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I'm asked to prove that the set consisting of all zero divisors in a commutative ring with unity contains at least one prime ideal.

I can't even start in the proof, I've just defined my set but cant move on construction the ideal !

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  • $\begingroup$ Nice Question.. Integral domain has $(??)$ as prime ideal... Can you see how and can you think of what would be the case of general commutative ring? $\endgroup$ – user87543 Jan 4 '14 at 16:59
  • $\begingroup$ @PraphullaKoushik for integral domains (p) is a prime ideal iff p is a prime number, then? $\endgroup$ – Enas Jan 4 '14 at 17:10
  • $\begingroup$ Check to see that your set is an ideal, then use the fact that $R/I$ is an integral domain if and only if $I$ is prime. By the way I think there are integral domains with prime ideals not generated by a single prime. It holds for principal ideal domains though. $\endgroup$ – user111013 Jan 4 '14 at 17:19
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One way to approach this is to combine these two lemmas:

  1. In a commutative ring, there exist minimal prime ideals.

  2. In a commutative ring with identity, minimal primes consist entirely of zero divisors (I use the convention that $0$ is a zero divisor here.)

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  • $\begingroup$ If you don't mind me asking a follow-up question: is it true or false that an ideal which is maximal with respect to being contained in the set of zero divisors is necessarily prime? That's what I started trying to prove... I can post a new question if that's preferred, but I figured I might as well ask you first. $\endgroup$ – Dustan Levenstein Jan 4 '14 at 17:33
  • $\begingroup$ @DustanLevenstein : I keep thinking about this extended question, but I don't have a conclusion either way. A counterexample would have to lie outside of reduced Noetherian rings. In those rings, the zero divisors are a union of prime ideals, and then prime avoidance tells you that an ideal maximal w.r.t. being all zero divisors is exactly one of these ideals. $\endgroup$ – rschwieb Jan 5 '14 at 0:21
  • $\begingroup$ Posted a question. $\endgroup$ – Dustan Levenstein Jan 5 '14 at 4:47
  • $\begingroup$ Another more detailed approach that I ran across later. $\endgroup$ – rschwieb Apr 27 '17 at 13:28
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Let $Z$ be the set of all zero divisors in a commutative ring $R$. Note that $R \setminus Z$ is closed under multiplication.

Indeed, take $a, b \in R \setminus Z$ and assume $ab \in Z$. Then there exists nonzero $c \in R$ such that $$0 = (ab)c = a(bc)$$

If $bc \ne 0$ then $a(bc) = 0$ implies that $a$ is a zero divisor. On the other hand, if $bc = 0$ then $b$ is a zero divisor. A contradiction.

Now consider the family of all ideals in $R$ contained in $Z$:$$\mathcal{S} = \{I \trianglelefteq R : I \subseteq Z\}$$

Order $\mathcal{S}$ by inclusion. We have $\{0\} \in \mathcal{S}$, and for every chain $\mathcal{L}$ in $\mathcal{S}$ we have that $\bigcup \mathcal{L}$ is an upper bound for $\mathcal{L}$ which is contained in $\mathcal{S}$. Zorn's lemma implies that $\mathcal{S}$ has a maximal element $P$. We claim that $P$ is prime.

Let $A, B$ be two ideals in $R$ such that $AB \subseteq P$. Assume $A, B \not\subseteq P$. Then $P + A$ and $P + B$ are two ideals in $R$ which strictly contain $P$. Therefore, $P + A$ and $P + B$ intersect $R \setminus Z$. Let $p_1 + a \in (P + A) \cap R \setminus Z$ and $p_2 + b \in (P + B) \cap R \setminus Z$.

Since $R \setminus Z$ is closed under multiplication we have:

$$R \setminus Z \ni (p_1 + a)(p_2 + b) = p_1p_2 + p_1b + ap_2 + ab \in P$$

which is a contradiction since $P \subseteq Z$.

Therefore $P$ is a prime ideal of $R$ contained in $Z$.

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