Proving Z[√2] is an Integral Domain

Abstract Algebra · Axiom Academy

EXAMPLE Proving Z[√2] is an Integral Domain Complete verification using ring axioms and the norm function N(a + b√2) = a² - 2b² Excellent work! You've proven that Z[√2] is an integral domain. Here's what we established: Ring Structure: Z[√2] is closed under addition and multiplication, with all ring axioms satisfied (associativity, distributivity, additive identity and inverses) Commutativity: Multiplication is commutative because it inherits this property from ℤ Unity: The element 1 = 1 + 0√2 serves as the multiplicative identity The Norm Function: N(a + b√2) = a² - 2b² is a powerful tool that maps to ℤ and is multiplicative: N(αβ) = N(α)N(β) No Zero Divisors: The multiplicative property of the norm proves that if αβ = 0, then α = 0 or β = 0, since N(α)N(β) = 0 in ℤ implies one factor is zero This proof technique—using a norm function to show no zero divisors—is fundamental in algebraic number theory. The same approach works for other quadratic integer rings like Z[i] and Z[ω], where similar norm functions reveal the integral domain structure!

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