Unraveling The Affine Mystery Is A Closed Injective Morphism Always Affine
Hey guys! Today, we're diving into a fascinating and somewhat tricky question in algebraic geometry: is a closed injective morphism always affine? This question, which pops up as Exercise 12.3 in Görtz and Wedhorn's renowned book Algebraic Geometry, has been lingering in the online forums for quite some time, with a five-year-old thread on Math Stack Exchange still lacking a definitive answer. This makes it a perfect topic for us to explore together, unraveling the concepts and nuances involved. My initial hunch? It might just be wrong, which makes the journey even more exciting!
The Core Question: Closed Injective Morphisms and Affineness
So, what exactly are we asking? Let's break it down. In the realm of algebraic geometry, we deal with morphisms, which are essentially maps between algebraic varieties or schemes. A closed morphism is one where the image of any closed set is also closed. An injective morphism is, as the name suggests, a map that is injective on points. Now, an affine morphism is a morphism that has a special relationship with affine schemes – schemes that are the spectrum of a ring. More precisely, a morphism f: X → Y is affine if, for every affine open subset V of Y, the preimage f⁻¹(V) is also an affine open subset of X.
The big question here is whether the properties of being closed and injective, when combined, automatically imply that a morphism is affine. This seems intuitive at first glance. After all, closed maps preserve topological structure, and injective maps ensure a one-to-one correspondence. But in the subtle world of schemes and algebraic geometry, intuition can often lead us astray. We need to rigorously examine the definitions and explore potential counterexamples.
To really get our hands dirty, let's consider some key concepts and theorems that might be relevant. We'll need to think about the properties of affine schemes, the behavior of closed immersions, and the relationship between morphisms and their underlying ring maps. We'll also need to be mindful of the different types of schemes – Noetherian, integral, etc. – as these conditions can significantly impact the validity of certain results. Throughout our exploration, we'll be keeping an eye out for potential pitfalls and edge cases that could invalidate a general statement.
Exploring the Key Concepts and Theorems
To tackle this problem effectively, we need to arm ourselves with the fundamental concepts and theorems that govern the behavior of morphisms, schemes, and affine varieties. First and foremost, let's delve deeper into the notion of affine schemes. An affine scheme is the prime spectrum of a commutative ring A, denoted as Spec(A). The points of Spec(A) are prime ideals of A, and the topology is defined by the Zariski topology, where closed sets correspond to ideals of A. Affine schemes serve as the building blocks of more general schemes, and their properties are crucial for understanding affine morphisms.
A critical aspect of affine schemes is their connection to ring homomorphisms. A morphism between affine schemes, say f: Spec(B) → Spec(A), is uniquely determined by a ring homomorphism φ: A → B. This correspondence allows us to translate geometric questions about morphisms into algebraic questions about ring maps, and vice versa. For instance, the injectivity of the morphism f can be related to the surjectivity of the ring map φ, and the closedness of f can be linked to the properties of the kernel of φ.
Another important concept to consider is that of a closed immersion. A closed immersion is a morphism f: X → Y that induces an isomorphism of X onto a closed subscheme of Y. In simpler terms, a closed immersion embeds X as a closed subset of Y, while preserving the scheme structure. Closed immersions are closely related to surjective morphisms of sheaves of rings, and they play a vital role in constructing and understanding subschemes.
Now, let's think about how these concepts might relate to our main question. If we have a closed injective morphism f: X → Y, we know that f maps X isomorphically onto a closed subscheme of Y. This suggests that we might be able to express X as the spectrum of a quotient ring of the ring associated with an affine open subset of Y. However, this is where the subtleties arise. We need to carefully analyze whether this quotient ring structure guarantees that f is indeed an affine morphism. We need to consider the implications of the injectivity and closedness conditions on the underlying ring maps and how they interact with the affine nature of schemes.
Furthermore, we should also consider the properties of Noetherian schemes and integral schemes. A Noetherian scheme is a scheme that can be covered by finitely many affine open subsets, where each corresponding ring is a Noetherian ring. Noetherian schemes often exhibit nicer properties than general schemes, and many theorems in algebraic geometry are stated with the assumption of being Noetherian. An integral scheme is a scheme where the structure sheaf has no zero divisors. Integral schemes are closely related to integral domains in ring theory, and they play a crucial role in studying the geometry of algebraic varieties.
Searching for a Counterexample: The Devil is in the Details
Given the complexity of the question, and the lack of a readily available answer, it's prudent to consider the possibility that the statement – a closed injective morphism is always affine – might be false. In mathematics, a single counterexample can shatter a seemingly plausible conjecture. So, let's put on our detective hats and start hunting for a potential counterexample.
To construct a counterexample, we need to find a closed injective morphism f: X → Y that is not affine. This means that there should exist an affine open subset V of Y such that the preimage f⁻¹(V) is not affine. This is where things get interesting. We need to carefully choose our schemes X and Y, and our morphism f, to satisfy these conditions.
One strategy we might employ is to look for a scheme X that is