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.

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    $\begingroup$ "If I hand you a constructible subset of a scheme, can you only find a monomorphism onto that set if it's locally closed?". I'm not so sure. Let me try this example: Let X be affine 2-space, let U be X minus the y-axis and let P be the origin. Then U union P is a nice example of a constructible set: it's the image of the map U disjoint P -->X (open immersion on U, closed immersion on P) and it's not locally closed. But I think the map (U disjoint union P) --> X is a monomorphism; if I have two maps T-->(U disjoint P) which become equal on X then write T=T_U disjoint T_P and follow your nose. $\endgroup$ Mar 30, 2010 at 20:23
  • $\begingroup$ re subject of the previous comment: for any constructible (indeed even pro- or ind-constructible) subset of a scheme X there is a monomorphism of schemes W->X whose image is that set. the reason is (see EGAIV.I.1.9, or 1971 EGAI.7.2) that for any scheme X there exists a scheme X^cons, a monomorphism X^cons -> X which is bijective on points and which has the property that the map induced on point sets gives a bijection between closed (resp. open, resp. open and closed) subsets of X^cons and pro-constructible (resp. ind-constructible, resp. constructible) subsets of X. $\endgroup$
    – LRG
    Mar 30, 2010 at 22:55
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    $\begingroup$ A definition of "image" whose formation is not visibly local on the target (i.e., formation does not visibly commute with base change alone open immersions) is unlikely to be of much use in proofs, let alone to do anything interesting. This is but one of the defects of the definition in terms of global minimal factorization through a closed immersion. Likewise, the reason not all monomorphisms are called subschemes is because a general monomorphism is a crazy thing. Remember that the subject is called algebraic geometry, not algebraic category theory. $\endgroup$
    – BCnrd
    Mar 31, 2010 at 3:08

2 Answers 2


$\newcommand\verts[1]{\lvert#1\rvert}$For any subset $\verts Y \subseteq \verts X$, there exists a monomorphism $\iota \colon Y \hookrightarrow X$ supported on $Y$. However, these do not have the property that any map landing in $\verts Y$ 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 $\verts Y \subseteq \verts X$, there exists a monomorphism $\iota \colon Y \hookrightarrow X$ supported on $Y$.

Proof. Indeed, let $Y = \coprod_{y \in \verts Y} \operatorname{Spec} \kappa(y)$. 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 $\verts Y = \verts{\operatorname{im}(f)}$ or $\verts Y = \verts{\operatorname{im}(f)} \cup \{\eta_{V(x)}\}$, and $Y$ is integral.

Proof. Clearly $\verts Y$ contains $\verts{\operatorname{im}(f)}$. 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 $\eta_{V(x)} \in Y$). $\square$


[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.

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    $\begingroup$ Numdam is a wonderful site, but they just can't leave their link format alone. Some time in the last 4 and a half years, they seem to have added an underscore to their volume links, so that the SAC link was broken. I have edited it, so that it will work until they change their links again, tomorrow or possibly the next day. $\endgroup$
    – LSpice
    Jun 23, 2023 at 13:38
  • $\begingroup$ While I was in the source, I noticed one other thing, that you very likely have figured out anyway in the years since this post: as you saw, something like $|\operatorname{im}(g)|$ |\operatorname{im}(g)| probably spaces too widely. You fixed this with $|\!\operatorname{im}(g)|$ |\!\operatorname{im}(g)|, but another way is to use braces to suppress undesired spacing: $|{\operatorname{im}(g)}|$ |{\operatorname{im}(g)}|. Even better, the semantically correct $\lvert\operatorname{im}(g)\rvert$ \lvert\operatorname{im}(g)\rvert takes care of it for you. I edited accordingly. $\endgroup$
    – LSpice
    Jun 23, 2023 at 13:47
  • $\begingroup$ I am re-visiting this question again, and I realised I don't think I understand: what does it mean to say that a monomorphism $\iota : Y \hookrightarrow X$ is supported on $Y$? Is it supposed to be something like "supported on $\lvert Y\rvert$", or am I just parsing incorrectly? $\endgroup$
    – LSpice
    Aug 1, 2023 at 20:22
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    $\begingroup$ @LSpice ah yes, that should be $\lvert Y\rvert$. $\endgroup$ Aug 1, 2023 at 21:58

$\DeclareMathOperator\Spec{Spec}$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 $\Spec \hat{A} \times_{\Spec A} \Spec \hat{A}$, 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.

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    $\begingroup$ Careful, $\widehat{A} \otimes_ A \widehat{A}$ is generally much much much bigger than $\widehat{A}$, and typically not even noetherian (and ${\rm{Spec}} \widehat{A} \rightarrow {\rm{Spec}} A$ is generally very far from a monomorphism, with gigantic geometric fibers). Due to its importance in fpqc descent, this was one of the key examples which motivated Grothendieck to change his mind about the necessity of noetherian hypotheses (as in his 1958 ICM talk). On the other hand, a localization at a multiplicative set provides a typical example of the sort I think you wanted to express. $\endgroup$
    – BCnrd
    Apr 2, 2010 at 7:49

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