# Zariski Topology (Tuesday, September 01)

## The Zariski Topology

Last time: 
\[  
V(I) &= \ts{x\in \AA^n \st f(x) = 0 \, \forall x\in I} \\
I(X) &= \ts{f\in k[x_1, \cdots, x_n] \st f(x) = 0\, \forall x\in X}
.\]

We proved the Nullstellensatz $I(V(J)) = \sqrt{J}$, defined the coordinate ring of an affine variety $X$ as 
\[
A(X) \da k[x_1, \cdots, x_n] / I(X)
\]
the ring of *regular* (polynomial) functions on $X$.
Recall that a *topology* on $X$ can be defined as a collection of closed subsets of $X$ that are closed under arbitrary intersections and finite unions.
A subset $Y\subset X$ inherits a subspace topology with closed sets of the form $Z\intersect Y$ for $Z\subset X$ closed in $X$.


:::{.definition title="Zariski Topology"}
Let $X$ be an affine variety.
The closed sets are affine subvarieties $Y\subset X$.
:::


:::{.proposition title="The Zariski topology is a topology"}
This satisfies the axioms for a topological space.
:::

:::{.proof title="?"}
\envlist

- We have $\emptyset, X$ closed, since 

  1. $V_X(1) = \emptyset$,
  2. $V_X(0) = X$

- Closure under finite unions:

  Let $V_X(I), V_X(J)$ be closed in $X$ with $I, J \subset A(X)$ ideals.
  Then $V_X(IJ) = V_X(I) \union V_X(J)$.

- Closure under intersections:
  
  We have $\bigcap_{i\in \sigma} V_X(J) = V_X\qty{ \sum_{i\in \sigma} J_i}$.

:::

:::{.remark}
There are few closed sets, so this is a "weak" topology.
:::


:::{.example title="Closedness differs in the analytic topology"}
Compare the classical topology on $\AA^1_{/\CC}$ to the Zariski topology.
Consider the set 
\[
A\da \ts{x\in \AA^1_{/\CC} \st \norm{x} \leq 1}
\]
which is closed in the classical topology.
However, $A$ is *not* closed in the Zariski topology, since the closed subsets are finite sets or the whole space.
In fact, the topology here is the cofinite topology.
:::

:::{.example}
Let $f: \AA^1_{/k} \to \AA^1_{/k}$ be any injective map.
Then $f$ is necessarily continuous wrt the Zariski topology.
Thus the notion of continuity is too weak in this situation.
:::


:::{.warnings title="The topology on the product is not the product topology"}
Consider $X\cross Y$ a product of affine varieties.
Then there is a product topology where open sets are of the form $\bigcup_{i=1}^n U_i \cross V_i$ with $U_i, V_i$ open in $X, Y$ respectively.
This is the wrong topology!
On $\AA^1 \cross \AA^1 = \AA^2$, the diagonal $\Delta \da V(x-y)$ is closed in the Zariski topology on $\AA^2$ but not in the product topology.
:::

## Irreducibility and Connectedness

:::{.example}
Consider $\AA^2_{/\CC}$, so the closed sets are curves and points.
Observation: $V(x_1 x_2 ) \subset \AA^2_{/\CC}$ decomposed into the union of the coordinate axes $X_1 \da V(x_1)$ and $X_2 \da V(x_2)$.
The Zariski topology can detect these decompositions.
:::

:::{.definition title="Irreducibility and Connectedness"}
Let $X$ be a topological space.

a. $X$ is **reducible** iff there exist nonempty proper closed subsets $X_1 ,X_2 \subset X$ such that $X = X_1 \union X_2$.
  Otherwise, $X$ is said to be *irreducible*.

b. $X$ is **disconnected** if there exist $X_1, X_2 \subset X$ such that $X = X_1 \disjoint X_2$.
  Otherwise, $X$ is said to be *connected*.
:::

:::{.example}
$V(x_1 x_2)$ is reducible but connected.
:::

:::{.example}
$\AA^1_{/\CC}$ is *not* irreducible, since we can write 
\[
\AA^1_{/\CC} = \ts{\norm{x} \leq 1} \union \ts{\norm{x} \geq 1}
\]
:::


