Exercise 15.5.6

Let α , β ∈ ℤ [ i ] be relatively prime. Prove the Chiese Remainder Theorem for ℤ [ i ] , which asserts that there is a ring isomorphism

ℤ [ i ] ∕ αβℤ [ i ] ≃ ℤ [ i ] ∕ αℤ [ i ] × ℤ [ i ] ∕ βℤ [ i ] .

Answers

Proof. Consider the map

ψ { ℤ [ i ] → ℤ [ i ] ∕ αℤ [ i ] × ℤ [ i ] ∕ βℤ [ i ] γ ↦ ( γ + αℤ [ i ] , γ + βℤ [ i ] ) .

Then φ is a ring homomorphism:

ψ ( γ ) ψ ( δ ) = ( γ + αℤ [ i ] , γ + βℤ [ i ] ) ⋅ ( δ + αℤ [ i ] , δ + βℤ [ i ] ) = ( ( γ + αℤ [ i ] ) ( δ + αℤ [ i ] ) , ( γ + βℤ [ i ] ) ( δ + βℤ [ i ] ) ) = ( γδ + αℤ [ i ] , γδ + βℤ [ i ] ) , = ψ ( γδ )

and similarly ψ ( γ ) + ψ ( δ ) = ψ ( γ + δ ) .

Moreover, since α , β are relatively prime, for all γ ∈ ℤ [ i ] ,

γ ∈ ker ⁡ ( ψ ) ⟺ γ ∈ αℤ [ i ]  and  γ ∈ βℤ [ i ] ⟺ α ∣ γ  and  β ∣ γ ⟺ αβ ∣ γ ⟺ γ ∈ αβℤ [ i ] .

Thus

ker ⁡ ( ψ ) = αβℤ [ i ] .

Therefore there is an injective ring homomorphism

ψ ¯ : ℤ [ i ] ∕ αβℤ [ i ] → ℤ [ i ] ∕ αℤ [ i ] × ℤ [ i ] ∕ βℤ [ i ]

such that ψ ¯ ( γ + αβℤ [ i ] ) = ψ ( γ ) .

Since

| ℤ [ i ] ∕ αβℤ [ i ] | = N ( αβ ) = N ( α ) N ( β ) = | ℤ [ i ] ∕ αℤ [ i ] × ℤ [ i ] ∕ βℤ [ i ] | ,

ψ ¯ is surjective, so ψ ¯ is a ring isomorphism.

ℤ [ i ] ∕ αβℤ [ i ] ≃ ℤ [ i ] ∕ αℤ [ i ] × ℤ [ i ] ∕ βℤ [ i ] .

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2022-07-19 00:00
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