Question

Let \(M_{1}\) and \(M_{2}\) be \(R\) -modules, and let \(N_{1}\) be a submodule of \(M_{1}\) and \(N_{2}\) a submodule of \(M_{2}\). Show that \(N_{1} \times N_{2}\) is a submodule of \(M_{1} \times M_{2}\)

Short answer

Question: Show that the cartesian product of two submodules of modules is itself a submodule of the cartesian product of the modules. Answer: To show that the cartesian product of two submodules, \(N_1 \times N_2\), is a submodule of the cartesian product of the modules, \(M_1 \times M_2\), we need to verify that \(N_1 \times N_2\) is closed under addition and scalar multiplication with respect to \(M_1 \times M_2\). By showing that \((n_{1_1} + n_{1_2}, n_{2_1} + n_{2_2}) \in N_1 \times N_2\) for all \((n_{1_1}, n_{2_1}), (n_{1_2}, n_{2_2}) \in N_1 \times N_2\) and \((r \cdot n_{1}, r \cdot n_{2}) \in N_1 \times N_2\) for all \(r \in R\) and \((n_{1}, n_{2}) \in N_1 \times N_2\), we conclude that \(N_1 \times N_2\) is a submodule of \(M_1 \times M_2\).

Step-by-step solution

Recall the submodule definition

A subset \(N\) of a module \(M\) is called a submodule if, for all \(n_1, n_2 \in N\) and \(r \in R\) (where \(R\) is a ring), we have the following properties: 1. \(n_1 + n_2 \in N\) 2. \(r \cdot n_1 \in N\)

Recall the cartesian product of modules

Given two \(R\)-modules \(M_1\) and \(M_2\), the cartesian product \(M_1 \times M_2\) is itself an \(R\)-module with addition and scalar multiplication defined componentwise. That is, for all \((m_1, m_2), (m_1', m_2') \in M_1 \times M_2\) and \(r \in R\), we have: 1. \((m_1, m_2) + (m_1', m_2') = (m_1 + m_1', m_2 + m_2')\) 2. \(r \cdot (m_1, m_2) = (r \cdot m_1, r \cdot m_2)\)

Show that the cartesian product of submodules is closed under addition

Take any two elements \((n_{1_1}, n_{2_1}), (n_{1_2}, n_{2_2}) \in N_1 \times N_2\). To show that \(N_1 \times N_2\) is closed under addition, we need to prove that \((n_{1_1}, n_{2_1}) + (n_{1_2}, n_{2_2}) \in N_1 \times N_2\). Since \(N_1\) and \(N_2\) are submodules, we have \(n_{1_1} + n_{1_2} \in N_1\) and \(n_{2_1} + n_{2_2} \in N_2\). Thus, \((n_{1_1}, n_{2_1}) + (n_{1_2}, n_{2_2}) = (n_{1_1} + n_{1_2}, n_{2_1} + n_{2_2}) \in N_1 \times N_2\).

Show that the cartesian product of submodules is closed under scalar multiplication

Take an element \((n_{1}, n_{2}) \in N_1 \times N_2\) and \(r \in R\). To show that \(N_1 \times N_2\) is closed under scalar multiplication, we need to prove that \(r \cdot (n_{1}, n_{2}) \in N_1 \times N_2\). Since \(N_1\) and \(N_2\) are submodules, we have \(r \cdot n_{1} \in N_1\) and \(r \cdot n_{2} \in N_2\). Thus, \(r \cdot (n_{1}, n_{2}) = (r \cdot n_{1}, r \cdot n_{2}) \in N_1 \times N_2\).

Conclusion

Since the cartesian product of submodules \(N_1 \times N_2\) is closed under both addition and scalar multiplication with respect to the cartesian product of modules \(M_1 \times M_2\), we can conclude that \(N_1 \times N_2\) is a submodule of \(M_1 \times M_2\).

Key concepts

Module Theory

Module theory is a fundamental concept in abstract algebra that extends the notion of vector spaces involving a field to a more general structure, which involves a ring. In essence, a module is similar to a vector space, but instead of a field, we have a ring that provides the scalars for scalar multiplication. This small change introduces significant complexity and richness to the structure.

For a subset of a module to qualify as a submodule, it must satisfy two main properties. First, it needs to be closed under addition. If you take any two elements within the subset and add them together, the result should still be within the subset. Second, it must be closed under scalar multiplication. When any element of the subset is multiplied by a scalar from the ring, the result must again fall within the subset.

These properties ensure that a submodule behaves like the larger module regarding addition and scalar multiplication. This ensures consistency in operations and allows us to study smaller pieces of modules (submodules) with similar properties to the entire module, making the analysis more manageable and modular.

Cartesian Product

The cartesian product in the context of module theory further advances the concept of structures that can be operated upon. Specifically, if you have two modules, you can form their cartesian product to create a new module. The cartesian product is simply a set of ordered pairs, where the first element of the pair comes from the first module, and the second element comes from the second module.

In our exercise, the cartesian product of the submodules plays a pivotal role. The componentwise operations on these pairs—where we add and multiply pairs by applying the operation to each component separately—are what give the cartesian product its module structure. This means that if you take two such pairs and add them, you get another pair in the product. Similarly, if you multiply a pair by a scalar, each component of the pair gets multiplied by the scalar, resulting in another pair that's still in the product.

Ring Theory

Ring theory is the study of rings - algebraic structures equipped with two binary operations: addition and multiplication. These operations within a ring must satisfy certain properties such as associativity, distributivity of multiplication over addition and the existence of an additive identity. Rings are generalizations of more familiar structures like the set of integers.

In the problem at hand, we're looking at a module over a ring, which means that the scalar multiplication in our modules and submodules uses elements from the ring. The properties and the behavior of this ring influence the structure of the modules and submodules in question. Understanding the nature of the ring is essential, as some rings have additional properties that could further dictate the uniqueness of operation within modules. For instance, if the ring is commutative (meaning that the multiplication operation is commutative), then the scalar multiplication in the module also behaves in a predictable, commutative manner with respect to the scalar from the ring.