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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?

edited May 21 2010 at 20:54

Simon Thomas
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asked May 21 2010 at 17:47

Akela
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The affine line over a field carries the cofinite topology, so its homeomorphism type depends only on its cardinality. – Qiaochu Yuan May 21 2010 at 17:52

Ahh, posted as an answer as you posted your comment! – Charles Siegel May 21 2010 at 17:53

Surely the answer is no (by which I mean I do not have a proof). At the very least, if $K$ is the algebraic closure of $\mathbb{Q}$ and $L$ is the algebraic closure of $\mathbb{Q}(t)$, then $\operatorname{Spec} K[t_1,\ldots,t_n]$ and $\operatorname{Spec} L[t_1,\ldots,t_n]$ should be homeomorphic for all $n$. It seems reasonable to conjecture that the only invariant of $K$ that these spectra see is the cardinality of $K$. – Pete L. Clark May 21 2010 at 17:59

Simon Thomas · Answer 1 · 2010-05-21 19:48:04Z UTC

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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

answered May 21 2010 at 19:48

Simon Thomas
6,009●1●18●36

Interesting. The relevant result seems to be Proposition 1.1. If you take $C$ to be the affine line over an algebraically closed field $F$ and $C'$ to be the affine line over an algebraically closed field $F'$, then the result says that an isomorphism of Zariski geometries from $C$ to $C'$ induces an isomorphism from $F$ to $F'$. It also says that a morphism of Zariski geometries is an isomorphism if it induces a homeomorphism on $C^n$ for all $n$.... – Pete L. Clark May 21 2010 at 20:17

...But what I wasn't immediately able to see was whether a morphism of Z.G.'s from $C$ to $C'$ involves more data than just a collection of such homeomorphisms -- e.g. whether various geometric compatibility conditions need to be satisfied. I would be very interested to see these details worked out. – Pete L. Clark May 21 2010 at 20:19

The experts in this area (such as Dave Marker) would know all about this. Hopefully one of them will stumble across this question and save us the trouble of working out the details ourselves! – Simon Thomas May 21 2010 at 20:32

In this type of argument you usually reconstruct the field from a 2-dimensional family of curves on $C\times C$. So probably you can get by knowing that the homeomorphism between $C$ and $C_1$ lifts to homeomorphisms of the Zariski topologies of $C^n$ and $C_1^n$ for $n=2,3,4$. – Dave Marker May 24 2010 at 6:56

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A simple version of this idea would be if $K$ and $L$ are algebraically closed fields and we had a homeomorphism between $P_K^2\timesP_K^2$' and $P_L^2\times P_L^2$. Let $V\subset P^2_K \times P^2_K$ be the incidence variety for lines in $P_2$. Desargues ideas allow us to reconstruct $K$ from $V$. A homeomorphism would give us an isomorphic variety in $P_L^2\times P_L^2$. This would allow us to interpret $K$ in $L$. But by a result of Poizat then $K$ and $L$ would be isomorphic. – Dave Marker May 24 2010 at 12:53

Charles Siegel · Answer 2 · 2010-05-21 17:53:22Z UTC

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.

Determining the field from the topology of the affine space over it.

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