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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