Prove every finite field is of cardinality $p^n$

Question

I realise that this question has been asked multiple times before, but I would like to ask about a specific detail in my proof.

Let F be a finite field.

I begin by showing that the characteristic of any finite integral domain must be a prime, say p.

This gives us that: $$ker(\phi) = pZ$$ where $\phi:Z \rightarrow F$ is a homomorphism.

By first isomorphism theorem, $$Z/pZ \cong Im(\phi) \subset F$$

Now the problem is prompting me to show that this means we can assume that $Z/pZ \subset F$.

From here it is easy to show that F is a field over $Z/pZ$ (which is also a field) and finish the proof.

From what I understand, we aren't really showing that $F$ is a field over $Z/pZ$, but rather over an isomorphic subfield of $F$. But the prompt explicitly asks us to show that F is a field over $Z/pZ$. This is where my confusion is coming from.

Any help would be appreciated

Answer 1

This is a common coercion of types. Note that $\phi(0) = 0$, $\phi(1) = 1$, $\phi(\underbrace{1 + \dots + 1}_{n \text{ times}}) = \underbrace{1 + \dots + 1}_{n \text{ times}}$. Yes, these arguments to $\phi$ are elements of $\mathbb{Z}/p\mathbb{Z}$ and the outputs of $\phi$ are elements of $\mathbb{F}_{p^n}$. But, once you have $0$ and $1$ in a structure with addition, you can always see how much of $\mathbb{Z}$ you get by iterated addition.

Answer 2

Well, if $F$ is a finite field of characteristic $p$, then $F$ contains a copy of ${\Bbb Z}_p$ given by the multiples of the unit element $1$. But each field extension $F$ of a field $K$ is a vector space over $K$. So $F$ is finite dimensional vector space over ${\Bbb Z}_p$. Let $\{b_1,\ldots,b_n\}$ be a basis of $F$ over ${\Bbb Z}_p$. Then each element $a$ of $F$ can be uniquely written as a linear combination $a= \sum_i k_i b_i$, where $k_i\in {\Bbb Z}_p$. Thus the field $F$ has $p^n$ elements, as required.

As a remark, the mapping ${\Bbb Z}_p \rightarrow F:a\mapsto a\cdot 1$ is a ring monomorphism, where $a\cdot 1 = 1+\ldots +1$, $a$ times. So as said, $F$ contains an isomorphic copy of ${\Bbb Z}_p$.