# The plane is not maximally affine

Over the last year, I have bugged a number of algebraic geometers with the following question:

Question 1: Say that an algebraic variety $X$ is maximally affine if, whenever $X$ occurs as an open subvariety of an affine variety $Y$, we have $X=Y$. Is the affine plane, $k^2$, maximally affine?

Here and throughout the post, $k$ 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 $n$ polynomials in two variables, $f_1(x,y)$, $f_2(x,y)$, …, $f_n(x,y)$ such that the resulting map $F: k^2 \to k^n$ is injective. Is the image necessarily closed? What if, instead of imposing that $F$ be injective, we impose that the Jacobian of $F$ have rank two everywhere on $k^2$?

Note that “Jacobian has rank $2$” 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 “$F$ is injective, the Jacobian of $F$ has rank two everywhere on $k^2$ and $F(k^2)$ is open in its closure.” If you don’t like me using topological language like “closure”, you can either take $k=\mathbb{C}$ 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 $f_1$, $f_2$, …, $f_n$ be $n$ polynomials in $n$-variables. If $F: k^n \to k^n$ 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 $f_1$, $f_2$, …, $f_n$ be $n$ polynomials in $n$-variables. If the Jacobian of $F$ has full rank everywhere on $k^n$, then the map $F$ is finite.

A few comments on this one. What does finite mean? Algebraically, it means that the ring $k[x_1, x_2, \ldots, x_n]$ is finitely generated as a module over the subring $k[f_1, f_2, \ldots, f_n]$. Topologically, if $k=\mathbb{C}$, it means that the map $F:\mathbb{C}^n\to \mathbb{C}^n$ 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 $xy=1$ onto the $x$-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 $k=\mathbb{C}$, and draws the conclusion that $F$ is an isomorphism. This would follow from the variant above: it is not hard to show that a finite map $\mathbb{C}^n \to \mathbb{C}^n$ whose Jacobian is everywhere nonvanishing is a covering map. But $\mathbb{C}^n$ is simply connected, so it has no nontrivial covers.

If $k$ has characteristic $p$, then $k^n$ is not “simply connected”. I haven’t told you what simply connected means for a variety over a field other than $\mathbb{C}$, but I’ll show you a nontrivial cover: map $k\to k$ by $x \mapsto x+x^p$. It is easy to see that every fiber has size $p$, and $(d/dx)(x+x^p)=1$ so there is no branching. This means that the Jacobian conjecture fails in characteristic $p$, 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 $p$ 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 $k^m \to k^n$ with $m. (They are not supposed to pair off, though; both questions were meant to address both results.) Obviously, you can’t ask that the map $k^m \to k^n$ 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 $F(k^2)$ not to be closed; say $a \in \overline{F(k^2)}$ but not in $F(k^2)$. Then there must be some curve $C$ in $k^2$ such that $a \in \overline{F(C)}$. What could $C$ be? We should try to make $C$ as simple as possible. A first thought might be to just take a line — say, the $x$-axis. But this can’t work. If we add another point to a line, we get $\mathbb{P}^1$, which is compact, and there are no positive-dimensional compact algebraic subvarieties of $k^n$. (For those who like analysis better than algebra, take $k = \mathbb{C}$. Then the restriction of any coordinate function on $\mathbb{C}^n$ to $\overline{F(C)}$ would be an analytic function on $\mathbb{CP}^1$, and hence constant, so $F(C)$ is a point and $F$ is not injective.)

Well, if a line doesn’t work, let’s try the next simplest thing. We’ll take $C$ to be the hyperbola $xy=1$. And we’ll try to arrange that $\lim_{t \to \infty} F(t,t^{-1})$ exist, this will give us our point $a$. Let’s just start writing down polynomials $f(x,y)$ such that $\lim_{t \to \infty} f(t,t^{-1})$ exists. Some things that work are $y$, $xy-1$ and $x(xy-1)$. These are the first three I thought of, and we need $n \geq 3$ to get interesting examples, so let’s see if this works. In other words, our putative counter-example will be $(x,y) \mapsto (y, xy-1, x(xy-1))$.

