# Definition of open set in Zariski Topology

I'm being confused by the definition of the open set in Shafarevich's Basic Algebraic Geometry I. He says that in p.23:

"Let $X\subset A^n$ be a closed subset of affine space. We say that $U\subset X$ is open if its complement $X∖U$ is closed."

how is this differ from the definition of usual Zariski Topology (taking complement in $A^n$ instead of $X$)?

Also note that using this definition of open set, Shafarevich proved that (p.36) the set of points for which a given rational map is regular forms a nonempty and open set.

How do we say about morphism using this definition?

## 1 Answer

It doesn't differ - this is just the subspace topology on $X$.

Given a topological space $A$ and a subset $B\subseteq A$, we can give $B$ the subspace topology.

Let us suppose $B$ is a closed subset of $A$.

Because we have made $B$ into a topological space via the subspace topology, we speak of subsets of $B$ being open or closed.

By the definition of the subspace topology, $C\subseteq B$ is open in $B$ if and only if there exists some $E\subseteq A$, that is open in $A$, such that $E\cap B = C$.

By the definition of "closed", the complement of $E$ in $A$, namely $A\setminus E$, is a closed subset of $A$.

Because $A\setminus E$ and $B$ are both closed subsets of $A$, their intersection $(A\setminus E)\cap B$ is also a closed subset of $A$.

Their intersection is equal to $B\setminus (B\cap E)=B\setminus C$.

Thus, $C\subseteq B$ is open in $B$ if and only if $B\setminus C$ is a closed subset of $A$.

In this case, $A=\mathbb{A}^n$ is affine space with the Zariski topology, $B=X$, and $C=U$.

In the quote from Shafarevich, at the end there appears "$X\setminus U$ is closed", which can be ambiguous when it is said without reference to which topological space it is closed in. However, it is implicitly being taken to mean in the big space, i.e. $A$.

• I can't understand your argument about existence of $D$. But, I think If we let $X$ to be subspace topology, then the set $U$ is open in $(X,\tau_X)$ if and only if $U=X \cap D$ for some $D$ open in $(\mathbb{A}^n,\tau)$. Since $X$ is closed in $\mathbb{A}^n$, we can't say that $U$ is open in $\mathbb{A}^n$ (unless there's special thing in our topological space) thanks for guidance. – Ajat Adriansyah Apr 24 '11 at 7:14
• @Ajat - I have added an alternative explanation, and tried to explain the steps. There is nothing that depends specially on $\mathbb{A}^n$ - this is a completely general argument. If you are still unsure let me know. – Zev Chonoles Apr 24 '11 at 7:35
• Dear Zev, Thanks for your explaination :) – Ajat Adriansyah Apr 24 '11 at 7:41