Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Ring $\mathbb{Z}$ has two units ($-1$ and $1$) and no zero divisors. –  drhab Nov 19 '13 at 18:51

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
    
Would the downvoter care to explain what is wrong with the answer? –  Brian M. Scott Nov 19 '13 at 19:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.