I encountered this statement in a book am reading (but without proof):
A finite field with $n$ elements exists iff $n=p^k$, where $p$ is a prime and $k$ is a positive integer. In this case, there is only one such field, up to isomorphism.
The statement came after a discussion on the characteristic of a field, i.e. a field $F$ is of positive characteristic if there exists an integer $n \geq 1$ s.t. $na=0$ for all $a \in F$, where $na$ denotes $a$ added to itself $n$ times, i.e. $na:=a+a+a+...+a$, where $a+a$ occurs $n$ times. However, I could not relate the discussion on field characteristics with the statement above... Any help ? ... or better yet is there a proof for this one to help me understand?
The connection is that the prime $p$ in $n = p^k$ is the characteristic of the field.
The proof that the order must be a power of characteristic is pretty easy, if $F$ is a finite field of characteristic $p$ then there is a homomorphism $\mathbb F_p \to F$ given by sending $x \in \mathbb F_p$ to $x\cdot 1$. Ring homomorphisms out of fields are injective so we can take $\mathbb F_p$ as a subfield of $F$. Then $F$ is a vector space over $\mathbb F_p$. As everything is finite $F$ must be finite dimensional over $\mathbb F_p$, say $\dim F = k$, then there's a vector space iso $F \simeq \mathbb F_p^k$, hence $|F| = p^k$.
The proof that there's only one field of that order up to isomorphism is a bit more involved, but you should be able to find it in any standard modern algebra text like Rotman or Dummit & Foote.