Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X = \mathrm{Spec}(A)$ be an affine scheme. Let $U$ be a quasi-compact open subset of $X$. Then there exist an affine scheme $Y$ and a morphism $f\colon Y \rightarrow X$ such that $f(Y) = U$, right? We are interested in a similar result(if any) on an arbitrary intersection of quasi-compact open subsets of $X$. Let $(U_i)_{i\in I}$ be a family of quasi-compact open subsets of $X$. Let $T = \bigcap_{i\in I} U_i$. Do there exist an affine scheme $Y$ and a morphism $f\colon Y \rightarrow X$ such that $f(Y) = T$?

share|cite|improve this question

2 Answers 2

up vote 4 down vote accepted

I don't know what is the motivation of the question. But the answer is yes.

In an affine (hence separated) scheme $X$, any affine open subset is retro-compact hence constructible. So any quasi-compact open subset of $X$ is constructible in $X$. Any intersection of constructible subsets of $X$ is pro-constructible (EGA IV.1.9.4), and, as $X$ is quasi-compact and quasi-separated, any pro-constructible subset of $X$ is the image of an affine scheme (EGA, IV.1.9.5(ix)).

Edit Sketch of the proof of EGA, IV.1.9.5(ix):

  1. First a constructible subset of $X$ is the image of an affine scheme $X'$ (EGA, IV.1.8.3). It is enough to prove it for a locally closed subset $U\cap (X\setminus V)$, so a quasi-compact open subset of an affine scheme $X\setminus V$, write it as a union of affine open subschemes $U_1, U_2...$ and take $X'$ the disjoint union of the $U_i$'s).

  2. (EGA IV. Now for a pro-constructible subset $\cap_i C_i$, if $C_i$ is the image of $f_i: X'_i=\mathrm{Spec}(A_i)\to C_i$, define $A'$ as the direct limit of tensor products of finitely many $A_i$'s and consider the canonical morphism $f: \mathrm{Spec}(A')\to X$ (each $A_i'$ is an $A$-algebra via $f_i$), show $f(X')=\cap_i C_i$. This part is harder.

share|cite|improve this answer
Thanks. I've just become interested in the set theoretic image of a morphism of schemes. Could you give us a hint so that we can prove it without referring to EGA? I think EGA is somewhat intimidating for most of us. – Makoto Kato Dec 22 '12 at 21:54
@MakotoKato: by the way, EGA IV.1.9.5(ix) is "if and only if". – user18119 Dec 23 '12 at 8:20
"EGA IV.1.9.5(ix) is "if and only if"" This is an interesting result. I searched for internet by the word "proconstructible", but very few was hit. – Makoto Kato Dec 23 '12 at 12:51
The title question is not so difficult to come up with. Let $C_1,\dots,C_n$ be quasi-compact open subsets(or constructible subsets). Let $f_i: X'_i=\mathrm{Spec}(A_i)\to C_i$ be a morhism such that $f_i(X'_i) = C_i$. Then $\bigcap C_i$ is the image of $\mathrm{Spec}(A_1\otimes\cdots\otimes A_n)$. So if $(C_i)_{i\in I}$ is a family of constructible subsets, it is natural to consider the infinite tensor product of $(A_i)_{i\in I}$. – Makoto Kato Dec 23 '12 at 13:56

No. Delete countably many points from the affine line over an uncountable field. This set is not constructible and cannot be the image of an affine.

share|cite|improve this answer
Could you explain why a non-constructible subset of an affine scheme cannot be the image of an affine scheme? – Makoto Kato Dec 22 '12 at 19:28
It can't be the image of a scheme by a locally finite presented morphism. – user18119 Dec 22 '12 at 21:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.