Problem 56 - A Computational Introduction to Number Theory and Algebra

Question

Let \(n\) be a positive integer, and consider the natural map that sends \(a \in \mathbb{Z}\) to \(\bar{a}:=[a]_{n} \in \mathbb{Z}_{n},\) which we may extend coefficient-wise to a ring homomorphism from \(\mathbb{Z}[X]\) to \(\mathbb{Z}_{n}[X],\) as in Example \(7.47 .\) Show that for every \(f \in \mathbb{Z}[X],\) we have a ring isomorphism \(\mathbb{Z}[X] /(f, n) \cong \mathbb{Z}_{n}[X] /(\bar{f})\).

Step-by-Step Solution

Answer

To prove that for every polynomial \(f \in \mathbb{Z}[X]\), there is a ring isomorphism between \(\mathbb{Z}[X]/(f,n)\) and \(\mathbb{Z}_n[X]/(\bar{f})\), we have done the following steps: 1. Defined a function \(\varphi\) that maps from \(\mathbb{Z}[X]/(f,n)\) to \(\mathbb{Z}_n[X]/(\bar{f})\) as \(\varphi([g]_{(f,n)}) = [\bar{g}]_{\bar{f}}\). 2. Proved that \(\varphi\) is well-defined, meaning if \([g]_{(f,n)} = [h]_{(f,n)}\), then \([\bar{g}]_{\bar{f}} = [\bar{h}]_{\bar{f}}\). 3. Shown that \(\varphi\) is a ring homomorphism, meaning it preserves addition and multiplication. 4. Proved that \(\varphi\) is bijective, which means it is both injective (one-to-one) and surjective (onto). As a result, we have shown that \(\mathbb{Z}[X]/(f,n) \cong \mathbb{Z}_n[X]/(\bar{f})\), indicating a ring isomorphism between the two quotient rings.

Step 1: Define the function \(\varphi\)

Let's define a function \(\varphi\) that maps from \(\mathbb{Z}[X]/(f,n)\) to \(\mathbb{Z}_n[X]/(\bar{f})\) as follows: $$\varphi \colon \mathbb{Z}[X]/(f,n) \to \mathbb{Z}_n[X]/(\bar{f}), \quad \varphi([g]_{(f,n)}) = [\bar{g}]_{\bar{f}}$$ where \([\bar{g}]_{\bar{f}}\) is reducing \(g\) modulo \(f\) and then modulo \(n\).

Step 2: Prove that \(\varphi\) is well-defined

To show that \(\varphi\) is well-defined, we need to prove that if \([g]_{(f,n)} = [h]_{(f,n)}\) in the domain \(\mathbb{Z}[X]/(f,n)\), then \([\bar{g}]_{\bar{f}} = [\bar{h}]_{\bar{f}}\) in the codomain \(\mathbb{Z}_n[X]/(\bar{f})\). If \([g]_{(f,n)} = [h]_{(f,n)}\), then \(g-h \in (f,n)\). That means there exist polynomials \(q_1, q_2 \in \mathbb{Z}[X]\) such that: $$g-h = q_1f + q_2n$$ Reducing both sides modulo \(n\) (i.e., replacing each coefficient by its remainder after division by \(n\)), we get: $$\bar{g}-\bar{h} = \bar{q_1}\bar{f}$$ This implies that \([\bar{g}]_{\bar{f}} = [\bar{h}]_{\bar{f}}\), so the function \(\varphi\) is well-defined.

Step 3: Prove that \(\varphi\) is a ring homomorphism

We need to show that \(\varphi\) preserves addition and multiplication: 1. \(\varphi([g]_{(f,n)} + [h]_{(f,n)}) = \varphi([g+h]_{(f,n)}) = [\bar{g+h}]_{\bar{f}}\). Since the reduction modulo \(n\) is a homomorphism, we have \([\bar{g+h}]_{\bar{f}} = [\bar{g}]_{\bar{f}} + [\bar{h}]_{\bar{f}} = \varphi([g]_{(f,n)}) + \varphi([h]_{(f,n)})\). 2. \(\varphi([g]_{(f,n)}\cdot [h]_{(f,n)}) = \varphi([gh]_{(f,n)}) = [\bar{gh}]_{\bar{f}}\). Again, since the reduction modulo \(n\) is a homomorphism, we have \([\bar{gh}]_{\bar{f}} = [\bar{g}]_{\bar{f}}\cdot [\bar{h}]_{\bar{f}} = \varphi([g]_{(f,n)})\cdot \varphi([h]_{(f,n)})\). Since \(\varphi\) preserves addition and multiplication, it is a ring homomorphism.

