Nontrivial Finite Rings with No Zero Divisors

[Bashyboy]

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?

 

[fresh_42]

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.

 

[Bashyboy]

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

 

[Bashyboy]

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?