Conditions on couples of elements to generate the same ideal Let $R$ be an integral domain. The following is a well known fact:

Let $a,b \in R$. Then $(a)=(b)$ if and only if there exist a unit $u \in R$ such that $a=ub$.

I would like to generalize this result for ideals which are generated by two elements. So the question is:

Let $a,a',b,b' \in R$. Does it exist some property for these four elements which is equivalent on having $(a,b)=(a',b')$?

I know that maybe this question is a bit vague, I hope that somebody could give me some reference or hint.
EDIT: As Mathmo123's comment suggests, maybe the condition could be

There exists $U \in GL_2(R)$ such that $$\left[ \begin{matrix} a \\ b\end{matrix} \right] = U \left[ \begin{matrix} a' \\ b'\end{matrix} \right]$$

Clearly, this is a sufficient condition. But is it necessary as well?
 A: We consider the following condition.

Condition. There exists $A \in \operatorname{GL}_2(R)$ such that
  $$\begin{pmatrix} a \\ b\end{pmatrix} = A \begin{pmatrix} a' \\ b'\end{pmatrix}.$$

The sufficiency of the condition has already been noted, so I will only consider its necessity.

Lemma. If $R$ is a Dedekind domain, then the condition is necessary.

Proof. Let $I = (a,b) = (a',b')$. If $I = 0$, we are trivially done. If $I \neq 0$, then $I$ is locally free of rank $1$: every localisation is a DVR, so $I_\mathfrak p \subseteq R_\mathfrak p$ is principal, hence free of rank 1. Thus, $I$ is projective, and the sequence
$$0 \to K \to R^2 \stackrel{(a,b)}\longrightarrow I \to 0$$
splits. Hence, $K$ is projective as well, and taking determinants (top wedges) we get $K \otimes I = R$. We get the same for $K'$ (replacing $(a,b)$ by $(a',b')$), so we deduce that $K \cong K'$ (abstractly, not necessarily as submodules of $R^2$), since $I$ is invertible. We get a commutative diagram
$$\begin{array}{ccccccccc} 0 & \xrightarrow{} & K & \xrightarrow{} & R^2 & \xrightarrow{} & I & \xrightarrow{} & 0\\
 & & \downarrow & &  & & \downarrow & & \\
0 & \xrightarrow{} & K' & \xrightarrow{} & R^2 & \xrightarrow{} & I & \xrightarrow{} & 0
\end{array}$$
where the two vertical arrows are isomorphisms. Choosing splittings of the top and bottom rows, we can construct an isomorphism $R^2 \to R^2$ making the diagram commutative.
Remark. Even for $R = k[x,y]$, the condition is not necessary! Indeed, consider the ideal $I = (x^2,y) = (x^2, (x+1)y)$. Then a choice of transition matrices (for the two directions) is given by
\begin{align*}
\begin{pmatrix}1 & 0 \\ 0 & x+1 \end{pmatrix}, & & \begin{pmatrix}1 & 0 \\ y & -x+1 \end{pmatrix}.
\end{align*}
We will show that there is no invertible matrix $A$ such that $A\ (x^2,y)^\top = (x^2, xy + y)^\top$. Indeed, any such matrix differs from the [first] matrix above by
$$\begin{pmatrix}-yP & x^2P \\ -yQ & x^2Q \end{pmatrix}$$
for some $P, Q \in k[x,y]$, because $k[x,y]$ is a UFD. We have to show that no such matrix can be invertible. But if we reduce mod $y$, we get the matrix
$$\begin{pmatrix}1 & x^2P \\ 0 & 1+x+x^2Q \end{pmatrix}.$$
The determinant is $1+x+x^2Q$, which can never be a unit in $k[x]$.
Remark. So we see that our condition is the correct one for Dedekind domains, and is too strong even for a UFD of dimension $2$. There are a few directions we could take from here:


*

*Try to find a better criterion. I'm not sure if we will be able to come up with one, though.

*Analyse the problem for more than $2$ generators. It seems that we have fundamentally used the fact that $K$ is locally free of rank $1$ to conclude that $K \cong K'$; thus one would imagine that there might exist counterexamples for Dedekind domains (maybe even for a PID) for more than $2$ generators.


Edit: Maybe I should comment a bit on where this example comes from. The ideals $(a,b), (a,c)$ coincide if and only if $\bar b, \bar c \in R/(a)$ differ by a unit. Thus, a lift of the unit and its inverse give lower triangular matrices taking one vector to the other. If $R$ is a UFD, then any two matrices taking $(a,b)$ to $(a,c)$ differ by a matrix of the form
$$\begin{pmatrix}-bP & aP \\ -bQ & aQ \end{pmatrix}.$$
Reducing mod $b$, the same method as above applies whenever the unit that $\bar b$ and $\bar c$ differ by does not lift to an element whose reduction mod $b$ is a unit. In the example above, the unit is $x+1 \in k[x,y]/(x^2)$; none of its lifts become constant mod $y$.
In summary, the obstruction is somehow related to liftability of units.
If you want a more arithmetic example, you can take $(5,X) = (5, 2X) \subseteq \mathbb Z[X]$. The unit $2 \in \mathbb F_5[X]$ does not lift to any polynomial whose constant term is a unit in $\mathbb Z$.
A: Concerning domains, there is a nice partial result covering the case where the ideal is principal.

Lemma. Let $R$ be a domain, and let $a,b,x\in R$ be elements of $R$. The following are equivalent :

*

*$\langle a,b\rangle = \langle x\rangle$

*There exists $P=\begin{pmatrix}u & v \\ -b' & a'\end{pmatrix}\in\mathrm{GL}_2(R)$ with $\det(P)=1$, such that $\begin{pmatrix}x\\0\end{pmatrix}=P\begin{pmatrix}a\\b\end{pmatrix}$

This is (part of) Theorem 1.1, chapter IV in Modules sur les anneaux commutatifs, G.-M. Diaz-Toca, H. Lombardi, C. Quitté.
Proof. The case where $x=0$ is trivial, so we assume $x\neq 0$.
(1)$\implies$(2). By hypothesis there exist $u,v\in R$ and $a',b'\in R$ such that $ua+vb=x$, $xa'=a$ and $xb'=b$. Then one also has $0=ab-ab=xa'b-xb'a=x(-b'a+a'b)$. Since $R$ is a domain and $x\neq 0$, this implies $-b'a+a'b=0$, thus
$$\begin{pmatrix}x\\0\end{pmatrix}=\begin{pmatrix}u & v \\ -b' & a'\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}$$
Furthermore $x=ua+vb=x(ua'+vb')$ and again, since $R$ is a domain and $x\neq 0$ we have $\det(P)=ua'+vb'=1$.
The converse implication (2)$\implies$(1) was already noted in previous posts. $\square$
