Every source I've looked at defines open subschemes and closed subschemes, but the definitions always look ad-hoc and not closely related to one another. Are there other kinds of subschemes? If not, can we give a definition for a "subscheme" and then deduce that the only ones that exist are open and closed subschemes?

For instance, EGA I, Definition 4.1.3, says (roughly):

We say a ringed space $(Y, \mathcal{O}_Y)$ is a subprescheme of a prescheme $(X, \mathcal{O}_X)$ if:

  1. $Y$ is a locally closed subset of $X$, and
  2. If $U$ denotes the largest open set of $X$ containing $Y$ and such that $X$ is closed in $U$, then $(Y, \mathcal{O}_Y)$ is a subprescheme of $(U, \mathcal{O}_X|_U)$ defined by a quasicoherent sheaf of ideals of $\mathcal{O}_X|_U$.

Is there some way to characterize this in terms of morphisms without having to build in the locally closed condition from the beginning?

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    $\begingroup$ To be short, a subscheme is a closed subscheme of an open subscheme. $\endgroup$
    – Cantlog
    May 31, 2014 at 19:22
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    $\begingroup$ @Cantlog: Yes, that's what the definition I included in my question says. But why? What motivates this seemingly arbitrary definition? $\endgroup$ May 31, 2014 at 19:25
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    $\begingroup$ @ZhenLin: Yes, but that's not a very satisfying definition. If I said something like "Permutation groups are reasonable, and cyclic groups are reasonable. Let's define a group to be either a permutation group or a cyclic group." then I think you'll agree most people wouldn't be satisfied with such a definition. $\endgroup$ Jun 1, 2014 at 2:09
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    $\begingroup$ I certainly didn't mean to imply that I thought a "subscheme" of $X$ ought to be any scheme homeomorphic to a subset of $X$, any more than I think a subgroup of $G$ is any group of the same cardinality as some subset of $G$. But it seems there ought to be some abstract characterizations of subschemes in terms of inclusion maps, rather than a characterization that explicitly forces us to build things in two steps by defining open subschemes and closed subschemes separately. $\endgroup$ Jun 1, 2014 at 4:53
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    $\begingroup$ The intelligent and courteous discussion between Daniel and @Zhen sets a standard of behaviour on this site. Kudos to both of them. $\endgroup$ Jun 1, 2014 at 7:56

2 Answers 2


I think a reasonable definition of an immersion of ringed topological spaces $f : X\to Y$ is :

  1. $f$ induces a homeomorphism $X\to f(X)$;

  2. For any $x\in X$, the canonical homomorphism $O_{Y, f(x)}\to O_{X,x}$ is surjective.

In the category of schemes, this implies that $f$ is a monomorphism (see http://stacks.math.columbia.edu/tag/01L6). Let us call such morphisms of schemes R-immersions (R for ringed topological spaces). Immersions of schemes in the standard sense (EGA, stacks project) are R-immersions.

Fact 1. If $X, Y$ are algebraic varieties over field $k$ (schemes of finite type over $k$), then any R-immersion $f: X\to Y$ is actually an immersion.

Fact 2. More generally, an R-immersion $f :X\to Y$ is an immersion if and only if $f$ is locally of finite type.

Now why immersions are prefered to R-immersions ? I don't really know. Fact 1 above could be an explanation. One question I don't know the answer is whether R-immersions are stable by base change (this is the case for immersions). If the answer is no, then this is one more reason to prefer immersions to R-immersions.

Example of an R-immersion which is not an immersion. Let $X$ be a scheme and let $x\in X$. Then the canonical morphism $i_x: \mathrm{Spec}(O_{X,x})\to X$ is an $R$-immersion. However, if e.g. $X$ is an infinite integral scheme, and if $\xi$ is its generic point, then $i_\xi$ is not an immersion because $\{ \xi\}$ is not locally closed in $X$.

Edit: afterthought.

The initial question can be interpreted as follows: given a subset $X$ in a scheme $(Y, O_Y)$, is it possible to endow $X$ with the structure of a scheme $(X, O_X)$, the latter being related in some manner to $(Y, O_Y)$ ? A first natural requirement is the underlying topological space of $X$ is given the induced topology, this is Condition (1) in my tentative definition of $R$-immersion. For the sheaf of regular functions, it is also natural to ask that the regular functions on $X$ are "induced" by regular functions on $Y$ and my Condition (2) is a natural candidate.

Then it turns out that the answer is positive for locally closed subsets of $Y$. I don't see other natural category of subsets for which the answer is positive. Anyway it seems hard to characterize subsets of $Y$ for which the answer is positive. A necessary condition is being pro-constructible (roughly speaking, possibly infinite intersection of constructible subsets), but even constructible is not enough to a get the structure of subscheme in the sense of $R$-immersion (one can show this for the classical example of constructible but non-locally closed subset of the affine plane: remove the $x$-axe and add the origin back).

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    $\begingroup$ Very good post, Cantlog. Indeed I have heard a mathematician I admire wonder why the canonical morphism $i_x: \mathrm{Spec}(O_{X,x})\to X$ is not considered an immersion. $\endgroup$ Jun 2, 2014 at 7:56
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    $\begingroup$ R-immersions are stable under base change. [But I'm too lazy to write down the proof, its quite long.] $\endgroup$ Jun 2, 2014 at 21:57
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    $\begingroup$ Fact 3: An R-immersion is an immersion iff $f(X) \subseteq Y$ is a locally closed subset. $\endgroup$ Jun 2, 2014 at 22:02

I think that there is no abstract nonsense answer. Locally closed immersions just arise naturally when one wants to study algebraic varieties. The local model looks as follows: We work in affine $n$-space $\mathbb{A}^n_k$ and consider not just closed subsets $\{x \in \mathbb{A}^n_k : p_1(x)=\dotsc=p_r(x) = 0\}$, but also allow inequalities by considering $\{x \in \mathbb{A}^n_k : p_1(x)=\dotsc=p_r(x) = 0 , f_1(x) \neq 0, \dotsc,f_s(x) \neq 0\}$. This is again an affine variety with coordinate ring $\{f_1,\dotsc,f_s\}^{-1} k[x_1,\dotsc,x_n]/(p_1,\dotsc,p_n)$. More generally, if $X$ is any scheme and $Z$ is a locally closed subset of the underlying set of $X$, then one may endow $Z$ with a scheme structure and the inclusion $Z \to X$ becomes a monomorphism of schemes. Of course one might try to characterize all those subsets with this property. Examples are infinite intersections of open affine schemes. Or one may try to characterize monomorphisms of schemes, see here for some information.


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