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I am wondering how to go about proving this,

Let $R$ be a commutative ring with identity such that not every ideal of $R$ is principal.

A) Use Zorn's lemma to show that $R$ has an ideal $J$ such that (i) $J$ is not principal ideal (ii) $J$ is not properly contained in any non-principal ideal.

B) Show that $R/J$ is a principal ideal ring (where $J$ is the ideal from part (a)).

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    $\begingroup$ Well, for (i) you use Zorn’s Lemma. And for (ii) you use the fact that every ideal that contains $J$ in $R$ is principal. This really is just a matter of “follow your nose.” $\endgroup$
    – Arturo Magidin
    Mar 8, 2019 at 5:41

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Follow the hint in (A) and use Zorn. You need the fact that the union of a chain of non-principal ideals is non-principal. If the union $J$ were principal, then it would have a generator, which would lie in some ideal $I$ of the chain, but then $I=J$ would then be principal.

in (B) all ideals containing $J$ (other than $J$) are principal, so reduce to principal ideals in $R/J$. Of course, $J$ also reduces to a principal ideal in $R/J$.

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  • $\begingroup$ For (B), Suppose $\tilde{I}$ is an ideal of $R/J$. By Lattice Ring Isomorphism Theorem, $\tilde{I} = I/J$ where $J \subset I$. Since $J$ is not properly contained in any non-principal ideal, $I$ is principal and $(a) = I$ for $a \in R$. And for $x \in I/J$, $x = ra + J$ for some $r \in R$. Therefore, $\tilde{I}$ is a principal ideal. Since $\tilde{I}$ is an arbitrary ideal, $R/J$ is principal ideal ring. $\endgroup$
    – user168915
    Mar 9, 2019 at 20:45

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