Chapter 8: Problem 6
Determine whether the indicate quotient rings are fields. Justify your answers. $$ \mathrm{Q}[x] /\left\langle x^{2}+3 x+2\right\rangle $$
Short Answer
Expert verified
The quotient ring is not a field since \(x^2 + 3x + 2\) is reducible over \(\mathbb{Q}\).
Step by step solution
01
Define the Quotient Ring
The quotient ring \(\mathbb{Q}[x] / \langle f(x) \rangle\) where \(f(x) = x^2 + 3x + 2\) is formed by the set of polynomials in \(\mathbb{Q}[x]\) modulo the ideal generated by \(f(x)\). This means any element in this quotient ring can be expressed as a polynomial \(g(x)\) of degree less than 2, because terms of degree 2 or higher can be reduced by the relation \(x^2 + 3x + 2 = 0\).
02
Check If the Ideal is Maximal
For \(\mathbb{Q}[x] / \langle f(x) \rangle\) to be a field, \(\langle f(x) \rangle\) must be a maximal ideal in \(\mathbb{Q}[x]\). A polynomial \(f(x)\) generates a maximal ideal if it is irreducible over \(\mathbb{Q}\). Check if \(x^2 + 3x + 2\) can be factored into polynomials of lower degree with rational coefficients.
03
Factor the Polynomial
Factor \(x^2 + 3x + 2\). The roots of this polynomial can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 3, c = 2\). Calculating this gives roots \(x = -1\) and \(x = -2\). Thus, \(x^2 + 3x + 2\) factors as \((x + 1)(x + 2)\) in \(\mathbb{Q}[x]\).
04
Determine if the Ring is a Field
Since \(x^2 + 3x + 2\) can be factored into \((x + 1)(x + 2)\) in \(\mathbb{Q}[x]\), the ideal \(\langle x^2 + 3x + 2 \rangle\) is not maximal, hence \(\mathbb{Q}[x] / \langle x^2 + 3x + 2 \rangle\) is not a field.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Factorization
Polynomial factorization is the process of expressing a polynomial as a product of its factors, which are simpler polynomials of lower degree. In the context of the given problem, the polynomial to be factored is \(x^2 + 3x + 2\). To factor this polynomial, we look for its roots using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = 3\), and \(c = 2\). Using these values, we calculate the discriminant \(b^2 - 4ac\) to be \(1\). The roots are:
Consequently, the factorization plays a crucial role in determining the nature of the quotient ring.
- \(x = -1\)
- \(x = -2\)
Consequently, the factorization plays a crucial role in determining the nature of the quotient ring.
Maximal Ideal
In ring theory, a maximal ideal is a very specific type of ideal that has great significance. An ideal \(I\) in a ring \(R\) is maximal if there exists no other ideal \(J\) such that \(I \subsetneq J \subsetneq R\). Maximal ideals are important because if a quotient ring formed from a ring \(R\) and an ideal \(I\) is a field, then \(I\) must be maximal.Checking whether the ideal \(\langle x^2 + 3x + 2 \rangle\) in \(\mathbb{Q}[x]\) is maximal involves checking the irreducibility of the polynomial \(f(x) = x^2 + 3x + 2\). This polynomial must be irreducible if the ideal it generates is to be maximal. A polynomial is irreducible over \(\mathbb{Q}\) if it cannot be expressed as a product of polynomials of lower degree with rational coefficients.In this case, since \(x^2 + 3x + 2\) factors as \((x + 1)(x + 2)\) with rational coefficients, \(\langle x^2 + 3x + 2 \rangle\) is not maximal. This lack of maximality shows that the quotient ring \(\mathbb{Q}[x] / \langle x^2 + 3x + 2 \rangle\) does not form a field.
Field Theory
Field theory is a branch of algebra that studies fields, which are mathematical structures with well-defined addition, subtraction, multiplication, and division operations. The concept of a field is critical for understanding how quotient rings may exhibit field-like properties.To determine if a quotient ring is a field, we must analyze the ideal used to create the quotient structure. In field theory, a ring \(R\) and an ideal \(I\) create a field \(R/I\) only if \(I\) is a maximal ideal. This is because fields are defined to have no nontrivial proper ideals, akin to having no smaller structures within them.In our case, we considered the quotient ring \(\mathbb{Q}[x] / \langle x^2 + 3x + 2 \rangle\). Based on field theory:
- The polynomial \(x^2 + 3x + 2\) was factored, and since it could be expressed in terms of lower degree polynomials, the ideal \(\langle x^2 + 3x + 2 \rangle\) is not maximal.
- Since the ideal is not maximal, the quotient ring cannot be a field.