A short proof for the open quadrant problem (Q321277)

It was proved in [\textit{J. F. Fernando} and \textit{J. M. Gamboa}, J. Pure Appl. Algebra 179, No. 3, 241--254 (2003; Zbl 1042.14035)] that given linearly independent linear forms \(\ell_1,\dots,\ell_r\) on \(\mathbb R^n\) there exists a polynomial map \(f:\mathbb R^n\to\mathbb R^n\) whose image is the open semialgebraic set NEWLINE\[NEWLINE \{x\in\mathbb R^n:\ell_1(x)>0,\dots, \ell_r(x)>0\}. NEWLINE\]NEWLINE As a particular case, which is the key to prove the general one, the authors constructed a polynomial map \(f:\mathbb R^2\to\mathbb R^2\) whose image is the open quadrant \(\mathbb Q:=\{x>0,\, y>0\}\subset\mathbb R^2\). The checking of the inclusion \(\mathbb Q\subset f(\mathbb R^2)\) makes use of Sturm's algorithm to show that some univariate polynomials have a real root in certain intervals. This required computer assistance using MAPLE, and it becomes a tedious task to verify that all computations are indeed correct. Thus, it was a challenge to find a more transparent proof avoiding the use of computers. The authors of the article under review have succeeded using a classical strategy: divide and conquer. They consider first the polynomial map NEWLINE\[NEWLINE F:\mathbb R^2\to\mathbb R^2,\, (x,y)\mapsto((xy-1)^2+x^2,(xy-1)^2+y^2) NEWLINE\]NEWLINE and easily prove the inclusions NEWLINE\[NEWLINE {\mathcal A}:=\{xy-1\geq0\}\subset F(\mathbb R^2)\subset\mathbb Q. NEWLINE\]NEWLINE They introduce next the subset \(\mathbb B={\mathcal A}\cup\{0<x\leq y\}\) and proved that the polynomial map NEWLINE\[NEWLINE G:\mathbb R^2\to\mathbb R^2,\, (x,y)\mapsto(x,y(xy-2)^2+x(xy-1)^2) NEWLINE\]NEWLINE satisfies the inclusions NEWLINE\[NEWLINE \mathbb B\subset G({\mathcal A})\subset G(\mathbb Q)\subset\mathbb Q. NEWLINE\]NEWLINE Finally they consider the polynomial map NEWLINE\[NEWLINE H:\mathbb R^2\to\mathbb R^2,\, (x,y)\mapsto\Big(x(xy-2)^2+\frac{xy^2}{2},y\Big) NEWLINE\]NEWLINE that satisfies the equalities \(H(\mathbb B)=H(\mathbb Q)=\mathbb Q\). From the above results it follows straightforwardly the equality \(\varphi(\mathbb R^2)=\mathbb Q\), where \(\varphi:=H\circ G\circ F\).NEWLINENEWLINEDespite the maps \(F\), \(G\) and \(H\) have few monomials and small degree it is worthwhile mentioning that the total degree of \(\varphi\) equals \(72\), whereas the total degree of the map introduced by Fernando and the reviewer is \(56\). It seems natural to ask which is the minimum total degree for the set of polynomial maps \(\mathbb R^2\to\mathbb R^2\) whose image is the open quadrant \(\mathbb Q\) and what is the sparsest polynomial map \(\mathbb R^2\to\mathbb R^2\) whose image is \(\mathbb Q\).NEWLINENEWLINEIn the recent paper [\textit{J. F. Fernando} et al., in: A mathematical tribute to Professor José María Montesinos Amilibia on the occasion of his seventieth birthday. Madrid: Universidad Complutense de Madrid, Facultad de Ciencias Matemáticas, Departamento de Geometría y Topología. 337--350 (2016; Zbl 1348.14135)] it is proved that \(\mathbb Q\) is the image of the polynomial map NEWLINE\[NEWLINE \mathbb R^2\to\mathbb R^2,\, (x,y)\mapsto((x^2y^4+x^4y^2-y^2-1)^2+x^6y^4,(x^6y^2+x^2y^2-x^2-1)^2+x^6y^4), NEWLINE\]NEWLINE whose total degree is \(28\). It seems plausible that the searched minimum total degree is \(16\), but a proof of this fact is not available nowadays.