Determining the field from the topology of the affine space over it. The inspiration is from another question in which it is remarked that the passage to Zariski topology is a very interesting functor from Commutative Rings to Topological spaces.
I am trying to understand how "faithful" this functor is. Any two fields have the same topology. But, with polynomial rings, we have more hope. So.

Let $K$, $L$ be two fields. Then if  $Spec\ K[X]$ and $Spec\ L[X]$ are homeomorphic, does it follow that $K$ and $L$ are isomorphic?

Similarly:

Let $K$, $L$ be two fields. Then if  $Spec\ K[X_1, \ldots , X_n]$ and $Spec\ L[X_1, \ldots , X_n]$ are homeomorphic, does it follow that $K$ and $L$ are isomorphic?

 A: The following paper of Hrushovski-Zilber shows that if we restrict our attention to algebraically closed fields $F$, then $F$ is uniquely determined up to isomorphism by its "Zariski geometry". Presumably an examination of the proof will show that an algebraically closed field
$F$ is determined by $Spec(F[x_{1}, \cdots F[x_{n}])$ for sufficiently large $n$. 
Hrushovski, Ehud; Zilber, Boris (1996). "Zariski Geometries". Journal of the American Mathematical Society 9: 1–56. 
Another "reference": http://en.wikipedia.org/wiki/Zariski_geometry
A: This is false.  Let $K=\bar{\mathbb{Q}}$ and $L=\bar{\mathbb{F}}_2$.  These are clearly both algebraically closed, of different characteristics, so $K\not\cong L$.  However, if we ONLY look at the topology, $\mathrm{Spec}(K[x])$ and $\mathrm{Spec}(L[x])$ will be be countable sets with the finite complement topology on the closed points, with a single generic point, so they're homeomorphic.  For algebraically closed fields, the TOPOLOGY on the affine line over the field is determined by the cardinality.
For higher dimensions, it's less clear to me, because you might be able to recover characteristic (I'm a char 0 kind of person, so I don't know) from how the various curves/hypersurfaces sit inside it.
