

$$B \otimes_A \text{Frac}(A) \cong \text{Frac}(A).$$

Since $B$ is a domain, the localization map $B \to B \otimes_A \text{Frac}(A) \cong \text{Frac}(A)$ is an embedding. This map must be the unique extension of the embedding $A \hookrightarrow \text{Frac}(A)$ of $A$ into its field of fractions, so it follows that $f$ exhibits $B$ as a localization of $A$.

But by Lemma 3, a field-local map can't invert any non-invertible elements of $A$. So $f$ is an isomorphism. $\Box$

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It should be possible to use the result for domains to prove the following (although I haven't checked this carefully): $f : A \to B$ is field-local iff it induces a bijection on prime ideals and moreover for each prime ideal $P$ of $A$ the induced map $A/P \to B/(f(P))$ is an isomorphism. At this point there is no need to consider arbitrary fields, and probably the result generalizes to non-affine schemes too. This seems like a very strong condition if $A$ and $B$ are also assumed to be reduced but somehow I still can't get it to imply that $f$ is an isomorphism.

Answer 2

Let's work with separated $k$-schemes of finite type throughout. We have the following theorem.

Theorem 1: Let $\phi : X \to Y$ be a morphism. Then the following are equivalent.

1. For each field $K/k$, $\phi(K) : X(K) \to Y(K)$ is a bijection.
2. There exist locally closed stratifications $X' \to X$ and $Y' \to Y$ and a factorization $\phi' : X' \to Y'$ with $\phi'$ an isomorphism.

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Recall a locally closed stratification $X' \to X$ is a morphism of the form $$ X' = \bigsqcup_{j = 1}^m X_j \to X $$ which is a bijection of underlying topological spaces and such that $X_j\to X$ is a locally closed embedding for each $j$. Note that $X_{red} \to X$ is a locally closed stratification. A morphism satisfying condition 2 is sometimes called a piecewise isomorphism. Here are some examples to show this is the best one can hope for in general.

Example:

1. Consider $\phi : \operatorname{Spec}k \sqcup \mathbb{G}_m \to \mathbb{A}^1$. Then $\phi(K)$ is a bijection for all $K/k$ but $\phi$ is not an isomorphism. More generally, $\phi(K)$ is a bijection for all $K/k$ whenever $\phi : X' \to X$ is a locally closed stratification.
2. Let $\phi : \mathbb{A}^1 \to \{y^2 = x^3\} \subset \mathbb{A}^2$ be the normalization of the cuspidal cubic given by $t \mapsto (t^2, t^3)$. Then $\phi(K)$ is a bijection for all $K/k$ but $\phi$ is not an isomorphism or a stratification but it is a piecewise isomorphism.

We say a morphism $\phi : X \to Y$ is radicial if $\phi(K) : X(K) \to Y(K)$ is injective for any field extension.

Lemma 1: (Mustaţă, Remark A.22) The following are equivalent:

1. $\phi$ is a radicial surjection.
2. $\phi(K) : X(K) \to Y(K)$ is a bijection for all algebraically closed fields $K$.

We say $\phi : X \to Y$ is a universal homeomorphism if for any $Y' \to Y$, the base change $X' \to Y'$ induces a homeomorphism of underlying topological spaces.

Lemma 2: (Stacks Project, Tag 04D5) The following are equivalent:

1. $\phi$ is a universal homeomorphism.
2. $\phi$ is radicial, surjective and proper.

For radicial surjections, we have the following analogous theorem.

Theorem 2: Let $\phi : X \to Y$ be a morphism. Then the following are equivalent.

1. $\phi$ is a radicial surjection.
2. There exist locally closed stratifications $X' \to X$ and $Y' \to Y$ and a factorization $\phi' : X' \to Y'$ with $\phi'$ a universal homeomorphism.

Proof. Universal homeomorphisms and locally closed stratifications are radicial surjections so 2 $\implies$ 1. For the converse note that radicial (resp. radicial and surjective) is stable under basechange. Up to stratifying $Y$ and restricting to irreducible components, we may assume $Y$ is integral. Similarly, up to stratifying $X$, we may assume $X$ is the union of its irreducible components $X_1, \ldots, X_m$. Let $\eta \in Y$ be the generic point. There exists a unique $i$ such that $X_i \to Y$ contains $\eta$ in its image. Since $X_i \to Y$ is radicial, it is quasi-finite. By Zariski's Main Theorem, There exists an open embedding $X_i \to \bar{X}_i$ and a finite dominant map $g: \bar{X}_i \to Y$. Let $Z = \bar{X}_i \setminus X_i$. Then $g(Z)$ is closed. Let $U = Y \setminus g(Z)$. Then $g^{-1}(U) \subset X_i$ so $g^{-1}(U) \to U$ is radicial, surjective and finite and thus a universal homeomorphism. Now we can replace $X$ with $X \setminus g^{-1}(U)$, $Y$ with $Y \setminus U$, and induct on the dimension of $Y$.

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We need one more result which tells us when a universal homeomorphism is a piecewise isomorphism.

Proposition 1: Let $\phi : X \to Y$ be a universal homeomorphism. Suppose either 1) $\mathrm{char} k = 0$, or 2) $\phi(K) : X(K) \to Y(K)$ is a bijection for all fields $K/k$. Then $\phi$ is a piecewise isomorphism.

Proof. As in the above proof, up to stratifying $Y$, restricting to components, and taking reduced subschemes, we can assume $X$ and $Y$ are integral. Let $\eta \in Y$ and $\xi \in X$ be the generic points. Under assumption 2), we can restrict to $\xi \to \eta$ and apply Lemma 1 in Qiaochu's answer to see that $\xi \to \eta$ is an isomorphism and in particular $\phi$ induces an isomorphism of function fields $k(Y) \to k(X)$. Since the smooth locus is open on the source, we get an open subset $U \subset X$ where $\phi|_U$ is smooth. In characteristic $0$, we have generic smoothness which yields such an open subset. In either case, $\phi(U) = V$ is open, $U = \phi^{-1}(V)$, and $U \to V$ is a smooth universal homeomorphism and in particular étale. By Stacks Project, Tag 02LC, $\phi : U \to V$ is an isomorphism. Now as before we replace $X$ and $Y$ by the complements of $U$ and $V$ and induct.

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Proof of Theorem 1. 2 $\implies$ 1 since locally closed stratifications and isomorphisms satisfy the property that $\phi(K)$ is an isomorphism for all $K/k$. For 1 $\implies$ 2, we first apply Lemma 1 to see $\phi$ is a radicial surjection. By Theorem 2, there exists a locally closed stratification after which $\phi$ becomes a universal homeomorphism. By Proposition 1, $\phi$ is a piecewise isomorphism.

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Finally, I want to give the following complement to the Theorem in Qiaochu's answer.

Theorem 3: Let $\phi : X \to Y$ be a morphism such that $\phi(K)$ is a bijection for all $K/k$. Suppose further that $X$ and $Y$ are integral and $Y$ is normal. Then $\phi$ is an isomorphism.

Proof. By the proof of Proposition 1, $\phi$ is birational and quasifinite. By Zariski's Main Theorem, we can factor $\phi$ as an open embedding $X \subset \bar{X}$ followed by a finite birational morphism $\bar{X} \to Y$ with $\bar{X}$ integral. By a a variant of Zariski's Main Theorem, $\bar{X} \to Y$ must be an isomorphism since $Y$ is normal. Then $X \subset \bar{X} \cong Y$ is an open immersion which is also a homeomorphism so $X = \bar{X}$.