# how do I prove that in a finite field, the ($p$th) power of a sum is the sum of the powers?

I did some search over the website and in google, but couldn't find the answer, even though I'm sure it exists, so maybe I did not look for the right key words.

Anyway, the question is how do I prove that in a finite field - GF(p), every exponent of a sum, is the sum of exponents?

Or, in a formula: $(a+b)^p=a^p+b^p$

I know it's proven with the binomial theorem, but I'm not sure why for example: $\binom{p}{0}$ is 1 (I do understand why it's 1 in a non-finite field). The way I see it is:

$\binom{p}{0}=\frac{p!}{(p-0)!0!}=\frac{p\cdot(p-1)!}{p!\cdot1}=\frac{0\cdot (p-1)!}{0\cdot(p-1)!}=\frac{0}{0}$

Now it seems that the result is not defined (or is it? in a finite field...).. while in a non finite field it goes like this:

$\binom{p}{0}=\frac{p!}{(p-0)!0!}=\frac{p!}{p!\cdot1}=\frac{1}{1}=1$

I would appreciate your help on the rest of the proof as well.. I know there is some main idea here..I'm new to finite fields, so maybe I'm missing something here...

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Your confusion arises because the binomial coefficient does not represent an element of the field in question.

If $R$ is a ring, then we let $\mathbb{Z}$ act on $R$ via addition: given an element $r\in R$, we define: \begin{align*} 0\cdot r &= 0_R\\ (n+1)\cdot r &= n\cdot r + r &\text{if }n\gt 0\\ (-n)\cdot r &= -(n\cdot r)&\text{if }n\gt 0. \end{align*} Note that $n\cdot r$ is not a product in $R$. Rather, it represents an "abbreviated sum".

In the binomial theorem in commutative rings, $$(a+b)^n = a^n + \sum_{k=1}^{n-1}\binom{n}{k}\cdot a^{n-k}b^k + b^n,$$ the binomial coefficients are not elements of the ring; they are integers, and the evaluation corresponds to the action of $\mathbb{Z}$ in $R$ described above.

If $R$ is a ring, and $1_R$ is a multiplicative identity, then it is important not to confuse the integer $n$ with the ring element $$n\cdot 1_R = \underbrace{1_R+\cdots+1_R}_{n\text{ summands}}.$$

So when you are computing $(a+b)^p$ in a field of characteristic $p$ you are not trying to compute $\binom{p}{k}$ in the field; you are computing this integer as an integer, and then letting that integer act on the elements of the field. So we compute $\binom{p}{0}$ as an integer (and get $1$), and then compute $1\cdot a^p = a^p$. You don't try to compute $$\binom{p}{k}=\frac{p!}{k!(p-k)!}$$ in $F$, you compute it in $\mathbb{Z}$ and then it acts on the elements of $F$.

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I loved your answer. I just want to point out that I didn't get to Rings in my class yet, so pretty much of your explanation doesn't say anything to me, but I did get the main idea here.. I'm so used to work with the binomial coefficients formula that I clearly forgot that it actually means how many times an element appears in the term, so it actually means integers like you said, and not an element in the field itself... thanks! – rboy Apr 19 '12 at 19:42
@RB14: In a ring, you have addition as usual (with a $0$ and with additive inverses), and you have a (not necessarily commutative) multiplication that is associative and distributes over the sum; you need not have a multiplicative identity, and nonzero elements may not have multiplicative inverses, and you may be unable to cancel nonzero factors. But the idea is the same for fields: given a field $F$, an element $\alpha\in F$, and an integer $n$, $n\alpha$ does not represent a product of elements of $F$, it represents a "sum in abbreviated notation". It is equal to $(n1_F)\alpha$ as well. – Arturo Magidin Apr 19 '12 at 19:57
Yeah, I get it now. thanks a lot! – rboy Apr 19 '12 at 20:12
But to master ring theory, it is essential to "confuse the integer $n$ with the ring element $n\cdot 1_R$" (but only after one can prove that one isn't confused by such notational abuse!) – Bill Dubuque Apr 19 '12 at 20:55

Remember that the binomial coefficients are integers. Their role in the binomial formula (that holds in all commutative rings, hence also in finite fields) is to tell how many times we add each term to itself, when building the sum. Because of this addition, all you need to do is to reduce the binomial coefficient modulo the prime $p$.

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Thanks, you did clarify some things too :) – rboy Apr 19 '12 at 19:46