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I'm looking at the exercise 14 (page 12) of Atiyah and Macdonald's Introduction to Commutative Algebra:

In a ring A, let $\Sigma$ be the set of all the ideals in which every element is a zero-divisor. Show that $\Sigma$ has maximal elements and that every maximal element of $\Sigma$ is a prime ideal.

Since one can apply Zorn's lemma to the set of all the ideals of $A$, one can apply it to the ideals in $\Sigma$ too, and prove that it has maximal elements. However I don't understand how these maximals ideals can be prime: every ideal in $A$ (and so every maximal element in $\Sigma$) is contained in a maximal ideal; it follows that every maximal element of $\Sigma$ is the intersection of a maximal ideal with the ideal of all the zero divisors of $A$. Since a prime ideal cannot be obtained as the intersection of two ideals, this is absurd. Where am I wrong in this reasoning? Thanks in advance

Bernard
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Dr. Scotti
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  • But a prime ideal can be the intersection of two ideals: Consider for example the intersection of such an ideal with itself. – asdq Sep 28 '20 at 14:59
  • @Dorian - As I understand it, the set of zero divisors is the union of the radicals of the annihilators of the nonzero elements of your ring (Prop. 1.15 of my edition of Atiyah-Macdonald). The maximal elements of the set (your $\Sigma$) are prime ideals. Let $I=ann(x)$ be a maximal element and $ab\in I.$ If $a$ is not in $I$, the $ax \neq 0.$ Since $ann(ax) \supseteq I,$ we must have $ann(x)=I,$ since $I$ is maximal. But $b(ax)=0,$ so $b \in I.$ This argument can be found in the proof of Theorem 6 in Kaplansky's "Commutative Rings." – Chris Leary Sep 28 '20 at 15:14

2 Answers2

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In ${\Bbb Z}/6{\Bbb Z}$, $2$ and $3$ are zero divisors. However, $3-2 = 1$ is not.

J.-E. Pin
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I don't believe the set of all zero-divisors is an ideal.

E.g. in $k[x,y]/(xy)$ this set is $(x) \cup (y)$ but for instance, $x + y$ is not a zero divisor.

Sera Gunn
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  • Thanks, so I reasoned like this. Let $R$ be a commmutative ring and $S\subset R$ be the subset of the zero-divisors. If $I\unlhd R$ and $I\subset S$, and $I$ is a maximal element in $\Sigma$, one must prove that $I$ is prime. So, suppose that $I$ is not prime, and that there are two elements $a,b\in R-I$ such that $ab \in I$. This means that there is a $c\in R$ such that $abc=0$, and so either $a$ or $b$ is a zero-divisor (suppose $b$). – Dr. Scotti Sep 29 '20 at 14:21
  • Consider the ideal $J$ generated by $I$ and $b$: I must prove that $J$ is contained in $S$. So, take any $i\in I$ and consider $b+i$: by the hypothesis $ab+ai\in I$, and since $a$ is not necessarily a zero-divisor, $b+i$ must be in $S$. It follows that $J$ is an ideal contained in $S$ which contains $I$, in contrast whith the hypothesis that $I$ is maximal in $\Sigma$. Is this right? Thanks again – Dr. Scotti Sep 29 '20 at 14:21
  • @Dorian It's not phrased perfectly. You want to assert that $c \neq 0$ for one. Then to show that $b + i$ is a zero divisor you need to know that whatever you're multiplying by is non-zero. So for instance if $zi = 0$ then $acz(b + i) = abci + aczi = 0$ but we don't know that $acz \neq 0$. I think this can be managed though if you work carefully. There are proofs here https://math.stackexchange.com/q/44481/437127 which are more straightforward I think. – Sera Gunn Sep 29 '20 at 16:46