A non-UFD such that $a^2 \mid b^2$ does not lead to $a\mid b$

Question

Is there any non-UFD that is a commutative ring such that $a^2 \mid b^2$ does not always lead to $a\mid b$?

It would be preferable if examples are something that does not involve monomials/polynomials, and there are some integer parts in the non-UFD (for example, gaussian integers have integer parts, though gaussian integer is UFD).

score 7 · Answer 1 · answered Aug 20 '14 at 04:42

7

Let $\,a = 2,\ b=\sqrt{12}\,$ in $\,\Bbb Z[\sqrt{12}]$

answered Aug 20 '14 at 04:42

Bill Dubuque

272,048

+1 a non-maximal order is surely the simplest way to go. At least if we insist that the ring should be a domain (without that requirement the question is rather dull). – Jyrki Lahtonen Aug 20 '14 at 06:16

score 3 · Answer 2 · answered Aug 20 '14 at 07:11

3

Let $a=4$, $b=8$ in $4{\bf Z}$.

answered Aug 20 '14 at 07:11

Gerry Myerson

179,216

That's as may be, but OP specifically asked for a non-UFD. – Gerry Myerson Aug 20 '14 at 23:02

A non-UFD such that $a^2 \mid b^2$ does not lead to $a\mid b$

2 Answers2