# How not to prove the Jacobian conjecture

A few years ago, I got curious to see how people attack the Jacobian conjecture. The Jacobian conjecture says the following:

Let $k$ be a field of characteristic zero, and let $F: k^n \to k^n$ be an unramified map. Then $F$ is an isomorphism.

The conjecture is usually stated algebraically, not geometrically. That statement goes as follows: Let the map $F$ be given by $\displaystyle{(x_1, \ldots, x_n) \mapsto \left( f_1(x_1, \ldots, x_n),\ \ldots,\ f_n(x_1, \ldots, x_n) \right)}$.
Define $J(F)$, the Jacobian of $F$, to be $\det(\partial f_i/\partial x_j)$. Then the algebraic statement is

Let $f_1$,$f_2$, …, $f_n$ are $n$ polynomials in $n$ variables such that $J(f_1, \ldots, f_n)$ is in $k^*$. Then the $f_i$ generate the ring $k[x_1, \ldots, x_n]$.

The geometric meaning of $J(F)$ is that $J(F)$ vanishes at the points where $F$ is ramified1.

I have noticed a pattern in many of the false proofs. Today, I’m going to tell you how to spot these proofs. The warning sign is the phrase “but $k[x_1, \ldots, x_n]$ has no nontrivial units.”

Disclaimer: I am not an expert on the Jacobian conjecture. I will not referee manuscripts on the subject unless they are relevant to my published work in some specific way. Please do not send me preprints on the subject (unless you are a personal friend).

It will be convenient to separate the domain and range of $F$. Let’s write $F: S \to T$. Assume for the sake of contradiction that $F$ is not an isomorphism. The proofs I am thinking of proceed as follows:

Step 1: We can factor $F$ as $S \hookrightarrow \overline{S} \xrightarrow{\overline{F}} T$ where $S \hookrightarrow \overline{S}$ is an open inclusion, $\overline{F}$ is finite, and $\overline{S}$ is affine and normal.

Step 2: $\overline{F}$ is ramified. (This is the part that uses characteristic zero and that uses the assumption that $F$ is not an isomorphism.)

Step 3: $\overline{F}$ is not ramified, a contradiction. (This is the step that uses the assumption that the Jacobian is nonvanishing.)

Step 4: Tenure!

The factorization in step 1 can be built; the easiest way is to construct the coordinate ring $\mathcal{O}(\overline{S})$ as the integral closure of $\mathcal{O}(T)$ in $\mathcal{O}(S)$. Step 2 can be made correct (although there are often errors here). The difficulty is in step 3.

Here is the argument that I think most people have in mind for step 3:

Completely Bogus Proof: Consider the Jacobian $J(\overline{F})$ of $\overline{F}$. The restriction of $J(\overline{F})$ to $S$ is $J(F)$, which is assumed to be a nonzero constant. Since $S$ is dense in $\overline{S}$, the function $J(\overline{F})$ is also a nonzero constant on $\overline{S}$, so $\overline{F}$ is unramified.

Now, this is completely bogus, because there is no such function as $J(\overline{F})$. The Jacobian is defined for maps from affine space to affine space, not for maps from an arbitrary variety to affine space. And no one writes this proof.

If I were trying to fix the bogus proof, I would think about how to define something like a Jacobian for a general map. Here’s how that would go:

False Proof: Let $\omega$ be the canonical module2 of $\overline{S}$. If $\overline{S}$ is smooth, this is $\bigwedge^n \Omega_{\overline{S}/k}$. Let $\eta$ be a nowhere vanishing section of $\omega$. Define $J(\overline{F})$ by
$\displaystyle{\overline{F}^* \left(d y_1 \wedge d y_2 \wedge \cdots \wedge d y_n \right) = J(\overline{F}) \eta}$.
Now, since $F$ is unramified, the function $J(\overline{F})$ must not vanish on $S$. But there are no nontrivial units in the coordinate ring of $S$, so $J(\overline{F})$ must be a nonzero constant. Thus, $\overline{F}^* \left(d y_1 \wedge d y_2 \wedge \cdots \wedge d y_n \right)$ is nowhere vanishing on $\overline{S}$, and the map $\overline{F}$ is nowhere ramified.

