Problem 3 - A First Course in Module Theory

Question

Let \(M_{1}, \ldots, M_{k}\) be a set of \(R\)-modules, and let \(L_{i}\) be a submodule of \(M_{i}\) for \(i=1, \ldots, k\) Show that there is an \(R\)-module isomorphism $$ \left(M_{1} \times \cdots \times M_{k}\right) /\left(L_{1} \times \cdots \times L_{k}\right) \cong M_{1} / L_{1} \times \cdots \times M_{k} / L_{k} $$

Step-by-Step Solution

Answer

Define the map \(\theta\), prove its well-definedness and homomorphism property, then show it is bijective to establish the isomorphism.

Step 1: Define the Map

Define the map \[ \theta : \frac{M_{1} \times \text{...} \times M_{k}} {L_{1} \times \text{...} \times L_{k}} \to \frac{M_{1}}{L_{1}} \times \text{...} \times \frac{M_{k}}{L_{k}} \] by \[ \theta((m_{1}, \text{...}, m_{k}) + (L_{1} \times \text{...} \times L_{k})) = (m_{1} + L_{1}, \text{...}, m_{k} + L_{k}) \] for every \((m_{1}, \text{...}, m_{k}) \text{ in } M_{1} \times \text{...} \times M_{k}.\) This map needs to be checked for well-definedness.

Step 2: Prove Well-definedness

To show that \(\theta\) is well-defined, consider two elements \((m_{1}, \text{...}, m_{k}) and (m'_{1}, \text{...}, m'_{k}) of M_{1} \times \text{...} \times M_{k}\) such that \((m_{1}, \text{...}, m_{k}) + (L_{1} \times \text{...} \times L_{k}) = (m'_{1}, \text{...}, m'_{k}) + (L_{1} \times \text{...} \times L_{k})\). This means \((m_{1}-m'_{1}, \text{...}, m_{k}-m'_{k}) \text{ belongs to } L_{1} \times \text{...} \times L_{k}\). Hence, \(m_{i} - m'_{i} \text{ belongs to } L_{i}\) for each \(i = 1, \text{...}, k\). Thus, \(\theta((m_{1}, \text{...}, m_{k}) + (L_{1} \times \text{...} \times L_{k})) \text{ is equal to } \theta((m'_{1}, \text{...}, m'_{k}) + (L_{1} \times \text{...} \times L_{k}))\), so \(\theta\) is well-defined.

Step 3: Prove Homomorphism Property

To show that \(\theta\) is a homomorphism, consider elements \((m_{1}, \text{...}, m_{k})\) and \((n_{1}, \text{...}, n_{k}) in M_{1} \times \text{...} \times M_{k}\). Then:\[ \theta(((m_{1}, \text{...}, m_{k}) + (n_{1}, \text{...}, n_{k})) + (L_{1} \times \text{...} \times L_{k})) \ \theta((m_{1} + n_{1}, \text{...}, m_{k} + n_{k}) + (L_{1} \times \text{...} \times L_{k})) = ((m_{1} + n_{1}) + L_{1}, \text{...}, (m_{k} + n_{k}) + L_{k}) \ = (m_{1} + L_{1}, \text{...}, m_{k} + L_{k}) + (n_{1} + L_{1}, \text{...}, n_{k} + L_{k}) = \theta((m_{1}, \text{...}, m_{k}) + (L_{1} \times \text{...} \times L_{k})) + \theta((n_{1}, \text{...}, n_{k}) + (L_{1} \times \text{...} \times L_{k}))\]. Hence, \(\theta\) is a homomorphism.

Step 4: Prove Injectivity

To show that \(\theta\) is injective, assume \(\theta((m_{1}, \text{...}, m_{k}) + (L_{1} \times \text{...} \times L_{k})) = 0\). This means \((m_{1} + L_{1}, \text{...}, m_{k} + L_{k}) = (0 + L_{1}, \text{...}, 0 + L_{k})\), i.e., \(m_{i} \text{ belongs to } L_{i}\) for each \(i = 1, \text{...}, k\). Hence \((m_{1}, \text{...}, m_{k}) \text{ belongs to} L_{1} \times \text{...} \times L_{k}\), implying \((m_{1}, \text{...}, m_{k}) + (L_{1} \times \text{...} \times L_{k}) = (0, \text{...}, 0) + (L_{1} \times \text{...} \times L_{k})\). Thus, \(\theta\) is injective.

