# Must this rng be a ring?

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?

• You have already asked the exact same question, buy you deleted it (which means a lot of people will not find anything at the end of that link). I guess my question is: why? – Arthur Jan 29 '16 at 0:33
• @Arthur No, I mistakenly forgot to add finite in my post - the question is trivial otherwise. I decided to delete and re-post as editing it would make the 10 comments irrelevant. – user111064 Jan 29 '16 at 0:35
• You are allowed to edit a question, rather than deleting it. @user111064 – Thomas Andrews Jan 29 '16 at 0:37
• So it is not the exact same question, then. Anyways, I made a comment on the other question that there is one statement that is closer to the existence of a $1$ than what you got, and it is this: "There exists a $b$ such that for any $a\neq 0$, $ab\neq0$". Of course, if your condition implies that there is a $1$ in the finite case, then so does mine, and they are equivalent. But if not, it might be worth looking into what finite rngs, if any, fulfill your statement but not mine, or my statement, but doesn't have a $1$. – Arthur Jan 29 '16 at 0:42
• @ThomasAndrews contributed the following statement that might also be worth a look: "Consider $f_a:R\to R$ given by $f_a(x)=ax$. Then $f_a=f_b$ implies $a=b$". – Arthur Jan 29 '16 at 0:49

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.

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.

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.

• Why is your map multiplicative? – Martin Brandenburg Jan 29 '16 at 21:27
• Because $(mx)(mx)c = (mx)(1-ny)c=(mx)c-(mnxy)c=(mx)c$. So $(mx)a(mx)b=(mx)(ab).$ @MartinBrandenburg – Thomas Andrews Jan 29 '16 at 21:35
• It's basically a usual Chinese Remainder theorem argument. It also why it is onto each component - you actually have $(mxm)c=mc$. – Thomas Andrews Jan 29 '16 at 21:38