Over the last year, I have bugged a number of algebraic geometers with the following question:
Question 1: Say that an algebraic variety is maximally affine if, whenever occurs as an open subvariety of an affine variety , we have . Is the affine plane, , maximally affine?
Here and throughout the post, is an algebraically closed field.
We can make the question less technical, at the expense of making it less likely to be true.
Question 2: Suppose that we have polynomials in two variables, , , …, such that the resulting map is injective. Is the image necessarily closed? What if, instead of imposing that be injective, we impose that the Jacobian of have rank two everywhere on ?
Note that “Jacobian has rank ” can be thought of as “infinitesimally injective”, although it is neither implies nor implied by “injective”, so it is reasonable to ask about both hypotheses.
To rephrase question 1 in the language of question 2, the hypotheses become “ is injective, the Jacobian of has rank two everywhere on and is open in its closure.” If you don’t like me using topological language like “closure”, you can either take or learn about the Zariski topology.
Recently, I’ve realized that the answer to both questions is “no”. In this post, I’ll explain why I wanted the answer to be “yes”, why I thought it might be, and why it isn’t.
Let me start by explaining why I thought this might be true, and why I wanted this to be true. First, we know:
Ax’s Theorem: Let , , …, be polynomials in -variables. If is injective, then it is surjective.
(Link requires academic access. If you can’t get in, the argument is repeated and generalized in another paper of Ax.)
Second, we want to know:
A variant of the Jacobian conjecture: Let , , …, be polynomials in -variables. If the Jacobian of has full rank everywhere on , then the map is finite.
A few comments on this one. What does finite mean? Algebraically, it means that the ring is finitely generated as a module over the subring . Topologically, if , it means that the map has finite fibers and is proper. The standard example of a map with finite fibers that is not proper is the projection of the hyperbola onto the -axis.
Do you think its amazing that this algebraic condition is equivalent to the topological one? You should study algebraic geometry!
How does this variant relate to the ordinary Jacobian conjecture? The standard Jacobian conjecture has the same hypotheses, and also imposes , and draws the conclusion that is an isomorphism. This would follow from the variant above: it is not hard to show that a finite map whose Jacobian is everywhere nonvanishing is a covering map. But is simply connected, so it has no nontrivial covers.
If has characteristic , then is not “simply connected”. I haven’t told you what simply connected means for a variety over a field other than , but I’ll show you a nontrivial cover: map by . It is easy to see that every fiber has size , and so there is no branching. This means that the Jacobian conjecture fails in characteristic , and this map is a counter-example. The variant I give above might still be true.
By the way, I spent an hour on mathscinet to see if someone had formulated this conjecture and I couldn’t find it, although I found lots of other characteristic variants of the Jacobian conjecture. I find it hard to believe that this is original to me; it seems very natural. (Of course, maybe its false!)
Questions 1 and 2 came about by wondering what the analogues of Ax’s theorem and the Jacobian conjecture might be for maps with . (They are not supposed to pair off, though; both questions were meant to address both results.) Obviously, you can’t ask that the map be a bijection, but you could ask that the image be closed, or that the map be finite.
Unfortunately, in math, wishing doesn’t make it so. I hope that you’ve been searching for a counterexample; if not, this is a good time to start looking because I’ll give it away in the next two paragraphs.
Let’s try to break Question 2 first. We want not to be closed; say but not in . Then there must be some curve in such that . What could be? We should try to make as simple as possible. A first thought might be to just take a line — say, the -axis. But this can’t work. If we add another point to a line, we get , which is compact, and there are no positive-dimensional compact algebraic subvarieties of . (For those who like analysis better than algebra, take . Then the restriction of any coordinate function on to would be an analytic function on , and hence constant, so is a point and is not injective.)
Well, if a line doesn’t work, let’s try the next simplest thing. We’ll take to be the hyperbola . And we’ll try to arrange that exist, this will give us our point . Let’s just start writing down polynomials such that exists. Some things that work are , and . These are the first three I thought of, and we need to get interesting examples, so let’s see if this works. In other words, our putative counter-example will be .
Well, the image isn’t closed. is in the closure of the image, but it can’t be in the image because, if then , not . Is the map injective? Yes. Let be in the image of . Then the preimage of is if , and if . So this is a counter-example to the first part of question 2.
The Jacobian of is
The first two columns are independent when , and the first and third are when .
Finally, is the image of open in its closure? I’ll leave the details to you, but the answer is yes. is and is the complement of the line .
In some sense, this is the end of a beautiful idea. In another, its a beginning. This gives us a great way to write down maps which can’t be isomorphisms. In theory, that could give us a way to break the Jacobian conjecture. I don’t expect it’s going to work, because the evidence for the Jacobian conjecture is very strong. So I’ll pose this as a problem:
Problem: Let and be two polynomials such that and exist. Show that the Jacobian of and vanishes somewhere on .