Step 4: Prove that \(\varphi\) is bijective

To show that \(\varphi\) is bijective, we need to prove that it is both injective (one-to-one) and surjective (onto): 1. Injectivity: Suppose \(\varphi([g]_{(f,n)}) = \varphi([h]_{(f,n)})\). This implies that \([\bar{g}]_{\bar{f}} = [\bar{h}]_{\bar{f}}\). We have already shown in Step 2 that this condition implies \([g]_{(f,n)} = [h]_{(f,n)}\). Therefore, \(\varphi\) is injective. 2. Surjectivity: Let \([\bar{k}]_{\bar{f}} \in \mathbb{Z}_n[X]/(\bar{f})\). Since \(k \in \mathbb{Z}_n[X]\), there must be a polynomial \(g \in \mathbb{Z}[X]\) such that \(\bar{g} = k\). We have \(\varphi([g]_{(f,n)}) = [\bar{g}]_{\bar{f}} = [\bar{k}]_{\bar{f}}\), so every element in \(\mathbb{Z}_n[X]/(\bar{f})\) has a preimage in \(\mathbb{Z}[X]/(f,n)\), meaning that \(\varphi\) is surjective. Since \(\varphi\) is both injective and surjective, it is bijective.

Step 5: Conclusion

We have shown that the function \(\varphi\) is a well-defined ring homomorphism between \(\mathbb{Z}[X]/(f,n)\) and \(\mathbb{Z}_n[X]/(\bar{f})\), and that it is bijective. Therefore, \(\varphi\) is a ring isomorphism, and the quotient rings are isomorphic: $$\mathbb{Z}[X]/(f,n) \cong \mathbb{Z}_n[X]/(\bar{f})$$.

Key Concepts

Polynomial Rings

Polynomial rings are algebraic structures consisting of polynomials with coefficients from a specific ring. A ring is a set equipped with two operations, addition and multiplication, that satisfy certain conditions, like associativity and distributivity.

When we talk about polynomial rings, we're essentially looking at expressions such as \( a_0 + a_1X + a_2X^2 + \ldots + a_nX^n \), where \( a_i \) are elements of a ring \( R \) and \( X \) is an indeterminate. This is denoted as \( R[X] \).

• **Addition and Multiplication:** The addition and multiplication of polynomials follow the usual algebraic rules, operating on coefficients directly.
• **Degree of a Polynomial:** The highest power of \( X \) with a non-zero coefficient is called the degree of the polynomial. It gives a measure of the polynomial's complexity.
• **Polynomial Modulo \( n \):** When dealing with polynomial rings over integers, \( \mathbb{Z}[X] \), we often extend these to include modular arithmetic, creating \( \mathbb{Z}_n[X] \).

For instance, \( \mathbb{Z}[X] \) includes all polynomials where each coefficient is an integer, while \( \mathbb{Z}_n[X] \) reduces these coefficients modulo \( n \), thus crafting a polynomial ring over a modular system.

Modular Arithmetic

Modular arithmetic deals with integers and revolves around the concept of equivalence classes, defined by a modulus \( n \). This arithmetic locally confines numbers within a limited set ranging from \( 0 \) to \( n-1 \).

Let's explore the key points:

• **Representatives and Equivalence Classes:** In modular arithmetic, integers can represent a set of numbers through equivalence classes. For instance, for a modulus 5, numbers like 0, 5, 10, and -5 all fall into the same class. This is written as \([0]_5\).
• **Operations Under Modulo \( n \):** Addition, subtraction, and multiplication in modular arithmetic are performed as usual, but results are adjusted using the modulus. For instance, \( 7 + 3 \equiv 0 \mod 5 \).
• **Polynomial Coefficients:** In polynomial rings like \( \mathbb{Z}_n[X] \), each polynomial coefficient operates under modular arithmetic, meaning they are taken modulo \( n \) during calculations.

Modular arithmetic plays a critical role in various fields like cryptography and computer science due to its property of wrapping around values, similar to how a clock's hour hand resets after hitting 12.

Ring Homomorphisms

A ring homomorphism is a map between two rings that preserves their structure, meaning it respects both addition and multiplication operations.

Here's what makes ring homomorphisms crucial:

• **Structure-Preserving Maps:** If \( \phi: R \to S \) is a ring homomorphism, then for any \( a, b \) in \( R \), \( \phi(a+b) = \phi(a) + \phi(b) \) and \( \phi(a \cdot b) = \phi(a) \cdot \phi(b) \). This ensures that the essential operations and relations in \( R \) are retained in \( S \) under the mapping.
• **Kernel and Image:** The kernel of a homomorphism consists of elements that map to the zero element in \( S \). This set helps in understanding the structure and characteristics of the homomorphism. The image is the set of all outputs and gives insight into how much of \( S \) can be represented by elements of \( R \).
• **Bijective Homomorphisms:** When a ring homomorphism is both injective (one-to-one) and surjective (onto), it becomes an isomorphism, meaning \( R \) and \( S \) are structurally identical as rings. This is pivotal in algebraic structures, ensuring equivalence between different mathematical frameworks.

In the context of the exercise, proving a ring map is an isomorphism shows how two ostensibly different algebraic structures are, in fact, perfectly aligned.