# Why Zariski topology is not Hausdorff

I am reading the book about Algebraic geometry. I am confused about the following two things the book mentioned:

Zariski topology is

1. different from the topology studied in real and complex analysis.
2. not Hausdorff.

Well, I roughly now about Hausdorff and some def. in real and complex analysis. However, I cannot know why still.

A hint or direction for thinking is ok.

Thanks,

## 2 Answers

The Zariski topology on $K$, where $K$ is a finite field, is indeed Hausdorff. However, as soon as $K$ is infinite, it is not Hausdorff. This might be interesting, if you are reading a book on algebraic geometry. The Zariski topology is the coarsest topology which satisfies the $T_1$ separation axiom, which means that singleton sets are closed.

So let $K$ be an infinite field. Then The Zariski topology over $K$ is not Hausdorff. Indeed, if we have any two nonempty open sets $U$ and $V$ , then $U^c$ and $V^c$ are finite, so $(U \cap V )^c = U^c \cup V^c$ is finite as well. In particular, $U \cap V$ is nonempty. Thus, open neighborhoods of any two points $x$ and $y$ will intersect.

• May I ask why Zarisky topology is $T_1$? Are there any conditions assumed here? A simple counterexample is the Zarisky topology of $\mathbb{Z}$. – Bingyan Liu Feb 1 at 16:38
• @BingyanLiu $K=\Bbb Z$ is not a field. – Dietrich Burde Feb 1 at 17:56
• this answer here: math.stackexchange.com/questions/1918669/…, fixes the argument for when we are dealing with $K^n$ with $n>1$. – John Doe May 14 at 19:08

Hint: What are the Zariski open subsets of $\mathbb C$?

• If its complement is Zariski closed. But in general case the complement of an open set is closed. – sleeve chen Dec 17 '14 at 9:38
• That's true for any topology. The question is more the following: What are the open sets in this particular topology? – Simon Rose Dec 17 '14 at 10:33