# Generators of Principal Ideals and Unit Association

Posted by Jason Polak on 16. August 2012 · 4 comments · Categories: commutative-algebra · Tags: , ,

Let us ponder a familiar property of any integral domain $R$. Suppose that $a,b\in R$ are such that $(a) = (b)$, where $(a)$ denotes the ideal generated by $a$. Then $a = rb$ and $b = sa$ for some elements $r,s\in R$. Combining these two facts together give $a = rsa$, or $(1 – rs)a = 0$. If $a\not = 0$, this means that $1 – rs = 0$ or that $r$ and $s$ are units. Thus $a$ is a unit multiple of $b$. Of course if $a = 0$ then $b= 0$ and $a$ is still a unit multiple of $b$.

Now, the question is, what happens if we drop the assumption that $R$ is an integral domain? If $(a) = (b)$, can we still prove that $a$ is a unit multiple of $b$? Let us still assume that $R$ is commutative. Of course, the above proof for integral domains can’t work because we used the defining property of integral domains to conclude that $1 – rs = 0$. In this post we shall see a counterexample to show that two generators of the same principal ideal do not necessarily have to be unit multiples of each other.

I first heard of this question in Hershy Kisilevsky‘s algebraic number theory class. I forgot about it for some time until I came across Question 101989 on MathOverflow that gave a reference for a paper that was purported to have a counterexample showing that one cannot prove this statement for arbitrary commutative rings.
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