This is a problem from Atiyah-MacDonald Introduction to Commutative Algebra and it goes as follows:
In a ring $A$, let $\Sigma$ be the set of all ideals in which every element is a zero-divisor. Show that the set $\Sigma$ has maximal elements and that every maximal element of $\Sigma$ is a prime ideal.
My proof:
By Zorn's Lemma, a maximal ideal exists and, in particular, it is the union of ideals in a chain. So, for example, let $I = \cup_{\alpha} I_{\alpha}$ where $(I_{\alpha})$ is a chain of ideals.
We assume for contradiction that while $xy \in I$, $x \notin I$ and $y \notin I$. In that case, $x$ and $y$ are non-zero-divisors, so their product cannot be a zero-divisor.
This assumption leads to a contradiction, therefore every maximal element of $\Sigma$ must be a prime ideal.
I saw elsewhere that the proof uses a different argument, by extending the ideal and deriving a contradicon by maximaility. Is my proof valid? Is there anything I should add or elaborate on ?
solution-verificationquestion to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Jun 19 '24 at 17:51