Must this rng be a ring?

Question

A rng is a ring without the assumption that the ring contains an identity. Consider a finite rng $\mathbf{R}$.

I am investigating conditions that get close forcing an identity but not quite. The closest condition I can think of is the following:

If $a\in \mathbf{R}$ is non-zero then there is $b\in\mathbf{R}$ such that $ab\neq 0$

I am finding myself unable to prove that $\mathbf{R}$ must/need-not have a multiplicative identity i.e. be a ring.

Are there well-known results/examples that deal with this sort of condition?

Answer 1

Not an answer, but a reduction to the prime power case.

It's pretty easy to reduce to when $|R|$ is the power of a prime.

If $|R|=mn$ with $\gcd(m,n)=1$, solve $mx+ny=1$. Show that $$R\to (mR)\times (nR); a\mapsto (mxa,nya)$$ is an isomorphism of rngs.

Now, if $ma\neq 0$, then $mab\neq 0$ for some $b\in R$. But note that $(ma)(mxb)=m(1-ny)(ab)$. So this means that $mR$ has the property we want, too. Similarly for $nR$.

If $mR$ and $nR$ have identities, then so does $R$. So one of $mR$ or $nR$ would have to not have an identity. We keep reducing until we find an example which is a prime power size.

Answer 2

In the commutative case (there are probably simple non-commutative counterexamples coming from matrix rings):

We say that a commutative rng $R$ has property $\mathcal{P}$ if, for all nonzero $a\in R$, there is some $b\in R$ with $ab\neq 0$. The zero ring has property $\mathcal{P}$, and it has a unit. Let $R$ be a finite nonzero commutative rng such that all smaller commutative rngs with property $\mathcal{P}$ have a unit.

Pick some $a\neq 0$. Then there is some $b$ with $ab\neq 0$, some $c$ with $abc\neq 0$, and so on. So there exist arbitrarily long nonzero products in $R$, which implies that there is some $x\in R$ that is not nilpotent.

Since $R$ is finite, there are $d,N$ such that $x^n = x^{n+d}$ for $n>N$. Choosing $M$ so that $Md>N$, we have $(x^{Md})^2 = x^{2Md} = x^{Md}$, so there is some non-zero idempotent $e=x^{Md}$.

Let $I = \{r\in R \mid er = 0\}$. $I$ has property $\mathcal{P}$: if $r\in I$ is nonzero, then there is some $s\in R$ with $rs\neq 0$. Then $r(s-es)=s(r-er)=sr\neq 0$, and $s-es\in I$.

$I$ is strictly smaller than $R$ ($e\notin I$, because $e$ is nonzero), so, by choice of $R$, $I$ has a unit $u$. But then $u+e$ is a unit of $R$: for any $t\in R$, $(u+e)t = ut + et=u(t-et) + et = (t-et) + et = t$.

By induction, we conclude that all finite commutative rngs with property $\mathcal{P}$ have a unit.

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In retrospect, the idea here is not so difficult: As in Thomas Andrews' answer, we are trying to write $R\cong R_1\bigoplus R_2$ for subrngs $R_1,R_2$. Direct summands inherit property $\mathcal{P}$, and $R$ has a unity if and only if both $R_1$ and $R_2$ do, so this lets us quickly reduce to the case of indecomposable rngs.

Furthermore, idempotents are one of the more natural ways to identify direct summands. Given an idempotent $e$, we can write $R\cong eR \bigoplus \operatorname{ann}(e)$, and $eR$ always has unity $e$. So the challenge is just to show that $R$ must contain a nonzero idempotent.