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If $p$ and $q$ are two maximal ideals in the set of zero-divisors in a ring $R$ with non-zero intersection between $p$ and $q$. does the set of all zero-divisors are a maximal ideal and equal the union of $p$ and $q$?

Different phrasing:

Let $Z(R)$ be the set of all zero-divisors of $R$. Let $p$ and $q$ be maximal ideals contained in $Z(R)$ with $p\cap q\neq\{0\}$. Show that $Z(R)=p\cup q$.

Davide Giraudo
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nesreen
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    The union of two ideals is only an ideal if one is contained in the other, so the answer to the last part is no. – Tobias Kildetoft Mar 08 '13 at 16:03
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    Do you mean "maximal ideal contained in the zero divisors" or do you really mean an ideal maximal with respect to being contained in the zero divisors? – rschwieb Mar 08 '13 at 16:28
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    @MaisamHedyelloo: Those are completely different questions. – Jim Mar 08 '13 at 17:29

2 Answers2

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Take any field $F$ and consider the ring $R=F^3$. It has three maximal ideals, corresponding to sets of elements where are zero on some fixed coordinate. Let's call them $I_1, I_2,I_3$ depending on if they contain elements that are zero on coordinate 1,2,3 respectively. Clearly all three consist of zero divisors. $I_1\cap I_2\neq \{0\}$ since it is the set of elements zero on both coordinates 1 and 2 (but they are nonzero on 3, often.)

Now every element of $I_1\cup I_2$ is zero on either coordinate 1 or coordinate 2. But there are elements of $I_3$ (necessarily zero divisors) which do not satisfy this. Thus $I_1\cup I_2\neq ZD(R)$.

In fact, $ZD(R)=I_1\cup I_2\cup I_3$. In any commutative Artinian ring, the zero divisors are the union of all the maximal ideals.

As for the title question/second part of your question about the zero divisors being an ideal, this is only the case in a commutative Artinian ring has exactly 1 maximal ideal.

rschwieb
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  • I'm studying the complement of the zero-divisor graph and I fined this: in a commutative ring R if ZD(R) contained only two maximal ideals p and q it is useful in my work to have ZD(R)=p∪q – nesreen Mar 09 '13 at 09:41
  • @nesreen OK, but that seems pretty strong. I'm not familiar enough with commutative rings to paint a broader picture. Good luck! – rschwieb Mar 09 '13 at 18:31
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Let B= be the union of all prime ideals which consisting elements of zero-divisors. Clearly, B⊆Z(R). Now let a∈Z(R). Then (a)⊆Z(R), and (a)∩S=∅. where S is a multiplicative closed set in R. So there exists a prime ideal P such that(a)⊆P and P∩S=∅. This implies a∈P⊆B. Hence B=Z(R). #

nesreen
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  • I used the Theorem: Let I be an ideal of R and let S be a multiplicative closed set in R such that I∩S=∅, then there is an ideal P of R which is maximal with respect to the properties I⊆P and P∩S=∅, furthermore, P is a prime ideal of R. – nesreen Mar 10 '13 at 10:46
  • I don't think anything you said establishes that P consists entirely of zero divisors, so the last line looks unjustified. – rschwieb Oct 01 '13 at 10:32
  • It's well known that the zero divisors are a union of prime ideals : http://math.stackexchange.com/q/44481/29335 – rschwieb Oct 01 '13 at 10:34