TITLE: Isomorphic factor rings of polynomial rings does imply isomorphic ideals?
QUESTION [9 upvotes]: Let $k$ be a field, $I$ and $J$ are ideals of $R=k[x_1,\dots,x_n]$. If $R/I\simeq R/J$ as rings, then $I \simeq J$ as $R$-modules holds?
Thanks in advance!

REPLY [8 votes]: The claim is not true.  A counterexample:
Let $R=k[w,x,y,z]$, $I=(w,x)$, and $J=(y,z)$.  Notice that $R/I\cong R/J$ as rings.  Now, suppose that $I\cong J$ as $R$-modules and let $f:I\to J$ be an isomorphism.
Then $0=f(xw-wx)=xf(w)-wf(x)$ so that $xf(w)=wf(x)$.  By unique factorization in $R$, it follows that $f(w)=wp$ and $f(x)=xp$ for some $p\in R$.  Since $f$ is an isomorphism, it is surjective, so there must exist $a,b\in R$ with:
$$y=f(aw+bx)=af(w)+bf(x)=(aw+bx)p$$
Looking at degrees, we arrive at a contradiction.