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Could someone explain to me, what is Zariski Topology? Under what condition a topology can be called Zariski Topology? Between the set $$V(E)=\{P \in \mathrm{Spec}(R)|E \subseteq P\}$$ and $$D(r)=\{P \in \mathrm{Spec}(R) | r \notin P \},$$ which one of them is Zariski Topology? Thank you.

Blackoffe
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    In the Zariski topology on Spec(R), closed subsets are of the form V(I) and open subsets are of the form D(I), where I is an ideal of the commutative ring R. Take a look at any reference... –  Jun 04 '13 at 19:08
  • Could you please change the tag? Even if this is a topology which arises in an algebraic contest, it's not algebraic topology at all; please put it under "algebraic geometry" – Edoardo Lanari Jun 04 '13 at 19:15
  • Thanks a lot for the help, I tried to look up in some lecture notes and journals, but it seems like I still don't quite understand.. Does it mean that the collection of V(E) is the Zariski Topology? – Blackoffe Jun 04 '13 at 19:25

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The Zariski topology can refer either to a topology on affine space $\mathbb{A}^n(k)$ over a field $k$ or to a topology on Spec$R$ for rings $R$.

The definition in affine space helps give geometric intuition about the Zariski topology. Recall that the affine space $\mathbb{A}^n(k)$ is defined to be the set of $n$-tuples $(a_1,...,a_n)$, where $a_i\in k$. An algebraic set in $\mathbb{A}^n(k)$ is defined to be a subset of $\mathbb{A}^n(k)$ which is the zero locus of a collection of polynomials $f_1,...f_m\in k[x_1,...,x_n]$. Equivalently, you could describe an algebraic set as the zero locus of polynomials in $I$, for an ideal $I\subset k[x_1,...,x_n]$. The zero locus of this ideal is denoted $V(I)$.

The Zariski topology on $\mathbb{A}^n(k)$ is then defined by letting the closed sets be precisely the algebraic sets. The open sets are then complements of algebraic sets. The idea behind the Zariski topology is to endow affine space with a topology that is described in purely algebraic terms. For example, if you're looking at the affine space $\mathbb{A}^1(\mathbb{C})$, you could just give it the topology of $\mathbb{C}$. But if you're working over a field $k$ other than $\mathbb{R}$ or $\mathbb{C}$, there may be no natural topology on $k$. Therefore we define the Zariski topology on affine space purely algebraically. Notice that closed sets in the Zariski topology of $\mathbb{A}^1(\mathbb{C})$ are also closed in the usual topology on $\mathbb{C}$, but the Zariski topology is coarser.

The Zariski topology on Spec$R$ is defined by analogy with the above definition for affine space. Now we define $V(I)$ as you indicated, $V(I)=\{P\in$Spec$R : I\subset P\}$. We define the Zariski topology on Spec$R$ by declaring the closed sets to be those sets of the form $V(I)$, where $I\subset R$ is an ideal. In particular, this tells us that the open sets are complements of sets of the form $V(I)$. We denote the complement of $V(I)$ by $D(I)$. The sets $D(r)$ which you mentioned are important, since if $I=(r_1,...,r_n)$, then $D(I)=D(r_1)\cup...\cup D(R_n)$.

Note: you shouldn't think of "Zariski" as a property of a topology in the same sense that "Hausdorff" is a property of a topology. Rather, the Zariski topology is a specific construction equipping affine space or Spec$R$ with a topology.

user64480
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    If one is discussing both schemes and classical varieties in the same text, it is probably better to write $\mathbb{A}^n (k)$ for the space of $n$-tuples of elements of $k$ and reserve $\mathbb{A}^n_k$ for the scheme $\operatorname{Spec} k[x_1, \ldots, x_n]$. – Zhen Lin Jun 04 '13 at 21:56
  • Thank you very much, this answer everything. :) – Blackoffe Jun 05 '13 at 03:29
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The first defines the closed sets, the second is the canonical base for the topology (thus in particular consists of zariski-open sets).

Edoardo Lanari
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