Isomorphism in UFD

Question

Is this statement true:

Let $R$ be a UFD and $p,q\in R$ are irreducible which are not associated. Prove that for all $n,m\in\mathbb N$: $$R/(p^mq^n)\cong R/(p^m)\times R/(q^n).$$

I don't think that the chinese remainder theorem works here, however, it is easy to construct a map $f$ from $R$ to $R/(p^m)\times R/(q^n)$ with $\ker f=(p^mq^n)$. The only problem is to prove the map is onto. I'll appreciate any help.

Answer 1

@JCAA probably gave the best answer to this question in comments. However if you want further elaboration read on:

Let $R=\mathbb{R}[x,y]$. Then $R$ is a UFD, and $x,y$ are irreducible, and not associates. If some element of $w\in R$ mapped to $(0,1)$ in $R/(x)\times R/(y)$ then $$w=ax=1+by$$ for some $a,b\in R$. Then we would have $ax-by=1$, which is impossible as $(x),(y)$ are not coprime.

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Edit:

The above just proves that your map is not an isomorphism. To show that the rings are not isomorphic, note that $R/(x)\times R/(y)$ contains the idempotent $(1,0)$, which is non-trivial (not $(0,0)$ or $(1,1)$). To see that $R/(xy)$ has no such idempotent, note that elements of $R/(xy)$ may be uniquely written in the form $\alpha=c+xp(x)+yq(y)$, where $p,q$ are polynomials over $\mathbb{R}$ and $c\in \mathbb{R}$. Let $$\alpha^2=c'+xp'(x)+yq'(y),$$ for a constant $c'$ and polynomials $p',q'$.

Then if $\alpha^2=\alpha$ we have $\deg(q')=\deg(q)$, and $\deg(p')=\deg(p)$. Thus $q=p=0$ and $\alpha=0$ or $1$.