Uniform Fiber Dimension In Sheaves On K-Schemes

Hey guys! Let's dive into an exciting topic in algebraic geometry: the dimensions of closed-point fibers in sheaves of modules on k-schemes. This is a fascinating area where algebraic geometry and commutative algebra meet, so buckle up, and let's get started!

The Quest for Uniform Dimension

In algebraic geometry, we often deal with schemes, which are geometric objects defined by commutative algebra. When we study sheaves of modules on these schemes, we encounter interesting questions about their properties. One such question is whether the fibers of these sheaves at closed points have the same dimension. Let's break down what this means and why it's important.

First, let's set the stage. Imagine we have an irreducible (or more generally, connected) scheme X over a field k. Think of X as our geometric playground. Now, let E be a sheaf of O-modules on X. A sheaf, in simple terms, is a way of assigning algebraic data (like modules) to open sets of our scheme. These modules tell us a lot about the local properties of our scheme. The question we're tackling today is all about understanding when these local properties behave nicely, specifically concerning the dimension of the fibers.

Now, what exactly are these fibers we keep mentioning? Well, if we pick a point x in our scheme X, the fiber of E at x, denoted as E(x), is essentially the module we get when we "specialize" E at the point x. More formally, it's the tensor product E ⊗O Ox/mx, where Ox is the local ring at x, and mx is its maximal ideal. This might sound a bit technical, but the key idea is that E(x) captures the behavior of E in the immediate vicinity of x. The dimension of this fiber, dimk E(x), is simply its dimension as a vector space over the field k. This dimension gives us a numerical measure of the "size" of the module at that point.

Defining Purity in Sheaves

To get a handle on this, we need a crucial definition: what does it mean for a sheaf to be "pure"? A sheaf E is said to be pure if for every O-module... Well, this is where it gets interesting, and where the original query leaves us hanging! To make this discussion complete, we need to clarify what conditions make a sheaf pure. Typically, purity is related to the concept of the support of a module. The support of a module M over a ring R, denoted SuppR(M), is the set of prime ideals p in R such that the localization Mp is nonzero. A sheaf E might be considered pure if, for example, the dimension of the support of E is constant across all points. Or, in the context of the fibers, one could define purity based on the uniformity of the dimensions of the fibers at closed points. In other words, E is pure if all the closed-point fibers E(x) have the same dimension.

The importance of this question becomes clear when we realize that schemes and sheaves are fundamental tools in algebraic geometry. Understanding the dimensions of fibers helps us classify and study the geometry of the schemes themselves. For instance, if we know that the fibers of a sheaf have the same dimension, it tells us that the sheaf behaves consistently across the scheme, which can simplify our analysis.

Let's delve deeper into the conditions that ensure the closed-point fibers of a sheaf of modules on a k-scheme share the same dimension.

Conditions for Uniform Fiber Dimension

Now, let's really get into the meat of the question: when can we guarantee that the closed-point fibers of a sheaf have the same dimension? This is a crucial question because it connects the abstract theory of sheaves with concrete geometric properties of schemes. We need to explore the conditions under which this uniformity holds.

One of the primary conditions that ensures uniform fiber dimension is the local freeness of the sheaf. Recall that a sheaf E of O-modules on a scheme X is locally free of rank r if there exists an open cover {Ui} of X such that for each i, the restriction E|Ui is isomorphic to the sheaf Ori, where Ori is the free module of rank r over the structure sheaf O restricted to Ui. In simpler terms, this means that locally, our sheaf looks like a free module, which is a very nice and well-behaved object. This local freeness is super important, guys.

If E is locally free of rank r, then for any point x in X, the fiber E(x) is a vector space over the residue field k(x) = Ox/mx of dimension r. This is a direct consequence of the properties of tensor products and the definition of local freeness. Essentially, since E looks like a free module locally, when we take the fiber at a point, we get a vector space whose dimension corresponds to the rank of the free module. Thus, if E is locally free, the dimension of the fiber is constant, and our question is answered in the affirmative!

