Zariski Topology is Topology

Theorem

Let $k$ be a field.

Let $n \in \N_{>0}$.

Let $\tau$ be the Zariski topology on $k^n$.

Then $\tau$ is indeed a topology on $k^n$.

Proof

Let $A := k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials.

Recall that by definition of Zariski topology:

$\forall U \in \tau: \exists T_U \subseteq A: U = X \setminus \map V {T_U}$

where $\map V {T_U}$ denotes the zero locus of $T_U$.

Each of the open set axioms is examined in turn:

Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets

For each $\UU \subseteq \tau$:

\(\ds \bigcup \UU\)

\(=\)

\(\ds \bigcup_{U \mathop \in \UU} k^n \setminus \map V {T_U}\)

\(\ds \)

\(=\)

\(\ds k^n \setminus \bigcap_{U \mathop \in \UU} \map V {T_U}\)

De Morgan's Laws

\(\ds \)

\(=\)

\(\ds k^n \setminus \map V {\bigcup_{U \mathop \in \UU} T_U}\)

Intersection of Zero Loci is Zero Locus

\(\ds \)

\(\in\)

\(\ds \tau\)

$\Box$

Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets

For all $U,V \in \tau$:

\(\ds U \cap V\)

\(=\)

\(\ds \paren {k^n \setminus \map V {T_U} } \cap \paren {k^n \setminus \map V {T_V} }\)

\(\ds \)

\(=\)

\(\ds k^n \setminus \paren { \map V {T_U} \cup \map V {T_V} }\)

De Morgan's Laws

\(\ds \)

\(=\)

\(\ds k^n \setminus \map V {T_U \cdot T_V}\)

Union of two Zero Loci is Zero Locus

\(\ds \)

\(\in\)

\(\ds \tau\)

$\Box$

Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

Since $\map V 1 = \O$:

$k^n = k^n \setminus \map V 1 \in \tau$

$\Box$

All the open set axioms are fulfilled, and the result follows.

$\blacksquare$