If $p$ is prime, prove that there exists an injective ring homomorphism $\phi:\mathbb{F}_{p^{n}}\longrightarrow\mathbb{F}_{p^{m}}$ iff $n|m$. This is what I have been able to conclude at the moment:
$\boxed{\Rightarrow}$ Since $\phi$ is a ring homomorphism, $Im_{\phi}(\mathbb{F}_{p^{n}})$ is a subring in $\mathbb{F}_{p^{m}}$. In particular, $(Im_{\phi}(\mathbb{F}_{p^{n}}),+)$ and $(\mathbb{F}_{p^{m}},+)$ are abelian groups, so $(Im_{\phi}(\mathbb{F}_{p^{n}}),+)$ is an abelian subgroup in $(\mathbb{F}_{p^{m}},+)$. Lagrange's theorem allow us to say that $|Im_{\phi}(\mathbb{F_{p^{n}}})|\,|\,|\mathbb{F}_{p^{m}}|=p^{m}$. As $\phi$ is an injection, $|Im_{\phi}(\mathbb{F}_{p^{n}})|=|\mathbb{F}_{p^{n}}|=p^{n}$, so we can write that $p^{n}|p^{m}$ (I don't know how to continue from here).
$\boxed{\Leftarrow}$ Since $n|m$, there exists an element $k$ in $\mathbb{N}$ such that $nk=m$. Therefore, $p^{nk}=p^{m}$, and so, $\mathbb{F}_{p^{nk}}=\mathbb{F}_{p^{m}}$. We can consider an embedding of the form $\mathbb{F}_{p^{n}}\hookrightarrow\mathbb{F}_{p^{m}}$, which is an injective ring homomorphism. $\square$
Any suggestion or correction will be welcome. Thanks!
 A: In the second part, I would like to know, how this embedding looks like. For the first part, you can use the degree formula. I will simplify this theorem as $\mathbb{F}_{p^m}<\mathbb{F}_{p^n}$ iff $m\mid n$, since according to Moore's theorem, two finite fields with the same number of elements are isomorphic (See here, Theorem 6).
$\Rightarrow$: If $\mathbb{F}_{p^m}<\mathbb{F}_{p^n}$, then because of the degree formula, we have:
\begin{equation}
\frac{n}{m}
=\frac{[\mathbb{F}_{p^n}:\mathbb{F}_p]}{[\mathbb{F}_{p^m}:\mathbb{F}_p]}
=[\mathbb{F}_{p^n}:\mathbb{F}_{p^m}]\in\mathbb{N}
\end{equation}
and therefore $m\mid n$.
$\Leftarrow$: Let $m\mid n$. We know that the Frobenius homomorphism $\operatorname{Frob}_p\colon\mathbb{F}_{p^n}\rightarrow\mathbb{F}_{p^n}.x\mapsto x^p$ is a generator of the cyclic Galois group $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$ with order $n$. Therefore $\operatorname{Frob}_p^m$ is a generator of a cyclic subgroup with order $n/m$, abbreviated as $H=\langle\operatorname{Frob}_p^m\rangle$, and we can consider its fixed field in $\mathbb{F}_{p^n}$, abbreviated as $\mathbb{K}=\operatorname{Fix}_{\mathbb{F}_{p^n}}(H)$. Using the degree formula for the field extensions $\mathbb{F}_p<\mathbb{K}<\mathbb{F}_{p^n}$ and using Artin's theorem, we get:
\begin{equation}
[\mathbb{K}:\mathbb{F}_p]
=\frac{[\mathbb{F}_{p^n}:\mathbb{F}_p]}{[\mathbb{F}_{p^n}:\mathbb{K}]}
=\frac{n}{|H|}=m.
\end{equation}
