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## Homework Statement

A finite ring with more than one element and no zero divisors is a division ring (Special case: a finite integral domain is a field)

## Homework Equations

## The Attempt at a Solution

Let ##r \in R \setminus \{0\}##, and define ##f : R \setminus \{0\} \to R \setminus \{0\}## by ##f(x) = rx##. Now suppose ##f(x) = f(y)##. Then ##rx = ry## or ##r(x-y)=0##, which implies ##x=y## since ##r \neq 0## and there can be no zero divisors. Since ##R \setminus \{0\}## is finite, we can conclude ##f## is, in addition to being injective (as we just demonstrated), surjective. Similarly, we can show map defined by right multiplication is bijective.

Given ##r \in R \setminus \{0\}##, there exists an ##x \in R \setminus \{0\}## such that ##f(x) = r## or ##rx=r##, which implies ##r## has a right identity. We can also deduce that ##r## has a left identity, and since left and right identities must coincide, ##x## is ##r##'s identity simpliciter. Therefore, every nonzero element has an identity. Now we show that these identities are in fact the same (this is where it gets a little hairy and uncertain).

Let ##r## and ##s## be nonzero elements and let ##x## and ##y## be their identity, respectively. Then ##rs = rs## or ##rs = rys## or ##r = ry## or ##rx = ry## or ##x=y##, where we used the cancellation law, which holds when there are no zero divisors, several times. There is probably a more direct way to conclude ##x=y##, but I can't see it at present.

Therefore, ##R## has a multiplicative identity, denote it as a ##1##. Thus given ##1##, there exists an ##x \in R \setminus \{0\}## such that ##f(x) = 1## or ##rx = 1##. Using the fact that left and right inverses coincide, ##x## is ##r##'s multiplicative identity.

How does this sound?