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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 Nov 19 '13 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 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.

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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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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? $\endgroup$ – Brian M. Scott Nov 19 '13 at 19:45

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