# Thread: Binomial Theorem and Commutative Rings

1. ## Binomial Theorem and Commutative Rings

If R is a ring of characteristic p, prove (x+y)^p = x^p + y^p for all x,y in R. We need to use the rule that for each n>= 1, (x+y)^n = (sigma, k=0 to infinity)(n choose k) x^k y^(n-k).

I truly don't know where to start other that maybe setting the summation = x^p + y^p

2. Originally Posted by KrazEcat01
If R is a ring of characteristic p, prove (x+y)^p = x^p + y^p for all x,y in R. We need to use the rule that for each n>= 1, (x+y)^n = (sigma, k=0 to infinity)(n choose k) x^k y^(n-k).

I truly don't know where to start other that maybe setting the summation = x^p + y^p
Indeed we have to use the binomial theorem. By the way, notice that for the binomial theorem to hold in a ring R, it is very important that the ring is commutative.

So (notice you wrote $\displaystyle \infty$ instead of p at the top of the sum):

$\displaystyle (x+y)^p = \sum_{k=0}^p \binom{p}{k} x^k y^{p-k}.$

Notice that in this sum, we get for $\displaystyle k=0$ the term $\displaystyle y^p$, and for $\displaystyle k=p$, we get the term $\displaystyle x^p$. So if all other terms vanish, then we are done.

The ring has characteristic $\displaystyle p$, so any term, which $\displaystyle p$ divides, will vanish. The following result holds true:

Lemma: If $\displaystyle 0<k<p$, then $\displaystyle p$ divides the binomial coefficient $\displaystyle \binom{p}{k}$.

Notice how this lemma implies that every term of the sum above vanishes, except for $\displaystyle x^p$ and $\displaystyle y^p$.

You might know this lemma from your book, otherwise let me know.

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### binomial theorem in a ring

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