When are two elements of $\mathbb{R}[x]$ “iso-associates”?

Question

Let $R$ denote a commutative ring. Then two elements of $R$ are said to be associates iff they generate the same ideal.

General question. What do we call two elements of $R$ satisfying the weaker condition that the rings obtained by quotienting out by the corresponding ideals are isomorphic?

Alternatively, what do we call two ideals that yield the same quotient ring up to isomorphism?

For the sake of readability, lets call any two such elements 'iso-associates.'

For example, in the ring $\mathbb{R}[x]$, we have the following.

• Given $a,b \in \mathbb{R}$, the polynomial $x-a$ is an associate of $x-b$ iff $a=b$. Yet $x-a$ and $x-b$ are always iso-associates, because such quotient rings are always isomorphic to $\mathbb{R}$.

• Given $a,b \in \mathbb{R}$, I think that the polynomial $x^2-a$ is an iso-associate of $x^2-b$ iff $\mathrm{sgn}(a) =\mathrm{sgn}(b)$. If I'm not mistaken, the three rings we can obtain in this way are the complex numbers, the hyperbolic numbers, and the dual numbers.

Specific question. When are two elements of $\mathbb{R}[x]$ iso-associates?

Answer 1

Any $0\neq f\in \mathbb{R}[x]$ can uniquely be written as $f=ap_1^{r_1}\cdots p_k^{r_k}q_1^{s_1}\cdots q_l^{s_l}$ where $0\neq a\in\mathbb{R}$, $p_i=x-a_i$ for distinct $a_i\in \mathbb{R}$, $q_i$s are irreducible monic polynomials of degree 2, all distinct, $r_i,s_i\geq 1$, integers. Thus, we can associate to such an $f$, the set $(k(f),l(f), p_i(f), q_i(f), r_i(f), s_i(f))$. Then, $f$ is iso-associate to $g$ if the corresponding set for $g$ is the same. The proof is a simple application of Chinese remainder theorem.