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Chapter 4: Problem 1

Give the details to show that any commutative ring with identity is a module over itself.

Short Answer

Expert verified
In summary, a commutative ring \(R\) with identity is a module over itself because addition in \(R\) forms an abelian group, the distributive laws \((a + b)*r = a*r + b*r\) and \(a*(r + s) = a*r + a*s\) hold for all elements in \(R\), compatibility with the ring operation is verified as \(a*(r*s) = (a*r)*s\), and there exists a scalar identity 1 such that for any element \(a\) in \(R\), \(1*a = a\). These properties fulfill the requirements for being a module, demonstrating that any commutative ring with identity is a module over itself.

Step by step solution

01

1. Reiterate Properties of Commutative Ring with Identity

The properties of a commutative ring with identity include having a set \(R\) where addition forms an abelian group, multiplication is commutative and associative, and there is a multiplicative identity, 1.
02

2. Establishing \(R\) under addition as an abelian group

Since \(R\) with addition already is an abelian group by definition of a ring, this fulfills the first condition of \(R\) being a module over itself.
03

3. Verifying the Distributive Laws

We will verify that for every \(a\), \(b\), and \(r\) in \(R\), \((a + b)*r = a*r + b*r\) and \(a*(r + s) = a*r + a*s\). These laws are easily verified because the distributive law holds in the ring \(R\) which is clear from the ring axioms.
04

4. Checking Compatibility with the Ring Operation

For this, we need to confirm that for any \(a\) in \(R\) and \(r\), \(s\) in \(R\), \(a*(r*s) = (a*r)*s\). Given that multiplication in \(R\) is associative, \(a*(r*s) = a*(r.s) = (a.r).s = (a*r)*s\), which is the required property.
05

5. Checking Identity Element

Finally, the condition that for any \(a\) in \(R\), \(1*a=a\) is also satisfied. Here \(1\) is the identity element of \(R\), and this is guaranteed by the properties of our ring, which we assumed to have an identity. Therefore, any commutative ring \(R\) with identity is a module over itself as it satisfies all the properties required for being a module.

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Key Concepts

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

Module Theory
In mathematics, module theory is an extension of linear algebra. It generalizes some of the ideas we see with vectors and scalars to more abstract algebraic structures. A module is a generalization of a vector space where the scalars come from a ring instead of a field. Modules are important because they allow exploration of these concepts with a wider variety of mathematical objects, like rings.
For a commutative ring with identity, such as \( R \), to be considered a module over itself, it must satisfy certain properties:
  • The ring \( R \) under addition forms an abelian group. This means that the sum of any two elements in \( R \) is still in \( R \), and the order of addition does not matter.
  • The distributive laws must hold; we'll discuss these further in a later section.
  • Scalars (elements of \( R \)) multiply elements from the module \( R \) in a way that respects the ring's multiplication structure (associativity, identity).
By satisfying these conditions, the ring \( R \) acts as a set of scalars, making \( R \) a module over itself.
Ring Identity
The identity of a ring is what makes it stand apart from other mathematical structures. In a ring \( R \), an identity is an element that, when multiplied by any element \( a \) in \( R \), leaves \( a \) unchanged. Mathematically, this is expressed as \( 1 \cdot a = a \), where \( 1 \) is the identity.
The presence of a multiplicative identity in a ring ensures the functionality of multiplication in a manner that's consistent and predictable. Not all rings have a multiplicative identity, but for a ring with identity, this ensures stability and a reference point for performing calculations.
When we consider a commutative ring with identity as a module over itself, the identity plays a crucial role. This condition must hold: for any element \( a \) in \( R \), multiplying by \( 1 \) yields \( a \). This simplifies interactions within the module and helps verify its compliance with module theory.
Distributive Laws
Distributive laws are essential in both ring and module theory. They describe how multiplication interacts with addition, both within the ring and the module. In any ring \( R \), these laws are written as:
  • \( a \cdot (b + c) = a \cdot b + a \cdot c \)
  • \((b + c) \cdot a = b \cdot a + c \cdot a \)
These properties tell us how the elements in \( R \) "distribute" over the elements they multiply by.
For a commutative ring considered as a module over itself, these laws must hold true to ensure consistency among the module operations. If you have any elements \( a, b, \) and \( r \) in \( R \), the act of multiplying and adding should follow these exact distributive laws.
The distributive laws guarantee that combining additions and multiplications within the module won't produce unexpected results. They are foundational for the module's structure, enforcing predictable outcomes for operations.

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Most popular questions from this chapter

Let \(M\) be an \(R\)-module and let \(I\) be an ideal in \(R\). Let \(I M\) be the set of all finite sums of the form $$ r_{1} v_{1}+\cdots+r_{n} v_{n} $$ where \(r_{i} \in I\) and \(v_{i} \in M\). Is \(I M\) a submodule of \(M\) ?

Show that if \(S\) and \(T\) are submodules of \(M\), then (with respect to set inclusion) $$ S \cap T=\operatorname{glb}\\{S, T\\} \text { and } S+T=\operatorname{lub}\\{S, T\\} $$

Prove that \(\operatorname{hom}_{R}(M, N)\) is an \(R\)-module under addition of functions and scalar multiplication defined by $$ (r \tau)(v)=r(\tau v)=\tau(r v) $$

Suppose that \(R\) is a commutative ring with identity. If \(\mathcal{I}\) and \(\mathcal{J}\) are ideals of \(R\) for which \(R / \mathcal{I} \approx R / \mathcal{J}\) as \(R\)-modules, then prove that \(\mathcal{I}=\mathcal{J}\). Is the result true if \(R / \mathcal{I} \approx R / \mathcal{J}\) as rings?

Give an example of a module \(M\) that has a finite basis but with the property that not every spanning set in \(M\) contains a basis and not every linearly independent set in \(M\) is contained in a basis.
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