abstract algebra - Inclusion of Fields whose order is a prime power

Blue was correct, I need to fix my understanding of this:

Finite fields have cardinality of a prime order because they have a prime subfield that has finite characteristic.

I do not completely understand why finite field of each characteristic is unique. Perhaps we look at a polynomial and express the finite field as a splitting field? And we use uniqueness of splitting fields?

Then, why is it that $\mathbb F_{p^m} \subseteq \mathbb F_{p^n}$ if and only if $m$ divides $n$?

Answer

The notation $\Bbb F_q$, where $q$ is a prime power, refers to the finite field of order $q$. If there were multiple finite fields of order $q$, then the notation $\Bbb F_q$ would be ambiguous! The existence and uniqueness of finite fields of given prime power order are one of the first things proven about them, and both parts can done by showing that $\Bbb F_q$ is the splitting field of $x^q-x$ over $\Bbb F_p$. This is because in general field theory we know that splitting fields exist and are unique up to isomorphism.

Another general fact from field theory we know is that finite groups of units within $F^\times$ for any field $F$ are always cyclic. Let $F$ be a field of order $q$. Then $F^\times$ is cyclic of order $q-1$, hence all of its elements satisfy $x^{q-1}=1$, hence satisfy $x^q=x$. But even $x=0$ satisfies the latter, so $x^q-x$ has every element of $F$ as a root, and since $|F|=q=\deg(x^q-x)$, these are precisely all of its roots, hence $F$ is the splitting field of $x^q-x$ over $\Bbb F_p$. Conversely if $F$ is the splitting field of that polynomial, then $F$ is generated over $\Bbb F_p$ by some primitive $(q-1)$th root of unity (since all of the nonzero roots of $x^q-x$ are powers of it), so $F^\times$ is generated by $\Bbb F_q^\times$ and this primitive root, which means that $F^\times$ must be cyclic of order $q-1$, hence $|F|=q$.

So we know for every prime power $p^m$ there exists a unique field $\Bbb F_{p^m}$ of order $p^m$, which is the splitting field of $x^{p^m}-x$. If $m\mid n$ then $(x^{q^m}-x)\mid(x^{q^n}-x)$ (can you prove this? what does the quotient look like?). Another fact from general field theory: if $f(x)\mid g(x)$ then $f$'s splitting field is contained in $g$'s splitting field (technical note: the splitting fields have to exist in a fixed algebraic closure). So $\Bbb F_{q^m}\subset \Bbb F_{q^n}$. Conversely, if $\Bbb F_{q^m}\subset\Bbb F_{q^n}$, then $q^n$ is a power of $q^m$, because the larger field is a vector space over the smaller field (general fact about vector spaces).