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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.

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    $\begingroup$ Use induction on $n$. The case $n=1$ follows from Newton Binomial formula, and the induction is easy : $(x+y)^{p^{n+1}} = ((x+y)^{p^n})^p$ ... $\endgroup$
    – user171326
    Jun 9, 2016 at 11:20
  • $\begingroup$ $p^n = 0$ additively or when you multiply it by an element of $F$ (which is really repeated addition), but here $p^n$ appears in an exponent. $\endgroup$
    – lhf
    Jun 9, 2016 at 11:37
  • $\begingroup$ It may be amusing to consider the case of $n = p-1$ and compute $(a+b)^p = (a+b)^{p-1} (a+b)$. $\endgroup$
    – user14972
    Jun 9, 2016 at 12:39
  • $\begingroup$ @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? $\endgroup$
    – rubik
    Jun 9, 2016 at 13:27
  • $\begingroup$ @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. $\endgroup$
    – rubik
    Jun 9, 2016 at 13:45

3 Answers 3

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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.

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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, $(\alpha + \beta)^{m} = \alpha + \beta = \alpha^{m} + \beta^{m}$.

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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 $0$.

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