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Let $R$ be a commutative ring with unity. Show that $Z(R)$, the set of all zero divisors of $R$, is an ideal if and only if $R$ is a local ring.

I have no idea for proving this.

Thanks in advance!

user26857
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Shivani Goel
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  • Can I know the reason for the down vote?? – Shivani Goel Jul 25 '16 at 21:08
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    It was not me, but no effort on your side seems to show up. You can start out from the definitions. – Berci Jul 25 '16 at 21:11
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    Don't know. I upvoted, in reaction. Is a local ring noetherian, in your conventions? – Bernard Jul 25 '16 at 21:11
  • no it is not noetherian @Bernard – Shivani Goel Jul 25 '16 at 21:13
  • If you put your cursor over the downvote arrow, part of the message that appears is "This question does not show any research effort". I think this is an adequate description of your post and hence the downvote, in my opinion, is justified. – Michael Albanese Jul 25 '16 at 21:15
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    Thus it is strictly forbidden to have no idea on how to tackle a problem? – Bernard Jul 25 '16 at 21:18
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    The problem is that neither implication is true, and no attempt at context was made. Perhaps you meant this question which is useful enough, but already asked? – rschwieb Jul 25 '16 at 21:38
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    @Bernard Counter upvotes (and counter downvotes) run contrary to what votes are supposed to achieve. Re your second comment, did you ever read some howtoask page? – Did Jul 25 '16 at 23:39
  • One direction here is trivial to disprove, a nice familiar ring whose zero divisors form an ideal, yet which is not a local ring (unique maximal ideal). Definitions are your friends. – hardmath Jul 26 '16 at 00:35

1 Answers1

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The if is false.

Counter-example:

Let $K$ be a field. consider the ring $R=\bigl(K[X,Y]/(XY)\bigr)_{(X,Y)}$. This ring is local by construction, has dimension $1$, and its zero divisors is the union of its two minimal prime ideals, generated by (the images of) $X$ and $Y$ respectively.

Added (thanks to an idea of @rschwieb):

The only if part is false too:

Consider a non-local domain $D$ and a non-zero torsion-free $D$-module $M$. The set $R=D\times M$, endowed with the ring structure defined by $$\begin{cases}(d,m)+(d',m')=(d+d',m+m')\\ (d,m)(d',m')=(dd',dm'+d'm) \end{cases}$$ is a counter-example.

For ease of notation, we identify the ideal $\{0\}\times M$ with $M$. It's straightforward to check that $M^2=\{0\}$ and $Z(R)=M$. However, as $M$ is nilpotent, the spectra of $R$ and of $D$ are in bijection, hence $R$ is non-local.

Bernard
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    +1 But you should also say that the only if is false too. In any nonlocal domain, the set of zero divisors is an ideal, and yet the ring isn't local. – rschwieb Jul 25 '16 at 21:36
  • @rschwied: Perhaps the O.P. means the case where there are non-trivial zero-divisors. – Bernard Jul 25 '16 at 21:43
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    Then you just take a nonzero ideal $M$ of a nonlocal domain $D$ and form the ring structure on $D\times M$ given by $(d,m)+(d',m')=(d+d',m+m')$ and $(d,m)(d',m')=(dd', dm'+md')$ whose ideal $(0, M)$ is the set of all zero divisors, and still isn't local. – rschwieb Jul 26 '16 at 00:12
  • Fine! I'll add it, if you're not opposed. – Bernard Jul 26 '16 at 00:19
  • Not opposed at all :) – rschwieb Jul 26 '16 at 01:57
  • @rschwieb: Updated. I've slightly generalised the construction. Thanks again for your idea! – Bernard Jul 26 '16 at 09:15
  • You need to add that $M$ is torsionfree (which is why I picked a nonzero ideal) or else it is not true that $ZD(R)=(0,M)$. – rschwieb Jul 26 '16 at 10:11
  • Oh yes! I forgot this point. Fixed.Thanks! – Bernard Jul 26 '16 at 10:13