Why the Zariski topology?
The Zariski topology is a topology that is well-suited for the study of polynomial equations in algebraic geometry, since a Zariski topology has many fewer open sets than in the usual metric topology. In fact, the only closed sets are the algebraic sets, which are the zeros of polynomials.
Is the Zariski topology Hausdorff?
The Zariski topology is not Hausdorff. In fact, any two open sets must intersect, and cannot be disjoint. Also, the open sets are dense, in the Zariski topology as well as in the usual metric topology.
What is Zariski problem?
The Zariski Cancellation Problem for Affine Spaces asks whether the affine space A k n is cancellative, i.e., if is an affine k-variety such that V × A k 1 ≅ A k n + 1 , does it follow that V ≅ A k n? Equivalently, if A is an affine k-algebra such that A [ 1 ] = k [ n + 1 ] , does it follow that A = k [ n ]?
Is the Zariski topology compact?
In particular, every open subset of an affine variety is compact in the Zariski topology. (Recall that by definition a topological space X is compact if every open cover of X has a finite subcover.)
Is the zariski topology compact?
Is the zariski topology path connected?
The Cremona group is topologically simple when endowed with the Zariski or Euclidean topology, in any dimension ≥ 2 and over any infinite field. Two elements are always connected by an affine line, so the group is path-connected.
What is zariski problem?
Are points closed in the Zariski topology?
Every regular map of varieties is continuous in the Zariski topology. In fact, the Zariski topology is the weakest topology (with the fewest open sets) in which this is true and in which points are closed.
Why is Zariski topology not hausdorff?
If the field k is not a finite field, then the Zariski topology on the affine space (def. 2.1) is not Hausdorff. This is because the solution set to a system of polynomials over an infinite polynomial is always a finite set. This means that in this case all the Zariski closed subsets V(ℱ) are finite sets.