5
$\begingroup$

First I apologize if this is elementary. I have just started looking at the basics of stacks and algebraic spaces so my understanding is lacking.

Let's work over an algebraically closed field $k$. Suppose I have an algebraic space $\mathcal{A}$ and two morphisms $f_i:S\rightarrow\mathcal{A}$ such that for every closed point $\operatorname{Spec}(k)\rightarrow S$ the induced maps $f_i|_{\operatorname{Spec}(k)}\rightarrow \mathcal{A}$ agree. Suppose $S$ is a reduced scheme with $\operatorname{Spec}(k)$ points which are dense in $S$. Must $f_1=f_2$? If not is there a condition we can impose on the algebraic space so this holds?

My intuition says since an algebraic space is fibered over sets instead of groupoids this should hold.

Thanks

$\endgroup$
9
  • 1
    $\begingroup$ This is already false for affine schemes. First learn schemes, then algebraic spaces. $\endgroup$
    – Martin Brandenburg
    Commented Mar 11, 2013 at 21:19
  • $\begingroup$ @Martin Sorry I suppose I need to assume the $\operatorname{Spec}(k)$ points are dense in $S$? What I have in mind is I have a functor which sends $S$ to the set of some sort of objects over $S$, and that functor is representable by an algebraic space. If the fibers over some $S$ (let's say a variety) are all the same object must it be the trivial family? $\endgroup$
    – user16544
    Commented Mar 11, 2013 at 21:42
  • $\begingroup$ Can't you just take a fiber-wise trivial family that isn't trivial to get a counterexample to your comment? $\endgroup$
    – Matt
    Commented Mar 11, 2013 at 22:10
  • $\begingroup$ @Matt Yes but if the functor is representable by an algebraic space can you do this? This means in particular that the objects being parameterized by $S$ have no automorphisms. $\endgroup$
    – user16544
    Commented Mar 11, 2013 at 22:15
  • 1
    $\begingroup$ While I think about that, we still need more conditions than $k$-points being dense. Can't you take two different tangent vectors at the same point? Then $f_i: Spec(k[\varepsilon]/(\varepsilon^2))\to \mathcal{A}$ agree on the $k$-point but shouldn't be considered "equal." $\endgroup$
    – Matt
    Commented Mar 11, 2013 at 22:37

1 Answer 1

Reset to default
0
$\begingroup$

It was difficult to capture the essence of what I was looking at, so I apologize that I never nailed down the correct hypotheses. This exercise from Hartshorne is what I was looking for an analog of in algebraic spaces:

II.4.2: Let $S$ be a scheme, let $X$ be a reduced scheme over $S$, and let $Y$ be a separated scheme over $S$. Let $f$ and $g$ be two $S$-morphisms which agree on an open dense subset of $X$. Then $f=g$.

$\endgroup$
2
  • $\begingroup$ I'm curious. Is this an "answer" because you've decided that this result is good enough (i.e. you figured out how to adapt the proof in the scheme case to the algebraic space case) or is this meant as an edit to the question and under these type of hypotheses you are still looking for an answer in the algebraic space case? $\endgroup$
    – Matt
    Commented May 13, 2013 at 23:48
  • $\begingroup$ It's not an answer, but it seemed a bit silly to me to edit the original question so the discussion several months ago makes no sense. I'm not sure what the proper policy is here so I erred on the side of not creating confusion. I haven't thought about this for a while but I happened to be leafing through Hartshorne today and noticed that this captured what I was after generalizing to algebraic spaces. $\endgroup$
    – user16544
    Commented May 14, 2013 at 1:51

You must log in to answer this question.