Question

Let \(\pi:=1+i \in \mathbb{Z}[i]\) (a) Show that \(2=\pi \bar{\pi}=-i \pi^{2},\) that \(N(\pi)=2,\) and that \(\pi\) is irreducible in \(\mathbb{Z}[i]\). \(458 \quad\) More rings (b) Let \(\alpha \in \mathbb{Z}[i],\) with \(\alpha=a+b i\) for \(a, b \in \mathbb{Z} .\) Show that \(\pi \mid \alpha\) if and only if \(a-b\) is even, in which case $$ \frac{\alpha}{\pi}=\frac{a+b}{2}+\frac{b-a}{2} i $$ (c) Show that for all \(\alpha \in \mathbb{Z}[i],\) we have \(\alpha \equiv 0(\bmod \pi)\) or \(\alpha \equiv 1(\bmod \pi)\). (d) Show that the quotient ring \(\mathbb{Z}[i] / \pi \mathbb{Z}[i]\) is isomorphic to the ring \(\mathbb{Z}_{2}\). (e) Show that for all \(\alpha \in \mathbb{Z}[i]\) with \(\alpha \equiv 1(\bmod \pi),\) there exists a unique \(\varepsilon \in\\{\pm 1, \pm i\\}\) such that \(\alpha \equiv \varepsilon(\bmod 2 \pi)\) (f) Show that for all \(\alpha, \beta \in \mathbb{Z}[i]\) with \(\alpha \equiv \beta \equiv 1(\bmod \pi),\) there exists a unique \(\varepsilon \in\\{\pm 1, \pm i\\}\) such that \(\alpha \equiv \varepsilon \beta(\bmod 2 \pi)\)

Short answer

Question: Show that a complex number α is divisible by π if and only if the difference between its real and imaginary parts is even. Solution: By our earlier analysis, we see that a complex number α is divisible by π if there exists x, y ∈ ℤ such that α = π(x + yi) = (x - y) + (x + y)i. Comparing the real and imaginary parts of α and π(x + yi), we have that a = x - y and b = x + y, where α = a + bi. Solving for x and y, we have x = (a + b)/2 and y = (b - a)/2. Thus, α is divisible by π if and only if a - b is even.

Step-by-step solution

Part (a): Expressions for 2 and norm of \(\pi\)

We need to show that \(2=\pi\bar{\pi}=-i\pi^2\). First, note that \(\pi=1+i\), so the conjugate of \(\pi\) is: \(\bar{\pi}=1-i\) Now, multiply \(\pi\) by \(\bar{\pi}\): \(\pi\bar{\pi}=(1+i)(1-i)=1-i+i-(-1)=2\) Next, square \(\pi\) and multiply by \(-i\): \(\pi^2=(1+i)^2=1+2i+i^2=1+2i-1=2i\) \(\implies -i\pi^2=-i(2i)=-i(-2) = 2\) Now, we compute the norm of \(\pi\): \(N(\pi)=\pi\bar{\pi}=2\) Finally, we show that \(\pi\) is irreducible in \(\mathbb{Z}[i]\). Since the norm of \(\pi\) is 2 (which is a prime number), any nontrivial factorization of \(\pi\) must have factors with norm \(1\) (as products of norms equal the norm of the product). But elements of norm 1 in \((\pm1,\pm i)\) are units in \(\mathbb{Z}[i]\), and therefore \(\pi\) cannot be nontrivially factored - making it irreducible.

Part (b): Condition for divisibility by \(\pi\) in \(\mathbb{Z}[i]\)

Let \(\alpha=a+bi \in \mathbb{Z}[i]\), where \(a, b \in \mathbb{Z}\). \(\pi\) divides \(\alpha\) if and only if \(\alpha=\pi\beta\) for some \(\beta \in \mathbb{Z}[i]\). Note: \(\pi\beta = (1+i)(x+yi)=(x-y)+(x+y)i\), for some \(x, y \in \mathbb{Z}\). Now equate the real and imaginary parts of \(\alpha\) and \(\pi\beta\): \(a=x-y\), \(b=x+y\). Solving these simultaneous equations, we obtain \(x=\frac{a+b}{2}\) and \(y=\frac{b-a}{2}\). If \(a-b\) is even, then \(x\) and \(y\) are integers, and \(\beta=x+yi=\frac{a+b}{2}+\frac{b-a}{2}i \in \mathbb{Z}[i]\). Conversely, if \(\pi\) divides \(\alpha\), then \(x,y \in \mathbb{Z}\) and this will always result in \(a-b\) being even.

Part (c): ModULO \(\pi\) in \(\mathbb{Z}[i]\)

For all \(\alpha \in \mathbb{Z}[i]\), we want to show that \(\alpha \equiv 0(\bmod \pi)\) or \(\alpha \equiv 1(\bmod \pi)\). Recall the conditions in part (b): \(\pi | \alpha \iff a-b\) is even, and \(\frac{\alpha}{\pi}=\frac{a+b}{2}+\frac{b-a}{2}i\). Now, consider the case when \(a-b\) is even. This means that \(\pi|\alpha\), so \(\alpha\equiv0(\bmod \pi)\). When \(a-b\) is odd, then \((a-1)-b\) is even. In this case, \(\pi\) divides \((\alpha-1)\), so \(\alpha \equiv 1(\bmod \pi)\). Hence, for any \(\alpha\in\mathbb{Z}[i]\), either \(\alpha\equiv 0(\bmod \pi)\) or \(\alpha\equiv 1(\bmod \pi)\).

