Problem 10 - A History of Mathematics: An Introduction

Question

Show that in the domain of integers of the form \(a+\) \(b \sqrt{-17}\), Liouville's factorization \(169=13 \cdot 13=(4+\) \(3 \sqrt{-17})(4-3 \sqrt{-17})\) in fact demonstrates that unique factorization into primes fails in that domain. (Hint: Use norms to show that each of the four factors is irreducible.)

Step-by-Step Solution

Answer

Based on the given factorization of 169 and the steps provided, demonstrate that the unique factorization into primes fails in the domain of integers of the form \(a+b\sqrt{-17}\).

Step 1: Define the norm function

For integers of the form \(a+b\sqrt{-17}\), the norm function is defined as: $$N(a+b\sqrt{-17}) = (a+b\sqrt{-17})(a-b\sqrt{-17}) = a^2 + 17b^2.$$

Step 2: Calculate the norms of the factors

We need to calculate the norms of each of the four factors (13, 13, \(4+3\sqrt{-17}\), and \(4-3\sqrt{-17}\)). $$N(13) = N(13)= 13^2 = 169.$$ $$N(4+3\sqrt{-17}) = (4^2 + 17\cdot(3^2)) = 16 + 153 = 169.$$ $$N(4-3\sqrt{-17}) = (4^2 + 17\cdot(-3^2)) = 16 + 153 = 169.$$

Step 3: Analyze the norms

Notice that all four factors have the same norm, which is 169. We now need to prove that these factors are irreducible. To do this, we will utilize the following property of norms: For any two integers \(x\) and \(y\) of the form \(a+b\sqrt{-17}\), we have \(N(xy) = N(x)N(y)\). Recall that an integer \(x\) of the form \(a+b\sqrt{-17}\) is irreducible if, whenever we can write \(x = y\cdot z\) for integers \(y\) and \(z\) of the form \(a+b\sqrt{-17}\), either \(N(y)=1\) or \(N(z)=1\) (which means either \(y\) or \(z\) is a unit).

Step 4: Prove the irreducibility of the given factors

Suppose one of the factors, say 13, can be factored further as \(13 = y\cdot z\), where \(y\) and \(z\) are integers of the form \(a+b\sqrt{-17}\). By the norm property mentioned above, we have: $$N(13) = N(y\cdot z) = N(y)N(z),$$ which implies that $$169 = N(y)N(z).$$ Since \(N(y), N(z) \ge 1\), the only possible ways to factor 169 in the domain of integers are either \((1,169)\) or \((13,13)\). However, we can see that neither of the elements 1 or 13 can be written in the form \(a+b\sqrt{-17}\). Therefore, 13 must be irreducible in the given domain. Apply the same reasoning to the factors \(4+3\sqrt{-17}\) and \(4-3\sqrt{-17}\), and you will also find them to be irreducible.

Step 5: Conclude the failure of unique factorization

Since all four factors (13, 13, \(4+3\sqrt{-17}\), and \(4-3\sqrt{-17}\)) have been shown to be irreducible, we can conclude that the unique factorization into primes fails in the domain of integers of the form \(a+b\sqrt{-17}\).

Key Concepts

Integers of the Form a+b√(-17)

In the realm of complex numbers, integers can take forms beyond the usual real numbers. One interesting form is \(a+b\sqrt{-17}\), where both \(a\) and \(b\) are regular integers (whole numbers). This form is fascinating because it extends the normal integer system into a complex number system, allowing us to explore more exotic mathematical properties.

Such integers are significant in certain algebraic structures, like rings, which allow for addition and multiplication similar to ordinary integers. In the mentioned problem, we delve into numbers of this type to explore properties like factorization, yet with influences of these additional complex components. The use of \(-17\) under the square root, a negative number, is fundamental in creating this complex number system, which behaves differently from the integers we usually handle.

Irreducibility

Irreducibility is a central concept in understanding factorization in different number systems. An integer of the form \(a+b\sqrt{-17}\) is considered irreducible if it cannot be factored into other non-unit integers of the same form. Essentially, you can think of irreducible numbers as the building blocks or prime numbers of this system, just as 2, 3, and 5 are in the integers you know.

If a number is irreducible, the only way it can be expressed as a product is as a product with units, which have a norm of 1 in these systems. The exercise demonstrates that certain numbers—specifically 13, \(4+3\sqrt{-17}\), and \(4-3\sqrt{-17}\)—are irreducible in this domain. This is crucial in determining whether unique factorization holds, as unique factorization relies on being able to break numbers down into prime factors, which in this case would be these irreducibles.

A key takeaway from the problem is that the failure of unique factorization indicates we're not simply in a straightforward generalization of the usual integers, meaning more complex behaviors and exceptions exist in these systems.

Norm Function in Algebra

The norm function is a useful tool in algebra that helps us understand various properties of numbers, especially in complex or extended integer systems. For integers of the form \(a+b\sqrt{-17}\), the norm is defined as \(N(a+b\sqrt{-17}) = a^2 + 17b^2\). This mathematical function gives us a non-negative integer that reflects the magnitude of the number in a certain sense.

The norm has a crucial property: it is multiplicative. This means that for two numbers \(x\) and \(y\), the norm of their product is equal to the product of their norms, i.e., \(N(xy) = N(x)\cdot N(y)\). This property is heavily used in the exercise to show the irreducibility of factors. If a factorization would imply a norm of 1 for one of the results, it would make that result a unit (a sort of 'one' in this system), indicating irreducibility if such a factorization does not occur without units.

Understanding norm functions not only helps in identifying irreducible elements but also in exploring concepts like the failure of unique factorization, as discussed in the exercise.