Step 5: Prove Surjectivity

To show that \(\theta\) is surjective, consider an element \((m'_{1} + L_{1}, \text{...}, m'_{k} + L_{k})\) in \( \frac{M_{1}}{L_{1}} \times \text{...} \times \frac{M_{k}}{L_{k}}\). Then we can map it to \((m'_{1}, \text{...}, m'_{k}) + (L_{1} \times \text{...} \times L_{k})\). Clearly, \(\theta((m'_{1}, \text{...}, m'_{k}) + (L_{1} \times \text{...} \times L_{k})) = (m'_{1} + L_{1}, \text{...}, m'_{k} + L_{k})\). Hence, \(\theta\) is surjective.

Step 6: Conclude the Isomorphism

Since \(\theta\) is a bijective homomorphism, it is an isomorphism. Therefore, \[ \frac{M_{1} \times \text{...} \times M_{k}} {L_{1} \times \text{...} \times L_{k}} \cong \frac{M_{1}}{L_{1}} \times \text{...} \times \frac{M_{k}}{L_{k}} \].

Key Concepts

Homomorphism

In algebra, a homomorphism is a structure-preserving map between two algebraic structures of the same type. For the context of R-modules, a homomorphism is a function that respects the operations of addition and scalar multiplication. When dealing with homomorphisms, it's important to check if the mapping holds these properties:

• Addition: If \(\theta(m + n) = \theta(m) + \theta(n)\), then it is additive.
• Scalar Multiplication: If \(\theta(r \times m) = r \times \theta(m)\), where \(r\) is a scalar, then it respects scalar multiplication.

The function \(\theta : \frac{M_1 \times ... \times M_k}{L_1 \times ... \times L_k} \to \frac{M_1}{L_1} \times ... \times \frac{M_k}{L_k}\) retains these properties, showing it is indeed a homomorphism. This step is crucial in proving the isomorphism as it ensures the structure of the module is preserved during mapping.

Injectivity

Injectivity, also known as one-to-one property, is an essential aspect to prove when determining an isomorphism. A map is injective if it maps distinct elements in the domain to distinct elements in the codomain.
For \(\theta\), to prove injectivity, one must show that if \(\theta((m_1, ..., m_k) + (L_1 \times ... \times L_k)) = 0\), then \(m_i \text{ belongs to } L_i\) for all \(i\). This step typically involves assuming the output is zero and showing that this implies a trivial element in the domain.
So, when proving injectivity:

• Assume \(\theta((m_1, ..., m_k) + (L_1 \times ... \times L_k)) = 0\).
• Show that this assumption leads to \(m_i\) being an element of \(L_i\) for all \(i\).
• This indicates the only preimage of zero is trivial, thereby proving the map is injective.

This injectivity ensures there are no overlapping images, solidifying \(\theta\)'s credibility as a one-to-one correspondence.

Surjectivity

Surjectivity, or onto property, is the other half of proving an isomorphism. A map is surjective if every element in the codomain has a preimage in the domain.
For \(\theta\), to show surjectivity, consider any element \((m'_1 + L_1, ..., m'_k + L_k)\) in the codomain \(\frac{M_1}{L_1} \times ... \times \frac{M_k}{L_k}\). The goal is to show there exists an element in the domain that \(\theta\) maps to it.
To achieve this, map each \(m'_i\) into the respective \(M_i\) and then construct the tuple \( (m'_1, ..., m'_k) + (L_1 \times ... \times L_k)\) in the domain:

• This element will map to \((m'_1 + L_1, ..., m'_k + L_k)\) under \(\theta\).
• Prove that every element in the codomain is representable this way, confirming surjectivity.

Ensuring surjectivity makes sure \(\theta\)'s map covers the entire codomain, positing that every element is reachable via \(\theta\) from the domain.

Submodules

Understanding submodules is foundational in module theory. A submodule is a subset of a module that is itself a module under the same operations.
In the context of this exercise, \(L_i\) is a submodule of \(M_i\) for all \(i = 1, ..., k\). The quotient \(M_i/L_i\) represents the set of cosets of \(L_i\) in \(M_i\).
When working with these submodules in proving the isomorphism, remember to:

• Recognize that the entire structure \((M_1 \times ... \times M_k)\) divided by \((L_1 \times ... \times L_k)\) maintains the submodule properties.
• Each \(m_i \text{ belongs to } M_i\) represents an element modulo \(L_i\).
• The map \(\theta\) preserves these submodule structures, ensuring consistency and coherence.

This foundational understanding ties in the modular arithmetic in modules, essential for grasping the isomorphism's thoroughness and integrity.