Can any smooth manifold be realized as the zero set of some polynomials? Is any real smooth manifold diffeomorphic to a real affine algebraic variety? (I.e. is there an "algebraic" Whitney embedding theorem?)
And are all possible ways of realizing a manifold $M$ as an algebraic variety equivalent? I.e. suppose $M$ is diffeomorphic to varieties $V_1$ and $V_2$, are these isomorphic in the algebraic category?
Admittely I'm just asking out of curiosity after reading this question: Can manifolds be uniformly approximated by varieties?
 A: For the sake of completeness, here is the answer  covering noncompact manifolds as well:
Definition. 1. A smooth manifold $M$ is tame if $M$ is diffeomorphic to the interior of a smooth compact manifold $N$ with (possibly empty) boundary.


*A smooth manifold $M$ is algebraic if it is diffeomorphic to a
nonsingular real-algebraic subset of ${\mathbb R}^n$ for some $n$.

Theorem. A smooth manifold is tame if and only if it is algebraic.
See Corollary 4.3 in
Akbulut, Selman; King, Henry C., The topology of real algebraic sets with isolated singularities, Ann. Math. (2) 113, 425-446 (1981). ZBL0494.57004.
A: A quick Google search found this paper, where it is stated that the answer to the first question is yes in the compact case (due to Tognoli): this result is called the Nash-Tognoli theorem. In general, the answer is no: a real affine variety has finite-rank homology groups, and it's easy to construct non-compact manifolds for which this is false (e.g. a surface of infinite genus). In fact, apparently there is a bound due to Milnor for the sum of the Betti numbers of a real variety. 
The answer to the second question is certainly not: just take two elliptic curves with slightly different $j$-invariants. 
