Problem 55 - Algebraic Curves: An Introduction to Algebraic Geometry

Question

Let \(F=X^{n}+a_{1} X^{n-1}+\cdots+a_{n}\) be a monic polynomial in \(R[X]\). Show that \(R[X] /(F)\) is a free \(R\)-module with basis \(\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\), where \(\bar{X}\) is the residue of \(X\).

Step-by-Step Solution

Answer

Question: Show that for any commutative ring \(R\), with identity and any monic polynomial \(F\) of degree \(n\), the quotient ring \(R[X]/(F)\) is a free \(R\)-module, with basis \(\{\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\}\). Answer: To show that \(R[X]/(F)\) is a free \(R\)-module with basis \(\{\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\}\), we proved the linear independence and spanning of these elements over \(R\). It has been shown that a linear combination of these elements equals the zero element in \(R[X]/(F)\) only when all coefficients are zero, proving their linear independence. Additionally, we showed that any element in \(R[X]/(F)\) can be expressed as a linear combination of these elements, proving that they span the entire \(R\)-module. Since the elements \(\{\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\}\) are linearly independent and span \(R[X]/(F)\), they form a basis for this \(R\)-module, and thus, \(R[X]/(F)\) is a free \(R\)-module.

Step 1: Definition of Free \(R\)-Module and Basis

A free \(R\)-module is an \(R\)-module that has a basis, which is a set of linearly independent elements that span the entire module. In other words, any element in the module can be expressed as a unique linear combination of the basis elements.

Step 2: Show Linear Independence

Assume that there exists \(r_0, r_1, \cdots, r_{n-1} \in R\) such that \(r_0\overline{1} + r_1\bar{X} + \cdots + r_{n-1}\bar{X}^{n-1} = \overline{0}\). This means that \(r_0 + r_1 X + \cdots + r_{n-1} X^{n-1} \in (F)\). Since \(F\) is monic and has degree \(n\), we have that \(r_0 + r_1 X + \cdots + r_{n-1} X^{n-1}\) must be a multiple of \(F\). However, the degree of the left-hand side is at most \(n-1\), which is less than the degree of \(F\). The only multiple of \(F\) with degree less than \(n\) is the zero polynomial. Therefore, \(r_0 + r_1 X + \cdots + r_{n-1} X^{n-1} = 0\). This implies that the coefficients are all zero: \(r_0 = r_1 = \cdots = r_{n-1} = 0\). Hence, the elements \(\{\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\}\) are linearly independent.

Step 3: Show Spanning

Let \(\bar{P} \in R[X]/(F)\). Then \(P \in R[X]\) is a polynomial of the form \(P = p_0 + p_1 X + \cdots + p_m X^m\), with \(m \geq 0\) and \(p_i \in R\). We can write \(P\) as $$ P = (p_0 + p_1 X + \cdots + p_{n-1} X^{n-1}) + (p_n X^n + p_{n+1} X^{n+1} + \cdots + p_m X^m). $$ Now, we can use polynomial division to express \(P\) as $$ P = QF + R $$ where \(Q\) is the quotient polynomial, \(F\) is the monic polynomial, and \(R\) is the remainder polynomial with the degree of \(R\) less than the degree of \(F\). Thus, \(R = r_0 + r_1 X + \cdots + r_{n-1} X^{n-1}\), with \(r_i \in R\). Then, $$ \overline{P} = \overline{Q}\overline{F} + \overline{R} = \overline{0} + \overline{R} = \overline{R}. $$ We can see that \(\overline{P} = \overline{R} = r_0\overline{1} + r_1\bar{X} + \cdots + r_{n-1}\bar{X}^{n-1}\) is a linear combination of the elements \(\{\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\}\). Therefore, these elements span \(R[X]/(F)\). Since the elements \(\{\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\}\) are linearly independent and span \(R[X]/(F)\), they form a basis for this \(R\)-module, and thus, \(R[X]/(F)\) is a free \(R\)-module.

Key Concepts

Polynomial Division

To truly understand how polynomial division contributes to the structure of an R-module, let's first look at the concept of polynomial division itself. When you divide polynomial P by polynomial F, where F is monic and has a higher degree, you obtain a quotient polynomial Q and a remainder R such that P = QF + R. Importantly, the degree of the remainder R is less than the degree of F.

In the context of an R-module, this operation allows us to represent any polynomial as a combination of the basis polynomials and multiples of F. Consequently, when we discuss the module R[X] / (F), any polynomial P in R[X] corresponds to the remainder R modulo F because the multiple of F would be considered as zero. Therefore, polynomial division plays a crucial role in identifying a spanning set for the module.

Linear Independence

Linear independence is a key concept in understanding a basis for an R-module. Elements \(\overline{1}, \bar{X}, \cdots, \bar{X}^{n-1}\) are said to be linearly independent if the only way to write the zero element of the module as a linear combination of these elements is by taking all scalar coefficients to be zero. What it means for our R-module is that no element in the set \(\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\) can be expressed as a linear combination of the others, which ensures that no element is 'redundant'.

To prove this, we assert that if a linear combination of these elements results in the zero element — here represented as the zero polynomial — then all the scalar coefficients must be zero. This strict requirement is what qualifies them to be the candidates for a basis of the module, as they contribute to its structure without overlapping functionality.

Module Spanning

For a set of elements to span an R-module R[X] / (F), they must be able to collectively express every element within the module. To show this for \(\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\), we take any polynomial \(\bar{P}\) in the module and demonstrate that it can be written as a linear combination of these elements.

By applying polynomial division to \(P\) within the polynomial ring R[X], and since F acts as zero within the module, every polynomial \(\bar{P}\) can be uniquely represented by the remainder term \(R\), which in turn is a linear combination of the spanning elements. Therefore, these elements reach out to cover every corner of the module, leaving no polynomial behind, and fulfilling the criterion of spanning the module.

Basis of a Module

A basis of a module unites the concepts of linear independence and spanning. For an R-module, the basis is the set of elements that both spans the module and is linearly independent. Our focus on \(\overline{1}, \bar{X}, \ldots, \bar{X}^{n-1}\) within the module R[X] / (F) demonstrates that these elements fulfill both conditions: they span the space by covering all elements in the module and are linearly independent because no one of them can be created by the others.

This duality ensures that the basis is an efficient and complete 'recipe' for reconstructing any element of the module. As the problem states, because these elements meet both criteria, they indeed form a basis for the free R-module R[X] / (F), which shows that any element of this module can be uniquely expressed through these basis elements.