explaing the following prove

Question

i need help with the proove of every finite integer domain is a field, the proof is as follows:

The main idea is that if $R$ is finite, then any map from $R$ to $R$ is injective iff it is surjective iff it is bijective.

We know $a\in R$ is a not a zero-divisor if $f_a$ is injective. Since $R$ is finite, then it is also surjective.

If $f_{a}$ is surjective then $a$ is a unit. Hence we have shown that any non-zero element in $R$ is a unit. Hence $R$ is a field.

I need to clarify why if $R$ is finite, then any map from $R$ to $R$ is injective iff it is surjective iff it is bijective.

and what is $f_a$ and why If $f_{a}$ is surjective then $a$ is a unit

Typos: "Proof" instead of "following prove" or "the proove". — Dietrich Burde, Apr 03 '19 at 19:06
Answers the first question completely: https://math.stackexchange.com/q/63072/29335 — rschwieb, Apr 03 '19 at 19:50

score 1 · Answer 1 · answered Apr 03 '19 at 18:56

First of all, if $f:X\to X$ is an injective map, then $f(X)\subset X$ and $|f(X)|=|X|$. If $X$ is finite, then it must be that $f(X)=X$ (this is false if $X$ is infinite, since $\{1,2,3,\ldots\}\subset\{0,1,2,3,\ldots\}$ have the same size but are not equal).

Now, the map $f_a:R\to R$ is defined by $f_a(x)=ax$. Since $a$ is not a zero divisor, the map $f_a$ is injective. Since $R$ is assumed to be finite, $f_a$ is surjective as well. Since $f_a$ is surjective, $f_a(x)=1$ for some $x$. This means that $ax=1$ for some $x$.

explaing the following prove

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