Let $\mathbb R[x,y]\_+$ denote the set of positive polynomials in two variables. My problem can be stated as follows: 



> 
> Does there exist a countable set $M\subseteq \mathbb R[x,y]\_+$ 
>  such that every $f\in \mathbb R[x,y]\_+$
>  can be represented as $$ f=\sum\_{i=1}^n p\_i^2\cdot h\_i,\ h\_i\in M,\ p\_i\in\mathbb R[x,y]\ ? $$
>  More concretely, can we take $M=\mathbb Q[x,y]\_+\ ?$
> 
> 
> 


Recall that a *quadratic module* in a commutative ring $A$ with $1\in A$ is a subset $Q\subseteq A$ such that 


$(i)\ 1\in Q,\ -1\notin Q$, $(ii)\ Q+Q\subseteq Q$, $(iii)\ p^2\cdot Q\subseteq Q,\ \forall p\in A.$


The problem can be now stated as follows: 



> 
> Is $\mathbb R[x,y]\_+$ countably generated as a quadratic module?
> 
> 
> 


The smallest quadratic module in $\mathbb R[x\_1,\dots,x\_d]$ is the set $\sum^2\mathbb R[x\_1,\dots,x\_d]$ of finite sums of squares. We have $\mathbb R[x\_1,\dots,x\_d]\_+=\sum^2\mathbb R[x\_1,\dots,x\_d]$ iff $d=1$ (Hilberts theorem). It is known that $\mathbb R[x\_1,\dots,x\_d]\_+$ is not finitely generated for $d\geq 2.$ One can also easily show that $\mathbb R[x\_1,\dots,x\_d]\_+$ is not countably generated for $d\geq 3.$



Is $\mathbb R[x,y]_+$ countably generated as a quadratic module?
Suppose that a countable set $M$ as required exists. (As you say, $M$ cannot be finite - in particular it cannot be empty.) Let $d$ be the minimal degree of all polynomials in $M$ and set $M\_d \subset M$ be the subset of elements of minimal degree. Since $M$ is not empty, $d\neq 0$, and the zero-set of any polynomial in $M\_d$ is a proper algebraic subset of $\mathbb R^2$.


Let $f \in \mathbb R[x,y]\_+$ be any element of minimal degree and without loss of generality, assume that there exists $(x\_0,y\_0) \in \mathbb R^2$, such that $f(x\_0,y\_0)=0$. (This needs a little additional argument.)


Now, for any $(\lambda,\mu) \in \mathbb R^2$, set 
$$f\_{\lambda,\mu}(x,y):= f(x-\lambda,y-\mu)$$ and note that $f\_{\lambda,\mu} \in \mathbb R[x,y]\_+$.


For any $(\lambda,\mu) \in \mathbb R^2$, the polynomial $f\_{\lambda,\mu}$ must be (for degree reasons) a positive linear combination of elements in $M\_d$. Thus, there is some $h \in M\_d$, which vanishes at $(x\_0+ \lambda,y\_0 + \mu)$. We conclude that the union of zero-sets of polynomials in $M\_d$ must be the whole of $\mathbb R^2$. This is a contradiction, since $M\_d$ is countable and this union is a countable union of proper algebraic subsets.


