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I am trying to understand the proof of Fleck congruence given in Binomial coefficients modulo prime powers, by A. Granville.

Fleck congruence states that for a prime $p$, integers $n \ge p$ and $0 \le r \le p-1$ and $q=\lfloor \frac{n-1}{p-1}\rfloor$ $$ \sum_{k \equiv r \bmod p}(-1)^k{n \choose k} \equiv 0 \pmod {p^q}.$$

The proof therein is very short and lean but it makes use of algebraic number theory, where my knowledge is very limited.

$\zeta$ being a $p$-th primitive root of $1$, I understand that $$ \sum_{k \equiv r \bmod p}(-1)^k{n \choose k} = \frac{1}{p}\sum_{i=0}^{p-1}\zeta^{-ir}(1-\zeta^i)^n .$$ I also understand that $(1-\zeta^i)^n$ belongs to the ideal generated by $(1-\zeta)^n$, since $\frac{1- \zeta^i}{1-\zeta}=(1+\zeta+\cdot\cdot+\zeta^{i-1})$ which is a unit (inversible) in the ring $\mathbb{Z}[\zeta]$, for $1\le i \le p-1$. It is also clear that $(1-\zeta)^{p-1}=p$. Then I would expect the quotient of $n$ by $p-1$ be involved but I fail to see why the quotient of $n-1$ by $p-1$ is eventually obtained, instead.

Thanks for any clarification.

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We have ($i=0$ can be dropped) $$ p\sum_{k\equiv r \ \mathrm{mod} \ p}(-1)^k\binom nk=(1-\zeta)^n\sum_{i=1}^{p-1} \zeta^{-ir}(1+\zeta+\cdots + \zeta^{i-1})^n. \ \ \ \ (1) $$ Writing $\zeta=\zeta-1+1$ and $\zeta^{-ir}=\zeta^{pM-ir}$ for some $M$ with $pM>ir$, we see that $$ \sum_{i=1}^{p-1}\zeta^{-ir}(1+\zeta+\cdots+\zeta^{i-1})^n \equiv \sum_{i=1}^{p-1} i^n \ \mathrm{mod} \ (1-\zeta). $$ The sum on the right is $0$ mod $p$ if $p-1\nmid n$, and $-1$ mod $p$ if $p-1|n$.

Letting $\sum_{k\equiv r \ \mathrm{mod} \ p}(-1)^k\binom nk=X$ and taking norm to (1), we see that $$ p^{p-1} X^{p-1} = p^{n+d}K, \ \ \ (2) $$ where $K\in\mathbb{Z}$, and $\begin{cases} d\geq 1 &\mbox{if } p-1\nmid n \\ d=0 &\mbox{if } p-1|n\end{cases}$.

If $p-1|n$, we have $\frac n{p-1}-1=\left\lfloor \frac{n-1}{p-1} \right\rfloor$ Thus, $\nu_p(X)\geq \left\lfloor \frac{n-1}{p-1} \right\rfloor$.

If $p-1\nmid n$, write (2) in the form of $$ p^{p-1}X^{p-1}=p^{n+d'}K' $$ where $d'\geq 1$ and $(K',p)=1$.

Considering prime factorization of $X$, we obtain $$ p-1+(p-1)\nu_p(X)=n+d', $$ and hence $p-1|n+d'$.

Then $\nu_p(X)=\frac{n+d'}{p-1}-1\geq\left\lfloor \frac{n-1}{p-1} \right\rfloor $.

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  • $\begingroup$ I am not familiar with the norm in $\mathbb {Q}(\zeta)$. May I kindly ask you to explain more? $\endgroup$ – René Gy May 9 '20 at 12:31
  • $\begingroup$ Let $c_0, c_1,\ldots, c_{p-1}$ be rational numbers and $\beta=c_0+c_1\zeta^1+\cdots c_{p-1}\zeta^{p-1}$. Then the norm of $\beta$ is denoted by $N(\beta)$ and $$ N(\beta)=\prod_{i=1}^{p-1} (c_0+c_1\zeta^{1i}+\cdots c_{p-1}\zeta^{(p-1)i}). $$ $\endgroup$ – Sungjin Kim May 9 '20 at 12:37
  • $\begingroup$ If $\beta\in\mathbb{Z}[\zeta]$ so that all coefficients $c_0, \ldots, c_{p-1}$ are integers, then $N(\beta)$ is also an integer. $\endgroup$ – Sungjin Kim May 9 '20 at 12:46
  • $\begingroup$ As you would already observed, $N(1-\zeta)=p$. Also, $N(z)=z^{p-1}$ if $z\in\mathbb{Z}$. $\endgroup$ – Sungjin Kim May 9 '20 at 12:49
  • $\begingroup$ Sorry, I need more help: how exactly do you obtain $ N(\sum_{i=1}^{p-1}\zeta^{-ir}(1+\zeta+\cdots+\zeta^{i-1})^n) = K p^d$ ? Also, for $\frac{n+d}{p-1} -1\geq\left\lfloor \frac{n-1}{p-1} \right\rfloor$ to hold when $d \neq 0$, don't we need $d\ge p-2$? $\endgroup$ – René Gy May 10 '20 at 9:14

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