Question

Let \(\rho: M \rightarrow M^{\prime}\) be an \(R\) -linear map with kernel \(K .\) Let \(N\) be a submodule of \(M\). Show that we have an \(R\) -module isomorphism \(M /(N+K) \cong\) \(\rho(\boldsymbol{M}) / \rho(N)\).

Short answer

In conclusion, we have shown that there is an R-module isomorphism between the quotient modules \(M / (N+K)\) and \(\rho(M) / \rho(N)\) by defining a map \(\phi\), proving that it is well-defined, R-linear, and a bijection. This result highlights the relationship between the quotient modules given the R-linear map \(\rho\), its kernel K, and a submodule N of M.

Step-by-step solution

Define the map

Define a map \(\phi: M / (N+K) \rightarrow \rho(M) / \rho(N)\) as follows: $$ \phi(m+(N+K))= \rho(m) + \rho(N) $$ where \(m \in M\).

Show that the map is well-defined

To show that \(\phi\) is well-defined, we need to show that if \(m + (N+K) = m' + (N+K)\) for some \(m, m' \in M\), then \(\rho(m) + \rho(N) = \rho(m') + \rho(N)\). This is equivalent to showing that if \(m - m' \in (N+K)\), then \(\rho(m - m') \in \rho(N)\). Suppose \(m - m' \in (N+K)\). Then there exists \(n \in N\) and \(k \in K\) such that \(m - m' = n + k\). Since \(\rho\) is an R-linear map, we have \(\rho(m - m') = \rho(n) + \rho(k)\). But \(\rho(k) = 0\) (as k belongs to the kernel of \(\rho\)), so we have \(\rho(m - m') = \rho(n)\). Since \(n \in N\), it follows that \(\rho(m - m') \in \rho(N)\), and hence \(\phi\) is well-defined.

Show that the map is R-linear

To show that \(\phi\) is R-linear, we need to show that it preserves addition and scalar multiplication. Let \(m_1 + (N+K), m_2 + (N+K) \in M / (N+K)\) and \(r \in R\). Then: 1. \(\phi((m_1 + (N+K)) + (m_2 + (N+K))) = \phi(m_1 + m_2 + (N+K)) = \rho(m_1 + m_2) + \rho(N) = (\rho(m_1) + \rho(N)) + (\rho(m_2) + \rho(N)) = \phi(m_1 + (N+K)) + \phi(m_2 + (N+K))\) 2. \(\phi(r(m_1 + (N+K))) = \phi(rm_1 + (N+K)) = \rho(rm_1) + \rho(N) = r(\rho(m_1) + \rho(N)) = r\phi(m_1 + (N+K))\) Thus, \(\phi\) is R-linear.

Show that the map is a bijection

First, we will show that \(\phi\) is injective. Suppose \(\phi(m + (N+K)) = \phi(m' + (N+K))\) for some \(m, m' \in M\). Then \(\rho(m) + \rho(N) = \rho(m') + \rho(N)\), which implies that \(\rho(m - m') \in \rho(N)\). Since \(\rho\) is R-linear, we have \(\rho(m - m') = \rho(n) + \rho(k)\) for some \(n \in N\) and \(k \in K\), where \(\rho(k) = 0\). Thus, \(m - m' - n = k\). Since \(k \in K\), it follows that \(m - m' \in N+K\), and thus, \(m + (N+K) = m' + (N+K)\). This establishes the injectivity of \(\phi\). Now, we will show that \(\phi\) is surjective. Let \(x + \rho(N) \in \rho(M) / \rho(N)\) for some \(x \in \rho(M)\). Then, there exists \(m \in M\) such that \(\rho(m) = x\). We have \(\phi(m + (N+K)) = \rho(m) + \rho(N) = x + \rho(N)\), and thus, \(\phi\) is surjective. Since \(\phi\) is R-linear, injective, and surjective, it is an R-module isomorphism between \(M / (N+K)\) and \(\rho(M) / \rho(N)\).

Key concepts

Linear Map

In mathematics, especially in the context of abstract algebra, a linear map is a fundamental concept that describes a function between two modules, say, M to M', that preserves addition and scalar multiplication. In simpler terms, it is a rule that connects elements of module M with those in M' in a way that is consistent with the structure of the modules involved.

Here, we encapsulate two properties that define a linear map:

• If you take two elements a and b from module M, and you map them to M', the result is the same as if you added them together first and then did the mapping. Mathematically, this means \( \rho(a + b) = \rho(a) + \rho(b) \).
• If you multiply an element from M by a scalar (let's say r from our ring R) and then map it, it's the same as if you mapped the element first and then multiplied the result by the scalar, which gives us \( \rho(ra) = r\rho(a) \).

This concept is vital in the proof of our R-module isomorphism. It is the linearity of the map that will ensure the structure of the module is preserved, making our mapping a true module homomorphism.

Submodule

The concept of a submodule is akin to that of a subset in set theory, but with additional structure. A submodule of module M is a subset N of M that itself forms a module with the same operations of addition and scalar multiplication as M. One can think of a submodule as a smaller 'universe' carved out from the bigger one, which still obeys the same mathematical 'laws'.

It's important to note a few crucial properties of submodules:

• The zero element of M is also in N.
• If you take any two elements from N, their sum is also in N.
• When you multiply any element of N by a scalar from R, the result is still in N.

These properties ensure that within the proof of our isomorphism, we can safely reason about operations within N and the larger module M in a parallel way.

Kernel of a Map

The kernel of a map is a concept that comes from the broader idea of functions and their properties. In the realm of R-modules and linear maps, the kernel of a map \rho from M to M' is the set of all elements in M that get sent to the zero element of M'.

Formally, the kernel is defined as \(K = \{m \in M | \rho(m) = 0'\}\), where 0' is the zero element in M'. The kernel captures the idea of 'loss of information'; it tells you what gets 'erased' when you apply the linear map \rho.

Interestingly, the kernel is not just any subset; it is actually a submodule of M. This fact is critical in our proof because it implies that everything in the kernel behaves nicely with respect to addition and scalar multiplication, which are pivotal properties when demonstrating that the proposed mapping is well-defined.

R-linear Map Properties

Now, let's delve into the R-linear map properties. These properties are what make linear maps between R-modules so special and structured. A map \rho: M \rightarrow M' is not only additive and compatible with scalar multiplication, as we discussed in the 'Linear Map' section, but it also must satisfy certain conditions to be considered an R-linear map.

For example, an R-linear map must send the zero element of the module M to the zero element of M' (a property arising from linearity). More formally, we say that for all r in R and m, n in M:

• \(\rho(0_M) = 0_{M'}\), where \(0_M\) is the zero element of M and \(0_{M'}\) is the zero element of M'.
• \(\rho(rm + n) = r\rho(m) + \rho(n)\), emphasizing the preserving of module operations.

In the context of our isomorphism proof, these properties ensure the map \phi we construct carries the R-module structure from M / (N+K) to \rho(M) / \rho(N) appropriately, solidifying it as an R-module homomorphism, hence allowing us to conclude that it is indeed an isomorphism between the given modules.