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If we identify \(\mathbf{A}^{2}\) with \(\mathbf{A}^{1} \times \mathbf{A}^{1}\) in the natural way, show that the Zariski topology on \(\mathbf{A}^{2}\) is not the product topology of the Zariski topologies on the two copies of \(\mathbf{A}^{1}\)

Short Answer

Expert verified
No, the Zariski topology on \(\mathbf{A}^{2}\) is not the product topology of the Zariski topologies on the two copies of \(\mathbf{A}^{1}\). This is evidenced by sets like the one defined by xy=1, which is closed in Zariski topology on \(\mathbf{A}^{2}\), but not in the product topology.

Step by step solution

01

Understanding Zariski topology

Zariski topology on an algebraic set \(\mathbf{A}^{n}\) is defined by taking all algebraic subsets of \(\mathbf{A}^{n}\) as closed sets. This is different from usual topologies such as product topology in which, conceptually, ‘closeness’ or ‘continuity’ applies on each coordinate individually.
02

Understanding product topology

Product topology on a Cartesian product of topological spaces is formed by taking as open sets all possible unions of Cartesian products of open sets from each space. This principle screams independence, as each space contributes to the product topology independently, unlike the Zariski topology where all coordinates collectively influence the topology.
03

Contrasting Zariski and product topology

To understand this difference, consider the subset of \(\mathbf{A}^{1} \times \mathbf{A}^{1}\) (i.e. \(\mathbf{A}^{2}\)) defined by the equation xy=1. In the product Zariski topology, this set wouldn't be closed; only the subsets where either \(x=0\) or \(y=0\) are closed, given the independent nature of product topology. However, in the Zariski topology on \(\mathbf{A}^{2}\), the set defined by xy=1 is indeed closed because Zariski topology works collectively on all coordinates.
04

Conclusion: Zariski topology is not product topology

Hence, the Zariski topology on \(\mathbf{A}^{2}\) is not the product topology of the Zariski topologies on the two copies of \(\mathbf{A}^{1}\). The intersecting set defined by xy=1 fails to be closed in the product topology, but is closed in the Zariski topology on \(\mathbf{A}^{2}\), showcasing the key difference.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Topology
When dealing with the concept of product topology, it's essential to understand how sets interact in the combined environment of multiple topological spaces. Imagine each space in your product as contributing its open sets independently. In the product topology, open sets in the resulting space are formed by assembling pieces of these open sets from each individual space.

This means you take all possible unions of these Cartesian products. For instance, given two topological spaces, say \(X\) and \(Y\), the product would look like all combinations of open sets \(U \subseteq X\) and \(V \subseteq Y\). This creates a scope where each dimension retains its characteristics, acting independently from others.

  • The independence ensures each dimension has no say over the other.
  • Open sets are formed from all possible combinations of existing open sets in each space.
This notion gives product topology a unique structure, ensuring that it respects the individuality of spaces, unlike more interconnected topologies like the Zariski.
Algebraic Sets
Algebraic sets are crucial in understanding the Zariski topology. They're composed of solutions to systems of polynomial equations. For example, consider \(xy = 1\) as an algebraic equation. The set of solutions to this equation, such as hyperbolas or lines, forms an algebraic set.

In the realm of Zariski topology, these sets become the closed ones. Unlike the product topology, where subsets are judged on their own merits as open or closed, algebraic sets gain their status because they satisfy particular polynomial relations. This interdependent nature is particularly significant.

  • Closed sets in Zariski topology come from equations like \(xy = 0\).
  • They cover complex patterns and shapes embedded within the algebraic structure.
In this way, Zariski topology appeals more to the collective behavior of coordinates, as opposed to the independence emphasized in the product topology.
Coordinate Spaces
Coordinate spaces, such as \(\mathbf{A}^n\) in algebraic geometry, represent a broad framework for solving equations and understanding geometric properties. These spaces consist of tuples, or coordinated sets of numbers, each element in the tuple drawn from a corresponding coordinate line \(\mathbf{A}^1\).

Consider \(\mathbf{A}^2\) as a combination of two \(\mathbf{A}^1\) spaces. It encompasses all pairs of points possible in the plane. In coordinate spaces, one can explore various properties and interactions of these points, especially when extended into topological concepts like Zariski or product topology.

