# Nontrivial Finite Rings with No Zero Divisors

## 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)

## 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?

fresh_42
Mentor
How does this sound?
Why do left and right identities have to be the same? I could follow you until - given an element ##r \neq 0## - there are elements ##x_r## and ##y_r## with ##rx_r =r = y_rr##. The index is necessary as up to this point, those elements still depend on the given ##r##. Also isn't clear whether we may assume the existence of a unity element ##1## or not. E.g., how about the ring of all polynomials over ##\mathbb{Z}_3## with zero absolute term and ##x^2=x##, i.e. ##x\,\mid \,p(x)\,##? And I haven't understood the part, where you show that ##x_r=x_s##. This leaves me with three questions:
1. ##1 \in R\,##?
2. ##x_r = y_r\,##?
3. ##x_r = x_s## and with it ##y_r=y_s\,##?
So
since left and right identities must coincide
isn't clear to me and the third part is a bit too hand waving. I think at some place associativity is needed (which also isn't mentioned as a given property). Without indexing the right identity ##x_r## and the left identity ##y_r## it is a bit too foggy for my taste.

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The ##x_r## and ##y_r## are guaranteed to exist because the map (and its right multiplication analogue) are surjective. I my problem lies in the fact that I confusingly calling ##x_r## the right identity of ##r## and ##y_r## the left identity of ##r##. Now, if I were able to show that ##x_r## is the right identity of EVERY element, as well as showing ##y_r## the left identity of EVERY element, then I could conclude ##x_r = y_r##; but really ##x_r## just an identity with respect to the single element ##r## (the same goes for ##y_r##). That's where I went wrong.

To answer your questions, (1) we are not assuming ##1 \in R##, but I am trying to deduce the existence of ##1##; (2) I was trying to deduce this, but, as I discussed above, I did it incorrectly; (3) I was also trying to deduce this, but I did it incorrectly.

So, I guess I could use a hint on how to solve this problem. I know that if ##1 \in R##, then we could use the surjectivity of ##f(x) = rx## to prove every nonzero element is invertible, but I am unsure of how to prove that ##R## is unital

I thought of a way to fix it. As before, let ##f_r : R_0 \to R_0## be defined by ##f_r(x) = rx##, where ##R_0 = R - \{0\}##. Since this is a bijection, given ##r \in R_0##, there exists an ##x \in R_0## such that ##f_r(x) = r## or ##rx= r##. Since ##x## is not zero, there exists a ##y## such that ##xy = x##. Multiplying both sides by ##r## gives ##rxy=rx## or ##ry = rx## or ##y=x##. Therefore ##x^2 = x##. Multiplying both sides by the arbitrary element ##a##, ##ax^2 = ax## or ##x(ax-a)=0## which implies ##ax=a##, because ##x## is not zero. Multiplying the same equation the right side and doing similar manipulations, we conclude that ##x## is a left and right identity of ##R##, which we denote as ##1##. Now ##f_r## is bijective, there exists a ##s \in R## such that ##rs=1##. All that remains to show is that ##s## is left inverse, which I won't do now.

How does this sound?