Show that a finite domain is a division ring

Question

Let $R$ be a finite ring. Show that the following are equivalent:
i. $R$ is a division ring.
ii. $R$ is nontrivial and if $r$,$s \in R$, with $rs=0$, then either $r=0$ or $s=0$.
$\textbf{NOTE:}$ A commutative ring that satisfies (ii) is called an integral domain or sometimes just a domain. The problem tells us that finite domains are fields.

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I am not sure if I have one direction done properly. The other direction stumps me.

$\textbf{Proof:}$ (i) $\implies$ (ii) Suppose $R$ is a division ring. Then every nonzero element is invertible. Let $r,s \in R$ where $r \neq 0$. Suppose $rs=0$. Then $r^{-1}(rs)=s \implies s=r^{-1}0=0 \implies s=0$.
$\therefore$ $rs=0 \implies r=0 \,\, \text{or} \,\, s=0$.

(ii) $\implies$ (i) Suppose $R$ is a finite ring that is nontrivial. Let $r,s \in R$ with $rs=0$ and either $r=0$ or $s=0$. [We want to show that $R$ is a division ring. So we want to show that every nonzero element is a unit. I know that a ring is a division ring if and only if the set of units denoted $U(R)=R-\{0\}$, where $R-\{0\}=R^\times$ is the multiplicative group of $R$.] How do I being to prove this direction?

Answer 1

(ii) $\Rightarrow$ (i): Let $r$ be a nonzero element of $R$. You want to show that $r$ has a multiplicative inverse. What can you say about the elements $r, r^2, r^3$ etc.?

Answer 2

Hint:

Let $r\in R,\;r\ne 0$. Consider the endomorphism of the underlying additive group: $\begin{aligned}[t] m_r:R&\longrightarrow R\\ s&\longmapsto rs \end{aligned}$

Show that $m_r$ is injective. As $R$ is finite, injective $\longleftrightarrow$ surjective $\longleftrightarrow$ bijective, and consequently, $1$ is attained.

Answer 3

Let $R=\{r_1=1,r_2,\dots,r_k\}$ be the distinct elements of $R.$

Assume $r\in R$ does no have an inverse. Then the set $rR=\{rr_1,rr_2,\dots,rr_k\}$ must have fewer than $k$ elements, since $1$ is not in $rR.$ So, by the pigeonhole principle, there must be some $i,j$ with $i\neq j$ Such that $rr_i=rr_j.$

But then $r(r_i-r_j)=0.$. So either $r=0$ or $r_i-r_j=0.$

But $i\neq j,$ so $r_i\neq r_j$ and thus by (ii) $r=0.$