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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 $d\geq 3.$

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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.

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  • $\begingroup$ I just realize that a positive polynomial does not need to realize its infimum, e.g., $f(x,y)=x^2 + (xy-1)^2$. Hence, the argument above is not complete in this form. $\endgroup$ – Andreas Thom Sep 22 '14 at 16:10

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