I have two (or more) polynomials $p(x,y)$, $q(x,y)$ going through $[0,0]$. Can I easily produce another irreducible polynomial having at the origin singularity having same qualities as $p(x,y)\cdot q(x,y)$? Nice candidate is $p(x,y)\cdot q(x,y) + x^{(deg_x p + deg_x q)}y^{(deg_y p + deg_y q)}$, but I cannot prove it.

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.