How to show that characteristic of the field $GF(p^n)$ is $p$?

I have come across this fact on Wikipedia webpage, but don't know how to prove it.



Let $r \in \mathbb{N}$ be the characteristic of $GF(p^n)$ .

We know that the characteristic of a domain is always $0$ or a prime number. In the case of a finite field, it can't be $0$ because otherwise the field would contain a copy of $\mathbb{Z}$. So the characteristic of $GF(p^n)$ is a prime.

Then we have $r \cdot 1 = 0 $, but a field is in particular an abelian group with respect to $+$ and so $$r \mid p^n \Rightarrow r = p$$

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    $\begingroup$ The characteristic of a domain is always a prime or zero. $\endgroup$ – Pedro Tamaroff May 18 '14 at 12:01
  • $\begingroup$ @PedroTamaroff: I edited $\endgroup$ – WLOG May 18 '14 at 12:02
  • $\begingroup$ Okay, but in your solution you're saying that if a field has characteristics 0 then it contains a copy of $\mathbb{Z}$. What do you mean by that? $\endgroup$ – michael May 18 '14 at 12:02
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    $\begingroup$ I mean that if the characteristic of $F$ is $0$, then we can define an homomoprhism $\phi : \mathbb{Z} \to F$ , with $\phi(1) = 1$ and this would be injective due to char$(F) = 0$ $\endgroup$ – WLOG May 18 '14 at 12:04

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