Zero Sets Closed? Sheaves Modules Schemes Explained

Hey guys! Let's dive deep into the fascinating world of algebraic geometry, commutative algebra, and schemes. Today, we're going to unravel a crucial concept: the closedness of zero sets in sections of sheaves of modules on schemes. This might sound like a mouthful, but trust me, we'll break it down piece by piece. So, grab your metaphorical hard hats, and let's get to work!

The Foundation: Schemes, Sheaves, and Modules

Before we can truly grasp the intricacies of closedness, we need to ensure we're all on the same page regarding the foundational concepts. Think of it like building a house; you can't put up the walls without a solid foundation, right? Similarly, we need a firm understanding of schemes, sheaves, and modules to tackle our main topic.

Schemes: The Geometric Landscape

First up are schemes. In algebraic geometry, schemes are the fundamental objects of study. They're like the geometric spaces where all the algebraic action happens. You can think of them as generalized manifolds, but instead of being built from smooth curves and surfaces, they're constructed from rings. Rings, in this context, are algebraic structures equipped with addition and multiplication operations. The connection between rings and geometry is what makes schemes so powerful. Schemes provide a framework for studying algebraic equations geometrically, and geometric objects algebraically. A scheme $(X, O)$ comprises a topological space $X$ and a structure sheaf $O$. The topological space gives us a notion of open sets and closeness, while the structure sheaf assigns a ring to each open set, encoding local algebraic information. It's like having a map (the topological space) overlaid with data tags (the sheaf) that tell you about the local algebraic properties.

Sheaves: Bundles of Information

Next, we have sheaves. Now, a sheaf is a tool for organizing algebraic data, such as rings or modules, over a topological space. Imagine a sheaf as a bundle of information attached to the open sets of our scheme. This information could be anything from functions to modules. The key idea is that the information varies continuously as we move across the space. A sheaf, in essence, is a way to keep track of how algebraic objects behave locally on a space and how these local behaviors glue together to give a global picture. Think of it as a sophisticated way of book-keeping in the world of algebraic geometry. The structure sheaf $O$ of a scheme $(X, O)$, which we mentioned earlier, is a prime example. It assigns to each open set $U$ in $X$ a ring $O(U)$, and these rings are compatible in a certain way. This compatibility is crucial for ensuring that our algebraic constructions behave consistently.

Modules: Algebraic Workhorses

Finally, let's talk about modules. In the context of ring theory, a module is a generalization of a vector space, where the scalars come from a ring instead of a field. Modules are the workhorses of commutative algebra and algebraic geometry. They allow us to study representations of rings and sheaves, and they pop up in various contexts, from solving systems of equations to understanding the geometry of schemes. A module over a ring is an algebraic object that behaves much like a vector space, but over a ring instead of a field. This flexibility allows us to capture more intricate algebraic structures. Sheaves of modules are a natural extension of this concept, where we have a module associated with each open set in our space, and these modules are compatible in a similar fashion to the rings in a structure sheaf.

The Central Question: Closedness of Zero Sets

Okay, with our foundation firmly in place, we can now tackle the central question: What does it mean for the zero set of a section of a sheaf of modules on a scheme to be closed? This might still sound abstract, but let's break it down step by step.

Setting the Stage

Let's set the stage with some notation. We have a scheme $(X, O)$, where $X$ is the topological space and $O$ is the structure sheaf. We're also given a sheaf of modules $E$ on $X$. Think of $E$ as a collection of modules, one for each open set in $X$, that are glued together in a compatible way. Now, consider a section $s$ of the sheaf $E$. A section is simply a choice of an element from the module $E(U)$ for each open set $U$ in $X$, such that these choices are compatible under restriction. It's like choosing a specific value from each module in our collection, making sure these values play nicely together.

Defining the Zero Set

The zero set of the section $s$, denoted as $Z(s)$, is the set of points $x$ in $X$ where the stalk $s_x$ of $s$ at $x$ is zero. The stalk $s_x$ is the localization of the module at the prime ideal corresponding to the point $x$. In simpler terms, the stalk is a way to zoom in on the behavior of the module at a particular point. So, the zero set consists of all points where our section vanishes locally. Imagine a function on a space; its zero set is the set of points where the function equals zero. In our case, the section $s$ plays the role of the function, and the zero set $Z(s)$ is the set of points where this section becomes zero in a local sense.

The Closedness Conjecture

Our main question boils down to this: Is the zero set $Z(s)$ a closed subset of $X$? In other words, can we be sure that the points where our section vanishes form a closed set in our topological space? This is a crucial question because closed sets have nice properties in topology and geometry. They often correspond to geometric objects like curves and surfaces, so understanding their behavior is essential.

The Proof: A Journey Through Localization and Affine Schemes

Now, let's embark on the journey of proving that the zero set $Z(s)$ is indeed closed. This proof involves some clever techniques from commutative algebra and algebraic geometry, particularly localization and the use of affine schemes.

The Power of Localization

Localization is a fundamental tool in commutative algebra. It allows us to focus on the behavior of a ring or module at a particular prime ideal. Imagine you have a map, and you want to study the details of a specific region. Localization is like zooming in on that region, allowing you to see the finer details. In our context, we use localization to study the behavior of the sheaf $E$ and the section $s$ at specific points in our scheme.

Affine Schemes: Our Local Laboratory

Affine schemes are schemes that are isomorphic to the spectrum of a ring, denoted as Spec $A$. These schemes are the building blocks of all schemes. Think of them as the local laboratories where we can perform algebraic experiments. They're particularly nice to work with because we can translate geometric questions into algebraic ones and vice versa. To prove that $Z(s)$ is closed, we can use the fact that closedness is a local property. This means that if $Z(s)$ is closed in a neighborhood of every point in $X$, then it is closed in $X$. So, we can cover $X$ by affine open sets and focus on proving closedness within each affine open set. This simplifies our problem considerably, as we can leverage the algebraic machinery available for affine schemes.

The Proof Unfolds

Let's dive into the nitty-gritty of the proof. Suppose we have an affine open subset $U$ of $X$. Since $U$ is affine, it is isomorphic to Spec $A$ for some ring $A$. Now, the restriction of the sheaf $E$ to $U$, denoted as $E|_U$, corresponds to a module $M$ over the ring $A$. Similarly, the restriction of the section $s$ to $U$ corresponds to an element $m$ in the module $M$. So, within our affine open set, we've translated our sheaf-theoretic problem into a module-theoretic one.

The zero set of $s$ in $U$, denoted as Z(s) igcap U, consists of all prime ideals $p$ in $A$ such that the localization of $m$ at $p$, denoted as $m_p$, is zero in the localization of $M$ at $p$, denoted as $M_p$. In other words, we're looking for all prime ideals where our element $m$ vanishes locally. Now, a crucial observation is that $m_p = 0$ in $M_p$ if and only if there exists an element $f$ in $A$ that is not in $p$ such that $fm = 0$ in $M$. This is a standard result from commutative algebra, and it's the key that unlocks our proof.

Consider the annihilator of $m$ in $M$, denoted as Ann$(m)$. This is the set of all elements $a$ in $A$ such that $am = 0$ in $M$. The annihilator is an ideal in $A$, and it captures all the elements that