# Ring such that $q^2\mid p^2$ does not imply $q\mid p$?

Let $R$ be a commutative ring with $1$ and suppose $q^2\mid p^2,$ for $p,q \in R$. Unless $R$ is a UFD, I don't believe I can conclude that $q\mid p,$ but I would like to know a concrete counterexample.

Does anyone know an example of such a ring, where there exist elements $p,q$ such that $q^2$ divides $p^2$ but $q$ does not divide $p$?

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$2^3\mid (2^2)^2$ but $2^3\nmid 2^2$ –  blue Jul 20 '14 at 7:30
It would be more reasonable to ask if $q^2|p^2$ implies $q|p$. (Namely, this holds in a UFD.) It doesn't hold in $\mathbb{C}[x,y,z]/(zx^2=y^2)$. There are also lots of more trivial examples, for example when $p^2=0$ and $p \neq 0$. –  Martin Brandenburg Jul 20 '14 at 7:43
@Martin Brandenburg: Yes, thanks. That's what I meant to ask. Apologies for the confusion, and thanks for the answer. (Question edited) –  user142700 Jul 20 '14 at 7:45

In $\mathbb{Z}/4$ take $q=0$ and $p=2$.

For an example of an integral domain, take $\mathbb{Z}[x,y]/(x y^2-4)$ and $p=y$, $q=2$.

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Just a remark: you need much less than a UFD to conclude that $p^2|q^2 \implies p \mid q$.

Namely, if $A$ is an integrally closed domain then this holds.

(If $q^2 = a p^2$ for some $a,p,q \in A$ with $p \neq 0$, then $(q/p)^2 = a$, so --- by integral closeness of $A$ --- we see $\alpha := q/p \in A$, so $q = p \alpha$ and hence $p | q$ in $A$.)

Note that UFD's are integrally closed, and so the case of a UFD follows from this one.

Note also that Martin Brandenburg's example in comments, namely $\mathbb C[x,y,z]/(zx^2 - y^2)$, corresponds to a surface in $\mathbb A^3$ whose singular locus is an entire curve (the line $x = y = 0$), which is the basic geometric cause of an affine ring being non-integrally closed. (Integrally closed implies that the singular locus has codimension at least two.)

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