The notion of a surjective map of schemes is somewhat confusing. For example, consider an elliptic curve {E} over {{\mathbb Q}} of positive rank. The multiplication-by-{m} map is obviously not surjective as a map {E({\mathbb Q}) \rightarrow E({\mathbb Q})}, but in fact it is surjective as a map of schemes, because schemes carry baggage associated to the points we haven’t yet `added in’ by passing to an algebraic closure. In this post we investigate the condition of a map being surjective on schemes as related to its inducing a surjection on the geometric points. I looked briefly in Hartshorne and did not seem to find anything about this, and I still do not have the whole picture (I have only considered what happens on closed points), so if readers can suggest a proper reference I would be grateful. Also, this post is likely to contain errors which might need fixing, so I urge readers not to assume I know what I’m doing.

Proposition 1 Let {\phi: Y \rightarrow X} be a map of schemes of finite type over {k}, and {k^a} an algebraic closure for {k}. The induced map {Hom_k(Spec \ k^a, Y) \rightarrow Hom_k(Spec \ k^a, X)} is surjective if and only if {Y \rightarrow X} is surjective on closed points.

I think one direction is reasonably straightforward. Given a closed point {x \in X}, take any geometric point {f:Spec\ k^a \rightarrow X} with image {x}, for example the point obtained from {O_U \rightarrow O_{U,x} \rightarrow k(x) \rightarrow k^a} for some affine {U} containing {x} (since we are of finite type over {k} and {x} is maximal, {k(x)} is a field of finite type over {k}, so finite over {k} by the Nullstellensatz). Then by assumption, there is a {g:Spec\ k^a \rightarrow Y} with {\phi g = f}, so in particular any closed point in the closure of {g(0)} is a closed point of {Y} lying above {x}.

Let’s try to prove the other direction. So now we are given a geometric point {f:Spec\ k^a \rightarrow X} and we want to lift it to such a point of {Y}. Well, such an {f} determines a unique pair {(x,\phi)} with {x\in X} and {\psi: k(x) \rightarrow k^a}, and {x} must be closed by a dimension argument, so let us choose a closed point {y \in Y} with {\phi(y)=x}. There is an induced map in the wrong direction {k(x) \rightarrow k(y)}. However, since {y} is closed, {k(y)} is a finite extension of {k}, so in particular we can generate a map {\psi'k(y) \rightarrow k^a}. Furthermore, it is possible to do so in such a way as to ensure that {k(x) \rightarrow k(y) \rightarrow k^a} is the map {\psi}.

But then the pair {(y,\psi')} determine a geometric point {g:Spec\ k^a \rightarrow Y} in the obvious fashion, and it is clear from our construction that {\phi g = f}, so we have proved that the map on geometric points is a surjection.

There are at least two things I still don’t know which would be good to know. Firstly, what conditions do I need to impose to guarantee that {k^a} can be replaced with {k^s}, a separable closure? There is some kind of condition needed to say that induced maps are separable (so I guess insisting that {\phi} is unramified) but also that the extensions {k(x)/k} are separable, which is probably some simple condition but I haven’t yet worked out what. Secondly, can I extend the result to surjectivity of the map {Y \rightarrow X} on all points? Again, hopefully this is fairly easy messing around with taking closures, etc. but I still don’t seem to have managed it.

It would also be nice to know if the above argument is actually correct and/or necessary. It’s been assembled in a very ad-hoc way to convince me that things I was doing with étale covers of elliptic curves weren’t nonsense, so I can believe my sketchy knowledge of scheme theory could have allowed a flaw to pass. Note that the result claimed about multiplication by {m} on an elliptic curve follows (on the closed points) immediately, because given any geometric point I can divide it by {m} by solving an appropriate degree {m^2} polynomial, hence I get a geometric point with co-ordinates in {k^a}.

Advertisements