For any subset $|Y| \subseteq |X|$, there exists a monomorphism $\iota \colon Y \hookrightarrow X$ supported on $Y$. However, these do not have the property that any map landing in $|Y|$ factors through $Y$.
In fact, I will show (using results from the Samuel seminar on epimorphisms of rings [Sam], [Laz]) that in the example $\mathbb A^2 \to \mathbb A^2$ given by $(x,y) \mapsto (x,xy)$ that you give, there is no minimal monomorphism it factors through.
Lemma 1. For any subset $|Y| \subseteq |X|$, there exists a monomorphism $\iota \colon Y \hookrightarrow X$ supported on $Y$.
Proof. Indeed, let $Y = \coprod_{y \in |Y|} \operatorname{Spec} \kappa(y)$. Then the natural map $\iota \colon Y \to X$ is a monomorphism, because $Y \times_X Y$ is just $Y$. (See [ML, Exc. III.4.4] for this criterion for monomorphism.) $\square$
Now we focus on the example $\mathbb A^2 \to \mathbb A^2$ given by $(x,y) \mapsto (x,xy)$. Write $f \colon Z \to X$ for this map, and note that $f$ is an isomorphism over $D(x) \subseteq X$. Assume $\iota \colon Y \to X$ is a minimal immersion such that there exists a factorisation of $f$ as
$$Z \stackrel g\to Y \stackrel \iota \to X.$$
For any irreducible scheme $S$, write $\eta_S$ for its generic point. We denote the origin of $\mathbb A^2$ by $0$.
Lemma 2. We must have $|Y| = |\!\operatorname{im}(f)|$ or $|Y| = |\!\operatorname{im}(f)| \cup \{\eta_{V(x)}\}$, and $Y$ is integral.
Proof. Clearly $|Y|$ contains $|\!\operatorname{im}(f)|$. If this inclusion were strict, then $|Y|$ contains some point $y$ that is not in the image of $f$. If $y$ is closed in $X$, then the open immersion $Y \setminus \{y\} \to Y$ gives a strictly smaller monomorphism that $f$ factors through, contradicting the choice of $Y$. Thus, $y$ has to be the generic point of $V(x)$. This proves the first statement.
For the second, note that the scheme-theoretic image $\operatorname{im}(g)$ of $g$ is $Y$. Indeed, if it weren't, then replacing $Y$ by $\operatorname{im}(g)$ would give a smaller monomorphism factoring $f$, contradicting minimality of $Y$. But the scheme-theoretic image of an integral scheme is integral, proving the second statement. $\square$
We now apply the following two results from the Samuel seminar on epimorphisms of rings [Sam]:
Theorem. Let $\iota \colon Y \to X$ be a quasi-compact birational monomorphism of integral schemes, with $X$ normal and locally Noetherian. Then $\iota$ is flat.
Proof. See [Sam, Lec. 7, Cor. 3.6]. $\square$
If $U \subseteq Y$ is an affine open neighbourhood of $0$ and $R = \Gamma(U,\mathcal O_U)$, then we get a flat epimorphism $\phi \colon k[x,y] \to R$ of $k$-algebras (not necessarily of finite type).
Theorem. Let $f \colon A \to B$ be a flat epimorphism of rings, and assume $A$ is normal and $\operatorname{Cl}(A)$ is torsion. Then $f$ is a localisation, i.e. $B = S^{-1}A$ for $S \subseteq A$ a multiplicative subset.
Proof. See [Laz, Prop. IV.4.5]. $\square$
Thus, $R = S^{-1}k[x,y]$ for some multiplicative set $S \subseteq k[x,y]$. This implies that $V = X \setminus U$ is a union of (possibly infinitely many) divisors. Moreover, $V \cap D(x) \subseteq D(x)$ has finitely many components since $D(x)$ is Noetherian (and $\iota$ is an isomorphism over $D(x)$). But then $V \cap V(x)$ is either finite or all of $V(x)$. This contradicts the fact that $U \cap V(x)$ equals $\{0\}$ or $\{0,\eta_{V(x)}\}$ (depending whether $\eta_{V(x)} \in Y$). $\square$
References.
[Laz] D. Lazard, Autour de la platitude, Bull. Soc. Math. Fr. 97, 81-128 (1969). ZBL0174.33301.
[ML] S. Mac Lane, Categories for the working mathematician. Graduate Texts in Mathematics 5. New York-Heidelberg-Berlin: Springer-Verlag (1971). ZBL0232.18001.
[Sam] P. Samuel et al., Séminaire d’algèbre commutative (1967/68): Les épimorphismes d’anneaux. Paris: École Normale Supérieure de Jeunes Filles (1968). ZBL0159.00101.
Assume $k$ is a field and $F$ is a contravariant functor from $k$-schemes of finite type to sets. For each $X/k$ and $x\in F(X)$ we get a "characteristic map" $X(k)\to F(k)$ by pulling back $x$. Now if $k$ is a topological field, we can define a topology on $F(k)$, which is the finest making all these maps (for all pairs $(X,x)$) continuous. Some facts are easy to check, for instance:
- if $F$ is representable we get the "usual" topology on $F(k)$,
- every morphism of functors $F\to G$ gives rise to a continuous map $F(k)\to G(k)$,
- restricting $X$ to affine $k$-schemes does not change anything.
With this generality, of course, one doesn't get very far without making some assumptions on $F$ and/or $k$. For instance, compatibility with products is not a formal consequence of the definition and therefore, probably, does not hold in general (example, please?). The nasty thing about this topology is that it is a quotient topology ($F(k)$ is a quotient space of the disjoint sum of all $X(k)$ indexed by pairs $(X,x)$) and general quotient topologies don't behave nicely.
Assume now that $k$ satisfies the following "weak henselian" property:
(WH) Every étale morphism $Y\to X$ gives rise to an open map $Y(k)\to X(k)$.
This is satisfied by local fields and more generally by henselian valued fields. It is of course particularly relevant if $F$ is an algebraic space of finite type over $k$. In this case, it is easy to check that if $U\to F$ is an étale presentation, then, assuming (WH), $U(k)\to F(k)$ is an open map. This (together with the existence of pointed étale neighborhoods) is enough to imply, for instance, that the topology on $F(k)$ is compatible with products of algebraic spaces. (The point is that we now have a quotient by an open equivalence relation, and these quotients do behave nicely). I have not checked the other conditions in detail but I am confident that it works.
Some remarks:
- I have used this trick in my paper "Problèmes de Skolem sur les champs algébriques" (Compositio 125 (2001), 1-30) to define open subcategories of $X(k)$ when $X$ is an algebraic stack of finite type over a local field $k$.
- I am currently working with P. Gille on the case where $F(X)=H^1_{fppf}(X, G)$, where $G$ is an affine $k$-group scheme of finite type.
- Condition (WH) seems to become even nicer when, in addition, every finite morphism gives rise to a closed map. This holds for local fields, and more generally for any field with a (rank 1?) valuation, which is algebraically closed in its completion, such as the quotient field of an excellent henselian DVR.
Best Answer
Hi Ravi,
There is an example in Tag 01QW in Johan's stacks project.
Jarod