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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?

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

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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 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.

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  • $\begingroup$ May I ask why $xr=1$ implies $rx=1$ also? Thanks $\endgroup$
    – user760
    Commented Apr 13, 2023 at 20:06
  • $\begingroup$ @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. $\endgroup$
    – rschwieb
    Commented Apr 13, 2023 at 20:51
  • $\begingroup$ Why is there a strictly descending chain, based on the fact that $rR$ is proper? $\endgroup$
    – user760
    Commented Apr 14, 2023 at 0:50
  • $\begingroup$ @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. $\endgroup$
    – rschwieb
    Commented Apr 14, 2023 at 3:27
  • $\begingroup$ 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? $\endgroup$
    – user760
    Commented Apr 14, 2023 at 5:17
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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.

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  • $\begingroup$ what about $0$? $\endgroup$ Commented May 2 at 1:19
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    $\begingroup$ @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. $\endgroup$
    – drhab
    Commented May 2 at 6:46
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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.

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

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