Coordinate Ring Direct Product And Variety Disconnectedness A Detailed Proof And Geometric Insights

Introduction

In the fascinating realm of algebraic geometry, a profound connection exists between the algebraic structure of a coordinate ring and the geometric properties of the variety it represents. Specifically, we delve into the relationship between a coordinate ring being a direct product of rings and the variety being disconnected. This exploration uncovers how the algebraic decomposition of the coordinate ring directly reflects the geometric separation of the variety into disjoint components. In this article, we aim to provide a comprehensive understanding of this relationship, offering a detailed proof and geometric insights. This concept is crucial for understanding the interplay between algebra and geometry, allowing us to translate algebraic properties into geometric interpretations and vice versa.

Coordinate Rings and Varieties

To begin, let's establish the foundational concepts. Consider an affine variety V in the affine n-space 𝔸ⁿ over an algebraically closed field k. The coordinate ring k[V] is defined as the quotient ring k[x₁, ..., x_n] / I(V), where I(V) is the ideal of all polynomials in k[x₁, ..., x_n] that vanish on V. The coordinate ring k[V] essentially captures the polynomial functions on the variety V. It serves as a bridge connecting the geometric object V with its algebraic representation. Understanding the structure of k[V] provides significant insights into the geometric properties of V.

Disconnected Varieties

A variety V is said to be disconnected if it can be expressed as a disjoint union of two non-empty, closed subsets. Formally, V = V₁ ∪ V₂, where V₁ and V₂ are closed subsets of V, V₁ ∩ V₂ = ∅, and both V₁ and V₂ are non-empty. Disconnectedness is a fundamental geometric property indicating that the variety consists of separate, non-intersecting components. Visualizing a disconnected variety helps in intuitively grasping the algebraic implications on its coordinate ring. For instance, consider two disjoint curves in the plane; their union forms a disconnected variety.

The Core Theorem: k[V] as a Direct Product Iff V is Disconnected

The central theorem we aim to explore and prove is: The coordinate ring k[V] is isomorphic to a direct product of rings if and only if the variety V is disconnected. This theorem provides a powerful link between algebraic structure and geometric connectivity. It states that if we can decompose the coordinate ring into a direct product, then the variety must consist of separate components, and conversely, if the variety is disconnected, its coordinate ring can be expressed as a direct product. This bidirectional relationship is crucial for translating algebraic properties into geometric interpretations and vice versa.

Proof: k[V] as a Direct Product Implies V is Disconnected

Let us first prove the forward direction: If k[V] is a direct product of rings, then V is disconnected. Assume that k[V] ≅ R₁ × R₂, where R₁ and R₂ are non-zero rings. This implies that there exist non-trivial idempotents in k[V]. An idempotent element e in a ring R is an element such that e² = e. In the direct product R₁ × R₂, the elements (1, 0) and (0, 1) are idempotents, where 1 and 0 are the multiplicative and additive identities, respectively.

Let e ∈ k[V] be a non-trivial idempotent, meaning e ≠ 0 and e ≠ 1. Since e is an element of the coordinate ring, it corresponds to a polynomial function on V. Consider the closed subsets V₁ = {P ∈ V | e(P) = 0} and V₂ = {P ∈ V | e(P) = 1}. These subsets are closed because they are the zero sets of the polynomials e and e - 1, respectively. Furthermore, V₁ ∩ V₂ = ∅ because if there were a point P in both V₁ and V₂, we would have e(P) = 0 and e(P) = 1 simultaneously, which is a contradiction.

Now, we show that V = V₁ ∪ V₂. For any point P ∈ V, we have e(P) ∈ k. Since e is an idempotent, e² = e, so e(P)² = e(P). This implies that e(P) must be either 0 or 1, because these are the only elements in a field that are equal to their squares. Thus, every point P in V must belong to either V₁ or V₂, meaning V = V₁ ∪ V₂.

Finally, we need to show that V₁ and V₂ are non-empty. If V₁ were empty, then e would be 0 on all of V, contradicting the non-triviality of e. Similarly, if V₂ were empty, then e would be 1 on all of V, again contradicting the non-triviality of e. Therefore, V₁ and V₂ are non-empty, closed, and disjoint subsets of V, which means that V is disconnected.

Proof: V Disconnected Implies k[V] as a Direct Product

Next, we prove the converse: If V is disconnected, then k[V] is a direct product of rings. Suppose V = V₁ ∪ V₂, where V₁ and V₂ are non-empty, closed, and disjoint subsets of V. Since V₁ and V₂ are closed, they are themselves varieties. Let I(V₁) and I(V₂) be the ideals of polynomials vanishing on V₁ and V₂, respectively.

Since V₁ and V₂ are disjoint, V₁ ∩ V₂ = ∅. By the Nullstellensatz, I(V₁ ∩ V₂) = I(∅) = (1) (the entire polynomial ring), and I(V₁ ∪ V₂) = I(V). Also, I(V₁ ∩ V₂) = √(I(V₁) + I(V₂)), so I(V₁) + I(V₂) = (1). This means there exist polynomials f ∈ I(V₁) and g ∈ I(V₂) such that f + g = 1. Consider the element e = g in k[V]; then e² = g² ≡ g (mod I(V)), because g² - g = g( g - 1) = g(-f) ∈ I(V). Thus, e is an idempotent in k[V].

Define a map φ: k[V] → k[V₁] × k[V₂] by φ(h) = (h|_V1, h|_V2), where h|_Vi denotes the restriction of the polynomial h to V_i. This map is a ring homomorphism. We claim that φ is an isomorphism. To show this, we need to prove that φ is both injective and surjective.

Injectivity: Suppose φ(h) = (0, 0). This means h vanishes on both V₁ and V₂, so h ∈ I(V₁) ∩ I(V₂). Since V = V₁ ∪ V₂, h vanishes on V, implying h ∈ I(V). Thus, h = 0 in k[V], and φ is injective.

Surjectivity: Let (h₁, h₂) ∈ k[V₁] × k[V₂]. We need to find a polynomial h ∈ k[V] such that h|_V1 = h₁ and h|_V2 = h₂. Consider the polynomial h = h₂ f + h₁ g, where f + g = 1 as before. Then, on V₁, f = 0, so h|_V1 = h₁ g|_V1 = h₁(1 - f)|_V1 = h₁. Similarly, on V₂, g = 0, so h|_V2 = h₂ f|_V2 = h₂(1 - g)|_V2 = h₂. Thus, φ(h) = (h₁, h₂), and φ is surjective.

Therefore, k[V] ≅ k[V₁] × k[V₂], demonstrating that the coordinate ring k[V] is a direct product of rings when V is disconnected. This completes the proof of the theorem.

Geometric Interpretation

The geometric intuition behind this theorem is quite elegant. When a variety V is disconnected, it means it comprises distinct, separate components. Each component corresponds to a factor in the direct product of the coordinate ring. The idempotents in k[V] act as characteristic functions for these components. For example, if V consists of two disjoint subvarieties, V₁ and V₂, then the idempotent e can be thought of as a function that is 0 on V₁ and 1 on V₂ (or vice versa). This function effectively