Are associates unit multiples (does $a\mid b\mid a \Rightarrow a = ub\,$ for a unit $u$)

Question

Let $R$ be a commutive ring with $1$. Recall the following relevant definitions.

• $b$ is divisible by $a$ in $R$, denotes $a\mid b$ in $R$, if $b = r a$ for some $r\in R$.

• $a$ and $b$ are associates in $R$ if $a\mid b$ and $b\mid a$, i.e. the ideals $aR = bR$.

• $u\in R$ is a unit if it is invertible ($uv=1$ for some $v\in R).$

• $a$ and $b$ are unit multiples in $R$ if $a = ub$ for some unit $u\in R$.

Given these definitions, my question is,

in $R,\,$ are associates $a,b$ unit multiples, i.e. does $a\mid b\mid a\Rightarrow a=ub$ for a unit $u$?

I was told that this not always true. But I encountered some difficulties in finding a counterexample.

Answer 1

See the following paper,

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.

It is mostly concerned with finding sufficient conditions on commutative rings that ensure that $Ra=Rb$ implies $a = bu$ for a unit $u$, but they do give some examples of $R$ where this fails. In particular this simple example of Kaplansky. Let $R=C[0,3]$, the set of continuous function from the interval $[0,3]$ to the reals. Let $f(t)$ and $g(t)$ equal $1-t$ on $[0,1]$, zero on $[1,2]$ but let $f(t)=t-2$ on $[2,3]$ and $g(t)=2-t$ on $[2,3]$. Then $f$ is not a unit multiple of $g$ in $R$ but each divides the other.

Answer 2

This is true if $R$ is a domain. More generally this is true if $a$ or $b$ is not a zero divisor in $R$. Suppose, $a$ is not a zero divisor and there exist $r,s\in R$ such that $a=rb$ and $b=sa$, then $a=rsa$, so $a(1-rs)=0$. Since $a$ is not a zero divisor, $1-rs=0$, so $rs=1$. So, $r,s$ are units. So $a$ and $b$ are associates.

Answer 3

It is a common oversight to assume that generally - as in domains - associates are unit multiples. To help remedy this, on earlier math forums I often cited the little-known below $2004$ survey $[1]\, $ (cf. sci.math $\text{Oct 15, 2008}$ in google groups and Ask an Algebraist 2008, and mathforum, etc). It contains a few simple counterexamples, excerpted below. The first two appeared in a seminal $1949$ paper $[2]$ by Kaplansky on Hermite rings. The third (mentioned by Fletcher $1969\ [3])$ employs a ring with a pair of universal associates $\,x\sim xy\:$ (so $\,xy\mid x,\,$ say $\,xyz = x)$. Fletcher gave no proof. Below we give a couple proofs (which easily extend to any commutative ring $D$, cf. 2nd $(2)$ below).

Theorem $ $ If $D$ is a domain, and $\,x,y,z\,$ are the images of $\,X,Y\!,Z\,$ in $D[X,\!Y\!,\!Z]/(X\!-\!XYZ)$ then $\,xy\sim x\,$ but $\,xy\neq xy_1\,$ for any unit $\:\!y_1,\,$ i.e. $\:\!xy\,$ is not a unit multiple of its associate $\:\!x$.

Proof $\ x = xyz,\,$ so $\,x\mid xy\mid x,\,$ so $\,x\sim xy.\,$ Assume $\,xy_1 = xy,\,$ so $X\!-\!XYZ\mid XY_1\!-\!XY\:\!$ so cancelling $X\Rightarrow\,1\!-\!YZ\mid Y_1\!-\!Y,\,$ which means $\,\color{#0a0}{Y_1 = Y + (1\!-\!Y\color{darkorange}Z)\:\!H},\,$ for some $\,H,\,$ so $$\begin{align} y_1\:\!\ {\rm unit}\iff (1)=(y_1) \iff&\ (1) = (\color{#0a0}{Y_1},\,X\!-\!XYZ)\ \ \ {\rm in}\,\ D[X,\!Y\!,\!Z]\\[.4em] {\rm If\:\!\ so,\ eval\ at}\,\ X\!=\!0,\,\color{darkorange}Z\!=\!\color{#0a0}Y\ \Rightarrow\: &\ \color{c00}{(1)} = (\color{#0a0}{Y\!+\!(1\!-\!Y^2)\:\!\color{#0cf}H})\,\ {\rm in}\:\!\ D[Y]\\[.4em] {\rm thus}\:\!\ {\color{#c00}{\deg}_{\:\!Y^{\phantom{|^.}}\!\!\!} \color{#0a0}{Y_1} = 0},\ {\rm so}\,\ \color{#0cf}H\!=\!0\, \Rightarrow\: &\ (1) = (0)\:\!\ {\rm by\ eval\ above\ at}\ \color{#0a0}Y\!=\!0\ \Rightarrow\!\Leftarrow \end{align}\qquad\quad\ $$

by $Y_1$ unit in $D[Y]\Rightarrow \color{#c00}{\deg} Y_1 = 0\,$ by here, by $D$ domain. [Alternatively: eval at $\:\!X\!=\!0\:\!$ shows $Y_1$ is a unit in $D[Y,Z]$ so constant, i.e. $Y_1\in D\,$ (by prior link). But $Y_1 = Y\!+\!(1\!-\!YZ)H\:\!$ is not constant since: $ $ if $\:\!H\neq 0\,$ then $\,\deg_Z Y_1 = 1\!+\!\deg_Z H \ge 1,\,$ else $H=0\,$ so $\,Y_1 = Y\not\in D\:\!$].

