Proof a Rng cannot have exactly five non-zero divisors.

Let $R$ be a Rng (a ring which does not necessarily have a $1$). We call an element $a$ regular if $xa=0$ implies $x=0$ and $ax=0$ implies $x=0$.

Prove $R$ cannot have exactly five regular elements.

This problem is from the Galois-Noether competition.

This is the progress so far. regular elements are closed under multiplication, their multiplications are cancellative. Therefore we can conceive the set $M$ of regular elements, if it had order $5$ then it would be a finite cancellative semigroup, hence a group. Since $5$ is prime it would have to be a cyclic group. Therefore $M$, with multiplication, would be isomorphic to $\mathbb Z_5$. I don't now if this is useful, but this is what I've found, I also have to thank Anon from chat for his help.

Solution 1:

Let $u$ be the unit in the group $M$ and let $S$ be the set of elements of $r\in R$ such that $ru=ur=r$. Then $S$ is a subrng of $R$ and contains all of $M$, and in fact $S$ is a ring (with unit $u$). Note furthermore that any unit of $S$ is in $M$, since $xy=yx=u\in M$ implies $x$ and $y$ are regular. (Edit: In fact, $S$ is all of $R$; see anon's comment below.)

We thus are reduced to showing that no ring $S$ can have exactly $5$ units. To show this, note that $-1$ is always a unit, but $\mathbb{Z}/5$ contains no nontrivial elements of order $2$, so we must have $1=-1$ in $S$. Thus $S$ is an algebra over $\mathbb{F}_2$. We may also assume that $S$ is generated by its units (otherwise, just consider the subring they generate). Thus we have that $S$ is isomorphic to some quotient of the group ring $\mathbb{F}_2[\mathbb{Z}/5]\cong \mathbb{F}_2\times \mathbb{F}_{16}$. The only such quotients are $0$, $\mathbb{F}_2$, $\mathbb{F}_{16}$, and $\mathbb{F}_2\times\mathbb{F}_{16}$, none of which have exactly $5$ units.