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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?

rschwieb
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sonicboom
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3 Answers3

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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 then 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.

rschwieb
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  • May I ask why $xr=1$ implies $rx=1$ also? Thanks – user760 Apr 13 '23 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. – rschwieb Apr 13 '23 at 20:51
  • Why is there a strictly descending chain, based on the fact that $rR$ is proper? – user760 Apr 14 '23 at 00:50
  • @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. – rschwieb Apr 14 '23 at 03:27
  • 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? – user760 Apr 14 '23 at 05:17
  • Perhaps it's better to start from the chain $rR\supseteq r^2R\supseteq r^3R\supseteq\dots$. Then, since the chain must stabilize after finitely many steps, there is a $k\in\mathbb{N}$ and $r'\in R$ such that $r^k=r^{k+1}r'$. Then we have $1=x^kr^k=x^kr^{k+1}r'=rr'$. But since $xr=1$, we have $rx=rxrr'=r1r'=rr'=1$. – user760 Apr 14 '23 at 06:19
  • Does that mean in a (left/right) Artinian ring, principal (left/right) ideals are always the whole ring? Of course it does not. When $r$ is not a right regular, one cannot expect the chain to keep shrinking. For example, if $r$ is an idempotent, then $r^2R\subseteq (rR)^2=rR$ and the chain is stable without being $R$. – rschwieb Apr 14 '23 at 11:21
  • @user760 Why do you say $1=x^kr^k$? It would be true if $x$ and $r$ commuted, but if that were the case it'd be trivial that $xr=rx=1$. – rschwieb Apr 14 '23 at 11:28
  • Right. I didn't realize $(xr)^k=x^k r^k$ only holds with commutativity. But for the chain of ideals, I still don't fully understand. Do you mean, since $xr=1$, then $xrR=1R=R$, and so $rR\simeq xr R=R$, where "$\simeq$" refers to bijectivity? And this bijectivity implies $rR\supseteq r^2R\supseteq r^3R\supseteq\dots$ descends indefinitely? – user760 Apr 14 '23 at 13:29
  • Let us continue this discussion in chat. – rschwieb Apr 14 '23 at 13:59
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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.

drhab
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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.

Brian M. Scott
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