Well, the image isn’t closed. $(0,0,0) = \lim_{t \to \infty} F(t, t^{-1})$ is in the closure of the image, but it can’t be in the image because, if $y=0$ then $xy-1=-1$, not ${0}$. Is the map injective? Yes. Let $(u,v,w)$ be in the image of $F$. Then the preimage of $(u,v,w)$ is $((v+1)/u,u)$ if $u \neq 0$, and $(-w,0)$ if $u=0$. So this is a counter-example to the first part of question 2.

The Jacobian of $F$ is

$\begin{pmatrix} 0 & y & 2xy-1 \\ 1 & x & x^2 \end{pmatrix}.$

The first two columns are independent when $y \neq 0$, and the first and third are when $y=0$.

Finally, is the image of $F$ open in its closure? I’ll leave the details to you, but the answer is yes. $\overline{F(k^2)}$ is $v(v+1)=uw$ and $F(k^2)$ is the complement of the line $u=v=0$.

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 $k^2 \to k^2$ 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 $f(x,y)$ and $g(x,y)$ be two polynomials such that $\lim_{t \to \infty} f(t,t^{-1})$ and $\lim_{t \to \infty} g(t,t^{-1})$ exist. Show that the Jacobian of $f$ and $g$ vanishes somewhere on $k^2$.

## 10 thoughts on “The plane is not maximally affine”

1. Are there *any* maximally affine varieties of dimension greater than one?

2. Ben Wieland says:

No, there are no maximally affine varieties of higher dimension. Consider U an affine variety, X the projective closure, Z the points at infinity. Blow up X in some point in Z. Delete the proper transform of Z.

To check this requires basic facts about blow-ups, which, when unwound into coordinates, are probably pretty close to David’s construction.

3. David Speyer says:

I don’t think that works. Applying your recipe to the plane, I get $\mathbb{P}^1 \times \mathbb{A}^1$. (Everything you do is a toric operation, so this is an easy computation.) To see that your object isn’t affine, let $z$ be the point which gets blown up and take $L$ a line in the plane, meeting $Z$ at $z$. Then your variety contains $L \cong \mathbb{P}^1$, and can’t be affine.

That said, my intuition now is that there may not be any maximal affine varieties other than curves.

4. Dear David,

Your construction has a more direct description.

Consider the projective surface P^1 x P^1. The diagonally
embedded P^1 is ample, and so removing it gives something
affine. (Incidentally, that is the problem with Ben’s putative
example: removing a divisor doesn’t necessarily give something
affine — you have to remove an ample divisor.)

Here is an open immersion from A^2 to the complement of
the diagonal:

(x,y) maps to (x, x + 1/y)

The image is equal to the complement of the diagonal in
A^1 x P^1.

What is the connection with your example?

Well P^1 x P^1 is an abstract model for a quadric (the closure
of the image of F in your example). In your example, the
identified with the diagonally embedded P^1 in my model.
So the closure of the image of your F is identified with the
complement of the diagonally embedded P^1 in my model.

The line u = v = 0 in your example is identified with the line

(point at infinity) x P^1

in my model.

Regards,

Matthew

5. Yup! And you can go further than this. If you take $\mathbb{P}^1 \times \mathbb{P}^1$ and remove the closure of $y=f(x)$ for $f(x)$ any polynomial of positive degree , you get something affine. If you then remove $\{ \infty \} \times \mathbb{P}^1$, you get $\mathbb{A}^2$.

6. Ben Wieland says:

oops
I think that there should be some way of expressing this in terms of points at infinity that makes it easy to go from A^n to general affines.

7. Attila Smith says:

Dear David,
another description of your example is that if you blow-up P^2 in z, you get the Hirzebruch surface \Sigma_1, a P^1 bundle over P^1.
Deleting the strict transform of the line through z leaves you with a P^1 bundle over A^1, necessarily trivial: it is your A^1 X P^1.
There is a beautiful picture of this on the cover of Shafarevich’s “Basic Algebraic Geometry ” volume 1 (Springer-Verlag).
The plane and its blow-up are shown in a magnificent
turquoise green.

8. ninguem says:

There are ruled varieties Y over a projective base X, with a section S, so that Y-S is affine. The corresponding rank 2 vector bundle on X giving rise to Y is said to be ample in the sense of Hartshorne. Anyway, Y-S is maximally affine for the same reason that the affine line is maximally affine. So there re maximally affine varieties of every dimension.