Question

Let \(n=p q,\) where \(p\) and \(q\) are distinct primes. Show that we have a ring isomorphism \(\mathbb{Z}_{n}[X] \cong \mathbb{Z}_{p}[X] \times \mathbb{Z}_{q}[X]\)

Short answer

Question: Given an integer \(n\) that is the product of two distinct prime numbers \(p\) and \(q\), prove that the polynomial ring \(\mathbb{Z}_{n}[X]\) is isomorphic to the direct product of polynomial rings \(\mathbb{Z}_{p}[X]\) and \(\mathbb{Z}_{q}[X]\). Solution: We defined a ring homomorphism \(\phi : \mathbb{Z}_{n}[X] \rightarrow \mathbb{Z}_{p}[X] \times \mathbb{Z}_{q}[X]\) as \(\phi(f(X)) = (f_p(X), f_q(X))\), where \(f_p(X)=f(X) \pmod{p}\) and \(f_q(X)=f(X) \pmod{q}\). Then, we showed that this homomorphism is injective and surjective. Therefore, there is a ring isomorphism \(\mathbb{Z}_{n}[X] \cong \mathbb{Z}_{p}[X] \times \mathbb{Z}_{q}[X]\).

Step-by-step solution

Define a homomorphism

Let's define a ring homomorphism \(\phi : \mathbb{Z}_{n}[X] \rightarrow \mathbb{Z}_{p}[X] \times \mathbb{Z}_{q}[X]\) as follows: For a polynomial \(f(X) \in \mathbb{Z}_{n}[X]\), let \(\phi(f(X)) = (f_p(X), f_q(X))\), where \(f_p(X)=f(X) \pmod{p}\) and \(f_q(X)=f(X) \pmod{q}\). We need to show that this is a ring homomorphism.

Show the functions preserve addition and multiplication

To prove that \(\phi\) is a homomorphism, we must show that it preserves addition and multiplication, i.e., \(\phi(f(X) + g(X)) = \phi(f(X)) + \phi(g(X))\) and \(\phi(f(X) \cdot g(X)) = \phi(f(X)) \cdot \phi(g(X))\). For addition, consider two polynomials \(f(X), g(X) \in \mathbb{Z}_{n}[X]\). Then \begin{align*} \phi(f(X) + g(X)) &= ((f(X) + g(X))\pmod{p},(f(X) + g(X))\pmod{q}) \\ &= (f(X)\pmod{p} + g(X)\pmod{p}, f(X)\pmod{q} + g(X)\pmod{q}) \\ &= (f_p(X) + g_p(X), f_q(X) + g_q(X)) \\ &= \phi(f(X)) + \phi(g(X)). \end{align*} Now, let's prove the property for multiplication: \begin{align*} \phi(f(X) \cdot g(X)) &= ((f(X) \cdot g(X))\pmod{p},(f(X) \cdot g(X))\pmod{q}) \\ &= (f(X)\pmod{p} \cdot g(X)\pmod{p}, f(X)\pmod{q} \cdot g(X)\pmod{q}) \\ &= (f_p(X) \cdot g_p(X), f_q(X) \cdot g_q(X)) \\ &= \phi(f(X)) \cdot \phi(g(X)). \end{align*} Since both addition and multiplication are preserved, \(\phi\) is a ring homomorphism.

Show that the homomorphism is bijective

Now we need to show that the defined homomorphism is bijective, meaning it is both injective (1-1) and surjective (onto). Injectivity: Suppose \(\phi(f(X)) = \phi(g(X))\), then \((f_p(X), f_q(X)) = (g_p(X), g_q(X))\). This implies \(f_p(X) = g_p(X)\) and \(f_q(X) = g_q(X)\). Therefore, \(f(X) \equiv g(X) \pmod{p}\) and \(f(X) \equiv g(X) \pmod{q}\), so \(f(X) \equiv g(X) \pmod{p q}\). Thus, \(f(X)=g(X)\), which means that \(\phi\) is injective. Surjectivity: Let \((h_p(X), h_q(X)) \in \mathbb{Z}_{p}[X] \times \mathbb{Z}_{q}[X]\). By the Chinese Remainder Theorem (CRT) for polynomials, there exists a polynomial \(h(X) \in \mathbb{Z}_{n}[X]\) such that \(h(X) \equiv h_p(X) \pmod{p}\) and \(h(X) \equiv h_q(X) \pmod{q}\). Then \(\phi(h(X)) = (h_p(X), h_q(X))\), which means that the homomorphism is surjective. Since \(\phi\) is injective and surjective, it is bijective.

