Ext⁰(M, N) = Hom(M, N): A Detailed Proof

by Omar Yusuf 41 views

Hey guys! Ever wondered how the Ext functor connects to the familiar Hom functor in the realm of homological algebra? Specifically, we're diving deep into proving that Ext⁰(M, N) is actually the same thing as Hom(M, N). This is a fundamental result, and while you might've heard about using the left-exactness of the Hom functor, we're going to tackle this head-on, directly from the definition using projective resolutions. So, buckle up, and let's get started!

Understanding the Basics: Hom and Ext

Before we jump into the proof, let's make sure we're all on the same page with the core concepts. The Hom functor, denoted as Hom(M, N), represents the set of all module homomorphisms (or linear transformations, depending on your context) from module M to module N. Think of it as the collection of all ways you can map elements from M to N while preserving the module structure (addition and scalar multiplication). This is a fundamental concept in module theory, and it provides a way to study the relationships between different modules.

Now, let's talk about Ext. The Ext functor, Extⁿ(M, N), is a bit more sophisticated. It arises from the study of extensions of modules and is a crucial tool in homological algebra for measuring the failure of exactness of the Hom functor. In simpler terms, Ext helps us understand how modules M and N can be "glued together" in various ways. To compute Extⁿ(M, N), we typically use a projective resolution of M (or an injective resolution of N). A projective resolution is an exact sequence of projective modules that "resolves" M. Projective modules are special because they have the property that any map from a projective module can be lifted through a surjective map, which makes them ideal for constructing resolutions. The Ext functor then emerges from taking the cohomology of a certain complex derived from this resolution. Don't worry if this sounds complicated; we'll break it down further as we go along.

So, the goal here is to show that when n = 0, Ext⁰(M, N) magically simplifies to Hom(M, N). This connection between Ext⁰ and Hom is a cornerstone in homological algebra, providing a bridge between the abstract machinery of Ext and the concrete world of module homomorphisms. It also highlights how Ext can be seen as a generalization of Hom, capturing more intricate relationships between modules.

The Proof: Unveiling the Equality

Okay, let's get our hands dirty with the proof! To show that Ext⁰(M, N) = Hom(M, N), we'll walk through the construction of Ext using a projective resolution and carefully examine what happens at the 0th level. Remember, the key idea is to build a projective resolution of M, apply the Hom functor, and then look at the resulting complex.

1. Start with a Projective Resolution:

We begin by choosing a projective resolution of M. This means we need an exact sequence of the form:

... → P₂ → P₁ → P₀ → M → 0

Where:

  • Each Pᵢ is a projective module. Remember, projective modules are "nice" modules that allow us to lift maps, which is crucial for constructing resolutions.
  • The sequence is exact. This means that the image of each map is equal to the kernel of the next map. Exactness is the cornerstone of homological algebra, ensuring that our resolution accurately represents M.
  • The map P₀ → M is surjective (onto). This means that every element in M has a pre-image in P₀, ensuring that our resolution "covers" all of M.

Constructing a projective resolution is always possible, although the specific construction can vary depending on the module M. For example, if M is a finitely generated module over a Noetherian ring, we can find a projective resolution where each Pᵢ is also finitely generated. The existence of projective resolutions is a fundamental result in homological algebra, and it allows us to define the Ext functor.

2. Apply the Hom Functor:

Now, we apply the contravariant Hom functor Hom(-, N) to our projective resolution. This means we replace each module Pᵢ in the resolution with Hom(Pᵢ, N) and reverse the direction of the arrows. Applying Hom(-, N) to the projective resolution, we obtain the following complex:

0 → Hom(M, N) → Hom(P₀, N) → Hom(P₁, N) → Hom(P₂, N) → ...

Notice a few key things here:

  • The arrows are reversed. This is because Hom is a contravariant functor, meaning it reverses the direction of maps.
  • We've added a 0 at the beginning. This is important for defining the cohomology of the complex.
  • Each term Hom(Pᵢ, N) is the set of module homomorphisms from Pᵢ to N. These are the "building blocks" of our Ext computation.

