Zariski cancellation and affine fibrations over the complex numbers

The explanation

Take five numbers p, s, u, F and J, and set each one to 1. The paper first forms x = s² + u³ + p²F, which here equals 3. Then it forms H = x²F − (1 + 2sx)J − p²J² − pu. With our numbers, H = 9 − 7 − 1 − 1 = 0. So the point (1, 1, 1, 1, 1) solves the equation H = 0. Complex numbers have the form a + bi, where i is a square root of −1. All complex solutions form a shape the paper calls X. One equation in five unknowns leaves four free directions, so X is four-dimensional.

Now give X one extra free direction, called w. Each point of X becomes a whole line, one point for each value of w. This new shape is the cylinder over X. The paper proves that this cylinder is ordinary five-dimensional spaceSource 1. Here two shapes count as the same when polynomial formulas match their points both ways. Yet X itself is not ordinary four-dimensional spaceSource 1. Adding one line hides a real difference.

Ordinary space of this kind is called affine space. The Zariski cancellation problem asks a simple question. If the cylinder over X is affine space, must X be affine space? For the line, the answer is yes, from work of Abhyankar, Heinzer and Eakin (1972). For surfaces over the complex numbers it is also yes, through work of Fujita and of Miyanishi and Sugie. Some number systems wrap around: adding 1 to itself enough times gives 0. Experts say these have positive characteristic. Over every such system, Gupta showed in 2014 that cancellation fails in every dimension from three up. Over the complex numbers, dimensions three and up stayed open. A July 2026 preprint by Gaifullin and Petrov still lists the question as open.

The hard part is showing that X is not four-space. The match with five-space does not help here. It can twist the extra line together with the old directions. So its existence alone does not give coordinates on XSource 2. The paper calls this its main workSource 3.

Its test uses sliding motions. Four-space has many of them: shift one coordinate and keep the others fixed. Makar-Limanov used such motions to show that a well-known threefold, the Russell cubic, is not three-space. A slice of X is the part where p is held at one value. The paper sorts the functions on X by how fast they vanish on the slice p = 0Source 4. This degenerates X into a simpler, graded model: a smooth shape with a simpler equation. If X were four-space, a sliding motion with a special property would pass to that model. The paper lifts that motion to a larger related space built over the model. It then argues that no such motion can exist. A classical inequality about polynomials, the Mason–Stothers abc inequality, is a key tool in that final argumentSource 5. Take two polynomials with no common factor and their sum. The inequality bounds their degrees by how many distinct roots the three have.

Go back to the numbers, but now set p = 0 and keep s = u = 1. Then x = 2, and H becomes 4F − 5J. On the slice p = 0, the equation is a straight-line condition on F and J. The paper claims more. Over every point of the p-line, the slice of X is three-dimensional affine spaceSource 6. This includes the extra abstract points that algebraists add to the ordinary values of p. Dolgachev and Weisfeiler predicted that such a family must be a product with three-space near each point. The paper claims that this family breaks the predictionSource 7. If it were locally a product, X would be four-space after allSource 8. A second family, from five-space to the plane, breaks it too.

The same H answers one more question. A coordinate is a polynomial that can serve as one of the axes after an invertible polynomial change of variables. The Stable Coordinate Conjecture says this: a polynomial that becomes a coordinate after adding one variable was already one. The paper claims that H is not a coordinate in five variables but becomes one in sixSource 9.

These results have limits. The cancellation example lives in dimension four. The paper does not treat complex dimension three or dimensions above four. Its families have three-dimensional slices. The paper says they do not address slices of dimension one or twoSource 10.

How sure is this? A computer proof checker called Lean has checked a formal version of the main theorem. It says that adding one variable to the polynomial functions on X gives polynomials in five variables. The functions on X themselves are not polynomials in four variables. The stable-coordinate consequence is outside that check. The Lean scope note does not list the fibration results either. No specialists have published a review yet.

What it doesn't show. This does not settle the cancellation question for complex dimension three.