Concluding the proof

As we have shown that \(\phi\) is a ring homomorphism and bijective, we can now conclude that there is a ring isomorphism \(\mathbb{Z}_{n}[X] \cong \mathbb{Z}_{p}[X] \times \mathbb{Z}_{q}[X]\).

Key concepts

Ring Homomorphism

A ring homomorphism is essentially a 'function' between two rings that respects the ring operations, which are addition and multiplication. By 'respects,' we mean that the homomorphism, let's call it \(\phi\), ensures that \(\phi(a + b) = \phi(a) + \phi(b)\) and \(\phi(a \cdot b) = \phi(a) \cdot \phi(b)\) for any two elements \(a\) and \(b\) from the first ring. These properties ensure that the ring structure is conserved under \(\phi\), thus allowing us to compare and link the algebraic properties of two distinct rings.

When solving textbook exercises regarding ring homomorphisms, it's important to verify both properties separately. This provides a clear and concise understanding that the required conditions are met, and they demonstrate the structuring respects the ring operations. Once you've proven a function is a homomorphism, you've established a strong foundational link between the two rings in question.

Polynomial modulo Primes

Dealing with polynomials modulo primes (\(\mathbb{Z}_{p}[X]\)) means working within the world of polynomial arithmetic under a prime modulus. In this setting, whenever we perform operations with polynomials—be that addition, subtraction, or multiplication—the coefficients are reduced modulo a prime number \(p\). This ensures all coefficients are within the set \(\{0, 1, ..., p-1\}\).

It's crucial to understand that when working modulo a prime number, we have a field structure, which implies that every non-zero element has a multiplicative inverse. This property is used extensively in theoretical and applied mathematics, including cryptography. It provides an extra layer of complexity and security in calculations—a fact that is just as fascinating as it is useful. When handling such problems, it’s good practice to systematically apply the modulo operation to each operation to prevent mistakes.

Chinese Remainder Theorem

The Chinese Remainder Theorem (CRT) is a beautiful piece of mathematics that applies to various areas of number theory and algebra. In essence, it gives us a way to understand and solve systems of simultaneous congruences with different moduli, under the condition that these moduli are pairwise coprime. The CRT says that for any set of integers \(a_1, a_2, ..., a_k\) and any set of pairwise coprime positive integers \(n_1, n_2, ..., n_k\), there is a unique integer x modulo \(N = n_1 \cdot n_2 \cdot ... \cdot n_k\) that solves the system
\[x \equiv a_i \pmod{n_i}\]
for \(i = 1, 2, ..., k\).

This theorem is not only theoretically interesting but also has very practical applications, including in algorithms and cryptography. When it comes to polynomial equations, the CRT provides a means to construct a unique solution from modular components. Students should aim to understand the practical construction of solutions using the CRT to fully appreciate its utility in exercises.

Bijective Function

A bijective function is one that is both injective (one-to-one) and surjective (onto). Injective means that different inputs map to different outputs (no two distinct elements map to the same element), while surjective means that every element in the target set is mapped by some element from the domain. In other words, for a function to be bijective, it must cover the entire target set, and it must be such that each element of the target is the image of precisely one element of the domain.

A bijective function allows us to establish a perfect 'pairing' between the elements of two sets, indicating the sets are equivalent in terms of size (cardinality). When discussing ring isomorphisms, the notion of a bijective homomorphism becomes crucial, as it signifies an exact correspondence between the ring elements while preserving the ring structure, which is necessary for two rings to be considered isomorphic. To master bijective functions, students should practice identifying whether a given function is one-to-one and onto, ensuring a deep understanding of this integral concept.