12

Suppose we have a ring $R$ and $(a),(b)$ are both ideals of $R$. Is it always true that $(a)=(b)$ if and only if there exists a unit $c$ such that $a=bc$ (i.e., $a$ and $b$ are associate)?

I have already verified that the forwards direction is true. But I have no idea on the backward direction. If it is true, can someone provide a proof to me?

Backward direction: Suppose that there exists a unit $c \in R$ such that $a=bc$. This implies that $(a)\subset (b)$. By using the same thing, (i.e., $b$ and $a$ are associate), there exists a unit $d \in R$ such that $b=ad$. This implies that $(b) \subset (a)$. is this correct?

Idonknow
  • 15,643
  • 2
    No. You need to prove it yourself. You have asked like 10 questions here, and have shown very little work. Let me provide a hint: (a)=(b) if and only if $a\mid b$ and $b\mid a$. – Alex Youcis Apr 09 '13 at 14:49
  • @AlexYoucis: So you mean the statement is false ? – Idonknow Apr 09 '13 at 14:52
  • 4
    @Idonknow No. The statement is true, I'm just saying that you need to take a more serious crack at if, or if you have, show us. I mean, come on, we aren't homework jockeys just sitting here waiting to answer other people's homework. We'd like to help you, but we'd also like to see that you're attempting to learn/improve in the process. – Alex Youcis Apr 09 '13 at 14:54

2 Answers2

21

This is easily proved true if $\rm\:a,b\:$ are not zero-divisors. It may fail otherwise, e.g. Kaplansky gave this counterexample: in the ring of continuous real functions on $[0,3],$ define piecewise $\rm\:a(t)\:$ and $\rm\:b(t)\:$ as follows: $\rm\: a(t) =1\!-\!t=b(t)\:$ on $[0,1];$ both $\,0\,$ on $[1,2];$ $\rm\:a(t)=t-2 = -b(t)$ on $[2,3].$

Kap remarked that the property is true for Artinian rings, principal ideal rings, and rings whose zero-divisors are contained in the Jacobson radical. For much further analysis see

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.

Math Gems
  • 19,574
10

This is not quite true over arbitrary (commutative) rings, although it holds over integral domains. From the given condition you get elements $u,v$ so that $b=ua$, $a=vb$ and so $b=uvb$ and $a=vua$. When $a$ or $b$ is a regular element, as always happens in an integral domain (with the exception of the trivial case $a=b=0$) then one can cancel it to conclude that $u$ and $v$ are inverses of each other.

This fails however in more general rings. As happens often with such statements, one can easily force an example by dividing out suitable things from a polynomial ring. Take $R=k[a,b,u,v]/(a-vb,b-ua)$ for some field (or ring) $k$, now we have forced $aR=bR$. However neither $u$ nor $v$ are invertible, as the evaluation $a:=0,b:=0$, which induces a morphism $R\to k[u,v]$ where the images of $u,v$ are non-invertible, shows.

Marc van Leeuwen
  • 115,048
  • A proof that the "generic" counterexample in your second paragraph works can be found in Theorem 8 of the paper I linked to (for any coefficient ring). They attribute to C.R. Fletcher (1970) the case when the coefficient ring is a field. But probably it is much older, since generic counterexamples are the first thing one tries (and such ideas are very old). – Math Gems Apr 09 '13 at 15:42
  • But your (Marc) example is different from the one in the paper (where as of now, it is Theorem 9(2) by the way). And forgive me for doubts on this one: Surely (the images of) $u$ and $v$ are not invertible in $R$, but we have to show more, namely, that there is no unit $w$ at all such that $a=wb$. (So far we only know that the obvious candidate is no unit.) In the example in the paper, there is a non-trivial proof for this, which does not immediately work for the $R$ here. – Torsten Schoeneberg Oct 25 '17 at 21:10
  • 1
    @Torsten Yes it does seem there is a bit more to prove. I'd need to think about this... – Marc van Leeuwen Nov 01 '17 at 14:08
  • @Marc Have you found something? Otherwise I would turn it into a new question. I would really like to know if this obvious "tailor-made" counterexample works or not. – Torsten Schoeneberg Jan 25 '18 at 07:39
  • 1
    Go ahaid, ask the question; this is not on on my mind at all right now. – Marc van Leeuwen Jan 25 '18 at 08:55
  • OK. I asked it, and user26857 pointed out that your example is actually isomorphic to the one in the paper. Which seems obvious to me now. Sorry to have bothered you with this. – Torsten Schoeneberg Jan 25 '18 at 23:59