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In my abstract algebra class we discussed the idea that for a finite field $F$, that the characteristic of $F$ is a prime number. The proof would go more or less like : Suppose that $char(F) = nm$ for $n,m \in \Bbb Z$ and $n,m \geq 2$. Then $nm(1_F)=n(1_F)m(1_F)=0$ implying that either $n(1_f)=0$ or $m(1_f)=0$ since $F$ is an integral domain, but this contradicts the minimality of $char(F)$. This lead me to two questions:

$1$)What effect does $char(F)$ have on the size of $F$

$2$)Are there finite fields of cardinality $k$ for any $k\in \Bbb Z, k \geq 2$

IntegrateThis
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Certainly if the characteristic of $F$ is $0$ then $F$ must be infinite. The converse is false.

If the characteristic of $F$ is the prime $p$, the field $F$ could be finite or infinite. In the finite case, it must have order $p^n$ for some $n$. This is most easily seen by viewing $F$ as a vector space over the prime field, which must be isomorphic to $\mathbb{Z}_p$. If this vector space has dimension $n$, a simple count shows that $F$ has $p^n$ elements. The infinite case is definitely a possibility, as the field $\mathbb{Z}_p(x)$ of rational functions shows.

Randall
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  • When you say $F$ as a vector space over the prime field, do you mean that the vectors are elements in $F$, and the scalars are linear combinations of $1_F$ ? – IntegrateThis Mar 13 '19 at 01:58
  • Yep. Actually, for any subfield $K$ of $F$, $F$ will be a vector space over $K$: this is just the definition of subring. This is frequently handy. I'm just applying it in the case where $K$ is as small as possible. – Randall Mar 13 '19 at 01:59
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Think about $F$ as a commutative group under addition (ignoring for the moment the multiplicative structure). You know that the order of every non-zero element of $F$ is $p$. What does that tell you about the order of $F$?

Robert Shore
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For any finite field $F$, take the subring given by integral linear combinations of $1$. This must be the field $F_p$, where $p$ is the characteristic of the field( you can take this to be a definition even). Hence $F_p\subseteq F$, and as a result $F$ must be an $F_p$-vector space. Consequentially $$|F|=p^{\operatorname{dim}_{F_p}F}$$

The converse also holds. There exists a field of cardinality $k\in \mathbb{N}$ if and only if $k=p^n$ for some $n\in \mathbb{N}$.

Noah Riggenbach
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