# The characteristic of the field $GF(p^n)$

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.

Thanks

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 $$r \mid p^n \Rightarrow r = p$$
• 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? – michael May 18 '14 at 12:02
• 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$(F) = 0$ – WLOG May 18 '14 at 12:04