Irreducible Disconnected Affine Scheme: Example & Explanation

Hey everyone! Let's explore a fascinating topic in algebraic geometry: affine schemes that are irreducible but not connected. It might sound like a contradiction at first, but trust me, it's a cool concept. We're diving deep into the world of schemes, specifically affine schemes, and tackling a question that often pops up when you're getting to grips with these abstract spaces. The question we're addressing is: Can we find an example of an affine scheme that, despite being irreducible, isn't connected? So, buckle up, and let's get started!

Understanding the Basics: Irreducibility and Connectedness

Before we jump into the nitty-gritty, let's refresh our understanding of what it means for a scheme to be irreducible and connected. These are fundamental topological properties, and grasping them is key to understanding our main question. In the realm of topological spaces, a space is considered irreducible if it cannot be expressed as the union of two proper closed subsets. Think of it like this: you can't break it down into two smaller, distinct pieces that are both closed. Imagine a line; you can't divide it into two closed intervals without including the point of division. This concept extends to schemes, where we consider closed subsets defined by ideals in the underlying ring.

Connectedness, on the other hand, is a more intuitive idea. A topological space is connected if it cannot be written as the disjoint union of two non-empty open sets. In simpler terms, it's all in one piece. Picture a figure eight; it's not connected because you can easily separate the two loops. Now, in the context of schemes, connectedness relates to the decomposition of the scheme into open subsets that don't overlap. The connection between irreducibility and connectedness is where things get interesting. Intuitively, you might think that if something is irreducible (can't be broken into closed pieces), it should also be connected (all in one piece). However, this intuition doesn't always hold, especially in the world of schemes!

The Affine Scheme Spec A: A Quick Recap

Now, let's zoom in on affine schemes. These are schemes that can be represented as the spectrum of a ring, denoted as Spec A, where A is a commutative ring. Spec A is a topological space whose points are prime ideals of A, and the topology is defined using ideals of A. This construction is the cornerstone of algebraic geometry, allowing us to translate algebraic properties of rings into geometric properties of spaces. The prime ideals of A play a crucial role in determining the topology and the structure of Spec A. They are the building blocks of the irreducible closed subsets, and their relationships dictate how Spec A is connected. For instance, the nilradical of A, which is the ideal of all nilpotent elements, is closely related to the irreducibility of Spec A. Remember, an affine scheme Spec A is irreducible if and only if the nilradical of A is a prime ideal. This is a key result that links the algebra (properties of the ring A) to the geometry (irreducibility of Spec A).

Understanding the relationship between the ring A and the scheme Spec A is crucial. Properties of the ring A directly translate into geometric properties of the scheme. For example, if A is an integral domain (has no zero divisors), then the zero ideal is a prime ideal, and Spec A is irreducible. However, the converse isn't always true, especially when we consider rings with nilpotents. Nilpotent elements, those that become zero when raised to some power, can create scenarios where Spec A is irreducible but not connected.

The Key Condition: Nilradical and Prime Ideals

To reiterate a crucial point, the irreducibility of an affine scheme Spec A hinges on the nilradical of A being a prime ideal. The nilradical, denoted as N(A), is the set of all nilpotent elements in A. A nilpotent element is one that, when raised to some positive integer power, equals zero. For example, in the ring Z/8Z (integers modulo 8), the element 2 is nilpotent because 2^3 = 8 ≡ 0 (mod 8). The nilradical N(A) forms an ideal in A, and its properties significantly influence the geometric nature of Spec A. If N(A) is a prime ideal, then Spec A is irreducible. This connection between the algebraic structure of A (specifically, the nilradical) and the geometric property of irreducibility is a cornerstone of the theory of schemes.

Now, why is this the case? Recall that a scheme is irreducible if it cannot be written as the union of two proper closed subsets. In Spec A, closed subsets correspond to ideals in A. If N(A) is prime, then any decomposition of Spec A into closed subsets must somehow reflect the structure of N(A). The primality of N(A) puts a strong constraint on how such a decomposition can occur, ultimately forcing Spec A to be irreducible. However, this doesn't automatically guarantee connectedness. We need to delve deeper to see how irreducibility and connectedness can diverge.

The Challenge: Irreducibility vs. Connectedness

The core of our challenge lies in finding an affine scheme that ticks the irreducibility box but fails to be connected. The irreducibility condition, as we've seen, is tied to the nilradical being a prime ideal. But what about connectedness? Connectedness is about whether the space can be separated into disjoint open sets. In the language of rings, this translates to the absence of nontrivial idempotents. An idempotent element in a ring A is an element e such that e^2 = e. If a ring A has a nontrivial idempotent (i.e., an idempotent other than 0 and 1), then Spec A is disconnected. Think of the idempotent as a switch that splits the ring, and hence the scheme, into two independent parts. The absence of such a switch ensures connectedness. So, our quest boils down to finding a ring where the nilradical is prime, but there's a