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Zariski Topology


An ideal $ P$ of a ring $ R$ is called prime if whenever $ ab\in P$ for $ a,b\in R$, then $ a\in P$ or $ b\in P$ (or both).

A variety $ X$ is called irreducible if for any decomposition $ X=X_1\cup X_2$ of $ X$ into a union of two subvarieties, either $ X_1=X$ or $ X_2=X$. In other words, there are no non-trivial decompositions of $ X$ into smaller varieties.

Proposition 2   A variety $ X$ is irreducible iff its ideal $ I(X)$ is prime.

Thus, there exists a one-to-one correspondence between the set of varieties $ \mathcal{X}$ and the set of prime (homogeneous if projective $ X$) ideals $ \mathcal{P}$ in the corresponding polynomial ring.

Theorem 4   Any radical ideal $ I\subset
K[x_1,...,x_n]$ is uniquely expressible as a finite intersection of prime ideals $ P_i$ with $ P_i\not\subset P_j$ for $ i\not = j$. Equivalently, any variety $ X$ can be uniquely expressed as a finite union of irreducible subvarieties $ X_i$ with $ X_i\not\subset X_j$ for $ i\not = j$.

The varieties $ X_i$ appearing in this unique decomposition are called the irreducible components of $ X$.

Definition. Let $ X$ be a set of points in some space. A topology on $ X$ is a set $ \mathcal{T}$ of designated subsets of $ X$, called the open sets of $ X$, so that the following axioms are satisfied:

The union of any collection of open sets is open.

The intersection of any finite collection of open sets is open.

$ X$ and $ \emptyset$ are open.

The closed sets in $ X$ are the complements of the open sets.

We define below the so-called Zariski topology on algebraic varieties. If we work over $ \mathbb{C}$, every variety can be roughly viewed as a complex manifold $ X$ (with the exception of a proper subset of its singular points). Through its embedding in, say, $ \mathbb{C}^n$, $ X$ will inherit the usual complex analytic topology from $ \mathbb{C}^n$ - a basis for the open sets on $ X$ will consist of the intersections of $ X$ with any finite balls in $ \mathbb{C}^n$.

The Zariski topology is a different kind of topology. A basis for the open sets in $ X$ is given by the sets

$\displaystyle U_f=\{p\in X\,\,\vert\,\,f(p)\not = 0\}$

where $ f$ ranges over polynomials (homogeneous if projective $ X$).

Lemma 2   The Zariski topology is indeed a topology on $ X$.

Exercise. Show that the Zariski topology on the projective line $ \mathbb{P}^1_{\mathbb{C}}$ is different from the analytic topology of $ \mathbb{P}^1_{\mathbb{C}}$.

Many statements in algebraic geometry are true for general points on varieties, i.e. if $ X$ is a variety and $ U$ is an open dense set of $ X$, then any point $ p\in U$ is called a general point on $ U$. (If $ X$ is irreducible, then any nonempty open set will be dense. This, in particular, makes Zariski topology a non-Housdorff topology - in the latter, one needs for any two points of $ X$ to have two nonintersecting open sets containing each one of the points. This confirms once again that the Zariski topology is much coarser than the analytic topology.)

next up previous
Next: Bezout's Theorem Up: Algebraic Geometry Previous: Examples of Varieties
Zvezdelina Stankova-Frenkel 2001-01-24