Problem 18 - Complex Analysis

Question

The factor ring \(K:=\mathbb{R}[X] /\left(X^{2}+1\right)\) is a field, since \(\mathbb{R}\) is a field and \(X^{2}+1\) is a prime prime element in \(\mathbb{R}[X]\). This must be proved. Let us denote by \(1_{K}\) the image of 1 , and by \(\mathrm{i}_{K}\) the image of \(X\) in \(K\). Then \(K=\mathbb{R} 1_{K}+\mathbb{R i}_{K}\). The axioms for a "field of complex numbers" are now easily checked, e.g. \(\mathrm{i}_{K}^{2}=-1_{K}\).

Step-by-Step Solution

Answer

The factor ring \(K = \mathbb{R}[X] / (X^2 + 1)\) is a field because it's isomorphic to complex numbers, ensuring component-wise closure and inversions.

Step 1: Understanding the Factor Ring

Starting with the definition of the factor ring \(K = \mathbb{R}[X] / (X^2 + 1)\), because \(\mathbb{R}\) is a field, all the nonzero polynomials in \(\mathbb{R}[X]\) are not zero divisors, making \(\mathbb{R}[X]\) a Principal Ideal Domain. Hence, factor rings built from prime ideals such as \((X^2 + 1)\) are fields.

Step 2: Polynomials in Quotient Ring

In \(K\), any element is a polynomial of degree less than 2, of the form \(a + bX\), where \(a, b \in \mathbb{R}\). This follows from reducing any polynomial in \(\mathbb{R}[X]\) by the modulus \(X^2 + 1\), where \(X^2 \equiv -1\).

Step 3: Establishing a Field Structure

To confirm that \(K\) is a field, we verify two things:1. Non-zero elements have inverses, ensuring division is possible (hence \(K\) is a field).2. The mapping \(a + bX \rightarrow a1_{K} + b\mathrm{i}_{K}\) satisfies field axioms.Since multiplication of polynomials reduces into the form \(a + bX\), and we can find an inverse for every non-zero polynomial, the structure of a field is achieved.

Step 4: Verification of Axioms

Identify \(1_{K}, \mathrm{i}_{K} \) such that \(1_{K}\) is the equivalence class represented by 1 and \(\mathrm{i}_{K}\) is represented by \(X\). Therefore, the relation \(\mathrm{i}_{K}^2 + 1_{K} = 0\) checks that \(\mathrm{i}_{K}^2 = -1_{K}\). Thus, the multiplication and addition operations coincide with the properties of complex numbers, completing the field verification.

Key Concepts

Field

In mathematics, a field is a set equipped with two operations: addition and multiplication. Fields provide a framework where one can add, subtract, multiply, and divide (except by zero), similar to how we handle rational numbers.

To classify a structure as a field, it must satisfy several properties:

• There is an additive identity, usually 0, where any number plus 0 remains unchanged.
• There is a multiplicative identity, often 1, where any number times 1 remains unchanged.
• Every element has an additive inverse, essentially meaning you can subtract any number using its negative.
• Every non-zero element has a multiplicative inverse, meaning you can divide by any non-zero number.

Real numbers (\(\mathbb{R}\)) form a familiar example of a field, but so do complex numbers, rational numbers, and certain polynomials under modulo conditions. In the exercise above, proving that the factor ring \(K = \mathbb{R}[X] / (X^2 + 1)\)is a field involves verifying these properties for polynomials modulo \(X^2 + 1\), demonstrating that non-zero elements have inverses.

Polynomial Ring

A polynomial ring consists of all polynomials with coefficients from a specific field, such as \(\mathbb{R}\), the field of real numbers. Polynomials are expressions like \(a_nX^n + a_{n-1}X^{n-1} + ... + a_0\), where \(a_i\) are coefficients in a field, and \(X\)is an indeterminant.

In the polynomial ring \(\mathbb{R}[X]\), we can perform operations such as addition, subtraction, and multiplication of polynomials. However, unlike a field, division is not always possible without a remainder, which is why not every polynomial ring is a field.

In our problem, \(\mathbb{R}[X]\)is described in the context of creating a factor ring. By taking the polynomial ring \(\mathbb{R}[X]\)and considering the ideal generated by \(X^2 + 1\), elements of the factor ring can be written as polynomials of degree less than 2, like \(a + bX\).

Principal Ideal Domain

A Principal Ideal Domain (PID) is a type of ring where every ideal can be generated by a single element. This means that for any ideal within the ring, there is some element in the ring such that the ideal comprises all the multiples of that element.

The polynomial ring \(\mathbb{R}[X]\)is an example of a PID, as every ideal can be generated by a polynomial. For instance, the ideal generated by the polynomial \(X^2 + 1\)is all the multiples of that polynomial within \(\mathbb{R}[X]\).

When the ideal is generated by a prime element, like in our factor ring \(\mathbb{R}[X] / (X^2 + 1)\), the factor ring becomes a field. This is because a prime element ensures no zero divisors are introduced, ensuring the invertibility necessary for a field.

Complex Numbers

Complex numbers extend the idea of one-dimensional real numbers to a two-dimensional plane using an imaginary unit, generally denoted as \(i\)satisfying \(i^2 = -1\). A complex number is expressed as \(a + bi\), where \(a\)and \(b\)are real numbers.

The factor ring \(K=\mathbb{R}[X] / (X^2 + 1)\)is demonstrated to exhibit the properties of complex numbers. In this setup, the element \(i_{K}\) corresponds to the imaginary unit, satisfying the relation \(i_{K}^2 = -1_{K}\).

Every element in \(K\)has the form \(a + bi_{K}\), mimicking the structure of complex numbers \(a + bi\)and confirming the field operations reflect the arithmetic of complex numbers. This allows polynomials modulo \(X^2 + 1\) to form a system analogous to complex numbers.