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Let $p(x,y),q(x,y)\in \mathbb{Q}[x,y]$. Assume that $p(0,0)=q(0,0)$. Is it generally possible to find a polynomial $r(x,y)\in\mathbb{Q}[x,y]$ irreducible in $\mathbb{Q}[x,y]$ such that $\mathbb{R}[x,y]/(r)\simeq \mathbb{R}[x,y]/(pq)$ as $\mathbb{R}$-algebras.

I need to generate polynomials having complicated singularities to test an algorithm. It will be nice if I can do it by prescribing the branches (polynomials $p,q$) and then to seek for an irreducible polynomial having the prescribed singularity.

I will be also grateful for any relevant reference to the literature. Many thanks.

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    $\begingroup$ What do you mean by "having same qualities as $p(x, y)\cdot q(x, y)$"? $\endgroup$
    – Michael Albanese
    Commented May 22, 2017 at 13:06
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    $\begingroup$ What do you mean by "going through $[0,0]$"? As these are polynomials in two variables, the graph of the corresponding function would be in 3D rather than 2D... Also, what domain are you considering? a field, a ring, the reels,...? And why do you assume that your given candidate is irreducible? I think it will not be in many cases, even in cases where $p$ and $q$ are both irreducible. All in all you might want to add quite some background on your question. $\endgroup$
    – Dirk
    Commented May 22, 2017 at 14:05
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    $\begingroup$ I presume that the OP means that $p(0,0)=q(0,0)=0$ and she seeks an irreducible $r(x,y)$ such that $\mathbb{R}[\![x,y]\!]/(r)\simeq\mathbb{R}[\![x,y]\!]/(pq)$ as $\mathbb{R}$-algebras, or something like that. For example you could have $p=x-y$ and $q=x+y$ and $r=pq-y^3=x^2-y^2-y^3$: this is the nodal cubic curve, which is globally irreducible, but reducible in a formal neighbourhood of the origin. $\endgroup$
    – Neil Strickland
    Commented May 22, 2017 at 17:10
  • $\begingroup$ I agree that this is a likely reading. Voting to reopen -- although this doesn't seem a very hard question, I think it is probably too difficult to be likely to get an answer at math.SE... $\endgroup$
    – R.P.
    Commented May 22, 2017 at 17:47

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