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This is a homework question I've been stuck on for the better chunk of a day, and I feel like I'm overlooking something obvious.

So far I've used the fact that in a finite field with characteristic p, we have that

$(1 + x^p)^n \equiv ((1 + x)^p)^n \equiv (1 + x)^{np} \space \text{(mod p)}$

and that when the left and right ends of this equivalence are expanded via the binomial theorem, I get

$(1 + x^p)^n = {n \choose 0}x^{0} + {n \choose 1}x^{p} + {n \choose 2}x^{2p} + {...} + {n \choose n}x^{np}$

$(1 + x)^{np} = {np \choose 0}x^0 + {np \choose 1}x^1 + {np \choose 2}x^2 + {...} + {np \choose np}x^{np}$

But here is where I'm stuck. By looking at the terms for both of these expressions, I can see that in order for the problem statement to be true, then we require

${np \choose k} = 0$

whenever k isn't a multiple of p. And I've read and understand the proof for

${p \choose k} \equiv 0$ (mod p)

which relies on the facts that k < p and p is prime. But here k need not be less than p, and np is definitely not prime. So I'm not sure how to proceed with the proof at this point.

*Edit: My confusion is on showing that the terms ${np \choose k}$ are all zero whenever k isn't a multiple of p, or at least their sum is. The argument here is that they all must be 0 because we already know that the two sums are congruent mod p, but that seems kind of circular to me? If they're congruent, does each coefficient with a k which isn't a multiple of p have to be exactly 0, or would it be better to say their sum is 0 (i.e. all zero implies sum zero, but sum zero doesn't necessarily imply all zero). Is it because by definition the binomial coefficient is a non-negative integer and thus a sum of non-negative integers summing to zero implies each one must itself be zero?

Any help would be greatly appreciated :)

drfpslegend
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  • Nitpick: The polynomial identity $(1+x)^p=1+x^p$ does not hold in a finite field, but rather in the polynomial ring $\Bbb{Z}_p[x]$. The formula (aka Freshman's dream) $(a+b)^p=a^p+b^p$ holds in any commutative ring of characteristic $p$. Commutativity is needed because without it we don't have the binomial formula. – Jyrki Lahtonen Apr 10 '22 at 09:55
  • Ah ok. But isn't $\mathbb{F}_p[x]$ a polynomial ring, with 1 and x being elements of the ring, meaning the congruence still holds? – drfpslegend Apr 10 '22 at 10:30
  • Absolutely. The proofs in the other threads rely on that identity. But it would not hold, for example, for all square matrices over $\Bbb{Z}_p$. It needs $ab=ba$, so if one of $a,b$ is equal to $1$, it still holds. – Jyrki Lahtonen Apr 10 '22 at 10:51

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