# Irreducible variety is connected in Zariski topology

I want to prove that an irreducible affine variety is connected in the Zariski topology. Actually this is an exercise from Dummit and Foote's Abstract Algebra. However, I have two questions:

• Is the nonempty open set mentioned in problem 12(also mentioned in problem 11) corresponding to the whole space $$\Bbb A^n$$ or the variety itself? It's hard for me to figure out what author mean.
• I have tried to think about it for a long while. My attempt was proof by contradiction(I assumed the nonempty "open" sets is open with respect to the whole space $$\Bbb A^n$$). Let $$V$$ be an irreducible affine variety, $$A,~B$$ be open in $$\Bbb A^n$$ and $$V\subseteq A\cup B$$, $$A\cap B=\emptyset$$, $$A\cap V\neq\emptyset$$, and $$B\cap V\neq\emptyset$$. I want to deduce a contradiction, but fail.

• The Zariski topology on an affine variety is the same as the subspace topology for the variety as a subset of $\mathbb{A}^n$ with the Zariski topology. I think this is why they are a little loose with saying where the opens are open. Commented May 14, 2019 at 15:50
• As user113102 pointed out, it doesn't really matter where the open sets live. However, to clarify the problem slightly, the fact you are given is that $A\cap V$ is dense in $V$, see problem 11. If you use problem 11, you are one sentence away from a contradiction.
– jgon
Commented May 14, 2019 at 18:53

$$\bullet$$ A topological space is said to be connected if it is impossible to write it as the union of two disjoint non-empty open subsets.
$$\bullet \bullet$$ A topological space is said to be irreducible if it is non-empty and if any two non-empty open subsets have non-empty intersection connected.
The simplest connected but not irreducible topological space is $$X=\{f,o,o'\}$$ endowed with the topology whose open sets are: $$X,\emptyset, \{o\},\{o'\},\{o,o'\}.$$