Most of us know the Jacobian conjecture. Here's a version below for fixed positive integers $d$ and $n$:


$J(d,n)$: If $f: C^n \rightarrow C^n$ is a polynomial map of degree $d$, and if the Jacobian determinant $\vert Jf \vert$ is nowhere vanishing (hence constant), then $f$ is injective (hence bijective).


We know that the sentence "For all $n$, $J(3,n)$" implies the sentence "For all $d,n$, $J(d,n)$." In other words, the Jacobian conjecture has been reduced to degree $3$.


We also know that, for any fixed $n$, $J(3,n)$ is provably true or provably false. This boils down to the completeness of the theory of algebraically closed fields of characteristic zero. 


But, do we know whether the sentence "For all $n$, $J(3,n)$" is provably true or false? In other words, might the Jacobian conjecture be... (oh no).. undecidable?!


In other words, I could theoretically program my computer to set out to prove the Jacobian conjecture $J(3,1)$ (easy) and $J(3,2)$ then $J(3,3)$, etc.., and my theoretical computer would keep on going for epochs and epochs. But would it ever halt? Might this be undecidable?



Could the Jacobian conjecture be undecidable?
There's no way to rule out *a priori* that the Jacobian conjecture is undecidable (in your favorite axiomatic system).


As I pointed out in my answer to [another MO question](https://mathoverflow.net/questions/74941/is-there-an-undecided-assertion-of-which-a-proof-that-its-not-undecidable-is-k), a proof that some statement is decidable would automatically mean that its status would be resolved up to a finite computation. You can be sure that if a conjecture as important as the Jacobian conjecture were reduced to a finite computation, then we would hear about it. This observation yields an informal "proof" that (almost) all the major open problems you care to name are not known to be decidable.


See also [this related MO question](https://mathoverflow.net/questions/79685/can-the-riemann-hypothesis-be-undecidable).


