TITLE: In the ring $\mathbb{Z}_p$, $p$ is prime, $(a+b)^p=a^p+b^p$ proof?
QUESTION [5 upvotes]: In the ring, $\mathbb{Z}_p$, $p$ a prime, prove that $(a+b)^p=a^p+b^p$.
The hint that is given to us says that the binomial expansion works in commutative rings, but I think I used something much more simple?
I said that $\mathbb{Z}_p$ of prime order is cyclic, so $\langle x\rangle = \mathbb{Z}_p$ is generated by $x$. So therefore $x^p \bmod p = x$. Thus in $\mathbb{Z}_p,(a+b)^p = a+b = a^p +b^p$
The only thing I am unsure of is if I have to prove that $x^p \bmod p = x$, and then if I can apply it to $(a+b)^p$.
If this way is super is super goofy and, even if it works somehow, requires a lot of proof, then how might I get started on the binomial theorem? Thanks team.

REPLY [2 votes]: You should use binomial theorem because the result will be more general and argument more elementary. The proof will work for any commutative ring with characteristic $p$ instead of just $\mathbb{Z}_p$.
The key comes down to one fundamental property of prime number.

If $p$ is a prime number and $p | ab$, then $p|a$ or $p|b$.

For any integer $k$ with $1 \le k \le p-1$, the binomical coefficients $\binom{p}{k}$ are given by following expression.
$$\binom{p}{k} = \frac{p!}{k!(p-k)!}\quad\iff\quad\binom{p}{k} k! (p-k)! = p!$$
Since $p$ divides RHS, $p$ divides one of the factors on LHS. Since $k!$ and $(p-k)!$ are products of integers smaller than $p$. $p$ doesn't divide $k!$ nor $(p-k)!$. This forces
$p$ divides $\binom{p}{k}$. In other words, $\frac1p \binom{p}{k}$ is an integer.
Apply binomial theorem to $(a+b)^p$, we get
$$(a+b)^p = \sum_{k=0}^p \binom{p}{k} a^k b^{p-k}
= a^p + b^p + p \sum_{k=1}^{p-1} \left[\frac1p \binom{p}{k}\right]a^kb^{p-k}
$$
In any commutative ring with characteristic $p$, multiply something by $p$ kills it.
This leaves us with $(a+b)^p = a^p + b^p$.
About the direction you are heading on, you can achieve what you want using following fact:

$\mathbb{Z}_p^{*} = \{ z \in \mathbb{Z}_p : z \ne 0 \}$, the non-zero elements of $\mathbb{Z}_p$, forms a group under multiplication.

For any $z \in \mathbb{Z}_p^{*}$, consider the sequence of elements $1, z, z^2, z^3, \cdots$. Since $\mathbb{Z}_p^{*}$ is finite, this sequence will repeat itself
somewhere. From that, you can deduce for some positive integer $d$, $z^{d} = 1$.
Furthermore, if $d$ is the smallest integer for $z^{d} = 1$, you can show the $d$ elements $1, z, z^2, \cdots, z^{d-1}$ forms a subgroup of $\mathbb{Z}^{*}$ with respect to multiplication.
By Lagrange theorem, $d$ divides $| \left|\mathbb{Z}_p^*\right| = p-1$. This implies $$z^{p-1} = \left(z^d\right)^{\frac{p-1}{d}} = 1^{\frac{p-1}{d}} = 1$$
Multiply $z$ by both side, you will find for any $z \in \mathbb{Z}_p$, one has $z^p = z$.
Please note that same argument works for any finite field. If $F$ is a finite field with
$n$ elements, then all its elements are roots of the polynomial $z^n - z = 0$.