Let $\mathbb{F}$ be a finite field. Consider the morphism $f: \mathbb{Z} \to \mathbb{F}, n \to n\cdot 1$. The kernel of $f$ is an ideal of $\mathbb{Z}$ which is an euclidean ring, thus it is principal and the kernel is of the form $p\mathbb{Z}$, where $p$ is the characteristic of the field $\mathbb{F}$, so $p$ is prime. We now define a new morphism by shifting to quotient:
$\tilde{f} : \mathbb{Z}/p\mathbb{Z} \to Im(f)$.
$\tilde{f}$ is an isomophism, so $\mathbb{Z}/p\mathbb{Z} \simeq Im(f)$. Therefore $Im(f)$ has $p$ elements. Now since $Im(f)$ is in bijection with $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z}$ is a field ($p$ prime), $Im(f)$ is a subfield of $\mathbb{F}$, so $\mathbb{F}$ is a vectorial space over $Im(f)$.
Why do we have that $\mathbb{F} \simeq Im(f)^{\dim\mathbb{F}}$?
