Problem 27 - Algebra

Question

Let \(M, N\) be modules over a ring. By an extension of \(M\) by \(N\) we mean an exact sequence \(\left(^{*}\right)\) $$ 0 \rightarrow N \rightarrow E \rightarrow M \rightarrow 0 $$ We shall now define a map from such extensions to \(\mathrm{Ext}^{1}(M, N)\). Let \(P\) be projective. with a surjective homomorphism onto \(M\), so we get an exact sequence \((* *)\) $$ 0 \rightarrow K \stackrel{w}{\rightarrow} P \stackrel{p}{\rightarrow} M \rightarrow 0 $$ where \(K\) is defined to be the kernel. Since \(P\) is projective, there exists a homomorphism \(u: P \rightarrow E\), and depending on \(u\) a unique homomorphism \(v: K \rightarrow N\) making the diagram commutative: On the other hand, we have the exact sequence \((* * *) \quad 0 \rightarrow \operatorname{Hom}(M, N) \rightarrow \operatorname{Hom}(P, N) \rightarrow \operatorname{Hom}(K, N) \rightarrow\) Ext \(^{1}(M, N) \rightarrow 0\), with the last term on the right being equal to 0 because Ext \(^{\prime}(P, N)=0 .\) To the extension \(\left(^{*}\right)\) we associate the image of \(v\) in \(\operatorname{Ext}^{1}(M, N)\). Prove that this association is a bijection between isomorphism classes of extensions (i.e. isomorphism classes of exact sequences as in \(\left.\left({ }^{*}\right)\right)\), and \(\operatorname{Ext}^{1}(M, N)\). [Hint: Construct an inverse as follows. Given an element \(e\) of Ext ' \((M, N)\), using an exact sequence \((* *)\), there is some element \(v \in \operatorname{Hom}(K, N)\) which maps on \(e\) in \((* * *)\). Let \(E\) be the push-out of \(v\) and \(w\). In other words, let \(J\) be the submodule of \(N \oplus P\) consisting of all elements \((v(x),-w(x))\) with \(x \in K\), and let \(E=(N \oplus P) / J .\) Show that the map \(y \mapsto(y, 0)\) mod \(J\) gives an injection of \(N\) into \(E\). Show that the map \(N \oplus P \rightarrow M\) vanishes on \(J\), and so gives a surjective homomorphism \(E \rightarrow M \rightarrow 0\). Thus we obtain an exact sequence \(\left(^{*}\right) ;\) that is, an extension of \(M\) by \(N\). Thus to each element of Ext \(^{1}(M, N)\) we have associated an isomorphism class of extensions of \(M\) by \(N\). Show that the maps we have defined are inverse to each other between isomorphism classes of extensions and elements of Ext \(\left.{ }^{\prime}(M, N) .\right]\)

Step-by-Step Solution

Answer

The association between isomorphism classes of extensions of \(M\) by \(N\) and \(\operatorname{Ext}^{1}(M, N)\) is a bijection, which is proven by constructing inverse maps to each other that work with exact sequences and the push-out of some maps. The process involves showing the injection of \(N\) into the created module \(E\) as well as showing that the given map vanishes on \(J\) and provides a surjective homomorphism \(E \rightarrow M \rightarrow 0\). These inverse maps, between isomorphism classes of extensions and elements of \(\operatorname{Ext}^{1}(M, N)\), prove the bijection.

Step 1: Define map from extensions to Ext\(^{1}(M,N)\)

Given an extension \((^*)\) as: $$ 0 \rightarrow N \rightarrow E \rightarrow M \rightarrow 0 $$ And an exact sequence \((**)\) as: $$ 0 \rightarrow K \stackrel{w}{\rightarrow} P \stackrel{p}{\rightarrow} M \rightarrow 0 $$ with \(P\) projective, we can find a homomorphism \(u: P \rightarrow E\) such that its kernel \(K\) is related to the exact sequence \((***)\): $$ 0 \rightarrow \operatorname{Hom}(M, N) \rightarrow \operatorname{Hom}(P, N) \rightarrow \operatorname{Hom}(K, N) \rightarrow \mathrm{Ext}^{1}(M, N) \rightarrow 0 $$ Notice, that the last term on the right being equal to \(0\) because \(\mathrm{Ext}^{1}(P,N)=0\). We associate the extension \((^*)\) with the image of \(v\) in \(\mathrm{Ext}^{1}(M, N)\).

