Problem 5 - A First Course in Module Theory

Question

Let \(P_{1} \times P_{2}\) be an external direct sum of \(R\)-modules. Define \(\omega: P_{1} \times\) \(P_{2} \rightarrow P_{2} \times P_{1}\) by \(\omega\left(p_{1}, p_{2}\right)=\left(p_{2}, p_{1}\right)\). Show that \(\omega\) is an isomorphism of \(R\)-modules. Given a set of modules \(\left\\{P_{1}, \ldots, P_{k}\right\\}, k>2\) and any permutation \(\alpha\) of the integers \(1, \ldots, k\), prove that $$ P_{1} \times \cdots \times P_{k} \cong P_{\alpha(1)} \times \cdots \times P_{\alpha(k)} $$

Step-by-Step Solution

Answer

Show that \( \omega((p_{1}, p_{2})) = (p_{2}, p_{1}) \) is a bijective homomorphism by proving linearity, injectivity, and surjectivity. Extend this method to permutations for \( k \) modules.

Step 1: Define homomorphism

Define the function \(\omega : P_{1} \times P_{2} \rightarrow P_{2} \times P_{1}\) by \(\omega(p_{1}, p_{2}) = (p_{2}, p_{1})\). This function swaps the elements of the pairs. We need to show that this function is a homomorphism.

Step 2: Verify linearity

To show \(\omega\) is an \(R\)-module homomorphism, verify that it preserves addition and scalar multiplication: \((p_{1}, p_{2}) + (q_{1}, q_{2}) = (p_{1} + q_{1}, p_{2} + q_{2})\) and \(r(p_{1}, p_{2}) = (rp_{1}, rp_{2})\). After applying \(\omega\), \(\omega((p_{1}, p_{2}) + (q_{1}, q_{2})) = (p_{2} + q_{2}, p_{1} + q_{1}) = \omega(p_{1}, p_{2}) + \omega(q_{1}, q_{2})\) and \(\omega(r(p_{1}, p_{2})) = (rp_{2}, rp_{1}) = r\omega(p_{1}, p_{2})\). Hence, \(\omega\) is an \(R\)-module homomorphism.

Step 3: Prove bijectivity

To show \(\omega\) is an isomorphism, we need to prove it is bijective. For injectivity: Assume \(\omega(p_{1}, p_{2}) = \omega(q_{1}, q_{2})\), then \((p_{2}, p_{1}) = (q_{2}, q_{1})\), implying \(p_{2} = q_{2}\) and \(p_{1} = q_{1}\). For surjectivity: For any \( (p_{2}, p_{1}) \in P_{2} \times P_{1} \)), there exists \( (p_{1}, p_{2}) \in P_{1} \times P_{2} \) such that \( \omega(p_{1}, p_{2}) = (p_{2}, p_{1}) \). Hence, \( \omega \) is bijective.

Step 4: Prove general case for any permutation

Given \( P_{1} \times \cdots \times P_{k} \) and any permutation \( \alpha \) of the integers \( 1, \ldots, k \), define \( \omega : P_{1} \times \cdots \times P_{k} \rightarrow P_{\alpha(1)} \times \cdots \times P_{\alpha(k)} \) by \( \omega((p_{1}, \ldots, p_{k})) = (p_{\alpha(1)}, \ldots, p_{\alpha(k)}) \). Show \( \omega \) is a bijective homomorphism similarly: verify it preserves addition and scalar multiplication, and show it's injective and surjective.

Key Concepts

R-module homomorphism

In the study of modules, an R-module homomorphism is a crucial concept. It refers to a function between two R-modules that preserves the module operations, meaning addition and scalar multiplication. Consider two R-modules, say, M and N, and a function \( f: M \rightarrow N \). For \( f \) to be an R-module homomorphism, it must satisfy both:

• Additivity: \( f(m_1 + m_2) = f(m_1) + f(m_2) \) for all \( m_1, m_2 \in M \).
• Scalar Multiplication: \( f(rm) = r f(m) \) for all \( r \in R \) and \( m \in M \).

When we consider the function \( \omega \) defined in the exercise by \( \omega : P_1 \times P_2 \rightarrow P_2 \times P_1 \), we see it swapping elements of pairs from two modules. To show that \( \omega \) is an R-module homomorphism, we check that it maintains both addition and scalar multiplication. By verifying the equations for additivity and scalar multiplication shared in step 2 of the solution, we solidify \( \omega \)'s status as an R-module homomorphism.

module isomorphism

A module isomorphism is a bijective R-module homomorphism, meaning it has both properties of a homomorphism and that it has an inverse. This notion is significant because isomorphic modules are essentially the same in the context of module theory—they have identical structure.
Demonstrating that \( \omega \) from the exercise is an isomorphism requires showing that \( \omega \) is bijective (one-to-one and onto). In step 3, the solution confirms injectivity by showing uniqueness—if \( \omega(p_1, p_2) = \omega(q_1, q_2) \), then \( p_1 = q_1 \) and \( p_2 = q_2 \). For surjectivity, it verifies that every element in the codomain (\( P_2 \times P_1 \)) can be traced back to an element in the domain (\( P_1 \times P_2 \)).
These properties together confirm that \( \omega \) is not only a homomorphism but an isomorphism.

permutation of modules

Permutations offer a unique way to re-order elements. When this concept is applied to modules, it can show the flexibility and inherent equivalence between different direct sums of modules. Given a set of modules \( \{ P_1, P_2, \ldots, P_k \} \) and considering any permutation \( \alpha \) of indices \( 1, 2, \ldots, k \), we aim to establish that the direct sum remains unchanged up to isomorphism.
That means, if we permute these modules, the resulting reordered modules will still be isomorphic to the original direct sum. This is presented in step 4 using the map \( \omega((p_1, p_2, \ldots, p_k)) = (p_{\alpha(1)}, p_{\alpha(2)}, \ldots, p_{\alpha(k)}) \). Verifying the homomorphism properties and bijectivity for this permutation function **\omega** confirms that regardless of the order, the direct sum of modules retain the same structure. This highlights the interchangeability and consistent properties of module sums under permutation.