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Chapter 16: Problem 31

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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. \((2)(3)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Let \(\rho: E \rightarrow E^{\prime}\) be an \(F\) -algebra homomorphism, let \(\alpha \in E,\) let \(\phi\) be the minimal polynomial of \(\alpha\) over \(F,\) and let \(\phi^{\prime}\) be the minimal polynomial of \(\rho(\alpha)\) over \(F\). Show that \(\phi^{\prime} \mid \phi,\) and that \(\phi^{\prime}=\phi\) if \(\rho\) is injective.

Show that the only \(\mathbb{R}\) -algebra homomorphisms from \(\mathbb{C}\) into itself are the identity map and the complex conjugation map.

Let \(a\) be a non-zero, square-free integer, with \(a \notin\\{\pm 1\\},\) and let \(n\) be a positive integer. Show that the polynomial \(X^{n}-a\) is irreducible in \(\mathbb{Q}[X]\)

Let \(g, h \in \mathbb{Z}[X]\) be non-zero polynomials with \(d:=\operatorname{gcd}(g, h) \in\) \(\mathbb{Z}[X]\). Show that for every prime \(p\) not dividing \(\operatorname{lc}(g) \operatorname{lc}(h),\) we have \(\bar{d} \mid \operatorname{gcd}(\bar{g}, \bar{h})\), and except for finitely many primes \(p,\) we have \(\bar{d}=\operatorname{gcd}(\bar{g}, \bar{h}) .\) Here, \(\bar{d}, \bar{g},\) and \(\bar{h}\) denote the images of \(d, g,\) and \(h\) in \(\mathbb{Z}_{p}[X]\) under the coefficient-wise extension of the natural map from \(\mathbb{Z}\) to \(\mathbb{Z}_{p}\) (see Example 7.47 ).

Consider the real numbers \(\sqrt{2}\) and \(\sqrt[3]{2}\). Show that \((\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]: \mathbb{Q})=(\mathbb{Q}[\sqrt{2}+\sqrt[3]{2}]: \mathbb{Q})=6\)
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