Images and Monomorphisms of Schemes

Question

If $X$ is an object in an arbitrary category, there is a natural definition of a subobject of $X$ as a monomorphism into $X$ (or really an equivalence class of monomorphisms). If $X$ is a scheme, however, the term 'subscheme' is conventionally reserved only for locally closed immersions (as in EGA I.4.1.2). There are certainly many monomorphisms of schemes that in this sense aren't subschemes, for example the inclusion of Spec of a local ring such as $Spec K[x]_{(x)} \to Spec K[x]$.

When we restrict 'subscheme' to mean 'locally closed immersion', defining images of schemes becomes problematic. A sensible definition, in any category, of the image of a morphism is the minimal subobject through which it factors. Using the above definition of subscheme, there are perfectly well-behaved examples of morphisms of schemes that don't have images in this sense. For example, consider the morphism $\mathbb A^2_K \to \mathbb A^2_K $ induced by the ring homomorphism $(x,y) \mapsto (x,xy)$; the set-theoretic image is the union of the origin and the complement of the $y$-axis, and there is no minimal locally closed set containing this.

There is, however, always a minimal closed immersion through which a given morphism factors, and so if one defines the scheme-theoretic image in this sense, it always exists. My question is that, if we let our notion of 'subscheme' include all monomorphisms, would the resulting notion of 'scheme-theoretic image' always exist? In other words, is there always a minimal monomorphism of schemes through which a given morphism factors? Say, in the above example? If I hand you a constructible subset of a scheme, can you only find a monomorphism onto that set if it's locally closed?

As a 'softer' question, can someone explain why we don't want to call general monomorphisms subschemes? In particular, suppose I have a morphism that is a submersion onto a locally-but-not-globally closed subscheme. It seems much more sensible to call that locally closed subscheme the image, rather than its global closure.

Answer 1

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 $\eta_{V(x)} \in Y$). $\square$

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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.

Answer 2

As a partial answer to your "softer" question, general monomorphisms (as Brian Conrad points out) can be more general than we really want subschemes to be. For instance, if $A$ is a Noetherian ring and $\hat{A}$ its completion with respect to some ideal, then $\hat{A}$ is isomorphic to $\hat{A} \otimes_A \hat{A}$. Consequently, Spec $\hat{A}$ is isomorphic to $\mathrm{Spec} \hat{A} \times_{\mathrm{Spec} A} \mathrm{Spec} \hat{A}$, and so Spec $\hat{A} \to $ Spec $A$ is a monomorphism of schemes. However, I do not think we want to call Spec $\hat{A}$ a subscheme of Spec $A$.

Edit: As the comment below points out, this example is incorrect. I had assumed that $\hat{A} \otimes_A \hat{A} = \hat{\hat{A}} = \hat{A}$, but the theorem I was using would require that $\hat{A}$ be a finitely generated $A$-module, which is typically not the case. However, the same reasoning shows that, for instance, Spec $k(t) \to$ Spec $k[t]$ is a monomorphism, whereas one would expect that any subscheme of a variety should contain at least one closed point.