if a ring is finite then the translation $x\rightarrow ax$ is surjective where $a\in A$ is regular

Question

In a proof of the inversibility of regular elements in a finite ring, there is the following argument:

let $A$ be a finite ring and $a\in A$ regular . the translation $A\rightarrow A: x\rightarrow ax$ is surjective because A is finite.

I don't see why it is the case, any help much appreciated.

Answer 1

If $a$ is regular, then $ax = 0$ only if $x=0$. Hence if $ax = ay$ we have $a(x-y) = 0 \implies x=y$.

So, the translation is an injective function.

We know that a function $f : A \to A$, for a finite set $A$, is injective iff it is surjective.