The maps in this complex are induced by the maps in the original projective resolution. Specifically, if dᵢ: Pᵢ → Pᵢ₋₁ is a map in the projective resolution, then the corresponding map in the complex is given by Hom(dᵢ, N): Hom(Pᵢ₋₁, N) → Hom(Pᵢ, N), where Hom(dᵢ, N)(f) = f ∘ dᵢ for any f ∈ Hom(Pᵢ₋₁, N). This composition is what makes the complex "work" and allows us to define cohomology.

3. Compute the Cohomology:

The Ext groups are defined as the cohomology of this complex. In general, the nth Ext group is given by:

Extⁿ(M, N) = ker(Hom(dₙ₊₁, N)) / im(Hom(dₙ, N))

Where ker denotes the kernel (the set of elements that map to zero) and im denotes the image (the set of elements that are the result of a map). Cohomology measures the "failure of exactness" of the complex. If the complex were exact, the kernel would be equal to the image, and the cohomology groups would be zero. However, in general, this is not the case, and the cohomology groups provide valuable information about the modules M and N.

For our specific case, we want to find Ext⁰(M, N). This corresponds to the 0th cohomology group, so we need to look at the maps at the beginning of the complex:

0 → Hom(M, N) → Hom(P₀, N) → Hom(P₁, N) → ...

Thus,

Ext⁰(M, N) = ker(Hom(d₁, N)) / im(Hom(d₀, N))

Where d₁: P₁ → P₀ and d₀: P₀ → M are the maps from our projective resolution. The map Hom(d₁, N): Hom(P₀, N) → Hom(P₁, N) is given by f ↦ f ∘ d₁, and the map Hom(d₀, N): Hom(M, N) → Hom(P₀, N) is given by g ↦ g ∘ d₀. Remember that d₀ is surjective because our resolution is exact.

4. The Key Insight: Left-Exactness at Play:

Here's where the magic happens! Let's carefully analyze the kernel and the image in our Ext⁰ formula. The kernel ker(Hom(d₁, N)) consists of all maps f: P₀ → N such that f ∘ d₁ = 0. This means that f vanishes on the image of d₁, which is the same as the kernel of d₀ (by exactness of the projective resolution). So, the maps in the kernel "factor through" M.

The image im(Hom(d₀, N)) consists of all maps of the form g ∘ d₀, where g: M → N. Now, consider the map Hom(d₀, N): Hom(M, N) → Hom(P₀, N) given by g ↦ g ∘ d₀. We want to show that this map is injective (one-to-one). Suppose g ∘ d₀ = 0 for some g: M → N. Since d₀ is surjective, this implies that g must be zero. Therefore, Hom(d₀, N) is injective.

Because d₀: P₀ → M is surjective, we can see that Hom(M, N) is precisely the kernel of Hom(P₀, N) → Hom(P₁, N). This is a manifestation of the left-exactness of the Hom functor! In fact, this is exactly what we wanted to show.

5. The Grand Finale:

Putting it all together, we have:

Ext⁰(M, N) = ker(Hom(d₁, N)) / im(Hom(d₀, N))

Since Hom(d₀, N) is injective, we have

im(Hom(d₀, N)) ≅ Hom(M, N)

Because of the exactness of the sequence

0 -> Hom(M, N) -> Hom(P₀, N) -> Hom(P₁, N)

We have

Hom(M, N) ≅ ker(Hom(P₀, N) -> Hom(P₁, N)) = ker(Hom(d₁, N))

Therefore,

Ext⁰(M, N) = ker(Hom(d₁, N)) / im(Hom(d₀, N)) ≅ Hom(M, N) / 0 ≅ Hom(M, N)

And there you have it! We've successfully proven that Ext⁰(M, N) is indeed isomorphic to Hom(M, N) using the definition of Ext via projective resolutions. High five!

Why This Matters: Applications and Implications

Okay, so we've proven a cool result, but why should we care? What's the big deal about Ext⁰(M, N) being equal to Hom(M, N)? Well, this result has several important applications and implications in homological algebra and beyond.

