Problem 7 - A First Course in Module Theory

Question

Let \(P=P_{1} \times P_{2}\) be an external direct sum of \(R\)-modules. Write \(i d_{i}\) for the identity map on \(P_{i}, i=1,2\) and define maps as follows: $$ \begin{aligned} &\pi_{1}: P \rightarrow P_{1}, \pi_{1}\left(p_{1}, p_{2}\right)=p_{1} \\ &\pi_{2}: P \rightarrow P_{2}, \pi_{2}\left(p_{1}, p_{2}\right)=p_{2} \\ &\epsilon_{1}: P_{1} \rightarrow P, \epsilon_{1}\left(p_{1}\right)=\left(p_{1}, 0\right) \\ &\epsilon_{2}: P_{2} \rightarrow P, \epsilon_{2}\left(p_{2}\right)=\left(0, p_{2}\right) \end{aligned} $$ Verify that these are all homomorphisms, and that the relations $$ \pi_{1} \epsilon_{1}=i d_{1}, \pi_{2} \epsilon_{2}=i d_{2}, \pi_{2} \epsilon_{1}=0, \pi_{1} \epsilon_{2}=0 $$ and $$ \epsilon_{1} \pi_{1}+\epsilon_{2} \pi_{2}=i d_{P} $$ hold. Conversely, suppose we are given a collection of modules \(P, P_{1}, P_{2}\) and homomorphisms as above. Show that \(P \cong P_{1} \times P_{2}\). Generalize this exercise from 2 to \(k\) terms.

Step-by-Step Solution

Answer

Maps \( \pi_1, \pi_2, \epsilon_1, \epsilon_2 \) are homomorphisms. Relations hold, so \( P \cong P_1 \times P_2 \). Generalize to k terms similarly.

Step 1 - Verify maps are homomorphisms

A map is a homomorphism if it preserves module operations (addition and scalar multiplication). Verify this for each map (\( \pi_{1}, \pi_{2}, \epsilon_{1}, \epsilon_{2} \)): For \( \pi_1 \), given \( p = (p_1, p_2) \) and \( q = (q_1, q_2) \), we have \( \pi_1(p+q) = \pi_1((p_1+q_1, p_2+q_2)) = p_1 + q_1 = \pi_1(p) + \pi_1(q) \), satisfying addition. For scalar multiplication by \( r \), \( \pi_1(rp) = \pi_1(r(p_1, p_2)) = r p_1 = r \pi_1(p) \).Analogous results hold for \( \pi_2 \).For \(\epsilon_1 \), given \( p_1, q_1 \in P_1 \), \(\epsilon_1 (p_1+q_1) = (p_1 + q_1, 0) = (p_1, 0) + (q_1, 0) = \epsilon_1(p_1) + \epsilon_1(q_1) \). For scalars, \( \epsilon_1(r p_1) = (r p_1, 0) = r (p_1, 0) = r \epsilon_1(p_1) \).Analogous results hold for \( \epsilon_2 \).

Step 2 - Verify the relation \( \pi_1 \epsilon_1 = i d_1 \)

Verification: for all elements \( p_1 \in P_1 \), we have \( \pi_1 \epsilon_1 (p_1) = \pi_1 ((p_1, 0)) = p_1 \). Thus, \( \pi_1 \epsilon_1 \) acts as the identity on \( P_1 \).

Step 3 - Verify the relation \( \pi_2 \epsilon_2 = i d_2 \)

Verification: for all elements \( p_2 \in P_2 \), we have \( \pi_2 \epsilon_2 (p_2) = \pi_2 ((0, p_2)) = p_2 \). Thus, \( \pi_2 \epsilon_2 \) acts as the identity on \( P_2 \).

Step 4 - Verify the relation \( \pi_2 \epsilon_1 = 0 \)

Verification: for all elements \( p_1 \in P_1 \), \( \pi_2 \epsilon_1 (p_1) = \pi_2 ((p_1, 0)) = 0 \). Thus, \( \pi_2 \epsilon_1 = 0 \).

Step 5 - Verify the relation \( \pi_1 \epsilon_2 = 0 \)

Verification: for all elements \( p_2 \in P_2 \), \( \pi_1 \epsilon_2 (p_2) = \pi_1 ((0, p_2)) = 0 \). Thus, \( \pi_1 \epsilon_2 = 0 \).

