Here is a precise statement of my question:

Let $p\in \mathbb{R}(x_1, \ldots, x_n)$ be a polynomial, and let $Z(p)\subset \mathbb{R}^n$ be the set of zeros of $p$. Must the singular homology $H_i (Z(p); \mathbb{Z})$ ($i\geq 0$) be finitely generated as an abelian group?

Here I really just mean the singular homology groups of this set as a topological space with the Euclidean topology.

It's an old theorem of Whitney that $Z(p)$ has finitely many connected components, so $H_0 (Z(p); \mathbb{Z})$ is finitely generated. Note that it is possible to triangulate $\mathbb{R}^n$ with $Z(p)$ as a subcomplex, so $H_i (Z(p)) = 0$ for $i>n$.

I'm guessing the answer is well-known (either a theorem or a counterexample), but I couldn't find an answer on Google or MathSciNet...

Here is a precise statement of my question:

Let $p\in \mathbb{R}[x_1, \ldots, x_n]$ be a polynomial, and let $Z(p)\subset \mathbb{R}^n$ be the set of zeros of $p$. Must the singular homology $H_i (Z(p); \mathbb{Z})$ ($i\geq 0$) be finitely generated as an abelian group?

Here I really just mean the singular homology groups of this set as a topological space with the Euclidean topology.

It's an old theorem of Whitney that $Z(p)$ has finitely many connected components, so $H_0 (Z(p); \mathbb{Z})$ is finitely generated. Note that it is possible to triangulate $\mathbb{R}^n$ with $Z(p)$ as a subcomplex, so $H_i (Z(p)) = 0$ for $i>n$.

I'm guessing the answer is well-known (either a theorem or a counterexample), but I couldn't find an answer on Google or MathSciNet...