# Why is $(a+b)^p = a^p+b^p$, where $a,b \in R$, a commutative ring with prime characteristic $p$?

Here is the answer from lecture notes. $$(a+b)^p = \sum {{p}\choose{k}}a^kb^{p-k}$$ and all the terms divide $$p$$ except $$a^p$$ and $$b^p$$ terms. So reducing (mod p) all terms are zero except the ones above.

So I don't understand how by reducing (mod p) makes those terms $$0$$? By definition of prime characteristic, $$p.1=0$$ where $$1$$ and $$0$$ are the multiplicative and additive identities. So I am not able to make the connection between the definition and the statement above. Thanks and appreciate a hint.

• By the definition of prime characteristic $p\cdot1=0$ (not $1^p=0$). – Lord Shark the Unknown Feb 10 at 22:13
• That's true, correcting that. – manifolded Feb 10 at 22:14
• $\binom{p}{k}$ is a multiple of $p$ if $k\in \{1,\dots, p-1\}.$ – mfl Feb 10 at 22:15
• Thanks, I understand that but don't understand how that would make the terms $0$ for $k\in \{1,..p-1\}$? – manifolded Feb 10 at 22:17
• $p\cdot 1=0\implies p\cdot x=0,\forall x\in R.$ – mfl Feb 10 at 22:18

$${{p}\choose{k}} = \frac{p\cdot(p-1)\cdot\ldots\cdot{(p-k+1)}}{k!}$$ As $$p$$ is prime and $$0 < k < p$$, $$p$$ doesn't divide $$k!$$, so this expression is a multiple of $$p$$ and thus it's zero in characteristic $$p$$.
Therefore, the sum $$\sum_{k=0}^p {{p}\choose{k}}a^kb^{p-k} = a^p + b^p + \sum_{k=1}^{p-1} {{p}\choose{k}}a^kb^{p-k} = a^p + b^p + \sum_{k=1}^{p-1}0\cdot a^kb^{p-k} = a^p+b^p$$
Because for all $$k$$ with $$0 the binomial coefficient $$\binom{p}{k}=\frac{p!}{k!(p-k)!}$$ is a multiple of $$p$$, because the numerator divisible by $$p$$, but the denominator isn't.