However, not all sheaves are locally free. So, what happens then? Are there other conditions we can impose to ensure uniform fiber dimension? Absolutely! Another key concept is the notion of a coherent sheaf. A sheaf E is coherent if it is of finite type (meaning it's locally finitely generated) and if, for any morphism On → E, the kernel is also of finite type. Coherent sheaves are a very important class of sheaves in algebraic geometry, and they often arise in natural contexts.

The Role of Coherent Sheaves

For coherent sheaves, the situation is a bit more nuanced. While local freeness directly implies uniform fiber dimension, coherence alone is not sufficient. We need an additional condition: flatness. A sheaf E is flat at a point x if the stalk Ex is a flat Ox-module. Flatness is a technical condition from commutative algebra that ensures that certain operations (like taking tensor products) behave well. In the context of sheaves, flatness at a point means that the local behavior of the sheaf is "smooth" in a certain sense.

If E is a coherent sheaf on X that is flat over k, then the function x → dimk(x) E(x) is upper semi-continuous. This means that the dimension of the fiber can "jump down" but not "jump up" as we move along the scheme. If X is integral (i.e., reduced and irreducible), then this function is constant on a dense open subset of X. In other words, for a flat coherent sheaf on an integral scheme, the fiber dimension is "generically" constant. This is a powerful result, guys!

Now, if we impose the additional condition that the scheme X is integral and the sheaf E is flat and coherent, we get closer to our goal. However, to guarantee that the fiber dimension is the same at all closed points, we need something more. One such condition is that E is universally catenary, which is a property related to the chain conditions on prime ideals in the local rings of X. If X is universally catenary, then the fiber dimension is constant on any irreducible component of X.

Let's recap. We've seen that local freeness is a strong condition that guarantees uniform fiber dimension. For coherent sheaves, we need flatness and possibly additional conditions on the scheme (like integrality and being universally catenary) to ensure that the fiber dimension is constant, at least generically or on irreducible components.

Examples and Counterexamples

Theory is great, but let's make things concrete with some examples and counterexamples. This will really help solidify our understanding.

Example 1: Locally Free Sheaves

Consider the projective space Pn over a field k. The structure sheaf OPn is locally free of rank 1, as it's just the sheaf of regular functions. The tangent sheaf TPn is also locally free. Therefore, at any closed point x in Pn, the fibers of OPn and TPn have constant dimensions. Specifically, the fiber of OPn at x is isomorphic to k, so its dimension is 1. The fiber of TPn at x is a vector space of dimension n, which is also constant.

Example 2: Coherent But Not Locally Free

Let's look at a more interesting case. Consider the affine plane A2 = Spec k[x, y]. Let I be the ideal (x, y) in k[x, y], and let E be the sheaf associated with the module k[x, y]/I. This sheaf is coherent because it corresponds to a finitely generated module. However, E is not locally free at the origin (0, 0). The fiber of E at the origin is isomorphic to k, which has dimension 1. At any other point, the fiber is 0. So, in this case, the fiber dimension is not constant, highlighting the necessity of additional conditions like flatness for coherent sheaves.

Counterexample: Non-Flat Sheaf

To illustrate the importance of flatness, consider the following example. Let X = Spec k[t], and let E be the sheaf associated with the module k[t]/(t2). This sheaf is coherent, but it's not flat at the point corresponding to the ideal (t) in k[t]. At this point, the fiber has dimension 2, while at any other point, the fiber has dimension 1. This shows that even coherence is not enough to guarantee constant fiber dimension without flatness.

Conclusion

So, guys, we've journeyed through the fascinating world of sheaves of modules on k-schemes and explored the conditions under which the closed-point fibers have the same dimension. We've seen that local freeness is a powerful condition that guarantees uniform fiber dimension, while for coherent sheaves, we need additional conditions like flatness and properties of the scheme itself to ensure constancy. By examining examples and counterexamples, we've gained a deeper understanding of this important topic in algebraic geometry.

This exploration is just the tip of the iceberg, and there's a whole universe of fascinating results and open questions in this area. Keep exploring, keep questioning, and most importantly, keep learning! Algebraic geometry is a beautiful field, and understanding these fundamental concepts will open doors to even more exciting discoveries. Keep up the great work, everyone!