Lech’s multiplicity conjecture

The explanation

Take the curve xy = 0: two lines crossing at the origin. Study it very close to the crossing. Write down the monomials in x and y, but drop any of degree N or more. Since xy = 0, only 1, x, …, x^(N−1) and y, …, y^(N−1) survive. That makes 2N − 1. For N = 100 it is 199. Divide by N and let N grow, and you get 2. A single smooth line keeps only N monomials, so its limit is 1.

That limit is the Hilbert–Samuel multiplicity, written e. It measures how singular a point is. A smooth point has e = 1, and the crossing has e = 2. In general, the count grows like a multiple of N^d, where d is the dimension. One multiplies the count by d! and divides by N^dSource 1. In dimension zero, e is simply the length, the number of independent quantitiesSource 2. The setting is a Noetherian local ring: the algebraic model of the neighbourhood of a single point, with a finiteness condition.

Now add a third variable z to the crossing lines. The new ring describes two planes crossing along a line. Monomials of degree below N now number N². For N = 100 that is 10,000. Multiply by 2! and divide by N², and e is again 2. Adding a variable is the simplest flat map. A flat map from one local ring R to another, S, makes S vary evenly over R, with no sudden jumps. Christer Lech asked whether multiplicity can ever go down along such a mapSource 3. His conjecture says e(R) ≤ e(S) always.

Lech proved it when R has dimension at most two, and in some special casesSource 4. Linquan Ma proved dimension three in equal characteristic, where the ring and its point share the same arithmeticSource 5. Ma also proved a weaker inequality, with a factor max{1, d!/2^d}Source 5. For d = 10 that factor is about 3,544. Ma also proved it in every dimension when R is a standard graded ring over a perfect fieldSource 6. One popular method needed special sequences of modules over every ring. Yhee showed some two-dimensional rings have noneSource 7.

This paper claims to prove Lech's conjecture with no restriction on dimension, residue field or characteristicSource 8.

The obstacle is a mismatch. Flatness controls powers of R's maximal ideal pushed into S. But e(S) uses powers of S's own maximal ideal. That difference blocks a direct comparison of the leading termsSource 9. Two problems remained: keeping the factor exactly 1 in every dimension, and estimating lengths in mixed characteristicSource 10. The paper claims to solve both with one estimate for chains of matrices, called free complexes. Its constants depend only on the number of parameters and the sizes of the matricesSource 11. In characteristic p, the Frobenius map raises matrix entries to high powers without changing sizesSource 12. In mixed characteristic, the paper uses perfectoid normalized lengths insteadSource 13. In residue characteristic zero, it shows a counterexample would give one in characteristic p, which cannot existSource 14.

The same estimate gives a side result. For a complete local domain of characteristic p, a quantity called the Dutta multiplicity is at least eSource 15.

How sure is this? A computer proof checker called Lean has checked a supporting statement. It proves that side result, the Dutta comparison for complete local domains of characteristic p. The full conjecture in every characteristic is outside that check. No specialist review of the paper has been published.

What it doesn't show. This does not say when equality e(R) = e(S) holds, and Lean has not checked the full theorem.