2
$\begingroup$

Let $x = \{x_{1},\dots, x_{n}\}$ be a set of variables and let $a = \{ a_{1}, \dots, a_{m}\}$ be a set of parameters. Let $\{f_{1}(a,x), \dots, f_{s}(a,x)\} \subset \mathbb{C}[a,x]$ be a set of polynomials and let $I$ be the ideal generated by this set. For $a \in \mathbb{C}^{m}$ let $I_{\overline{a}}$ denote the evaluation of the ideal $I$ at $\overline{a}$, i.e. $I_{\overline{a}} = \{f_{1}(\overline{a},x), \dots, f_{s}(\overline{a},x)\} \cdot \mathbb{C}[x]$ and let $J_{\overline{a}}= \sqrt{I_{\overline{a}}}$

Problem: Describe the prime decomposition of each $J_{\overline{a}}$ by splitting the parameter space in suitable subsets $A_{i}$ so that for each $A_{i}$ the generators of the different prime components can be obtained by evaluating at the parameter $\overline{a}$.

Question: Is there a finite decomposition of the parameter space $\mathbb{C}^{m}$ into constructible sets $A_{1},\dots, A_{l}$ so that for each $A_{i}$ a finite number of finite sets of polynomials $Q_{i,j}= \{h_{1,i}(a,x), \dots, h_{l_{i},i}(a,x)\}$, $j = \{1, \dots, t_{i}\}$ exists, with the property that for every $\overline{a} \in A_{i}$ the evaluation (specialization) of the $(Q_{i,j}(\overline{a},x))_{j=1,\dots, t_{i}}$ gives a prime decomposition of the ideal $J_{\overline{a}}$?

The motivation (and possibly the solution) comes of the theory of comprehensive Groebner Basis: Given a parametric ideal as above there exists a decomposition of the parameter space as above so that for every $A_{i}$ a finite set of polynomials $G_{i} \subset \mathbb{C}[a,x]$ can be found so that the specialization (=evaluation) at $\overline{a}$ gives a Groebner basis for $I_{\overline{a}}$ for every $\overline{a} \in A_{i}$.

Actually the case I am really interest in is when $\mathbb{C}[a,x]$ is replaced by the ring of convergent power series $\mathbb{C}\{a,x\}$, so when $I$ is an ideal of convergent power series. But since I think that the polynomial case is propably not so far from the power series one and more material on polynomials is available I posed the question formulated for polynomials. Thanks for all responses!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.