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When $p$ is a prime number and $x$ and $y$ are members of a commutative ring of characteristic $p$, then $$(x+y)^p=x^p+y^p.$$ This can be seen by examining the prime factors of the binomial coefficients: the $n$th binomial coefficient is

$$\binom{p}{n} = \frac{p!}{n!(p-n)!}.$$

The numerator is $p$ factorial, which is divisible by $p$. However, when $0 < n < p$, neither $n!$ nor $(p - n)!$ is divisible by $p$ since all the terms are less than $p$ and $p$ is prime. Since a binomial coefficient is always an integer, the $n$th binomial coefficient is divisible by $p$ and hence equal to 0 in the ring. We are left with the zeroth and $p$th coefficients, which both equal 1, yielding the desired equation.

1) My immediate thought is, so when we say binomial coefficient we mean addition forms also: for example, $3x^2y = x^2y+x^2y+x^2y$. But can this be generalized to all rings of characteristic $p$?

2) Also, how is coefficient being divided by $p$ related to coefficient becoming zero in characteristic $p$?

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  • $\begingroup$ Note this is additivity of the Frobenius endomorphism. $\endgroup$ – Julien Apr 9 '13 at 13:03
  • $\begingroup$ What's the point of the double dollar signs? $\endgroup$ – Joe Z. Apr 9 '13 at 13:04
  • $\begingroup$ @julien Of course, the ring might not have an identity, depending on the definition of ring that OPs instructor uses. (Personally, I hate any book that defines rings without identities, but there are such books.) $\endgroup$ – Thomas Andrews Apr 9 '13 at 13:15
  • $\begingroup$ @ThomasAndrews Right. Even more: $p$ need not be prime here, which seems to be assumed to talk about the Frobenius endomorphism. $\endgroup$ – Julien Apr 9 '13 at 13:18
  • $\begingroup$ @julien Yes, $p$ being prime is only needed for $(x+y)^p\equiv x^p+y^p$. We can certainly define $n\cdot r$ with $n$ an integer and $r$ an element of a ring of any characteristic. $\endgroup$ – Thomas Andrews Apr 9 '13 at 13:21
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This should answer both of your questions.

Characteristic $p > 0$ means (informally, see the post by Thomas Andrews for a formal definition) that the multiple $$ p \cdot 1 = \underbrace{1+ \dots + 1}_{\text{$p$ times}} $$ is zero. It follows $p \cdot a = 0$ for all $a$ in the ring. This is because $$p \cdot a = \underbrace{a + \dots + a}_{\text{$p$ times}} = (\underbrace{1+ \dots + 1}_{\text{$p$ times}}) a = (p \cdot 1) a = 0 a = 0.$$

Also, if $p$ divides the integer $n$, so that $n = m p$ for some $m$, then $$n \cdot a = (m p) \cdot a = m \cdot (p \cdot a) = m \cdot 0 = 0.$$

Now the binomial theorem in a commutative ring is $$ (x + y)^{n} = \sum_{i=0}^{n} \binom{n}{i} \cdot x^{i} y^{n-i}, $$ where note that the binomal coefficients act as multiples.

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  • $\begingroup$ This requires $p$ to be prime to hold. E.g $\binom{4}{2}=6$ is not divisible by $4$, the characteristic of $\mathbb{Z}/4\mathbb{Z}$. There is a little arithmetic argument to show that $\binom{p}{k}$ is divisible by $p$ for $1\leq k\leq p-1$ when $p$ is prime. $\endgroup$ – Julien Apr 9 '13 at 13:36
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    $\begingroup$ @julien, $p$ was taken as a prime in OP, and the little arithmetic argument was given in OP. $\endgroup$ – Andreas Caranti Apr 9 '13 at 13:38
  • $\begingroup$ Sorry. How could I not see the word "prime"...? $\endgroup$ – Julien Apr 9 '13 at 13:40
  • $\begingroup$ @julien, I wish I had an answer to that, because it happens to me all the time to read through a text (usually hastily, in my case), only to discover later that I have missed some bits. $\endgroup$ – Andreas Caranti Apr 9 '13 at 13:44
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1) You need to deal with the order of multiplication in non-commutative rings, so it's no longer $3x^2y$ in the binomial theorem but $xxy + xyx + yxx$. I'm not entirely sure if it breaks down in this case, but it seems like it would.

2) If the coefficient is divisible by $p$, then it is equal to $pq$ for some $q$. If the ring has characteristic $p$, then $pq$ = $0q$ = $0$.

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  • $\begingroup$ I'm pretty sure he didn't ask about non-commutative rings, so it seems a bit odd to jump to that with point (1). This theorem is not true for non-commutative rings, but that isn't the question. $\endgroup$ – Thomas Andrews Apr 9 '13 at 13:26
  • $\begingroup$ Oh, did he mean when $p$ is not prime rather than when the ring is not commutative? (When he said "all rings of characteristic $p$", I mean.) $\endgroup$ – Joe Z. Apr 9 '13 at 13:29
  • $\begingroup$ I took that to mean "can I general the idea of $n\cdot r$ for all rings with characteristic $p$?" where $n$ is an integer and $r$ is in your ring. But I can see how you could read it the other way. $\endgroup$ – Thomas Andrews Apr 9 '13 at 13:32

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