Let $f$ and $g$ be positive definite forms in the polynomial ring ${\mathbb{R}}[x\_0,\ldots, x\_n]$ such that $\deg(g)$ divides $\deg(f)$. A generalization of a theorem by Reznick is that $g^N f$ is a sum of squares for large $N$. 


A corollary of the Representation Theorem says that if $V$ is an affine real variety with compact set $V({\mathbb{R}})$ of real points, then every $f\in {\mathbb{R}}[V]$ that is strictly positive on $V(\mathbb{R})$ is a sum of squares in ${\mathbb{R}}[V]$. 


A proof of the theorem by Reznick using this corollary goes like this: Let $V$ be the complement of the hypersurface $g=0$ in real projective space ${\mathbb{P}}^{n-1}$. This variety $V$ is affine and has a compact set of real points. Let $r=\deg f/\deg g$. By assumption $f/g^r>0$ and thus a sum of squares. 


My question is: Why is there a need to remove the hypersurface $g=0$. In the first place, if $g$ is positive-definite, isn't this hypersurface $g=0$ an empty set?



Removing a hypersurface when applying the Representation theorem to prove Positivstellensatz with uniform denominators 
The real point of $g = 0$ are the empty set, but $g = 0$ is still a non-trivial hypersurface over $\mathbb R$, just with no real points. (The fact that is has no real points is the reason why the real points of $V$ are compact.)


