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 $p$?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $pq$ = $0q$ = $0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.