Zariski topology
The Zariski topology is a topological space
With the Zariski topology,
Proof this is a topology
The set of vanishing ideals must satisfy the axioms of a topological space.
Let
-
and . -
Given ideals
,
- Given ideals
we have
Why this topology
Given a continuous function
This means that since given the Euclidean topology polynomials are continuous, that the Zariski topology has far fewer open (or closed) sets and is much more restrictive.
Zariski topology on prime spectrum.
For a ring
Note that if
Thus, the sets
form the closed sets of a topology on
Proof
First,
Next, if
Proof:
If
therefore, we have
This gives
Say
Lastly, if
Proof: Say
Say
However,
Induced topology on
\text{MSpec }R
.Say
. Then the subspace topology on is the “normal” Zariski topology described above.
Open sets of Zariski topology
For
are an open topological basis for the Zariski topology.