9
$\begingroup$

Let $d\ge2$, and let $q$ be a power of a prime. As usual, define $N(d,q)=\sum_{k=0}^d{d\choose k}_q$.

I wonder if there are $d$ and $q$ as above such that $1+N(d,q)=q^{d+1}$.

(If the answer is positive then this helps me give an example of a finite commutative local ring which is not principal and having exactly as many ideals as elements; see The number of ideals in a ring.)

$\endgroup$
1

1 Answer 1

8
+50
$\begingroup$

There are no solutions to this equation -- the point is that for $d\ge 5$ the term $1+N(d,q)$ is (much) larger than $q^{d+1}$, while for $d=2$, $3$ or $4$ we get a polynomial equation whose solutions are easily determined. To see this, first note that $$ \binom{d}{k}_q =\frac{(1-q^d)(1-q^{d-1})\cdots (1-q^{d-k+1})}{(1-q)(1-q^2)\cdots(1-q^k)} = \frac{(q^d-1)}{(q^k-1)}\frac{(q^{d-1}-1)}{(q^{k-1}-1)}\cdots \frac{(q^{d-k+1}-1)}{(q-1)}, $$ and since $(x-1)/(y-1) \ge x/y$ if $x \ge y >1$, it follows that $$ \binom{d}{k}_q \ge \frac{q^d}{q^k} \frac{q^{d-1}}{q^{k-1}}\cdots \frac{q^{d-k+1}}{q} = q^{k(d-k)}. $$ Thus if $d\ge 5$, then $$ N(d,q) \ge \binom{d}{2}_q \ge q^{2(d-2)} \ge q^{d+1}, $$ and there are no solutions to the given equation. (In fact, much more is known about $\binom{d}{k}_q$ -- namely this is a polynomial in $q$ of degree $k(d-k)$ with leading coefficient $1$, and with all other coefficients being non-negative integers. See for example Amritanshu Prasad's answer to the linked MO question Sum of Gaussian binomial coefficients. This gives an alternative proof that $\binom{d}{k}_q \ge q^{k(d-k)}$ for all non-negative $q$.)

This leaves us with the cases $d=2$, $3$ and $4$. When $d=2$, we see that $N(2,q)= 3+q$ and for all prime powers $q$ we see that $1+N(2,q) < q^3$. If $d=3$ then $N(3,q)=2+2(1+q+q^2)$ so that the equation is $5+2(q+q^2)=q^4$ which can readily be checked to have no prime power solutions. Finally if $d=4$ then $N(4,q) = 2+2(1+q+q^2+q^3)+(1+q^2)(1+q+q^2) = q^4+3q^3+4q^2+3q+5$, and the equation is $$ q^4+3q^3+4q^2+3q+6=q^5, $$ which once again doesn't have any prime solutions (just need to check $q=2$ and $3$).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.