Vtgyjfy

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

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 $mathbb{Z}/pmathbb{Z}$ 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.

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 $k_iin {Bbb Z}_p$. Thus the field $F$ has $p^n$ elements, as required.

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