# Thread: Show (a + b) ^ p = a^p + b^p

1. ## Show (a + b) ^ p = a^p + b^p

Hi I'm trying to show that for a commutative ring with characteristic p that
(a + b)^p = a^p + b^p

I have verified this with a few examples but can't seem to prove that the binomial coefficients C(p,r) = 0 (modp) where r /= 0 or p (i.e. the first or last term)

Can anyone help me out? Cheers =)

2. Originally Posted by liedora
Hi I'm trying to show that for a commutative ring with characteristic p that
(a + b)^p = a^p + b^p

I have verified this with a few examples but can't seem to prove that the binomial coefficients C(p,r) = 0 (modp) where r /= 0 or p (i.e. the first or last term)

Can anyone help me out? Cheers =)

Hints:

1) Since there's commutativity we can use Newton's binomial theorem: $\displaystyle \displaystyle{(a+b)^p=\sum\imits_{k=0}^p\binom{p}{ k}a^{p-k}b^k$

2) Now prove that $\displaystyle \displaystyle{\forall 1\leq k\leq p-1\,,\,\,p\mid\binom{p}{k}$ .

Tonio

3. Hi Tonio, the part 2) was where I was having the difficulty, I can't seem to prove that. Any advice of how I might get there?

Thanks

4. Originally Posted by liedora
Hi Tonio, the part 2) was where I was having the difficulty, I can't seem to prove that. Any advice of how I might get there?

Thanks

$\displaystyle \displaystyle{\binom{p}{k}=\frac{p!}{k!(p-k)!}=\frac{(k+1)(k+2)\cdot\ldots\cdot p}{(p-k+1)(p-k+2)\cdot\ldots\cdot (p-k)}}$ , and now

check that no factor in the denominator divides p for $\displaystyle 1\leq k\leq p-1$ , and since

the binomial coefficient is always an integer then we're done.

Tonio

5. Cheers Tonio, got it now =)