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Chapter 4: Problem 2

Let \(V=\cdot \sum_{k=1}^{n} V_{k}\) be a direct sum of \(R\)-modules. Prove that the endomorphism ring \(\operatorname{End}_{R}(V)\) is isomorphic to the checkered matrix ring $$ S=\left\\{\left(\alpha_{i, j}\right) \mid \alpha_{i, j} \in \operatorname{Hom}_{R}\left(V_{j}, V_{i}\right)\right\\} $$ Here multiplication of entries is given by function composition.

Short Answer

Expert verified
The ring \(\operatorname{End}_{R}(V)\) is isomorphic to the matrix ring \(S\).

Step by step solution

01

Define the Homomorphism

Consider the map \( \phi: \operatorname{End}_{R}(V) \to S \) that sends each endomorphism \( f: V \to V \) to the matrix \( (\alpha_{i,j}) \), where each \( \alpha_{i,j} = \pi_i \circ f \circ \iota_j \). Here, \( \pi_i: V \to V_i \) is the canonical projection, and \( \iota_j: V_j \to V \) is the canonical inclusion.
02

Show \(\phi\) is a Ring Homomorphism

We verify that \( \phi(f+g) = \phi(f) + \phi(g) \) and \( \phi(fg) = \phi(f) \phi(g) \) for any \( f, g \in \operatorname{End}_{R}(V) \), considering the properties of matrix addition and multiplication with function composition. This shows \( \phi \) is indeed a ring homomorphism.
03

Demonstrate \(\phi\) is Surjective

For any matrix \((\alpha_{i,j})\) in \(S\), define an endomorphism \( f \in \operatorname{End}_{R}(V)\) composed of \( \sum_{i,j} \iota_i \circ \alpha_{i,j} \circ \pi_j \). Verify \( \phi(f) = (\alpha_{i,j}) \), ensuring that \( \phi \) can obtain any element of \( S \), thus proving surjectivity.
04

Prove \(\phi\) is Injective

Suppose \( \phi(f) = (0) \), meaning \( \pi_i \circ f \circ \iota_j = 0 \) for all \(i, j\). Hence, \( f = 0 \) because the restriction to each summand square to zero enforces \( f\equiv 0\) on all of \( V \). This establishes injectivity.
05

