Image of ring morphism is a subfield Let $\Bbb F$ a finite field and $\varphi$ the map defined by
$$\varphi: \Bbb Z\to \Bbb F,\; n\mapsto n\cdot 1_{\Bbb F}$$
I want to prove that $Im \varphi$ is a subfield of $\Bbb F$. So far I proved that $\varphi$ is a ring morphism so $Im \varphi$ is a subring. What I need to prove now is that all element of $Im \varphi\setminus\{0_{\Bbb F}\}$ are invertible and I'm stuck on it. 
 A: If $R\to \mathbb F$ is any nonzero ring homomorphism, then $Im(f)$ is obviously a domain, and of course finite domains are fields.
This is easy to see, and contained in many questions on the site. For a nonzero element $a$, you can consider the powers $a^n$. This can't be infinte, so there exists $n, m$ integers such that $a^n=a^m$. Cancelling away the smaller power, you get a power of $a$ equal to $1$. So in fact, $a^{-1}$ is a power of $a$, and that's in the image too, of course.

Now that I actually go back and see how you're defining the map from $\mathbb Z\to \mathbb F$, I see other possiblities too. The fact that the codomain of the map is an integral domain implies that $\ker(f)=(p)$ for some prime $p$ or $\{0\}$, but the latter case is ruled out since the codomain is finite. By the first isomorphism theorem, $Im(f)\cong \mathbb Z/(p)$.
A: Another proof:
There's a very simple reason for this: $\operatorname{Im f}$ is a subring of the finite field $F$, hence is an integral domain. So it results from the following result:

A finite integral domain is a field.

Indeed, if $A$ is this domain, for any $a\neq 0$ in $A$, the map $\;\begin{aligned}[t]m_a\colon A&\longrightarrow A\\x&\longmapsto ax\end{aligned}\;$ is injective since $A$ is an  integral domain. Now on a finite set, injective means surjective, so $1$ is attained, i.e. there exists $x\in A$ such that $ax=1$: this proves any non-zero $a$ is invertible.