  • Conceptual Clarity: This result provides a crucial link between the abstract world of Ext functors and the more concrete world of module homomorphisms. It shows that Ext is not just some esoteric construction but rather a generalization of the familiar Hom functor. This understanding can help you build intuition for working with Ext and other derived functors.
  • Foundation for Higher Ext Groups: The fact that Ext⁰(M, N) = Hom(M, N) serves as a base case for understanding the higher Ext groups (Extⁿ(M, N) for n > 0). These higher Ext groups capture more subtle information about the relationships between modules and play a crucial role in classifying extensions of modules. By understanding the base case, we can better grasp the meaning and significance of the higher Ext groups.
  • Applications in Algebraic Topology: Homological algebra, including the Ext functor, has deep connections to algebraic topology. For example, Ext groups can be used to compute cohomology groups of topological spaces. The result Ext⁰(M, N) = Hom(M, N) can be used to relate topological concepts to algebraic ones, providing a powerful tool for studying topological spaces.
  • Applications in Ring Theory: Ext functors are also used extensively in ring theory to study the structure of rings and modules. For example, Ext groups can be used to classify modules over a ring and to understand the homological properties of rings. The result Ext⁰(M, N) = Hom(M, N) provides a starting point for these investigations.

In short, proving Ext⁰(M, N) = Hom(M, N) is not just an exercise in homological algebra; it's a fundamental result that underpins much of the theory and has wide-ranging applications in various areas of mathematics. It’s like understanding the ABCs before writing a novel – crucial for the bigger picture!

Common Pitfalls and How to Avoid Them

Alright, before we wrap things up, let's talk about some common pitfalls that people often encounter when proving Ext⁰(M, N) = Hom(M, N). Being aware of these pitfalls can save you a lot of headaches and help you truly master the concept.

  • Forgetting the Projective Resolution: The entire proof hinges on using a projective resolution. If you try to prove this result without explicitly constructing or referring to a projective resolution, you're going to get lost in the weeds. Always start by writing down the projective resolution and clearly identifying the maps involved. It’s the map that gets you to the treasure!
  • Misunderstanding Exactness: Exactness is the lifeblood of homological algebra. A sequence is exact if the image of each map is equal to the kernel of the next map. If you don't fully grasp this concept, you'll struggle to understand why the projective resolution works and how it leads to the desired result. Make sure you can clearly explain what exactness means in the context of a projective resolution. Think of it as the perfect flow of information – nothing lost, nothing gained.
  • Getting Lost in the Notation: Homological algebra is notorious for its heavy notation. It's easy to get bogged down in symbols and lose sight of the underlying concepts. When working with Ext, Hom, kernels, images, and complexes, take your time and carefully write out each step. Don't be afraid to use diagrams and visual aids to help you keep track of the maps and modules involved. Visualizing it is half the battle!
  • Ignoring Left-Exactness: While we aimed for a direct proof, the left-exactness of the Hom functor is lurking in the background. Understanding this property can provide valuable intuition and help you see why Ext⁰(M, N) = Hom(M, N) makes sense. If you're struggling with the proof, try thinking about how left-exactness plays a role. It’s the secret ingredient that makes the dish perfect!
  • Not Checking Injectivity: A crucial step in the proof is showing that the map Hom(d₀, N): Hom(M, N) → Hom(P₀, N) is injective. If you skip this step, your proof will be incomplete. Make sure you explicitly demonstrate why this map is injective. This is where the surjectivity of d₀ comes into play, so be sure to connect these ideas. Injectivity is the key to unlocking the final result!

By being mindful of these pitfalls, you can navigate the proof of Ext⁰(M, N) = Hom(M, N) with confidence and gain a deeper understanding of the concepts involved. Remember, practice makes perfect, so don't hesitate to work through examples and ask questions whenever you're unsure about something.

Conclusion: Mastering the Connection

Alright, guys, we've reached the end of our journey! We've successfully proven that Ext⁰(M, N) = Hom(M, N) by diving deep into the construction of Ext using projective resolutions. We've explored the significance of this result, its applications in various areas of mathematics, and the common pitfalls to avoid. Hopefully, this comprehensive guide has equipped you with the knowledge and understanding you need to confidently tackle this fundamental concept in homological algebra.

Remember, the key to mastering homological algebra is to build a strong foundation in the basic concepts, such as projective resolutions, exactness, and the Hom and Ext functors. Don't be afraid to ask questions, work through examples, and explore the connections between different ideas. The more you engage with the material, the deeper your understanding will become.

So, go forth and conquer the world of homological algebra! And remember, Ext⁰(M, N) will always be there for you, reminding you of the beautiful connection between the abstract and the concrete in mathematics. Keep exploring, keep learning, and keep having fun!