Part (d): Quotient Ring Isomorphism

We want to show that \(\mathbb{Z}[i]/\pi\mathbb{Z}[i]\cong \mathbb{Z}_2\). Define a surjective ring homomorphism \(\phi:\mathbb{Z}[i] \to \mathbb{Z}_2\) by \(\phi(a+bi)=a-b \pmod 2\) and consider its kernel, which is the preimage of 0 in \(\mathbb{Z}_2\): \(\mathrm{Ker}(\phi)=\{\alpha \in \mathbb{Z}[i] \mid \alpha=a+bi, a-b\equiv 0\pmod 2\}\) By part (b), this kernel is exactly the ideal \(\pi\mathbb{Z}[i]\), and by the first isomorphism theorem for rings: \(\mathbb{Z}[i]/\mathrm{Ker}(\phi) \cong \mathrm{Im}(\phi)\) Therefore, \(\mathbb{Z}[i]/\pi\mathbb{Z}[i]\cong\mathbb{Z}_2\).

Part (e): Modulo \(2\pi\) condition

For all \(\alpha\equiv1(\bmod\pi)\) in \(\mathbb{Z}[i]\), we need to find a unique \(\varepsilon \in \{\pm1,\pm i\}\) such that \(\alpha \equiv \varepsilon(\bmod 2\pi)\). Note that \(\alpha \equiv 1(\bmod\pi)\) implies that \(\pi\) divides \((\alpha-1)\), so \(\alpha-1=\pi\beta\) for some \(\beta\in\mathbb{Z}[i]\). Then: \(\alpha=\pi\beta+1=(1+i)(x+yi)+1=(x-y+1)+(x+y)i\) For \(\alpha \equiv \varepsilon (\bmod 2\pi)\), we need to verify that \(\alpha-2\pi\varepsilon=\alpha-(2+2i)\varepsilon \in \pi \mathbb{Z}[i]\). Simplify the expressions for \(\alpha-2\pi\varepsilon\) for each \(\varepsilon \in \{\pm1,\pm i\}\): 1. \(\alpha-2\pi=\left(x-y-1\right)+\left(x+y-2\right)i\) 2. \(\alpha+2\pi=\left(x-y+3\right)+\left(x+y+2\right)i\) 3. \(\alpha-2\pi i=\left(x-y+1+2\right)+\left(x+y-1-2\right)i\) 4. \(\alpha+2\pi i=\left(x-y+1-2\right)+\left(x+y+1+2\right)i\) Since \(\alpha\equiv1(\bmod\pi)\), \(x-y+1\) is odd and \(x+y\) is even. This is consistent only with the first case, where \(\alpha\equiv-1(\bmod 2\pi)\). Hence, \(\varepsilon=-1\) is the unique choice satisfying the desired congruence.

Part (f): Modulo \(2\pi\) condition for products

For all \(\alpha, \beta\equiv1(\bmod\pi)\) in \(\mathbb{Z}[i]\), we need to find a unique \(\varepsilon \in \{\pm1,\pm i\}\) such that \(\alpha\equiv\varepsilon\beta(\bmod 2\pi)\). Let \(\alpha=a+bi\) and \(\beta=c+di\). Then \(\alpha\beta=(ac-bd)+(ad+bc)i\). From part (b), we know that \(\frac{\alpha}{\pi}=a'b'\) and \(\frac{\beta}{\pi}=c'd'\), where \(a',b',c',d' \in \mathbb{Z}\). Hence, \(\frac{\alpha\beta}{\pi}=\frac{1}{\pi}(ac-bd+(ad+bc)i)=a'd'+b'c'\). Now consider congruence modulo 2. The right side of the above equation is either \(0\) (even) or \(1\) (odd), since \(\alpha, \beta\equiv1(\bmod\pi)\). We conclude that: \((ac-bd) \equiv a'd'(\bmod 2)\) and \((ad+bc) \equiv b'c'(\bmod 2)\) Observe that \((ad+bc)-(ac-bd)=ad+bc-ac+bd=(b+a)(c+d)\) is always even, which implies that \(\alpha\beta\equiv\alpha+\beta(\bmod 2\pi)\). By part (e) for \(\alpha\equiv\varepsilon_1(\bmod 2\pi)\) and \(\beta\equiv\varepsilon_2(\bmod 2\pi)\), we have \(\varepsilon_1=-1\) and \(\varepsilon_2=-1\). Thus, \(\alpha\equiv-1(\bmod 2\pi)\) and \(\beta\equiv-1(\bmod 2\pi)\) \(\implies \alpha\beta\equiv1(\bmod 2\pi)\) From part (e) for \(\alpha\beta\), we get \(\alpha\beta\equiv-1(\bmod 2\pi)\). Thus, \(\varepsilon=-1\) is the unique choice satisfying the desired congruence.

