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Chapter 1: Problem 3

If \(V\) is a right \(R\)-module, prove that \(v 0=0=0 r\) for all \(v \in V\) and \(r \in R\). Furthermore, show that \((-v) r=-(v r)=v(-r)\).

Short Answer

Expert verified
The properties are verified using distribution and module axioms: \(v0 = 0\), \(0r = 0\), \((-v)r = -(vr) = v(-r)\).

Step by step solution

01

Prove that \(v 0 = 0\) for all \(v \in V\)

By the definition of a right module over a ring \(R\), we have an operation such that for any \(v \in V\): \[ v(r_1 + r_2) = vr_1 + vr_2 \] Substitute \(r_1 = 0\) and \(r_2 = 0\) into the equation:\[ v(0 + 0) = v0 \] which implies \[ v0 = v0 + v0 \]. Subtract \(v0\) from both sides to get \(v0 = 0\).
02

Prove that \(0r = 0\) for all \(r \in R\)

By module axioms, in particular the scalar multiplication axiom, multiplying zero (which is in \(V\)) by any scalar \(r \in R\) yields:\[ 0r = 0 \]. This follows directly from the property of zero in vector spaces/modules which states multiplying a zero vector with any scalar still gives the zero vector.
03

Prove that \((-v)r = -(vr)\)

To prove \((-v)r = -(vr)\), note using module properties: Add \((vr)\) to both sides of the equation and use distributive properties:\[ (-v)r + vr = (-v+v)r = 0r = 0 \].Thus, \((-v)r+ vr = 0\) implies \((-v)r = -(vr)\).
04

Prove that \((-v)r = v(-r)\)

Using distribution and associativity over ring elements, we have:\[ (-1)v = -v \], and consistently:\[ v(-r) = -(vr) \].This equality confirms that \((-v)r = v(-r)\). Additionally, since \(-1 \cdot v \cdot r = (-v)r = v(-1 \cdot r)\), the relation holds, showing that dealing with negatives in modules aligns with basic algebraic operations.

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Key Concepts

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

Scalar Multiplication
Scalar multiplication is a key operation in modules and vector spaces. In the context of a right \( R \)-module, it involves multiplying elements of the module by scalars from the ring \( R \). Scalar multiplication must satisfy specific properties to maintain the structure of the module.

  • Closed under Scalar Multiplication: If \( v \) is an element of the module \( V \) and \( r \) is a scalar from \( R \), then \( vr \) is also an element of \( V \).
  • Distributive with Respect to Module Addition: Scalar multiplication distributes over module addition, meaning \( v(r_1 + r_2) = vr_1 + vr_2 \) for any module element \( v \) and scalars \( r_1, r_2 \).
  • Associative with Ring Multiplication: For any \( v \in V \) and scalars \( r_1, r_2 \), \( v(r_1r_2) = (vr_1)r_2 \). This associativity ensures scalar multiplication behaves like multiplication in familiar algebraic structures.
Distributive Property
The distributive property is a fundamental rule in algebra that applies to many mathematical structures, including right \( R \)-modules. In the context of our exercise, we can see how it governs the behavior of scalar multiplication.

The property states that for any element \( v \) in a module and any scalars \( r_1, r_2 \) from the ring \( R \):
  • Right Distributive Law: \( v(r_1 + r_2) = vr_1 + vr_2 \). This depicts how a single module element multiplied by a sum of scalars results in the sum of the individual products.
This property ensures that multiplication is linear in nature, providing a consistent framework for combining operations involving scalars and module elements. It's crucial for verifying statements like \( v0 = 0 \) and for proving results involving negative elements, as discussed in the solution.
Module Axioms
Module axioms form the foundational rules that define the structure of a module. These axioms ensure the module behaves in a predictable and structured way, similar to vector spaces but with more generalized scalar fields.

Let's focus on some crucial axioms that were highlighted in the exercise:
  • Zero Element: There's an element \( 0 \in V \) such that for all \( v \in V \), \( v + 0 = v \) and \( 0v = 0 \). This axiom was used in proving \( v0 = 0 \) and \( 0r = 0 \).
  • Existence of Negatives: For every \( v \in V \), there is an element \( -v \) in \( V \) such that \( v + (-v) = 0 \). This ensures we can deal with subtractions and relates to our solution involving \((-v)r = -(vr) \).
  • Compatibility with Ring Operations: The operations of scalar multiplication are consistent with the ring operations, adhering to distributive and associative laws. This axiom helps ensure the module behaves like a familiar algebraic system, consistent with our understanding of addition and multiplication.

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Most popular questions from this chapter

Let \(W\) be a subset of \(V\). Prove that \(W\) is a submodule if and only if it is nonempty, closed under \(+\) and closed under multiplication by \(R\). If \(X\) and \(Y\) are submodules of \(V\), show that \(X \cap Y\) and \(X+Y\) are also submodules.

If \(V\) and \(W\) are vector spaces over the field \(K\), observe that they are \(K\)-modules and describe \(\operatorname{Hom}_{K}(V, W)\) and \(\operatorname{End}_{K}(V)\). Be more specific in case \(\operatorname{dim}_{K} V=n<\infty\) and \(\operatorname{dim}_{K} W=m<\infty\).

Let \(V\) and \(W\) be additive abelian groups. Verify that \(\operatorname{Hom}(V, W)\) is an additive abelian group and that \(\operatorname{End}(V)\) is a ring. If \(V\) and \(W\) are \(R\)-modules, verify that \(\operatorname{Hom}_{R}(V, W)\) is a subgroup of \(\operatorname{Hom}(V, W)\) and that \(\operatorname{End}_{R}(V)\) is a subring of \(\operatorname{End}(V)\).

Let \(V\) be a nonzero finitely generated \(R\)-module. Prove that \(V\) has a maximal submodule \(W\). By this we mean that \(W \neq V\) and that there are no submodules contained properly between \(W\) and \(V\). Show by example that this result is false for arbitrary nonzero \(V\).

Suppose \(\begin{aligned} A & \stackrel{\sigma}{\longrightarrow} \quad B \\ \alpha \downarrow & \downarrow \beta \\ A^{\prime} & \stackrel{\sigma^{\prime}}{\longrightarrow} \quad B^{\prime} \end{aligned}\) is a commutative diagram of \(R\)-modules and of \(R\)-homomorphisms. Prove that \(\alpha: \operatorname{Ker}(\sigma) \rightarrow \operatorname{Ker}\left(\sigma^{\prime}\right)\) and furthermore that \(\beta\) induces a \(\operatorname{map} \tilde{\beta}: B / \operatorname{Im}(\sigma) \rightarrow B^{\prime} / \operatorname{Im}\left(\sigma^{\prime}\right) .\) Here \(B / \operatorname{Im}(\sigma)\) is called the cokernel of \(\sigma\).
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