Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

let $X : Ring \to Set$ be a functor (a Z-functor in the language of demazure, gabriel) and $V \subseteq X$ a locally closed subfunctor. assume that $U \subseteq V$ is an open subfunctor. does then exist an open subfunctor $W \subseteq X$ such that $U = V \cap W$? if $X$ and $V$ are schemes, this should be true.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

I don't think this is true even when $X$ and $V$ are schemes if you only require that the map $V\to X$ is an embedding of functors (rather than a locally closed embedding). Example: Take $X=Spec(k[x])$, $V=Spec(k[x,x^{-1}]\times k)$, $U=Spec(k)$. That is, $X$ is an affine line, $V$ is the disjoint union of a punctured line and origin, and $U$ is a point, which we view as the connected component of $V$. The natural map $V\to X$ (corresponding to stratification of $X$) is a categorical monomorphism: it induces an embedding of functors.

EDIT:

Now we add the assumption that $V\hookrightarrow X$ is locally closed. I think the statement still fails, here's a counterexample:

$X$ will be an ind-scheme: so there is a sequence of schemes $X_n$ related by closed embeddings $X_n\hookrightarrow X_{n+1}$ and $$X(A)=\lim_{\to} X_n(A).$$ In other words, every point of $X(A)$ factors through one of $X_n$'s.

For ind-schemes, a locally closed subfunctor $V\subset X$ is given by a compatible family of locally closed $V_n\subset X_n$ (so that it is a locally closed sub-ind-scheme). Problem is, the statement fails for ind-schemes.

Let's take $X_n={\mathbb A}^n$, with the embedding $X_n\hookrightarrow X_{n+1}$ being the coordinate embedding. Now let $V_n$ be the union of the origin $0$ and $n-1$ punctured lines $$l_k:=\lbrace(k,0,\dots,0,x,0,\dots,0)|x\ne 0\rbrace,$$ where $x$ is in the $k$-th position, and $k$ varies from $2$ to $n$. (Let's say I work over a field of characteristic $0$, so all integers are distinct.)

Finally, $U_n\subset V_n$ is the origin, which is a component of each $V_n$, so it is both open and closed.

It is easy to see that it is impossible to find a compatible family of open subsets $W_n\subset X_n$ such that $U_n=V_n\cap W_n$. Indeed, $W_1$ contains $0$, so it is non-empty, and thus contain $n\in{\mathbb A}^1=X_1$ for some $n$. But then $W_n$ must contain $(n,0,\dots,0)$ without meeting $l_n$.

share|improve this answer
    
that's right. I've edited the question. –  Martin Brandenburg Dec 30 '09 at 16:31

Your Answer

 
discard

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.