Is a prime subfield a set of integers? By definition: If $F$ is a field and $K\subset F$ is the smallest field contained in $F$, we call $K$ the prime subfield of $F$. Denote the prime subfield of $F$ by $P(F)$ (hope you don't mind my introducing this notation).
We know that for any field $F$ with characteristic $p$, $P(F)=\{a\cdot b^{-1}, \text{where } a\in\{0,1,...,p-1\},b\in\{1,2,...,p-1\}\}$. I'd like to know if $P(F)$ is a set of integers, but it's not clear to me that it should be so (e.g. $2(p-1)^{-1}$ an integer?)
For some context: I am trying to prove that if the characteristic of $F$ is $p$ for some $p$ prime, $P(F)$ is isomorphic to $\mathbb{F}_p$. I'm convinced that $\sigma:P(F)\to \mathbb{Z}_p$ with $\sigma(f)=\overline{f}\equiv f$ mod $p$ will give a homomorphism, and hence $\sigma$ is an isomorphism between $P(F)$ and $P(\mathbb{Z}_p)=\mathbb{Z}_p$. [I haven't yet verified that $P(\mathbb{Z}_p)=\mathbb{Z}_p$; seems $\mathbb{Z}_2\subset \mathbb{Z}_p$, but $\mathbb{Z}_2$ is not a field with the addition inherited by $\mathbb{Z}_p$?] I know that the desired homomorphism properties - $\sigma(a+b)=\sigma(a)+\sigma(b); \sigma(ab)=\sigma(a)+\sigma(b); \sigma(1)=1$ - hold if $a,b$ are integers, hence the initial question of the post.
If my wishful thinking is off and $P(F)$ is not a set of integers, I was thinking $P(F)$ is isomorphic to $\{0,1,...,M\}$ for some $M\le p^p$, and then we can define the above homomorphism from $\{0,1,...,M\} \to \mathbb{Z}_p$. But we no longer have the guarantee that $\{0,1,...,M\}$ is a field. So I'm not sure where I would go from there.
Thanks a bunch in advance! And please let me know if I should filter my posts more before I answer questions. I usually don't include much "thought-process" text - maybe I should not do so.
 A: 
If my wishful thinking is off and P(F) is not a set of integers,

You're right to question your wishful thinking. The prime subfield is not a set of integers. 
Think about the $p$-element field $\mathbb{Z}_p$, which is its own prime subfield. It has $p$ elements that it's convenient to name with the names $0, 1, \ldots , p-1$ of the first $p$ nonnegative integers, but it doesn't contain those integers, since their arithmetic isn't ordinary integer arithmetic, it's arithmetic mod $p$. Your question suggests that you sort of understand this.
In answer to your last paragraph: you should think a question through as best you can before asking it here - but you need not arrive at perfect clarity. If you could, you'd probably have the answer.
A: In characteristic $p$, the prime subfield is just $\mathbf Z/p\mathbf Z$.
This is because for any commutative ring $R$, there's a canonical ring homorphism which maps each $n\in \mathbf Z$ onto $n\cdot 1_R$, and if it is not injective, its kernel is generated by an integer $a>0$,  whence by the 1st isomorphism theorem an injective ring homomorphism $\mathbf Z/a\mathbf Z\hookrightarrow R$.
Now if $R$ is an integral domain (e.g. a field $F$),  the kernel is a prime ideal, generated by a prime number $p$. You now have your injection from  $\mathbf Z/p\mathbf Z$ into the field $F$.
