TITLE: Quick way to show a finite field always has prime power order?
QUESTION [7 upvotes]: Possible Duplicate:
Order of finite fields is $p^n$ 

Is there a quick way to prove that finite fields always have cardinality that is the power of a prime p?

REPLY [27 votes]: Here is a quick way to derive the characteristic of a finite field with pretty low level concepts. Consider the ring homomorphism
$$
f\colon\mathbb{Z}\to F:n\mapsto n\cdot 1_F
$$
where $1_F$ is the multiplicative identity in $F$. Now
$$
\mathbb{Z}/\ker(f)\simeq\operatorname{im}(f)
$$
but $\operatorname{im}(f)$ is a subring of a field, hence an integral domain. So necessarily $\ker(f)$ is a prime ideal, thus of form $p\mathbb{Z}$ for some prime $p$. Then $\operatorname{im}(f)$ is a subfield of order $p$ of $F$, so viewing $F$ as a vector space over this field immediately implies $|F|=p^n$ for some $n$.