:::{.proposition title="?"}
Let $X$ be a disconnected affine variety with $X = X_1 \disjoint X_2$.
Then $A(X) \cong A(X_1) \cross A(X_2)$.
:::

:::{.proof}
We have 
\[
X_1 \union X_2 = X \implies I(X_1) \intersect I(X_2) = I(X) = (0) \in A(X)
,\]
recalling that the coordinate ring $A(X)$ is a quotient by $I(X)$.
Since $X_1 \intersect X_1 = \emptyset$, we have 
\[  
I(X_1 \intersect X_2) = \sqrt{I(X_1) + I(X_2) } = I(\emptyset) = \gens{1}
.\]

Thus $I(X_1) + I(X_2) = \gens{1}$, and by the Chinese Remainder Theorem, the following map is an isomorphism:
\[  
A(X) \to A(X)/I(X_1) \cross A(X) / I(X_2)
.\]

However, the codomain is precisely $A(X_1) \cross A(X_2)$.
:::

### Irreducibility on the Algebra Side

:::{.proposition title="?"}
An affine variety $X$ is irreducible $\iff$ $A(X)$ is an integral domain.
:::

:::{.proof}
$\implies$:
By contrapositive, suppose $f_1, f_2 \in A(X)$ are nonzero with $f_1 f_2 = 0$.
Let $X_i \da V(f_i)$, then 
\[
X= V(0) = V(f_1 f_2) = X_1 \union X_2
\]
which are closed and proper since $f_i \neq 0$.

\

$\impliedby$:
Suppose $X$ is reducible with $X = X_1 \union X_2$ with $X_i$ proper and closed.
Define $J_i \da I(X_i)$, 
then by part (a) of the Nullstellensatz.
\[  
V(J_i) = V(I(X_i)) = X_i \implies J_i \neq 0
.\]
So there exists a nonzero $f_i \in J_i = I(X_i)$, so $f_i$ vanishes on $X_i$.
But then 
\[
V(f_1) \union V(f_2) \supset X_1 \union X_2 = X
,\]
so $X= V(f_1 f_2)$ and $f_1 f_2 \in I(X) = \gens{0}$ and $f_1 f_2 = 0$, and $A(X)$ is thus not a domain. 
:::

:::{.example}
Let $X = \ts{\vector p^1, \cdots, \vector p^d}$ be a finite set in $\AA^n$.
The Zariski topology on $X$ is the discrete topology, and $X = \disjoint_{i=1}^d \ts{\vector p^i}$.
So 
\[  
A(X) = A\qty{\Disjoint_{i=1}^d \ts{\vector p^i}} = \prod_{i=1}^d A\qty{\ts{\vector p^i}} = \prod_{i=1}^d \frac{ k[x_1, \cdots, x_n]} {\gens{x_1 - p^i_1, \cdots, x_n - p^i_n} }
\]
where $p_j^i$ is the $j$th component of $\vector p^i$.
:::

:::{.example}
Set $V(x_1 x_2) = X$, then $A(X) = k[x_1, x_2]/ \gens{x_1 x_2}$.
This not being a domain (since $x_1 x_2 = 0$) corresponds to $X = V(x_1) \union V(x_2)$ not being irreducible.
:::

:::{.example}
Let $X_1$ be the $xy\dash$plane and $X_2$ be the line parallel to the $y\dash$axis through $\thevector{0,0,1}$, and let $X= X_1 \disjoint X_2$:

![Union of a plane and a parallel line.](figures/image_2020-09-01-10-43-00.png){width=350px}

Then $X_1 = V(z)$ and $X_2 = V(x, z-1)$, and 
\[
I(X) = \gens{z} \cdot \gens{x, z-1}=  \gens{xz, z^2 - z}
\]
The coordinate ring is then given by 
\[
A(X) = { \CC[x, y, z] \over \gens{xz, z^2 - z} } = { \CC[x, y, z] \over  \gens{z} } \oplus { \CC[x, y,z] \over \gens{x, z-1} }
\]
:::