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
![Rendered by QuickLaTeX.com \[ \frac{1}{n!(q-1)^n}\prod_{i=1}^n (q^i-1) \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-26e85e26dd00656bcff2e6606f530bf7_l3.png)
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
![Rendered by QuickLaTeX.com \[ \frac{|G|}{|N|}=q^{|\Phi^+|}\prod_i \frac{q^{d_i}-1}{d_i(q-1)}. \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-b216f63471917ce0214cd7d080c3f511_l3.png)
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
![Rendered by QuickLaTeX.com \[|G|=q^{\frac{n(n-1)}{2}}\prod_{d=1}^n q^d-1 \qquad \mbox{and}\qquad |N|=n!(q-1)^n.\]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-c5112c5a49a3ed8b1de5418ae3254e86_l3.png)
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
![Rendered by QuickLaTeX.com \[ \frac{|G|}{q^{\frac{n(n-1)}{2}}|N|}=\frac{1}{n!(q-1)^n}\prod_{d=1}^n (q^d-1) \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-95bc147a5806e8df19cba7e984846134_l3.png)
is an integer.
Now let
be a prime dividing
and let
be a positive integer. To conclude, it suffices to show that the fraction
![Rendered by QuickLaTeX.com \[ \frac{q^d-1}{d(q-1)} \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-3961f776dadaad7c871dc8841d79916c_l3.png)
has zero
-adic valuation. Write
, then by the binomial theorem,
![Rendered by QuickLaTeX.com \[ \frac{q^d-1}{d(q-1)}=\frac{1}{2dm}\sum_{i=1}^d {d \choose i}(2n)^i. \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-2aa84d78c540fc645843d69e37ee6b9a_l3.png)
Let
and
. Since
, we get
![Rendered by QuickLaTeX.com \[ v_p\left( {d \choose i}(2n)^i \right)\geq v_p\left(\frac{d(2n)^i}{i!}\right)>a-\frac{i}{p-1}+i(b+v_p(2)). \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-6d0f6a9558bf6167ad01b7c1769c45eb_l3.png)
We have the inequality
, so
![Rendered by QuickLaTeX.com \[ v_p\left( {d \choose i}(2n)^i \right)>a+i\left(b+\frac{v_p(2)-1}{2}\right). \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-90d52f8d4b379f54f8a477e446921274_l3.png)
For
, we therefore get
![Rendered by QuickLaTeX.com \[ v_p\left( {d \choose i}(2n)^i \right)>a+2(b-\frac{1}{2})=a+2b+v_p(2)-1\geq a+b+v_p(2)=v_p(2dm) \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-4f0a630137991b30d360cfd080f89706_l3.png)
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
![Rendered by QuickLaTeX.com \[ v_p \left( \frac{q^d-1}{d(q-1)} \right)=v_p\left(\frac{2md}{2md}\right)=0, \]](https://petermc.net/blog/wp-content/ql-cache/quicklatex.com-c7db2b4abc73fbebbca9f70e501e9b00_l3.png)
completing the proof.