Key concepts

Irreducible Elements in Number Theory

In number theory and algebra, understanding the concept of irreducible elements is crucial. An element in a ring is called irreducible if it is not a unit (an element that has a multiplicative inverse) and cannot be factored into a product of two non-unit elements. In simpler terms, irreducible elements are the 'primes' of the ring: they cannot be broken down into smaller components that are also non-units.

For example, in the set of Gaussian integers (\(mathbb{Z}[i]\)), which are complex numbers of the form (\(a + bi\)), where (\(a\)) and (\(b\)) are integers and (\(i\)) is the imaginary unit, an element is irreducible if its norm (the product of the element and its complex conjugate) is a prime number in the standard integers (\(mathbb{Z}\)). This property makes irreducible elements vital in factorization within rings, highlighting the behavior of the element in multiplication and division scenarios much like prime numbers do in the set of whole numbers.

In the exercise provided, (\(pi = 1 + i\)) was demonstrated to be irreducible in (\(mathbb{Z}[i]\)) since its norm, (\(N(pi) = 2\)), is prime. As irreducibility is a key property for many concepts in number theory, such as factorization and the study of prime elements, it is a fundamental building block for understanding more complex algebraic structures.

Gaussian Integers and Their Properties

The Gaussian integers constitute a fascinating subset of complex numbers. Specifically, they are numbers that can be expressed in the form (\(a + bi\)), where (\(a\)) and (\(b\)) are integers and (\(i\)) represents the imaginary unit, satisfying (\(i^2 = -1\)). This set forms a mathematical structure known as a ring, denoted (\(mathbb{Z}[i]\)), where addition, subtraction, and multiplication follow the conventional rules of arithmetic and maintain the integrity of the Gaussian integers.

An essential characteristic of Gaussian integers is their divisibility properties, which mirror that of the integers in many ways but come with the added complexity due to their two-dimensional nature. Within this ring, divisibility by a Gaussian integer involves finding another Gaussian integer that can be multiplied to give the original number, much in line with divisibility in the familiar integers. This facet is showcased in part (b) of the exercise, where the condition for a Gaussian integer to be divisible by (\(pi = 1 + i\)) was determined by an equivalence regarding the parity of the difference between its real and imaginary parts.

Ring Homomorphism and Its Role in Algebra

A ring homomorphism is a function between two rings that respects the operations of addition and multiplication inherent to ring structures. This means that for a function (\(f: R \to S\)), where (\(R\)) and (\(S\)) are rings, it must satisfy (\(f(x + y) = f(x) + f(y)\)) and (\(f(x \times y) = f(x) \times f(y)\)) for all (\(x, y \)) in (\(R\)). Ring homomorphisms are analogous to 'translators' that convert between different ring languages while preserving their essential arithmetic relationships.

In our exercise's part (d), we see a homomorphism being employed to establish an isomorphism between a quotient ring and (\(mathbb{Z}_2\)). This homomorphism provides a systematic method to simplify the problem, proving congruences and establishing the fundamental theorem of isomorphism for rings, key to understanding the structure-preserving nature of certain algebraic operations. Moreover, it integrates the quotient ring and ring isomorphism concepts, offering a bridge between these intertwined algebraic ideas.

Quotient Ring and Its Algebraic Implications

A quotient ring is formed when a ring (\(R\)) is partitioned by an ideal (\(I\)), written as (\(R/I\)). You can think of it as a way to simplify the ring by 'collapsing' all the elements that are differ by an element of (\(I\)) into a single entity. This process creates a new ring that encapsulates the structure of (\(R\)) but within a simplified context. One can say it offers a 'scaled-down' version of the original ring, illustrating what happens when all elements of an ideal are considered indistinguishable.

The quotient ring concept was illustrated in part (d) of the given exercise by showing that the Gaussian integers modulo the ideal generated by (\(pi = 1 + i\)) are isomorphic to the ring of integers modulo 2, (\(mathbb{Z}[i]/pimathbb{Z}[i] cong mathbb{Z}_2\)). This result connects to the broader themes in algebra of reducing complexity and finding fundamental structures.

Ring Isomorphism Unveiled

When two rings are said to be isomorphic, they have a mutually inverse pair of homomorphisms between them, indicating a perfect correspondence between their structure and operations. One can transform one ring into the other via these mappings without losing any algebraic properties, which reveals that they are, in essence, the same from an algebraic perspective—even if they may look entirely different superficially.

This means they have an equal number of elements corresponding to each other, and their addition and multiplication tables match. It is an equivalence relationship in the realm of algebra that signifies two rings share the same abstract structure. In the solution to part (d) of the exercise, we witnessed the establishment of such a ring isomorphism between the quotient ring (\(mathbb{Z}[i]/pimathbb{Z}[i]\)) and the binary field (\(mathbb{Z}_2\)), thereby demonstrating that despite their apparent differences, these two rings are algebraically identical when viewed through the correct 'lens'.