The part of this proof which is true is that $\overline{F}^* \left(d y_1 \wedge d y_2 \wedge \cdots \wedge d y_n \right)$ really is a well defined section of $\omega$ and this section vanihses exactly where $\overline{F}$ is ramified. Can you spot the lie in this proof? Here it is: “Let $\eta$ be a nowhere vanishing section of $\omega$.” There is no reason to believe that $\omega$ will be a trivial line bundle. The reader may take it as a challenge to construct an embedding $S \hookrightarrow \overline{S}$ where $S \cong k^n$, $\overline{S}$ is affine and $\omega$ is nontrivial. The construction in my previous post doesn’t quite work for this purpose, but it may be a good starting point.

Authors don’t explicitly write the false proof. What they do is to define a function $j$, usually by some explicit formula. This function is often equal in many special cases to the function $J(\overline{F})$ defined above, so in those cases $j$ vanishes precisely where $\overline{F}$ is ramified; the authors tend to believe that their $j$ will always capture the ramification behavior of $\overline{F}$. But in reality, there may be no function $j$ whose vanishing encodes the ramification of $\overline{F}$. For number theorists, this is related to the possibility that the different ideal need not be principal.

Authors who take this approach will often phrase step 2 as “$j$ vanishes somewhere on $\overline{S}$” and step 3 as “$j$ vanishes nowhere on $\overline{S}$.” This is why the proof of step 2 is often broken, because the definition of $j$ may make this restatement is false.

But the real problem is in step 3. The authors want to say “$F$ is unramified, so $j$ does not vanish on $S$, so since the coordinate ring of $S$ has no nontrivial units, $j$ does not vanish anywhere.” But, because $j$ doesn’t truly control the vanishing of $\overline{F}^* \left(d y_1 \wedge d y_2 \wedge \cdots \wedge d y_n \right)$, it is quite possible that $j$ does vanish on $S$. This is why the observation about nontrivial units is a warning flag to me.

I would like to build a counter-example to defeat this line of attack once and for all. Namely, I would like to build spaces and maps $S \hookrightarrow \overline{S} \to T$ such that

(1) $T \cong k^n$;
(2) $S \to T$ is unramified;
(3) the coordinate ring of $S$ has no nontrivial units;
BUT (4) $\overline{S} \to T$ is ramified.

I suspect that I have such a counter-example, as described above. It’s pretty darn messy though, and I haven’t locked down the proof of (3). Better ideas are welcome!

1 When $k$ is not algebraically closed, this statement has to be interpreted with some care.
2 There are some technical issues defining $\omega$ when $\overline{S}$ is not Cohen-Macaulay, but I don’t think that they are crucial.

## 9 thoughts on “How not to prove the Jacobian conjecture”

1. Sorry, can you please explain why Step 2 is “$\overline{F}$ is ramified” and Step 3 is “$\overline{F}$ is not ramified” and why those two statements don’t contradict one another?

And if you don’t mind me asking, is this post inspired by the recent preprint on math.AG claiming to prove the Jacobian conjecture?

2. “Sorry, can you please explain why Step 2 is “\overline{F} is ramified” and Step 3 is “\overline{F} is not ramified” and why those two statements don’t contradict one another?”

These do contradict each other. Hence, tenure!

It is partly inspired by that preprint. (Which I do believe is broken. The author and I have exchanged several e-mails, and I think he now agrees with me.) But it is also inspired by reading several preprints of S. Oda and others. I thought about giving particular examples of the errors I mention, but I decided it was best not to get into it.

3. Edited post to try to avoid the confusion.

4. I enjoyed this post. This conjecture is so wonderful, but so hard. Cannot believe still open even for n=2.

5. My understanding is that, even worse, the n=2 case implies the general case. Don’t have a reference for this, though.

6. rjlipton says:

Do not believe that two dimensional case implies full case. What is true is that cubic in any dimension does imply full case. There are other reductions also known.

7. How about Dang Vu Giang’s very recent attempt in

It may be true and it is short proof (there is one inessential typo where it says 4 for 2, and there are about 2-3 places which need a little nonobvious but elementary elaboration, but a colleague of mine checked those; and there is one essential earlier result used, cited in the literature). What do you think ?

8. rjlipton says:

The proof has many holes. It is not correct.

9. David Speyer says:

The first error I spotted was that H and $H_1 := \partial H/\partial z_1$ need not commute, so $\partial H^n/\partial z_1$ need not be $n H^{n-1} H_1$ (bottom of page 1, top of page 2). This causes problems with all the future formulas involving derivatives of matrices. I sent Giang a note about this and he seemed to agree.