  • Coordinate spaces allow examining equations such as \(xy = 1\) across multiple dimensions.
  • They form the foundation in algebraic geometry for exploration of shapes and positions.
This structure plays a pivotal role, enabling the transition from individual algebraic equations to an all-encompassing view across larger dimensional spaces.

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Most popular questions from this chapter

Projective Closure of an Affine Varicty. If \(Y \subseteq \mathbf{A}^{n}\) is an affine variety, we identify \(\mathbf{A}^{n}\) with an open set \(U_{0} \subseteq \mathbf{P}^{n}\) by the homeomorphism \(\varphi_{0} .\) Then we can speak of \(\bar{Y},\) the closure of \(Y\) in \(\mathbf{P}^{n}\), which is called the projectice closure of \(Y\) (a) Show that \(I(\bar{Y})\) is the ideal generated by \(\beta(I(Y)),\) using the notation of the proof of (2.2) (b) Let \(Y \subseteq \mathbf{A}^{3}\) be the twisted cubic of (Ex. 1.2 ). Its projective closure \(\bar{Y} \subseteq \mathbf{P}^{3}\) is called the twisted cubic curce in \(\mathbf{P}^{3}\). Find generators for \(I(Y)\) and \(I(\bar{Y})\), and use this example to show that if \(f_{1}, \ldots, f_{r}\) generate \(I(Y),\) then \(\beta\left(f_{1}\right), \ldots, \beta\left(f_{r}\right)\) do not necessarily generate \(I(\bar{Y})\)

Let \(Y \subseteq \mathbf{P}^{n}\) be a projective variety of dimension \(r .\) Let \(f_{1}, \ldots, f_{t} \in S=\) \(k\left[x_{0}, \ldots, x_{n}\right]\) be homogeneous polynomials which generate the ideal of \(Y\). Let \(P \in Y\) be a point, with homogeneous coordinates \(P=\left(a_{0}, \ldots, a_{n}\right) .\) Show that \(P\) is nonsingular on \(Y\) if and only if the rank of the matrix \(\left\|\left(\partial f_{i} / \partial x_{j}\right)\left(a_{0}, \ldots, a_{n}\right)\right\|\) is \(n-r .\) [Hint: (a) Show that this rank is independent of the homogeneous coordinates chosen for \(P ;\) (b) pass to an open affine \(U_{i} \subseteq \mathbf{P}^{n}\) containing \(P\) and use the affine Jacobian matrix; (c) you will need Euler's lemma, which says that if \(\left.f \text { is a homogeneous polynomial of degree } d, \text { then } \sum x_{i}\left(\partial f / \partial x_{i}\right)=d \cdot f .\right]\)

(a) Show that the following conditions are equivalent for a topological space \(X:\) (i) \(X\) is noetherian; (ii) every nonempty family of closed subsets has a minimal element; (iii) \(X\) satisfies the ascending chain condition for open subsets; (iv) every nonempty family of open subsets has a maximal element. (b) A noetherian topological space is quasi-compact, i.e., every open cover has a finite subcover. (c) Any subset of a noetherian topological space is noetherian in its induced topology. (d) A noetherian space which is also Hausdorff must be a finite set with the discrete topology.

There are quasi-affine varieties which are not affine. For example, show that \(\mathrm{N}=\mathbf{A}^{2}-\\{(0,0)\\}\) is not affine. \([\text {Hint}: \text { Show that }((X) \cong h[x, 1] \text { and use }(3.5)\) See (III, Ex. 4.3 for another proof.

(a) Let \(Y\) be the plane curve \(y=x^{2}\) (i.e., \(Y\) is the zero set of the polynomial \(f=\) \(y-x^{2}\) ). Show that \(A(Y)\) is isomorphic to a polynomial ring in one variable over \(k\) (b) Let \(Z\) be the plane curve \(x y=1 .\) Show that \(A(Z)\) is not isomorphic to a polynomial ring in one variable over \(k\) (c) Let \(f\) be any irreducible quadratic polynomial in \(k[x, y],\) and let \(W\) be the conic defined by \(f .\) Show that \(A(W)\) is isomorphic to \(A(Y)\) or \(A(Z) .\) Which one is it when?

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