A counterexample to Kurosh’s division-ring problem

The explanation

Start with the quaternions, numbers like q = 1 + 2i + 3j. Here i² = j² = −1, and ij = −ji, so order matters. Square q. The cross terms 6ij and 6ji cancel, and q² = −12 + 4i + 6j. Then q² − 2q = −14. So q solves the equation q² − 2q + 14 = 0, whose coefficients are ordinary real numbers.

Every quaternion solves such an equation. The 2 is twice the real part, and the 14 is 1 + 4 + 9. The real numbers are the center of the quaternions: the numbers that commute with everything. A ring is algebraic over its center when every element solves a nonzero polynomial with central coefficients. The quaternions are also small: every quaternion is built from 1, i, j and ij, so they have dimension 4 over the reals.

A division ring is a number system where every nonzero element can be divided by, though multiplication need not commute. The Kurosh problem asks whether algebraic over the center forces local finitenessSource 1. Locally finite means anything generated by finitely many elements has finite dimension over the center. Alexander Kurosh asked this in 1941, inspired by Burnside's problem about groupsSource 2.

Positive results came under extra conditions. If one degree bound works for every element, the ring satisfies a polynomial identity, and theorems of Jacobson and Kaplansky then give local finitenessSource 3. Bell, Drensky and Sharifi sharpened this to dimension at most d² over the centerSource 4. Golod built infinite-dimensional algebras of a related kind in 1964. But they contain nonzero elements whose powers vanish, so they are not division ringsSource 5.

This paper gives a negative answerSource 6. It builds a countable division ring of characteristic zero, generated over its center by two elements x and ySource 7. Every element is algebraic over the center, yet the ring has infinite dimension over itSource 7. The trick is that each element may need its own polynomial, of growing degreeSource 8. That is exactly what the uniform-degree results forbid.

The construction is a long iteration. Countability matters here: it lets the paper list every polynomial expression in x and y and treat them one at a timeSource 9. At each stage a lifting theorem makes the next listed expression algebraic over a larger central fieldSource 10. The same step enlarges the division algebra and keeps every earlier equation trueSource 11. A limiting construction, called an ultraproduct, then keeps all these equations at onceSource 12.

The one obstacle is keeping the process alive. Each step must leave two independent directions free and keep x and y generating everythingSource 10. Otherwise the next step would have nothing to work with. Most of the paper proves that these properties survive every step, using matrix calculations and formal power series. A last argument shows the dimension is infinite over the true center, which the construction does not fix in advanceSource 13.

How sure is this? No Lean formalization is listed for this paper. The claim rests on the paper's own long argument. No specialist review of the paper has been published.

What it doesn't show. This does not give a ring where one polynomial degree works for every element; that case was already known to force finite size.