Questions: The Norm in Algebraic Number Fields

Question 1

An element α ∈ ℤ[i] has norm N(α) = 13, where 13 is a prime integer. What can you conclude about α?

• A. α must be a unit in ℤ[i]
• B. α is irreducible in ℤ[i]
• C. α = 2 + 3i specifically
• D. α must be a real integer

Answer: (B)

If α = βγ for non-units β, γ ∈ ℤ[i], then multiplicativity gives N(β)·N(γ) = N(α) = 13. Since N takes positive integer values for Gaussian integers, this forces one factor to be 1, meaning one of β or γ is a unit. So α cannot factor into two non-units — it is irreducible. This is the direct payoff of multiplicativity: prime norm implies irreducibility.

Question 2

A number theorist wants to find all integer solutions to x² + y² = 5. How does the norm in ℤ[i] turn this into a factorization problem?

• A. By finding the prime factorization of 5 in ℤ and lifting it
• B. By expressing the equation as N(x + yi) = 5 and finding Gaussian integers with that norm
• C. By counting lattice points on the circle of radius √5
• D. By computing 5 mod 4 to determine representability

Answer: (B)

The equation x² + y² = 5 is exactly N(x + yi) = 5 in ℤ[i], since N(a+bi) = a²+b². Solutions correspond to Gaussian integers with norm 5. Since N(2+i) = 4+1 = 5 and 5 is prime, 2+i is a Gaussian prime, and the integer solutions follow from its factorizations (up to units). The norm translates a Diophantine equation into an algebraic factorization question.

Question 3

The multiplicativity of the norm, N(αβ) = N(α)N(β), follows from the fact that each field embedding σᵢ: K → ℂ is a ring homomorphism.

• T. True
• F. False

Answer: True

Since σᵢ is a ring homomorphism, σᵢ(αβ) = σᵢ(α)·σᵢ(β). Taking the product over all embeddings: N(αβ) = ∏ σᵢ(αβ) = ∏ σᵢ(α)·σᵢ(β) = (∏ σᵢ(α))·(∏ σᵢ(β)) = N(α)·N(β). The multiplicativity is a direct algebraic consequence of how embeddings interact with multiplication.

Question 4

If α, β ∈ ℤ[i] each have norm 5, then N(αβ) = 10.

• T. True
• F. False

Answer: False

By multiplicativity, N(αβ) = N(α)·N(β) = 5·5 = 25, not 10. The norm multiplies — it does not add. A common error is to treat norm like a linear function. For instance, N(2+i) = 5 and N(1+2i) = 5, so N((2+i)(1+2i)) = N(2+i)·N(1+2i) = 25.

Question 5

Why does an element with prime norm have to be irreducible in the ring of integers O_K?

Model answer: If α = βγ with β, γ non-units, then N(α) = N(β)·N(γ) by multiplicativity. Both N(β) and N(γ) are positive integers greater than 1 (since units have norm 1). But then N(α) = N(β)·N(γ) is a product of two integers each ≥ 2, so N(α) is composite — contradicting the assumption that N(α) is prime. Therefore no such factorization into non-units exists, and α is irreducible.

This argument works directly because norm is multiplicative and norms of units equal 1. The key logical move is: prime norm → can't be written as a product of two integers both > 1 → can't factor into two non-units. This makes prime-norm-checking a practical tool for identifying irreducibles without constructing all possible factorizations.