Step 6 - Verify relation \( \epsilon_1 \pi_1 + \epsilon_2 \pi_2 = i d_P \)

For any \( p = (p_1, p_2) \in P \), \( ( \epsilon_1 \pi_1 + \epsilon_2 \pi_2 ) (p) = \epsilon_1 \pi_1(p) + \epsilon_2 \pi_2(p) = \epsilon_1(p_1) + \epsilon_2(p_2) = (p_1, 0) + (0, p_2) = (p_1, p_2) = p \). Thus, \( \epsilon_1 \pi_1 + \epsilon_2 \pi_2 = i d_P \).

Step 7 - Show \( P \cong P_1 \times P_2 \)

Given a collection of modules and homomorphisms satisfying the given relations, define a map \( \phi : P_1 \times P_2 \rightarrow P \) by \( (p_1, p_2) \mapsto \epsilon_1(p_1) + \epsilon_2(p_2) \) and its inverse \( \psi : P \rightarrow P_1 \times P_2 \) by \( p \mapsto (\pi_1(p), \pi_2(p)) \). Verify these are indeed inverses using the given relations, concluding \( P \cong P_1 \times P_2 \).

Step 8 - Generalize to k terms

Consider homomorphisms \( \pi_i : P \rightarrow P_i \) and \( \epsilon_i : P_i \rightarrow P \) for \( i = 1, 2, \ldots, k \) defined similarly. Verify relations: \( \pi_i \epsilon_i = i d_i \), \( \pi_j \epsilon_i = 0 \) for \( j \e i \), \( \sum_{i=1}^{k} \epsilon_i \pi_i = i d_P \).

Key Concepts

R-Modules

In mathematics, an R-module is a generalization of the concept of vector spaces. The key difference is that the scalars, instead of being from a field, come from a ring, which can have more complex structure and fewer restrictions than a field. Modules help extend linear algebra concepts to more advanced structures.

• Definition: An R-module is essentially a set equipped with an operation of addition and a compatible action by the ring R.
• Operations: The addition operation must satisfy properties like associativity, commutativity, and the existence of an additive identity. Scalar multiplication must distribute over both module addition and ring addition, and it must associate with the ring multiplication.

For example, consider the module P = P_1 x P_2 where P_1 and P_2 are R-modules. This structure allows us to form tuples (pairs here) with elements from P_1 and P_2, enabling combined operations and properties from both modules. This sets the stage for defining more complex mappings and relationships between such modules.

Homomorphisms

In the context of R-modules, homomorphisms play a crucial role. A homomorphism is a map between two modules that preserves the module structure. This means that it respects both addition and scalar multiplication. Let's break down why this is important.
Module Homomorphism Properties:

• Additivity: For a homomorphism φ: P → Q and for all elements u, v in P, φ(u + v) = φ(u) + φ(v) holds, ensuring the preservation of module addition.
• Compatibility with Scalar Multiplication: For all r in R and for all elements u in P, φ(r * u) = r * φ(u) ensures that scalar multiplication within the module is preserved by the homomorphism.

For instance, in the given solution, maps such as π_1, π_2, ε_1, and ε_2 are shown to be homomorphisms by verifying these properties. Showing that these properties hold helps in proving the identities like π_1(ε_1(p_1)) = p_1, confirming that such maps maintain the integrity of module operations.

Direct Sum

The concept of the direct sum is fundamental in module theory and many other areas of algebra. When we say that P is the direct sum of modules P_1 and P_2, written P = P_1 ⊕ P_2, it means that every element of P can be uniquely expressed as a sum of elements from P_1 and P_2. Here are key points:

• Unique Representation: Each element p in P has a unique pair (p_1, p_2) from P_1 and P_2 such that p = (p_1, p_2).
• Projection and Inclusion Maps: Maps like π_1, π_2, ε_1, and ε_2 help in projecting and including elements between the modules, preserving their structures.

For example, the projection maps π_1 and π_2 help retrieve components from the direct sum—π_1 extracts the P_1 component, while π_2 extracts the P_2 component. In contrast, the inclusion maps ε_1 and ε_2 embed elements from P_1 and P_2 into the direct sum in their respective positions. Verification of the relations π_1(ε_1) = id_1 and π_2(ε_2) = id_2 ensures that these maps interact correctly, marking the concept of direct sums essential in understanding module distributions and compositions.