A Quick Note on Rings with Mostly Zero Divisors

Niven Achenjang bio photo By Niven Achenjang Comment

I apologize for the title of this post, but I couldn’t think of anything good. I had some random mathematical thoughts this morning that led into to a couple (hopefully interesting) questions. Before I forget, I want to write them down along with their resolutions.

Sadly, I don’t remember why, but I was thinking about conditions on when a localization is 0 this morning. Put more formally

Let $R$ be a commutative ring, and let $S\subset R$ be a multiplicative set. What sufficient and necessary conditions for $\sinv R=0$?

The natural thing to expect is that this means exactly when $0\in S$, and indeed, one can try to prove this. It’s clear that $\sinv R=0$ when $0\in S$ since you’ve inverted $0$, so we only need to worry about the other direction. Suppose that $\sinv R=0$. Then, $a/s=0/1$ for all $a\in R$ and $s\in S$. For a fixed $a$ (and $s$), this means that there exists some $u\in S$ such that

so $u=0$ or $a$ is a zero divisor. Well, let $a=1$; then it certinaly isn’t a zero divisor, so $u=0\in S$ and we win, except not quite. Our proof made use of $1\in R$, so we’ve really only showed the following.

Let $R$ be a commutative ring with unity, and let $S\subset R$ be a multiplicative set. Then, $\sinv R=0\iff0\in S$.

However, the original question was phrased for all commutative rings, not just those with unity. This led me to wonder the following:

Does there exist a (commutative) ring $R$ (without unity) such that every element is a zero divisor?

At this point, I don’t know a good way to detail my thought process in thinking about this, so I’ll just skip to the answer: yes. Specifically, take (the abelian group) $R=\zmod2$ and give it a ring structure by setting $0^2=1^2=0$. At first glance, one may worry that this doesn’t actually give a (consistent, well-defined, whatever) ring, but there’s nothing to worry since this is just the ideal $\{0,2\}$ generated by $2$ in the ring $\zmod4$ 1.

Sweet, got that figured out. After this, the next natural thing to ask was if we can get an analagous phenomenon in the presence of units?

Does there exist a commutative ring $R$ with unity such that every element is a unit or a zero divisor?

I hope the last condition seemed weird when you first read the question, and I hope even more that this one seems strange now; I know they both did to me 2. Because of that, I suspected that the answer to this question was no, and started thinking about contradictions I could reach from such a ring. This led me to the following.

Let $R$ be a commutative ring with unity such that every element is a unit or a zero divisor. Then, $R$ is a local ring.
Let $\mfm=\brackets{r\in R:r\text{ is a zero divisor}}\subset R$. We claim that $\mfm$ is an ideal. Since it is literally the set of all non-units, this is enough to prove that $R$ is local. Fix any $r\in R$ and $a,b\in\mfm$. Then, there exist $x,y\in R$ (even in $\mfm$) s.t. $ax=0=by$. Hence, $rax=0$ so $(ra)\in\mfm$ and $xy(a+b)=(ax)y+(by)x=0$ so $a+b\in\mfm$.

I’m gonna be honest; I’m not sure if this helps answer the question or not, but it’s something. Since I had no luck deriving a contradiction from this, I decided to instead try and construct such an $R$, and as it turns out, they do exist. Take $R=\F_2[x]/(x^2)$. As a set, $R=\brackets{0,1,x,1+x}$; we see that $0,x$ are zero divisors while $1,1+x$ are units as $(1+x)^2=x^2+2x+1=1$.

Finally, I never actually answered my original question. That is, does $\sinv R=0\implies0\in S$ where $R$ is a commutative ring? The $\brackets{0,2}\subset\zmod4$ from the second question doesn’t give a counterexample to this because all of its multiplicative subsets contain 0. As it turns out, the answer to this question is yes. There are no counterexamples. To see this we only need to modify our original proof slightly.

Let $R$ be a commutative ring, and let $S\subset R$ be a multiplicative set such that $\sinv R=0$. Then, $0\in S$.
Fix any $s\in S$. Then, $s/s=0/1$ so there exists some $u\in S$ such that $us=0$. However, $S$ is multiplicative, so since $s,u\in S$ we must also have $us=0\in S$.

Luckily for me, I didn’t think to make the numerator an element of $S$ when I first thought about this question; if I had, I would have missed out on the fun of the other two questions.

  1. ideals are (sub)rings when you don’t require rings to have unity 

  2. I really like examples of strange phenomona in math 

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