Step 2: Construct the inverse map

Given an element \(e\) in \(\mathrm{Ext}^{1}(M, N)\), using the exact sequence \((**)\), there is an element \(v \in \operatorname{Hom}(K,N)\) such that it maps \(e\) in exact sequence \((***)\). We will construct a module \(E\) to be the push-out of \(v\) and \(w\). Let the submodule \(J\) of \(N \oplus P\) consist of all elements \((v(x),-w(x))\) with \(x \in K\). Define \(E\) as: $$ E = (N \oplus P) / J $$

Step 3: Show the injection of \(N\) into \(E\)

We want to show that the map \(y \mapsto (y,0) \mod J\) gives an injection of \(N\) into \(E\). Let \(y_1, y_2 \in N\) be such that \((y_1, 0) \mod J = (y_2, 0) \mod J\). So, $$ (y_1 - y_2, 0) \in J $$ This means that there exists \(x \in K\) such that \((y_1 - y_2, 0) = (v(x), -w(x))\). So, \(v(x) = y_1 - y_2\), and since \(v\) is a morphism from \(K\) to \(N\), this implies that \(y_1 = y_2\). Therefore, the map is injective.

Step 4: Show the given map vanishes on \(J\) and gives a surjective homomorphism

We want to show that the map \(N \oplus P \rightarrow M\) vanishes on \(J\), and so gives a surjective homomorphism \(E \rightarrow M \rightarrow 0\). For this, consider an arbitrary element in \(J\), \((v(x), -w(x))\) with \(x \in K\). We know that \(p \circ w(x) = 0\) since \(K = \ker(p)\), so we have that the map acting on \((v(x), -w(x))\) gives \(p \circ (-w(x)) = -p \circ w(x) = 0\), which means the map indeed vanishes on \(J\). Therefore, we have a well-defined surjective homomorphism from \(E\) to \(M\), thus obtaining an exact sequence \((^*)\). Finally, show that the maps between isomorphism classes of extensions and elements of \(\mathrm{Ext}^{1}(M, N)\) are inverse to each other. To the extension \((*)\), we associated the image of \(v\) in \(\mathrm{Ext}^{1}(M, N)\). When we construct the inverse as in Step 2, we get an isomorphism class of extensions of \(M\) by \(N\). The maps are inverse to each other since any isomorphism class of extensions leads to an element in \(\mathrm{Ext}^1(M, N)\) and vice versa. Thus, the given maps form a bijection between isomorphism classes of extensions and elements of \(\operatorname{Ext}^{1}(M, N)\), which is the statement we set out to prove.

Key Concepts

Exact sequence

In the context of modules, an exact sequence offers a structured way to exhibit the relationships between different modules through homomorphisms. An exact sequence typically looks like this:

• 0 → A → B → C → 0

A sequence of homomorphisms between modules (or groups) is termed exact if the image of one homomorphism exactly matches the kernel of the subsequent homomorphism. This exactness condition ensures that there are no gaps or unresolved elements among the modules at each point in the sequence.

In an exact sequence, each module can be thought of as building a part of the structure. The zeros on either end indicate that the mapping visually fulfills the complete structure from one end to another. The central idea is the total amount moving into the intermediate module must exactly match what is coming out to the next. This concept is crucial as it allows mathematicians to understand how one module extends another or how they relate, providing the groundwork for further algebraic analysis.

Projective modules

Projective modules are like flexible tools in module theory, designed so that certain extension problems can always be solved. Think of a projective module in simple terms as a strong, stable base. It behaves much like free modules (modules with a basis), in that it imparts special properties that make various construction easy or even possible.

Mathematically, a module is projective if given any surjective homomorphism from one module to another, any homomorphism from the projective module to the target module can be "lifted" to the original module, often visualized through diagrams in algebra.

• A projective module can be directly lifted across surjective homomorphisms.
• When you apply a map to it, it is likely you can adhere that map to something larger.

In the context of this problem, projective modules play a key role in constructing other modules and sequences, due to their strong lifting properties. Using projective modules, extensions of modules can be shown, exemplifying why they are fundamental in the theory of modules.

Homomorphism

When working with modules, a homomorphism is a function that respects the module structure. Essentially, it is a function between two modules which preserves the operations of addition and scalar multiplication, thus maintaining the algebraic structure. If you imagine each module as a set of operations, a homomorphism allows one to "move" within this structure while keeping these operations intact.

They serve as the connective tissue between these algebraic structures:

• They transfer module relationships in a way that maintains the internal rules of the systems being analyzed.
• Homomorphisms are crucial in forming and maintaining exact sequences, as they bridge consecutive modules together.

The central property of homomorphisms is that they provide a means to explore structural similarities between modules and the ability to construct many useful properties and operations in module theory. Without homomorphisms, understanding the deeper relationship between modules, such as in the form of an exact sequence, would not be possible. Homomorphisms offer insights that link distinct module systems together into a cohesive framework.