# Rings whose elements are partitioned between units and zero-divisors.

In $\mathbb{Z}_n$ the elements are fully partitioned between the units and the zero-divisors. I believe this is the case, am I correct?

Now, I take it this does not hold true in general, there may be rings with elements that are neither units nor zero divisors?

• Ring $\mathbb{Z}$ has two units ($-1$ and $1$) and no zero divisors. Commented Nov 19, 2013 at 18:51

You're correct in this case, and more generally elements in Artinian rings are either units or are zero divisors. It's not hard to prove: basically you can show that if $$x$$ isn't a zero divisor, then the chain $$xR\supseteq x^2R\supseteq\dots$$ has to stabilize, whence there will be an $$r$$ such that $$x^n=x^{n+1}r$$. Rewriting that, you get $$x^n(xr-1)=0$$. If $$x$$ isn't a zero divisor, then the $$x^n$$ can be cancelled, resulting in $$xr=1$$, so that $$x$$ is a unit.

Any commutative domain which isn't a field has LOTS of nonunits which aren't zero divisors. So for example $$\Bbb Z$$ has two units $$\{\pm1\}$$, zero, and the rest of the elements are not zero divisors.

• May I ask why $xr=1$ implies $rx=1$ also? Thanks Commented Apr 13, 2023 at 20:06
• @user760 Let's suppose $R$ is simply right Artinian. Now $xr=1$ tells us that $rR$ is an isomorphic copy of $R$ inside of $R$. If $rx\neq 1$ then it is a nontrivial idempotent, and $rxR\subseteq rR=rxrR\subseteq rxR$ shows that $rR$ would be a proper right ideal of $R$. But now you have a strictly descending chain $rR\supseteq r^2R\supseteq r^3R\supseteq\ldots$. This cannot happen, so $rx=1$ after all. Obviously it works if you just assumed "left Artinian" symmetrically. Commented Apr 13, 2023 at 20:51
• Why is there a strictly descending chain, based on the fact that $rR$ is proper? Commented Apr 14, 2023 at 0:50
• @user760 If $R$ contains a copy of itself properly, then that copy contains another proper copy, and so on. That's what the chain is doing. Commented Apr 14, 2023 at 3:27
• Hmmm. Does that mean in a (left/right) Artinian ring, principal (left/right) ideals are always the whole ring? Otherwise, $Ra$ or $aR$ or $RaR$ are always a copy of $R$, right? Or am I missing something? Commented Apr 14, 2023 at 5:17

Ring $\mathbb{Z}$ has two units ($-1$ and $1$) and no zero divisors.

So actually every element $n$ in it with $n\notin\left\{ -1,1\right\}$ is neither a unit nor a zero-divisor.

• what about $0$? Commented May 2 at 1:19
• @J.W.Tanner I confess that it depends on definition. I have seen both: $0$ is a trivial zero-divisor or $0$ is no zero-divisor. Commented May 2 at 6:46

An example is the ring $\Bbb R[[X]]$ of formal power series over $\Bbb R$: it has no zero divisors, but an element $\sum_{n\ge 0}a_nX^n$ is invertible if and only if $a_0\ne 0$. Thus, every non-zero power series of the form $\sum_{n\ge 1}a_nX^n$ is neither a unit nor a zero divisor.

• Would the downvoter care to explain what is wrong with the answer? Commented Nov 19, 2013 at 19:45