Polynomial Reducibility Analysis

Let's analyze the properties of the polynomial \(f(x) = x^2 + 3x - 1\) concerning its reducibility over different algebraic structures and the unique factorization property of \(\mathbb{Z}[\sqrt{13}]\).

A polynomial is said to be reducible over a field or a ring if it can be factored into two non-constant polynomials with coefficients in that field or ring. For a monic quadratic polynomial like \(f(x)\), it is reducible over a field if and only if it has roots in that field. For an integral domain like \(\mathbb{Z}[\sqrt{13}]\), a monic quadratic polynomial is reducible if and only if it has roots in that domain.

Polynomial Reducibility over \(\mathbb{Q}\)

To check the reducibility of \(f(x) = x^2 + 3x - 1\) over the field of rational numbers \(\mathbb{Q}\), we can find its roots using the quadratic formula:

The roots are given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for a quadratic equation \(ax^2 + bx + c = 0\).

For \(f(x) = x^2 + 3x - 1\), we have \(a=1\), \(b=3\), and \(c=-1\). The roots are:

\(x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)}\)

\(x = \frac{-3 \pm \sqrt{9 + 4}}{2}\)

\(x = \frac{-3 \pm \sqrt{13}}{2}\)

The roots are \(\frac{-3 + \sqrt{13}}{2}\) and \(\frac{-3 - \sqrt{13}}{2}\). The number \(\sqrt{13}\) is irrational, so \(\frac{-3 + \sqrt{13}}{2}\) and \(\frac{-3 - \sqrt{13}}{2}\) are irrational numbers. Since the roots are not in \(\mathbb{Q}\), the polynomial \(f(x)\) cannot be factored into linear factors with coefficients in \(\mathbb{Q}\). Therefore, \(f(x)\) is irreducible over \(\mathbb{Q}\).

• Statement: f is irreducible over \(\mathbb{Q}\) - This statement is true.

Polynomial Reducibility over \(\mathbb{Q}[\sqrt{13}]\)

The field \(\mathbb{Q}[\sqrt{13}]\) consists of numbers of the form \(a + b\sqrt{13}\), where \(a, b \in \mathbb{Q}\).

The roots of \(f(x)\) are \(\frac{-3 + \sqrt{13}}{2}\) and \(\frac{-3 - \sqrt{13}}{2}\). Let's rewrite these roots in the form \(a + b\sqrt{13}\):

• \(\frac{-3 + \sqrt{13}}{2} = -\frac{3}{2} + \frac{1}{2}\sqrt{13}\). Here, \(a = -3/2\) and \(b = 1/2\). Since \(a, b \in \mathbb{Q}\), this root is in \(\mathbb{Q}[\sqrt{13}]\).
• \(\frac{-3 - \sqrt{13}}{2} = -\frac{3}{2} - \frac{1}{2}\sqrt{13}\). Here, \(a = -3/2\) and \(b = -1/2\). Since \(a, b \in \mathbb{Q}\), this root is in \(\mathbb{Q}[\sqrt{13}]\).

Since both roots of \(f(x)\) are in the field \(\mathbb{Q}[\sqrt{13}]\), the polynomial \(f(x)\) can be factored into linear factors over \(\mathbb{Q}[\sqrt{13}]\):

\(f(x) = (x - \frac{-3 + \sqrt{13}}{2})(x - \frac{-3 - \sqrt{13}}{2})\)

Thus, \(f(x)\) is reducible over \(\mathbb{Q}[\sqrt{13}]\).

• Statement: f is reducible over \(\mathbb{Q}\)[\(\sqrt{13}\)] - This statement is true.

Polynomial Reducibility over \(\mathbb{Z}[\sqrt{13}]\)

The ring \(\mathbb{Z}[\sqrt{13}]\) consists of numbers of the form \(a + b\sqrt{13}\), where \(a, b \in \mathbb{Z}\) (integers).

For a monic quadratic polynomial to be reducible over an integral domain like \(\mathbb{Z}[\sqrt{13}]\), it must have roots in that domain.

The roots of \(f(x)\) are \(\frac{-3}{2} + \frac{1}{2}\sqrt{13}\) and \(\frac{-3}{2} - \frac{1}{2}\sqrt{13}\). For these roots to be in \(\mathbb{Z}[\sqrt{13}]\), the coefficients \(a = -3/2\) and \(b = \pm 1/2\) must be integers. However, \(-3/2\) and \(\pm 1/2\) are not integers.

