# Definition of Prevariety

1. Oct 5, 2005

### dg0666

The book I am reading, Linear Algebraic Groups by Humphreys defines a prevariety X in projective space P^n to be a noetherian topological space endowed with a sheaf of functions such that X is the union of finitely many open subsets, each isomorphic to an affine variety.

This confuses me because I do not understand how affine varieties, which are closed sets in the Zariski topology, can be isomorphic to open sets in P^n.

Daniel

2. Oct 5, 2005

### Hurkyl

Staff Emeritus
Ooh, is it the closed vs open bit that's confusing you?

Don't forget that openness and closedness are properties of a subspace relative to the whole, and not properties of the subspace itself.

For example, (0, 1) is a closed subset of (0, 1), despite the fact it's an open subset of [0, 1].

3. Oct 5, 2005

### dg0666

In fact I understand that distinction. What is confusing me is that the affine varieties are closed in one topology and they are mapped isomorphically onto open subsets in P^n. It is just this bijection between open and closed subsets which is confusing me.

Should I be confused by this or is this just something counter-intuitive? I always thought that a continuous map sends closed sets to closed sets. Do you have to hand a simple elementary example to clarify how this situation is possible.

Dan

4. Oct 5, 2005

### Hurkyl

Staff Emeritus
A continuous map reflects closed sets -- that is if f(T) is closed, then T is closed.

The other way is not guaranteed. For example, consider the map RR:x→arctan x. This is an isomorphism of a closed set onto an open set.

Another popular example is to take the curve in R² given by xy = 1, and project it onto the x-axis, giving an isomorphism of a closed subset of R² onto an open subset of R.

But, continuous maps do preserve compact sets. One of the oddities of Zariski topologies is that compact sets are not necessarily closed.

For example, C - {0} is an open, compact subset of C with the Zariski topology.

Incidentally, the embedding (0, 1)→[0, 1] is representative of what's going on here. If we identify the endpoints of [0, 1], then it is exactly analogous to embedding the real line into the projective real line.

Last edited: Oct 5, 2005
5. Oct 6, 2005

### dg0666

Great. I think that has answered my question very well. My confusion arosed because of a misunderstanding between isomorphisms and continuous maps but thanks very much for your help.

Dan

6. Oct 6, 2005

### Hurkyl

Staff Emeritus
It strikes me that I should be more explicit about which topology I was using, and in which category I was claiming isomorphism in my examples. (Though I presume you figured out my meaning, I still feel the need to elaborate)

x→arctan x was using the Euclidean topology, since it's clearly not a continuous map under the Zariski topology. It is an isomorphism onto its image, that is an isomorphism of R and (-π/2, π/2), as a map of topological spaces. (i.e. a homeomorphism)

The second example, the projection of the hyperbola onto the punctured x-axis is continuous in both the Zariski and the Euclidean topologies. The domain is closed in both topologies, and the image is open in both topologies. This one is actually a regular map of quasiaffine varieties, and thus an isomorphism in that category, as well as being a homeomorphism of topological spaces.

7. Nov 1, 2005

### mathwonk

I am puzzled about your saying that the prevariety is "in projective space" as then you give a completely abstract definition.

the distinction between a "pre"variety and a variety, is usually the lack of a separation axiom, which is always satisfied for such things that occur in projective space. are you sure you read him correctly?

8. Nov 1, 2005

### mathwonk

I am puzzled about your saying that the prevariety is "in projective space" as then you give a completely abstract definition.

the distinction between a "pre"variety and a variety, is usually the lack of a separation axiom, which is always satisfied for such things that occur in projective space. are you sure you read him correctly?