SMMC 2022 A4

The Simon Marais Mathematics Competition happened last weekend. It is a maths competition for undergraduate students across Europe, Asia, Africa and Oceania. This post is about problem A4, which I submitted. I’ll talk a bit about where the problem came from, a generalisation, a conjecture and also provide a solution. The entire paper is available on the Marais website, and solutions should be put up there at some time in the near future.

Problem (SMMC 2022 A4)
Let be a positive integer, and let be an odd integer such that every prime factor of is larger than . Prove that

is an integer that has no prime factor in common with .

Origins

Let and let be the subgroup of monomial matrices (a matrix is a monomial matrix if and only if it has exactly one nonzero entry in each row and column). I show below in my solution that this question is equivalent to the fact that the integer is coprime to . Now why would I ever care about that?

This coprimality fact implies that the cohomology of with mod coefficients is isomorphic to the cohomology of with mod coefficients. And I was interested in these cohomology groups because the second cohomology group classifies central extensions, which is what I used to think about back in my PhD days. The group feels somewhat more “combinatorial” than , so it is nice to be able to pass information from to for free.

Generalisations (known and conjectural)

Let be a split reductive group over , which I conflate with its -points below in an abuse of notation. Let be a maximal split torus and its normaliser in . Then

Here is the set of positive roots and the collection of integers are the exponents of the Weyl group. Then the same argument as in my proof below shows that this fraction is an integer, relatively prime to .

If we remove the assumption that is split, then I suspect the same conclusion is satisfied, but there is an additional argument needed as the formula for the quotient has additional factors. I have not worked out this argument and really don’t want to resort to case by case arguments, so there is your conjecture (I expect we now need to say is a maximal torus containing a maximal split torus).

Solution

First we show that the fraction in the question is an integer. Since divides as a polynomial for all , the statement only depends on the residue class of modulo . Since every prime factor of is greater than , is relatively prime to . So by Dirichlet’s theorem on primes in arithmetic progressions, we may assume without loss of generality that is prime.

Let and let be the subgroup of monomial matrices. Then

By Lagrange’s theorem is an integer. Since is relatively prime to , we can further divide by the largest power of in and deduce that

is an integer.

Now let be a prime dividing and let be a positive integer. To conclude, it suffices to show that the fraction

has zero -adic valuation. Write , then by the binomial theorem,

Let and . Since , we get

We have the inequality , so

For , we therefore get

as from our assumption that divides .
Thus in our sum, the term with has a strictly smaller -adic valuation than every other term, so determines the -adic valuation of the sum, and we get

completing the proof.