Let $F$ be a finite field and $\varphi:\mathbb{Z}\to F$ a homomorphism such that $\varphi(1)=1_F$. Is $Im\varphi=\{\varphi(x):x\in \mathbb{Z}\} $ a subfield of $F$?
I know that $1_F\in Im\varphi$, $0_F\in Im\varphi$ and $(Im\varphi ,+)$ is a group, but I couldn't prove that given $x\in Im\varphi\setminus\{0_F\}$, then $x^{-1}\in Im\varphi$ in which $x^{-1}$ is the multiplicative inverse of $x$.
Another thing I know but I couldn't use to answer the question is that since $F$ is finite, then $\ker\varphi\neq\{0\}$. And since $\mathbb{Z}$ is a principal ideal domain and $\ker\varphi$ is a ideal, then exists $p\in\mathbb{Z}\setminus\{0\}$ such that $\ker\varphi=p\cdot \mathbb{Z}$.
I saw a teacher stating in a video lesson that $Im\varphi$ is a subfield of $F$ to prove that $p$ is prime. But unfortunately I couldn't understand why $Im\varphi$ is a subfield of $F$.
The image of $\phi$ is a subring of the field $F$, hence is an integral domain. This implies that $\ker(\phi)$ is a prime ideal of $\mathbb{Z}$.
But as you noted, $\ker(\phi)\neq \{0\}$ since $F$ is finite, which implies that $\ker(\phi)=p\mathbb{Z}$ for some prime $p$. Hence $\mathrm{Im}(\phi)\simeq \mathbb{Z}/p\mathbb{Z}$ is a field.