Let $(R, +, \cdot)$ be a finite ring without zero divisors, show that $R$ has a neutral element for $\cdot$.

Question

I have to prove the question in the title, but I am having some difficulties.

Here's a sketch what I've already tried:

Choose $a \in R$. Because $R$ is finite, there exist positive integers $i$ and $j$ ($i\neq j$) so that $a^i = a^j \Leftrightarrow a^ia = a^ja \Leftrightarrow a^ia - a^ja = 0 \Leftrightarrow a^j(a^{i-j}a - a) = 0$.

Because $R$ doesn't have any zero divisors and $a^j \neq 0$, we must have $a^{i-j}a - a = 0$ and thus, $a^{i-j}a = a$.

So now I think that $a^{i-j}$ is the neutral element for $\cdot$ but I'm not sure how to proceed.

Could anyone help me?

Thanks!!

Answer 1

So at this point, you have $a^{i-j}a = a$. We want to show that $a^{i-j}$ is in fact the identity element for $\cdot$. Now, this amounts to saying that for any $b \in R$, $a^{i-j}\cdot b = b$. So we have the following:

$a^{i-j} \cdot b = c \iff a \cdot a^{i-j} \cdot b = a \cdot c \iff a^{i-j}a \cdot b = a \cdot c \iff ab = ac $.

But since we have no zero divisors, this means that $b = c$.