Conclude the Isomorphism

Since \( \phi \) is both injective and surjective, it is bijective. Therefore, by the isomorphism theorem for rings, \( \operatorname{End}_{R}(V) \cong S \), concluding the problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Endomorphism Ring
The endomorphism ring, denoted as \( \operatorname{End}_R(V) \), is a fascinating concept in ring theory. It consists of all \( R \)-module homomorphisms from a module \( V \) to itself.
  • An endomorphism is essentially a map \( f: V \rightarrow V \) that respects the operations of addition and scalar multiplication in the module \( V \).
  • Endomorphisms can be added together and composed, giving the structure of a ring.
  • The identity map \( \operatorname{id}_V \), which simply returns any input as itself, acts as the multiplicative identity in the endomorphism ring.
This structure is useful for understanding the transformations and symmetries within the module \( V \). It is especially powerful because it allows us to categorize such maps with ring properties鈥攖his makes theoretical predictions about the module's behavior possible.
Homomorphism
A homomorphism is a map between two algebraic structures that preserves their defining operations. In the context of modules, an \( R \)-module homomorphism \( g: M \to N \) satisfies two core properties:
  • It respects addition: \( g(m_1 + m_2) = g(m_1) + g(m_2) \) for all \( m_1, m_2 \in M \).
  • It is compatible with scalar multiplication: \( g(rm) = rg(m) \) for any \( r \in R \) and \( m \in M \).
This preservation ensures that the essential structure of modules is maintained when transferring information from one to another. In our exercise, the homomorphisms \( \pi_i: V \to V_i \) and \( \iota_j: V_j \to V \) play a role in mapping between different parts of a direct sum to facilitate the construction of the checkered matrix descriptions in \( S \).
Direct Sum of Modules
The direct sum of modules, \( V = \sum_{k=1}^{n} V_k \), is a way to combine several modules into a single framework. Each module \( V_k \) represents a component of the entire structure, and elements of \( V \) can be expressed uniquely as a sum of elements from each component module. Key properties include:
  • Every element \( v \in V \) can be uniquely written as \( v = v_1 + v_2 + \ldots + v_n \), with each \( v_k \) from the respective \( V_k \).
  • This uniqueness enables the splitting of complex transformations over \( V \) into simpler transformations over each component \( V_k \).
Such decomposition plays a crucial role in understanding how actions on \( V \) can separate into actions on each \( V_k \). In our problem, it helps to visualize \( V \) as sections being organized in a checkered pattern for matrix representation.
Isomorphism Proof
An isomorphism between two algebraic structures establishes a perfect correspondence between them. In ring theory, showing that two rings are isomorphic involves demonstrating an isomorphism鈥攁 bijective homomorphism鈥攅xists between them. The process breaks down into:
  • Showing surjectivity: Demonstrating every element of the target ring is the image of some element from the source ring.
  • Showing injectivity: Proving that no two distinct elements in the source map to the same element in the target.
For the exercise, a map \( \phi: \operatorname{End}_R(V) \to S \) was constructed to represent every endomorphism through matrix entries. It was shown to be both surjective and injective, thus proving \( \phi \) is an isomorphism.This means the endomorphism ring of the direct sum \( V \) corresponds directly to the checkered matrix ring \( S \); hence, they are structurally identical as rings.

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Most popular questions from this chapter

Let \(R=\mathrm{M}_{n}(D)\) with \(D\) a division ring. Show that \(R\) has finitely many right ideals if and only if either \(n=1\) or \(D\) is finite.

Let \(K\) be a field and let \(R \subseteq M_{2}(K)\) be given by \(R=\left(\begin{array}{cc}K & K \\ 0 & K\end{array}\right)\). If \(e=e_{1,1}\), show that \(\operatorname{End}_{R}(e R)\) is a field even though \(e R\) is not an irreducible \(R\)-module.

Suppose the ring \(R\) has elements \(e_{i, j}\) for \(i, j=1,2, \ldots, n\) that satisfy \(e_{i, j} e_{j^{\prime}, k}=0\) if \(j \neq j^{\prime}, e_{i, j} e_{j, k}=e_{i, k}\) and \(1=e_{1,1}+e_{2,2}+\cdots+e_{n, n} .\) If \(S\) is the centralizer in \(R\) of these \(n^{2}\) elements, prove that \(R \cong \mathrm{M}_{n}(S)\) and that \(S \cong e_{1,1} R e_{1,1} .\) To start with, define a map \(\sigma: \mathrm{M}_{n}(S) \rightarrow R\) by \(\left(s_{i, j}\right) \mapsto \sum_{i, j} s_{i, j} e_{i, j} .\) Then show that \(\sigma\) is a ring isomorphism.

Let \(D\) be a division ring and let \(R\) be a ring of \(D\)-linear transformations on \(_{D} V\). Assume that \(R\) is doubly transitive on \(V\) so that, by definition, if \(\left\\{v_{1}, v_{2}\right\\}\) is a \(D\)-linearly independent subset of \(V\) and if \(\left\\{w_{1}, w_{2}\right\\} \subseteq\) \(V\) is arbitrary, then there exists \(r \in R\) with \(v_{1} r=w_{1}\) and \(v_{2} r=w_{2}\). Show that \(\operatorname{End}_{R}(V)=D\) and conclude that \(R\) is a dense ring of linear transformations on \(V .\) What happens if we merely assume that \(R\) is transitive on \(V ?\)

Prove that an Artinian ring with no nonzero nilpotent elements and no nontrivial central idempotents is a division ring.
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