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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?

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  • $\begingroup$ Feels like for $\mathbb{C}[x]$ we would have $p(x)=\prod\limits_i (x-a_i)^{n_i}$ and $q(x)=\prod\limits_i (x-b_i)^{m_i}$ are iso-associates if they have isomorphic decompositions into irreducibles, that is if there exists a permutation $\pi$ such that $n_i = m_{\pi_i}$. In particular, the powers of $p$ and $q$ must coincide. For $\mathbb{R}[x]$ we'd have to consider irreducibles of the form $x^2+ax+b$ with negative determinant. $\endgroup$
    – lisyarus
    Commented Aug 8, 2017 at 14:44

1 Answer 1

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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.

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    $\begingroup$ The $p_i$ and $q_i$ should not be part of the data, since they don't matter; all that matters is how many there are of each degree and multiplicity. $\endgroup$ Commented Aug 8, 2017 at 20:27
  • $\begingroup$ @EricWofsey You are right, I put too many things there. $\endgroup$
    – Mohan
    Commented Aug 8, 2017 at 21:20

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