Therefore, the polynomial \(f(x)\) has no roots in \(\mathbb{Z}[\sqrt{13}]\). Since it is a monic quadratic polynomial with no roots in the domain, it is irreducible over \(\mathbb{Z}[\sqrt{13}]\).

• Statement: f is irreducible over \(\mathbb{Z}\) [ \(\sqrt{13}\) ] - This statement is true.

Unique Factorization Domain \(\mathbb{Z}[\sqrt{13}]\)

A unique factorization domain (UFD) is an integral domain where every non-zero, non-unit element can be written as a product of irreducible elements, uniquely up to order and associates.

Let's examine the element \(-4\) in \(\mathbb{Z}[\sqrt{13}]\).

We can factor \(-4\) in two ways:

1. \(-4 = 2 \times (-2)\)
2. \(-4 = (3 + \sqrt{13})(3 - \sqrt{13})\) because \((3 + \sqrt{13})(3 - \sqrt{13}) = 3^2 - (\sqrt{13})^2 = 9 - 13 = -4\).

Let's check if the factors \(2\), \(-2\), \(3 + \sqrt{13}\), and \(3 - \sqrt{13}\) are irreducible in \(\mathbb{Z}[\sqrt{13}]\). An element \(a+b\sqrt{13}\) is irreducible if its norm, \(N(a+b\sqrt{13}) = a^2 - 13b^2\), cannot be factored into norms of non-unit elements.

• \(N(2) = 2^2 - 13(0)^2 = 4\). For 2 to be reducible, it must have factors with norms \(\pm 2\). We check if \(a^2 - 13b^2 = \pm 2\) has integer solutions. Modulo 13, this congruence is \(a^2 \equiv \pm 2 \pmod{13}\). The squares modulo 13 are \{0, 1, 4, 9, 3, 12, 10\}. Neither 2 nor -2 (which is 11 mod 13) is in this set. So there are no elements with norm \(\pm 2\). Thus, 2 is irreducible in \(\mathbb{Z}[\sqrt{13}]\). Similarly, \(-2\) is irreducible as \(N(-2)=4\).
• \(N(3 + \sqrt{13}) = 3^2 - 13(1)^2 = 9 - 13 = -4\). For \(3 + \sqrt{13}\) to be reducible, it must have factors with norms \(\pm 2\). As shown above, there are no elements with norm \(\pm 2\). Thus, \(3 + \sqrt{13}\) is irreducible in \(\mathbb{Z}[\sqrt{13}]\). Similarly, \(3 - \sqrt{13}\) is irreducible as \(N(3-\sqrt{13}) = 3^2 - 13(-1)^2 = 9 - 13 = -4\).

Now we check if 2 is an associate of \(3 + \sqrt{13}\) or \(3 - \sqrt{13}\). This would mean \((3 + \sqrt{13})/2\) or \((3 - \sqrt{13})/2\) is a unit in \(\mathbb{Z}[\sqrt{13}]\). A unit is an element \(a+b\sqrt{13}\) with norm \(\pm 1\). \((3 + \sqrt{13})/2 = 3/2 + (1/2)\sqrt{13}\). This is not in \(\mathbb{Z}[\sqrt{13}]\). So 2 is not an associate of \(3 + \sqrt{13}\). Similarly, 2 is not an associate of \(3 - \sqrt{13}\).

We have found two distinct factorizations of \(-4\) into irreducibles in \(\mathbb{Z}[\sqrt{13}]\): \(-4 = 2 \times (-2)\) and \(-4 = (3 + \sqrt{13})(3 - \sqrt{13})\). Since these factorizations are distinct (up to order and associates), \(\mathbb{Z}[\sqrt{13}]\) is not a unique factorization domain.

• Statement: \(\mathbb{Z}\) [ \(\sqrt{13}\) ] is a unique factorization domain - This statement is false.

Summary of Statements

Based on our analysis:

• f is irreducible over \(\mathbb{Z}\) [ \(\sqrt{13}\) ] - True
• f is irreducible over \(\mathbb{Q}\) - True
• f is reducible over \(\mathbb{Q}\)[\(\sqrt{13}\)] - True
• \(\mathbb{Z}\) [ \(\sqrt{13}\) ] is a unique factorization domain - False

The statements that are true are: f is irreducible over \(\mathbb{Z}\) [ \(\sqrt{13}\) ], f is irreducible over \(\mathbb{Q}\), and f is reducible over \(\mathbb{Q}\)[\(\sqrt{13}\)].