In the ring $\mathbb{Z}_p$, $p$ is prime, $(a+b)^p=a^p+b^p$ proof?

Question

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.

Answer 1

Your argument seems perfectly fine to me!

The reason the question gives this hint is that the same result holds in any commutative ring of characteristic $p$, and you obviously can't use the same proof for rings of order $\neq p$.

So to prove the result in general, let $R$ be a commutative ring of characteristic $p$, and let $a, b \in R$. Then by the Binomial Theorem, we have $$ (a+b)^p = a^p + \binom{p}{1}a^{p-1}b + \binom{p}{2}a^{p-2}b^2 + \ldots + \binom{p}{p-1}ab^{p-1} + b^p. $$

Now, for $1 \leq k \leq p - 1$, we have $$ \binom{p}{k} = \frac{p!}{k!(p-k)!}, $$ and $p$ divides the numerator but not the denominator (since $p$ is prime), so $\binom{p}{k}$ is a multiple of $p$, and hence it is zero in $R$. Therefore, all the terms in our big sum are zero, except for the first and last ones, which gives us $$ (a+b)^p = a^p + b^p. $$

Answer 2

The Binomial Theorem says:

$(a + b)^{p} = a^{p} + \binom{p}{1}a^{p-1}b + \binom{p}{2}a^{p-2}b^{2} + ... + \binom{p}{p - 1}ab^{p-1} + b^{p}$

Where $\binom{n}{k}$, the binomial coefficients, $ = \frac{n!}{k!(n-k)!} = \frac{n(n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...(1)}$. (Intuitively, $\binom{n}{k}$ is defined as the number of ways to pick $k$ objects out a group of $n$, which is why it is read "$n$ choose $k$". This definition may also give you some insight into why the Binomial Theorem works.) Note that if $n$ is prime and $1\leq k \leq n-1$, then $n\vert \binom{n}{k}$ because there are no factors of $n$ in the denominator. Then, because all terms except $a^{p}$ and $b^{p}$ contain binomial coefficients divisible by $p$, we have $\boxed{(a + b)^{p} = a^{p} + b^{p}}$

You used Fermat's Little Theorem instead, which also works.

Answer 3

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 $z^n - z = 0$.