The proof of the day is: Finite integral domain is Galois field
Theorem: Every finite integral domain is a Galois field.
Proof:
By the definition of field, for a integral domain $\mathcal{D}=(D,+,\cdot)$ to be
a field, it must exist $a^{-1}$ such that $a\cdot a^{-1} = 1, \forall a\in \mathcal{D}^{*}$.
Consider the homomorphism $\psi:\mathcal{D}\to\mathcal{D}$ such that $\psi(x)=\alpha\cdot x$,
$\alpha \neq 0$.
We have that $ker(\psi)=\{x\in\mathcal{D}\mid\alpha\cdot x = 0\} = \{0\}$, since by definition,
$\mathcal{D}$ doesn't have zero divisors. So $\psi$ is one-to-one, which implies that it is surjective too,
since $\mathcal{D}$ is finite (pigeonhole princple). So, $\exists x\in\mathcal{D} \mid \alpha\cdot x=1$.
Then $x=a^{-1}$, and $\mathcal{D}$ is a finite field, as we wanted.