# Proving $(\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n}$ in a finite field [duplicate]

Prove that if $F$ is a field with $p^n$ elements and $\alpha,\beta \in F$, then $$(\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n}$$

From Newton identity, we have that $$(a + b)^n = \sum_{i = 0}^n \binom ni a^i b^{n - i}$$ However, I'm already stuck because in a finite field we cannot necessarily compute the binomial, since we may divide by zero. How do I get around this?

Furthermore, since $F$ has characteristic $p$, isn't it $p^n = 0$? How does it make sense in the first place? I'm a bit confused.

• Use induction on $n$. The case $n=1$ follows from Newton Binomial formula, and the induction is easy : $(x+y)^{p^{n+1}} = ((x+y)^{p^n})^p$ ...
– user171326
Jun 9, 2016 at 11:20
• $p^n = 0$ additively or when you multiply it by an element of $F$ (which is really repeated addition), but here $p^n$ appears in an exponent.
– lhf
Jun 9, 2016 at 11:37
• It may be amusing to consider the case of $n = p-1$ and compute $(a+b)^p = (a+b)^{p-1} (a+b)$.
– user14972
Jun 9, 2016 at 12:39
• @lhf I still don't understand. By definition of characteristic, $\underbrace{1 + \cdots + 1}_{p\text{ times}} = 0$. So $p = p\cdot 1 = 0$. But then $p^n = 0$ and we are exponentiating by zero. Does that make sense? Jun 9, 2016 at 13:27
• @lhf Oh no, exponentiation is defined by $a^n$ with $n \in \mathbb Z$, so I have to consider $p^n \in \mathbb Z$. Ok, nevermind. Jun 9, 2016 at 13:45

First consider the case $n=1$. We take the $p$th power as follows: $(a+b)^p = a^p + {p \choose 1} a^{p-1} b + \cdots + b^p$. Observe that except for the first and last term in this sum, the coefficient of each term is a multiple of $p$. Since the field has characteristic $p$, these are zero. Hence, $(a+b)^p=a^p+b^p$. Now take the $p$th power again, and repeat this process $n$ times.

Let $F$ be a finite field with $m$ elements.

By Lagrange's theorem of group theory applied to $F^\times$, we have $z^{m-1}=1$ for all $z \in F$, $z\ne0$.

Therefore, we have $z^m=z$ for all $z \in F$. In particular, $(\alpha + \beta)^{m} = \alpha + \beta = \alpha^{m} + \beta^{m}$.

## It's fine to leave it as an integer

Every ring admits scalar multiplication by integers; you do not need to consider $\binom{n}{i}$ as an element of the finite field.

## It's fine to map it into the finite field

We can always uniquely interpret an integer as an element of any ring by applying the unique map $\mathbb{Z} \to R$. If you want to interpret $\binom{n}{i}$ as an element of the ring, that is how you should do so.

## Either way, study the coefficient first as an integer

As seen in this answer, there is an explicit formula for how many times $p$ divides $\binom{p^a}{b}$ in the integers:

$$v_p\left( \binom{p^a}{b} \right) = a - v_p(b)$$

which we can use to show that all of the coefficients except the first and last are divisible by $p$ over the integers.

Then we either argue that we are multiplying by an integer multiple of $p$, or we are multiplying by a finite field element that is $0$.