Cancel a line?

0:00Cancelling a line

Suppose a variety X, times a line, is affine space. Must X itself be affine space? This is Zariski’s cancellation problem. In topology the answer can be no: the Whitehead manifold is not ordinary three-space, yet its product with a line is four-space. In algebra, the question is whether a ring that becomes a polynomial ring after adjoining one variable was already a polynomial ring. For curves, Abhyankar, Heinzer and Eakin showed that cancellation holds. For the plane over the complex numbers, it holds too: if a surface times a line is three-space, the surface is the plane. That is work of Fujita, Miyanishi and Sugie, around 1980. In 2014, Neena Gupta showed that it fails in every dimension from three up, over every field of positive characteristic. Over the complex numbers, dimensions three and higher stayed open.

0:57One equation

The paper writes down one equation in five variables, p, s, u, F and J. Put x = s² + u³ + p²F. Then H = x²F − (1 + 2sx)J − p²J² − pu. The fourfold X is the zero set of H. Slice X along p. Away from p = 0, two more combinations, y and z, satisfy xy − z(z + 1) = p²(H + pu). There, x, y, z and u can replace s, u, F and J as coordinates. On X, u is then fixed by x, y and z, so each slice is affine three-space. At p = 0, H becomes linear in F and J, and a change of coordinates with determinant one shows that slice is affine three-space too. Now add one new variable, w. A derivation that moves u while fixing x, y and z sends H to p³, so its flow turns H into H + p³w. The paper then writes five explicit coordinates on the result: X times a line is affine five-space.

2:11The claim

Result 047 is a 29-page OpenAI preprint, dated September 23. It claims that X is not affine four-space, so cancellation fails over the complex numbers in dimension four. Two consequences follow. H becomes a coordinate after adding one variable without being one before, which disproves the stable coordinate conjecture in five variables. And the maps from X to the line, and from affine five-space to the plane given by p and H, have every fiber affine three-space, but are not products even over small Zariski-open pieces of the base. Over the complex numbers, that goes against a conjecture of Dolgachev and Weisfeiler.

2:55Why X is not affine space

Affine space has many additive symmetries, like sliding along a coordinate. Makar-Limanov used such actions to show that the Russell cubic is not affine three-space. Here the paper measures how fast each function blows up as p goes to zero, using a simpler model of X near that slice. If X were affine four-space, one coordinate would have positive degree. Its leading terms would give a homogeneous additive action on a simpler, degenerate ring, with an invariant of positive degree. That degenerate ring is built from the smooth quadric xy = z(z + 1). Pulling back along the torus bundle from 2-by-2 matrices of determinant one makes everything explicit. It yields a derivation on the matrix ring, of weight zero or negative, that kills one polynomial when applied twice. After one more grading step, the polynomial to kill becomes d² + a²u³. The last step is the Mason–Stothers theorem, the polynomial version of the abc conjecture, together with the fact that −u³ is not a square. These force the action to fix a, d and u, and the determinant relation then gives the wrong sign. So, the paper concludes, no such action exists, and X is not affine four-space.

4:15Where it stands

This is a preprint, not yet peer reviewed, and it has not been revised. OpenAI also published a Lean formalization of the main theorem, about 5,600 lines, not counting Mathlib. We haven’t run the check. The stable coordinate and fibration consequences are outside the Lean scope. The paper cites no other release paper. We found no public assessment. If it holds, one extra line can erase the difference between a strange fourfold and affine space, and Zariski’s question has a negative answer over the complex numbers.