Question about the ring of polynomials bounded on a real variety Suppose that $I$ is a  prime real radical ideal in the polynomial ring $\mathbb{R}[x_1,\ldots,x_n]$. "Real radical" means that if a sum of squares $a^2+b^2+\ldots$ is an element of $I$ then so are the polynomials $a, b,\ldots$. Let  $V\subseteq \mathbb{R}^n$ be the zero-set of $I$, and assume that $V$ is unbounded.  Obviously the ring $B:=\mathbb{R}[I+\mathbb{R}]$ consists entirely of of polynomials bounded on $V$. Is every polynomial bounded on $V$ necessarily an element of the ring $B$?
 A: I wish I had thought more about this before I posted. Anyway, here is a counterexample. Let $f\in \mathbb{R}[x,y]$ be the polynomial $y(x^2+1)-1$. Let $I$ be the ideal generated by $f$. Let $V$ be the zero-set of $f$ in $\mathbb{R}^2$. Writing the equation $f=0$ in the form $$y=\frac{1}{1+x^2},$$ it is clear that the polynomial $y$ is bounded on $V$. Moreover $y$ is not an element of the ring $\mathbb{R}[\mathbb{R}+I]$ because $y$ is not constant as a function on $V$. We must verify that $I$ is prime and real radical.


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*$I$ is prime: To check this, we note that the quotient ring $\mathbb{R}[x,y]/I$ is isomorphic to the ring $\mathbb{R}[x,1/(1+x^2)]$, which is an integral domain, being a subring of the field $\mathbb{R}(x)$. 

*I is real radical: It is enough to find an order on the ring $\mathbb{R}[x,1/(1+x^2)]$. But the field $\mathbb{R}(x)$ can be ordered: For example think of the elements as functions on $\mathbb{R}$ ordered by eventual domination.
It seems that the general problem of calculating the ring of bounded functions on a real variety is much much harder than I realized. If anyone has any references or thoughts on the matter please post.
