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Let \(\mathcal{I}\) be an ideal of \(R\). Prove that if \(R / \mathcal{I}\) is a free \(R\)-module, then \(\mathcal{I}\) is the zero ideal.

Short Answer

Expert verified
If \(R / \mathcal{I}\) is a free R-module with basis \(\{x_i + \mathcal{I}\}\), then a linear combination of basis elements equaling the zero element in \(R / \mathcal{I}\) (i.e., \(\mathcal{I}\)) can be written as \(\sum_{i} r_i (x_i + \mathcal{I}) = \mathcal{I}\), with \(r_i \in R\). This implies that \(\sum_{i} r_i x_i = \mathcal{I}\). Due to the linear independence of basis elements, we have \(r_i = 0\) for all \(i\), which means that \(\mathcal{I}\) must be the zero ideal.

Step by step solution

01

Understand Free Modules

First, let's recall the definition of a free module. A module M over a ring R is said to be free if it has a basis, which is a set of linearly independent elements that span M. In our case, the module in consideration is \(R / \mathcal{I}\), and we are given that it is free.
02

Identify the Basis of R/\(\mathcal{I}\)

Since \(R / \mathcal{I}\) is a free R-module, it has a basis. Let's denote this basis by \(\{x_i + \mathcal{I}\}\), where \(i\) runs over some index set, and \(x_i \in R\). Since the elements \(\{x_i + \mathcal{I}\}\) form a basis, they are linearly independent.
03

Consider Linear Combination of Basis Elements

Let's consider a linear combination of basis elements that equals the zero element in \(R / \mathcal{I}\), i.e., \(\mathcal{I}\) itself. We have: \[\sum_{i} r_i (x_i + \mathcal{I}) = \mathcal{I},\] where \(r_i \in R\). Now, using the properties of the quotient module and ideals, we can rewrite this as: \[\sum_{i} r_i x_i + \mathcal{I} = \mathcal{I}.\]
04

Linear Independence and Zero Ideal

Since the basis elements are linearly independent: \[\sum_{i} r_i x_i = \mathcal{I}.\] Since the left-hand side is an element of the ideal \(\mathcal{I}\) and the right-hand side is the ideal \(\mathcal{I}\) itself, this implies that the only possibility for this linear combination to equal the ideal is if for all \(i\), \(r_i x_i = 0\). But since \(\{x_i\}\) are linearly independent, we must have \(r_i = 0\) for all \(i\), meaning that the linear combination consists only of zero elements. Now, we can rewrite the sum as: \[\sum_{i} 0 \cdot x_i = 0,\] which means that the ideal \(\mathcal{I}\) must be the zero ideal. Therefore, we have proven that if \(R / \mathcal{I}\) is a free R-module, then \(\mathcal{I}\) is the zero ideal.

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Key Concepts

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

Free Module
A free module is an important concept within the field of abstract algebra. For instance, consider an R-module M, where R is a ring. We call M free if it enjoys a property similar to that of a vector space having a basis. More specifically, M has a set of generators that are linearly independent, and every element of M can be uniquely expressed as a finite linear combination of these generators.

This means, given a ring R and a free module M over it, we can find elements \(x_1, x_2, ..., x_n\) such that any element m in M can be written uniquely as \(r_1x_1 + r_2x_2 + ... + r_nx_n\) where \(r_1, r_2, ..., r_n\) are elements of R. This concept is crucial as it provides an easy understanding of the module's structure, which can parallel that of a vector space, making computations and understanding relations within the module much more manageable.

For the free module \(R / \mathcal{I}\), if we can show it has a basis, then this implies that \(R / \mathcal{I}\) behaves much like the ring R does in the absence of the ideal \mathcal{I}. If \mathcal{I} contributes nothing (is the zero ideal), then the module \(R / \mathcal{I}\) is essentially 'free' of any relations imposed by \mathcal{I}, preserving the same algebraic structure as R.
Linear Independence
Linear independence is a key concept when discussing free modules and vector spaces. A set of elements within a module or a vector space is said to be linearly independent if none of the elements can be expressed as a linear combination of the others. In other words, for any finite set of elements \(\{x_1, x_2, ..., x_n\}\), the only solution to the equation \(r_1x_1 + r_2x_2 + ... + r_nx_n = 0\), where zero represents the additive identity in the module, is for all the coefficients \(r_i\) to be zero.

This property ensures that no element in the set is 'redundant'; the set is a minimal generating set for the module. It has a strong implication for our exercise: Since we have established that \(R / \mathcal{I}\) is a free module with a basis, each element of the basis of \(R / \mathcal{I}\) is such that it cannot be written as a combination of other basis elements, reinforcing the conclusion that the ideal \mathcal{I} must indeed be the zero ideal for this to hold.
Quotient Module
In abstract algebra, a quotient module arises when you form a modular congruence class. This is an analogy to quotient groups in group theory. For a module M and a submodule N in M, we can define a new module called the quotient module, denoted as \(M/N\), consisting of the set of cosets of N in M. Each coset is a subcollection of M and looks like \(m + N\) for some \(m \in M\), which represents all the elements m could be when paired with any element of N.

A coset simply encapsulates the idea of adding a fixed element to every element of the submodule N. The quotient module \(M/N\) itself turns out to have a module structure with addition and scalar multiplication defined 'coset-wise'. The exercise at hand centers on the quotient module \(R / \mathcal{I}\), which is formed by taking the ring R and modding out by the ideal \mathcal{I}. It's a special case because ideals in rings behave well enough that the quotient actually forms a ring (and hence a module) in its own right.
Basis of Module
The concept of a basis of a module generalizes the idea of a basis in linear algebra. For a free module M over a ring R, a basis is a set of elements in M that are linearly independent (as we discussed earlier) and such that every element in M can be uniquely represented as a linear combination of these basis elements.

In our context, if the quotient module \(R / \mathcal{I}\) is said to have a basis, it means there is a set of cosets \(\{x_i + \mathcal{I}\}\) that satisfies these conditions. The unique representation part is crucial here. If \(\mathcal{I}\) is not the zero ideal, then the representation wouldn't be unique, violating the very definition of a basis. Hence, the fact that \(R / \mathcal{I}\) has a basis forces \mathcal{I} to be the zero ideal, leading to the resolution of our exercise. This underlines why a basis is critical for understanding the structure and properties of modules, particularly free modules.

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