Prove an isomorphism between a quotient ring and a direct product of integers | Step-by-Step Solution

What We're Solving:

We need to prove that the quotient ring Z[x]/(x² + x) is isomorphic to Z × Z. This means we need to show these two rings have the same algebraic structure by finding a bijective ring homomorphism between them.

The Approach:

The key insight here is to use the Chinese Remainder Theorem for rings! Notice that x² + x = x(x + 1), and since gcd(x, x+1) = 1 in Z[x], we can "split" our quotient ring. We'll construct an explicit isomorphism using evaluation maps at specific points.

Step-by-Step Solution:

Step 1: Factor the ideal generator First, let's observe that x² + x = x(x + 1). This factorization is crucial because x and (x + 1) are coprime in Z[x].

Step 2: Set up the Chinese Remainder Theorem Since x and (x + 1) are coprime, we can apply CRT: Z[x]/(x(x + 1)) ≅ Z[x]/(x) × Z[x]/(x + 1)

Step 3: Simplify each quotient ring

• Z[x]/(x): When we set x = 0, any polynomial f(x) = a₀ + a₁x + a₂x² + ... becomes just a₀. So Z[x]/(x) ≅ Z.
• Z[x]/(x + 1): When we set x = -1, any polynomial f(x) becomes f(-1). So Z[x]/(x + 1) ≅ Z.

Step 4: Construct the explicit isomorphism Define φ: Z[x]/(x² + x) → Z × Z by: φ([f(x)]) = (f(0), f(-1))

where [f(x)] represents the equivalence class of f(x) in the quotient ring.

Step 5: Verify φ is well-defined If f(x) ≡ g(x) (mod x² + x), then f(x) - g(x) = (x² + x)h(x) for some h(x) ∈ Z[x]. Evaluating at x = 0: f(0) - g(0) = 0, so f(0) = g(0). Evaluating at x = -1: f(-1) - g(-1) = 0, so f(-1) = g(-1). Therefore, φ is well-defined.

Step 6: Verify φ is a ring homomorphism

• φ([f] + [g]) = φ([f + g]) = ((f + g)(0), (f + g)(-1)) = (f(0) + g(0), f(-1) + g(-1)) = φ([f]) + φ([g])
• φ([f] · [g]) = φ([fg]) = ((fg)(0), (fg)(-1)) = (f(0)g(0), f(-1)g(-1)) = φ([f]) · φ([g])

Step 7: Show φ is bijective

• Injective: If φ([f]) = (0, 0), then f(0) = 0 and f(-1) = 0. This means f(x) is divisible by both x and (x + 1), hence by x(x + 1) = x² + x. So [f] = [0].
• Surjective: For any (a, b) ∈ Z × Z, we can find f(x) = a + (b - a)x. Then φ([f]) = (a, a + (b - a)(-1)) = (a, b).

The Answer:

The isomorphism φ: Z[x]/(x² + x) → Z × Z given by φ([f(x)]) = (f(0), f(-1)) proves that Y ≅ Z × Z. This works because x² + x = x(x + 1) factors into coprime elements, allowing us to use the Chinese Remainder Theorem.

Memory Tip:

Remember "Evaluate at the roots"! Since x² + x = x(x + 1) has roots 0 and -1, evaluating polynomials at these points gives us the natural map to Z × Z. The Chinese Remainder Theorem is your friend whenever you see a product of coprime ideals!