Problem 31 - A Computational Introduction to Number Theory and Algebra

Question

Show that the ring \(\mathbb{Z}[\sqrt{-5}]\) is not a UFD.

Step-by-Step Solution

Answer

Explain your answer. No, the ring \(\mathbb{Z}[\sqrt{-5}]\) is not a UFD. We found an element (6) that has more than one distinct factorizations into irreducible elements: \((1+\sqrt{-5})(1-\sqrt{-5})\) and \((2)(3)\). These factorizations cannot be made multiples of one another by the units in \(\mathbb{Z}[\sqrt{-5}]\), which demonstrates that the ring is not a UFD.

Step 1: Find distinct factorizations of 6 in \(\mathbb{Z}[\sqrt{-5}]\)

First, let's find distinct factorizations of 6 in \(\mathbb{Z}[\sqrt{-5}]\). We can have two possible factorizations for 6: 1. \((1+\sqrt{-5})(1-\sqrt{-5})\) 2. \((2)(3)\) Now, we need to show that these factorizations are in terms of irreducible elements.

Step 2: Show the elements involved in the factorizations are irreducible

Recall that an element is irreducible if it cannot be written as the product of two non-unit elements. 1. \((1+\sqrt{-5})\) and \((1-\sqrt{-5})\): For both of these elements, their norms are given by \(N(a+b\sqrt{-5}) = a^2 + 5b^2\). So, \(N(1+\sqrt{-5}) = 1^2 + 5(1)^2 = 6\) and \(N(1-\sqrt{-5}) = 1^2 + 5(1)^2 = 6\). Both elements have a norm of 6, which is a prime number in \(\mathbb{Z}\). Thus, neither of them can be factored as a product of two non-unit elements in \(\mathbb{Z}[\sqrt{-5}]\), making them irreducible. 2. \((2)\) and \((3)\): These are prime integers in \(\mathbb{Z}\), and hence cannot be factored as a product of two elements in \(\mathbb{Z}[\sqrt{-5}]\) with smaller distinct norms, making them irreducible in \(\mathbb{Z}[\sqrt{-5}]\) as well.

Step 3: Show that the factorizations are not multiples of one another by units in \(\mathbb{Z}[\sqrt{-5}]\)

Now, let's consider the units of \(\mathbb{Z}[\sqrt{-5}]\). An element is a unit if its norm is 1. In \(\mathbb{Z}[\sqrt{-5}]\), the only elements with a norm of 1 are 1 and -1 (considering that the norm is given by \(N(a+b\sqrt{-5}) = a^2 + 5b^2\)). So, there's no possible way to make the factorizations multiples of one another by using the units 1 or -1. Thus, the ring \(\mathbb{Z}[\sqrt{-5}]\) is not a UFD since the element 6 has two distinct factorizations into irreducible elements: 1. \((1+\sqrt{-5})(1-\sqrt{-5})\) 2. \((2)(3)\)

Key Concepts

Ring Theory

Ring theory is a fundamental branch of abstract algebra focusing on rings and their properties. A ring is a collection of elements equipped with two binary operations: addition and multiplication. These operations must satisfy several axioms, such as associativity and distributivity. This means:

• Associativity: For all elements \(a, b, \) and \(c\) in the ring, \((a + b) + c = a + (b + c)\) and \((a \cdot b) \cdot c = a \cdot (b \cdot c)\).
• Distributivity: Multiplication distributes over addition, \(a \cdot (b + c) = (a \cdot b) + (a \cdot c)\).

Rings often have additional structures, like unity (identity elements) or being commutative. For a ring to be a Unique Factorization Domain (UFD), every element in the ring should be expressible as a product of irreducible elements uniquely, except for order and units. The challenge sometimes lies in confirming whether specific number rings, such as \(\mathbb{Z}[\sqrt{-5}]\), fulfill these conditions.

Irreducible Elements

Irreducible elements are crucial in understanding the structure of rings. An element \(r\) in a ring is irreducible if not only is it non-zero and not a unit, but it also cannot be factored into a product of two non-unit elements within that ring. However, not every irreducible element is prime.
Consider the ring \(\mathbb{Z}[\sqrt{-5}]\). Elements like \(1+\sqrt{-5}\) and \(1-\sqrt{-5}\) are irreducible here because their norms are 6, a prime number in the integers \(\mathbb{Z}\).
They cannot be split into products of other non-unit elements with smaller norms. Recognizing irreducible elements helps assess the factorization properties of a ring, informing us about its nature as either a UFD or not.

Norms in Quadratic Rings

Norm functions are pivotal when working with quadratic rings and extended numbers. In a quadratic ring of the form \(\mathbb{Z}[\sqrt{d}]\), the norm of an element \(a + b\sqrt{d}\) is calculated as \(a^2 - db^2\). This measure helps determine properties like factorization and irreducibility.
For \(\mathbb{Z}[\sqrt{-5}]\), the norm is given by the formula \(N(a+b\sqrt{-5}) = a^2 + 5b^2\). This allows mathematicians to ascribe each element a positive integer value, guiding the factorization into irreducible components.

• If the norm is prime in \(\mathbb{Z}\), then the element may be irreducible.
• Norms help in verifying non-equivalent factorizations within a ring.

By studying policy norms, we identify that norms provide a way to unveil deeper insights into factorization behaviors within algebraic rings.