# $R$ commutative ring with 1 and not every ideal is principal. Prove $R$ has ideal that is not principal.

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)).

• 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.” Mar 8, 2019 at 5:41

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 $$R/J$$.
• 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. Mar 9, 2019 at 20:45