Exercise 1.14

Proof. We order $\mathrm{\Sigma }$ by inclusion; $\mathrm{\Sigma }\ne \varnothing$ since $(0)\in \mathrm{\Sigma }$. By Zorn’s lemma, to show $\mathrm{\Sigma }$ has maximal elements it suffices to show for every ascending chain $\{{𝔞}_{\alpha }\}$ in $\mathrm{\Sigma }$, the upper bound $𝔞={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha }{𝔞}_{\alpha }$ is in $\mathrm{\Sigma }$. $𝔞$ is an ideal since if $x,y\in 𝔞$, they are both in some ${𝔞}_{\alpha }$, and so $x+y\in {𝔞}_{\alpha }\subseteq 𝔞$; similarly, if $x\in 𝔞$, it is in some ${𝔞}_{\alpha }$, and so $\mathit{ax}\in {𝔞}_{\alpha }\subseteq 𝔞$ for any $a\in A$. Moreover, $𝔞\in \mathrm{\Sigma }$ since every element $x\in 𝔞$ must be contained in some ${𝔞}_{\alpha }$ and so is a zero-divisor.

We now show these maximal $𝔞\in \mathrm{\Sigma }$ are prime. So, suppose $\mathit{xy}\in 𝔞$ but $x,y\notin 𝔞$. Then, $𝔞⊊(𝔞:x)$ since $y\notin 𝔞$. By maximality, $(𝔞:x)$ contains some $z$ which is not a zero-divisor. We now claim that $(𝔞:z)\in \mathrm{\Sigma }$. If $0\ne w\in (𝔞:z)$, then $\mathit{wz}\in 𝔞$, hence there exists $v\in A$ such that $\mathit{vwz}=0$. Since $z$ is not a zero-divisor, $\mathit{vz}\ne 0$, hence $w$ is a zero-divisor. But $𝔞⊊(𝔞:z)$ since $x\in (𝔞:z)$, contradicting the maximality of $𝔞$.

Now we show that the set of zero-divisors in $A$ is a union of prime ideals. If $x\in A$ is a zero-divisor, then $(x)\in \mathrm{\Sigma }$. By repeating the argument in the first paragraph to the subset of $\mathrm{\Sigma }$ of ideals containing $(x)$, we see that there is a maximal $𝔞\in \mathrm{\Sigma }$ such that $x\in 𝔞$. Thus, the set of zero-divisors in $A$ is the union of maximal elements in $\mathrm{\Sigma }$, which are prime ideals. □