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Below are said three counterexamples excerpted from $[1]$.
Notation: $\,a\sim b\,$ means $\,a,b\,$ are associates, and $\,a\approx b\,$ means they are unit multiples.

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Beware in general rings there are a few different interesting notions of "associate", e.g.

• $\ a\sim b\ $ are $ $ associates $ $ if $\, a\mid b\,$ and $\,b\mid a$
• $\ a\approx b\ $ are $ $ strong associates $ $ if $\, a = ub\,$ for some unit $\,u\ \,$ (a.k.a. unit multiples)
• $\ a \cong b\ $ are $ $ very strong associates $ $ if $\,a\sim b\,$ and $\,a\ne 0,\ a = rb\,\Rightarrow\, r\,$ unit

See $[1]$ for much further discussion. See also the survey linked here for the effect that this bifurcation has on the notion of unique factorization ring and related matters.

$[1]\,$ D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
When are Associates Unit Multiples?
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.

$[2]\,$ I. Kaplansky. Elementary divisors and modules, Trans. Amer. Math. Soc. 66 (1949), 464-491

$[3]\,$ C.R. Fletcher, Unique factorization rings, Proc. Cambridge Philos. Soc. 65 (1969), 579-583

Answer 4

For a counterexample, I will use the ring $R := \mathbb{Z}[a, b, r, s] / \langle b - ra, a - sb \rangle$. In this ring, we certainly have $b = ra$ and $a = sb$. It remains to show there is no unit $u \in R^*$ such that $b = ua$.

Thus, suppose we did have $b = ua$. Then $a(u-r) = 0$. On the other hand (and here is where the argument gets somewhat technically involved), it can be shown that the kernel of the $R$-module homomorphism $a\cdot : R \to R$ is generated by $rs - 1$. Therefore, there exists $p \in R$ such that $u = r + p(rs - 1)$. Also, if $u$ is a unit, then the image $\bar u$ of $u$ in $R / \langle a, b \rangle \simeq \mathbb{Z}[a,b,r,s] / \langle b - ar, a - sb, a, b \rangle = \mathbb{Z}[a,b,r,s] / \langle a, b \rangle \simeq \mathbb{Z}[r,s]$ would also have to be a unit. But the only units of $\mathbb{Z}[r,s]$ are the constant polynomials $\pm 1$, whereas with $\bar u = r + \bar p (rs - 1)$ we would need to have $\bar u(1, 1) = 1 \ne -1 = \bar u(-1, -1)$ if we consider $\bar u$ as inducing a function $\mathbb{Z}^2 \to \mathbb{Z}$. This gives a contradiction, so the assumption that $u$ is a unit must have been false.

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The basic idea here is that we are in some sense taking the "universal example" of a ring with two associate elements $a,b$ (though with the caveat that the elements $u,v$ making the elements associate are also distinguished, rather than simply assumed to exist somewhere in the ring; and also that the actual universal example would be $\mathbb{Z}[a,b,r,s] / \langle b-ra, a-sb \rangle$ whereas I used $\mathbb{Q}$ for a slightly easier time with the technical details I omitted). If we had found a unit within $\mathbb{Q}[a,b,r,s] / \langle b-ra, a-sb \rangle$ making $a$ and $b$ associates, then that expression would have worked in particular for any $\mathbb{Q}$-algebra. On the other hand, once we instead found that there is no unit in this universal example, we already had our counterexample that we needed.

(On the other hand, this shows that trying to work directly with the universal example might not always be the best or easiest approach. In this example, we had some mildly hairy calculations, ultimately based on Groebner basis theory, that were needed. However, in the counterexample based on $R = C[0,3]$ in another answer, it was more straightforward to check the details.)

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Addendum on the technical detail omitted above, verifying the fact in Macaulay2:

Macaulay2, version 1.16
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases, MinimalPrimes, PrimaryDecomposition, ReesAlgebra, TangentCone, Truncations
i1 : R = ZZ[a,b,r,s] / (b - ra, a - sb)
o1 = R
o1 : QuotientRing
i2 : mult_a = map(R^1, R^1, matrix {{ a }})
o2 = | a |
         1       1

o2 : Matrix R  <--- R
i3 : kernel(mult_a)
o3 = image | rs-1 |
                         1

o3 : R-module, submodule of R

Answer 5

Let me add this historic counterexample from I. Kaplansky: Elementary divisors and modules. Trans. Amer. Math. Soc. 66 (1949), 464-491 (page 466, example (b)).

Take the (Noetherian!) subring $R \subset \mathbb Z \times \mathbb F_5[x]$ consisting of all $(z,f)$ with $f(0) = z\bmod 5$. The elements $a = (0,x)$ and $b = (0,2x)$ are associates by $b = 2a$ and $a = 3b$. The unit group of $R$ consists only of the elements $\pm 1$, so $a$ and $b$ are no unit multiples.