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