A finite integral domain with more than one element is a field (Proof explanation)

Question

Here is the question I want to answer:

Let $R$ be a finite commutative ring with more than one element and no zero-divisors. Show that $R$ is a field.

I do not understand some ideas in the proof given here How to show that a finite commutative ring without zero divisors is a field? , here is the proof:

It follows from the pigeon-hole principle.

First, we show that $R$ has an identity. (Sometimes the existence of an identity is included in the definition of a ring, but you do not need it here.)

For each $0 \ne a \in R$, consider the map $$ \phi_{a} : R \to R, \qquad x \mapsto a x. $$ Because $a$ is not a zero-divisor, the map $\phi_{a}$ is injective, thus surjective. So there is $e \in R$ such that $$a = \phi_{a}(e) = e a.\tag{1}$$ Again because the map $\phi_{a}$ is surjective, every $b \in R$ is of the form $$b = \phi_{a}(x_{b}) = a x_{b}$$ for some $x_{b}$, so multiplying (1) by $x_{b}$ you find $$ b = e b $$ for all $b \in R$, so $e$ is the identity.

Now use once more the fact that $\phi_{a}$ is surjective, to show that there is $c \in R$ such that $$a c = \phi_{a}(c) = e,$$ so $c$ is an inverse of $a$, $a$ being an arbitrary non-zero element.

My questions are:

1 - what is the intuition for defining the map $\phi_a$ in this form?

2- Why because $a$ is not a zero divisor, we can conclude that the map $\phi_a$ is injective and why this implies that this map $\phi_a$ is also surjective?

Could someone explain to me these questions please?

Answer 1

Generally when working with rings the cool thing you have is the multiplication. If you want to show something, the easiest non-trivial map one can think of is a simple multiplication map (as you would look for an addition map if working with groups). Most of the times (as in this case) that is already enough to prove it. And this is exactly what happens here with $\phi_a$.

For your second question: A map is injective iff the kernel of the map is just $\{0\}$. If you calculate the kernel of this map, you get $\ker \phi_a = \{x: ax=0\} $. But as $a$ is not a zero divisor, we have $ax\neq 0$ for all $x\neq 0$, thus only $0$ is part of the kernel.

As we know that the map between $R$ and itself is injective (i.e. all inputs give distinctive outputs) and that the ring is finite (this is a necessary condition here!), we can conclude set-theoretically that the number of elements in the image of $\phi_a$ must be exactly $|R|$. But the codomain of $\phi_a$ is $R$, so in order to have $|\text{im } \phi_a|=|R|$ we need to have $R=\text{im }\phi_a$, so $\phi_a$ is also surjective.

I hope these explanations were helpful.