This seems to be a fairly classic problem in algebra:
Let $R$ be a nonzero finite commutative ring with no zero divisors. Prove that $R$ is a field.
Here is the a solution I came across:
It has previously been shown that if $R$ is a finite integral domain, then it is a field. Thus we show that $R$ is an integral domain.
If $ab=0$ we know since $R$ has no zero divisors that $a=0$ or $b=0$, so the only thing that remains to show is that there exists an identity $1_R\neq 0_R$.
Since $R$ is finite we can write $R=\{0_R,r_1,r_2, ...\, r_n\}$. Consider the set $R_i=\{r_ir_1, r_ir_2, ..., r_ir_n\}$, where $1\leq i \leq n$. For each $i$, if $r_ir_j=r_ir_k$, then $r_ir_j-r_ir_k=r_i(r_j-r_k)=0_R$ and since $r_i \neq 0_R$ we have $r_j=r_k$ (i.e. cancellation holds). This means the elements in $R_i$ are distinct.
It follows that for each $i$ there exists some $1\leq t_i \leq n$ such that $r_ir_{t_i}=r_i$.
Particularly $r_{t_i}r_{t_{t_i}}=r_{t_i}$. Thus $r_ir_{t_i}=r_ir_{t_i}r_{t_{t_i}}=r_ir_{t_{t_i}}$. Now by cancellation $r_{t_i}=r_{t_{t_i}}$.
That is, $r_{t_i}r_i=r_i=r_ir_{t_i}$.
Now we claim that $r_{t_1}$ is the identity in $R$. It is enough to show that for each $r_i\in R$, $r_{t_1}r_i=r_i$. Given $i$ there exists a $k$ such that $r_{t_1}r_k=r_i$, and thus $r_1r_k=r_1r_{t_1}r_k=r_1r_i$. Now by cancellation $r_k=r_i$ and thus $r_{t_1}r_i=r_i$. We have shown that $R$ contains a nonzero identity.
I don't understand why the bold part is needed. Can't we just say directly that $r_{t_i}r_i=r_i=r_ir_{t_i}$ since $R$ is commutative? I don't understand why $r_{t_i}=r_{t_{t_i}}$ is needed. Any help in clearing this up for me will be greatly appreciated!
Awful lot of subscripts, not needed. Multiplication by a nonzero element permutes the nonzero elements, the outcome set is everything. With no zero divisors, multiplication by a fixed nonzero element is injective, and with finiteness it is also surjective.
Take your favorite nonzero element $r.$ There is some element, call it $e,$ such that $re=r.$ Take another nonzero element $s.$ $$ res = (re)s = rs, $$ $$ r(es) = rs, $$ $$ r(es) - rs = 0, $$ $$ r(es-s) = 0. $$ No zero divisors, so $$ es = s. $$ We may now say $e=1,$ because it is a multiplicative identity.