Questions: Gaussian Integers

Question 1

Why is 2 + i a Gaussian prime?

• A. Its real and imaginary parts share no common integer factors
• B. Its norm N(2+i) = 5 is a rational prime, so it cannot factor further in ℤ[i]
• C. It lies on the Gaussian unit circle and is therefore irreducible
• D. It corresponds to the prime 2 in ℤ, which remains prime in all extensions

Answer: (B)

The key tool is the multiplicativity of the norm: N(αβ) = N(α)N(β). If 2+i = βγ in ℤ[i], then N(2+i) = N(β)N(γ), i.e., 5 = N(β)N(γ). Since 5 is a rational prime, the only factorizations in ℕ are 1×5 and 5×1. An element with norm 1 is a unit, so this means one factor is a unit — making 2+i irreducible (a Gaussian prime). The same argument applies to any Gaussian integer whose norm is a rational prime.

Question 2

A student argues that the rational prime p = 7 must remain prime in ℤ[i] because '7 cannot be split into two pieces.' What is the correct analysis?

• A. The student is right; 7 is prime in ℤ and therefore prime in every ring containing ℤ
• B. The student reaches the right conclusion for the wrong reason; 7 remains prime in ℤ[i] specifically because 7 ≡ 3 (mod 4)
• C. The student is wrong; 7 splits in ℤ[i] as a product of two conjugate Gaussian primes
• D. The student is wrong; 7 ramifies in ℤ[i] like the prime 2

Answer: (B)

The correct conclusion is that 7 remains prime (inert) in ℤ[i], but the reason is the mod 4 classification: a rational prime p remains prime in ℤ[i] if and only if p ≡ 3 (mod 4). Since 7 ≡ 3 (mod 4), it stays prime. The reason this works is Fermat's theorem on sums of squares: p ≡ 3 (mod 4) cannot be written as a² + b² = N(a+bi), so p cannot be a norm of a non-unit Gaussian integer, hence it cannot split. The student's informal argument doesn't capture this; a prime like 5 ≡ 1 (mod 4) splits as 5 = (2+i)(2-i) even though '5 cannot be split into two pieces' by ordinary integers.

Question 3

The norm function N(a+bi) = a² + b² satisfies N(αβ) = N(α)N(β) for all Gaussian integers α and β.

• T. True
• F. False

Answer: True

Multiplicativity of the norm is the engine of Gaussian integer factorization theory. Because N(αβ) = N(α)N(β), any factorization in ℤ[i] produces a corresponding factorization of norms in ℤ. This means divisibility and primality questions in ℤ[i] can be partially reduced to ordinary integer arithmetic. In particular, an element with prime norm must be a Gaussian prime, because the only way to factor its norm in ℤ would require one factor to be 1 (a unit).

Question 4

Most rational prime remains prime in ℤ[i] because ℤ[i] contains ℤ as a subring, and divisibility is preserved under ring extensions.

• T. True
• F. False

Answer: False

This is false. Rational primes fall into three categories in ℤ[i]: primes p ≡ 3 (mod 4) remain prime (inert); primes p ≡ 1 (mod 4) split into a product of two distinct Gaussian primes — for example, 5 = (2+i)(2−i); and p = 2 ramifies as 2 = −i(1+i)². Being prime in a subring does not guarantee primality in an extension ring; it is a property that must be checked in the new ring.

Question 5

Why does the behavior of rational primes in ℤ[i] depend on their residue mod 4, and what role does the norm play in proving this?

Model answer: A rational prime p splits in ℤ[i] precisely when it can be written as a sum of two squares, p = a² + b² = N(a+bi). Fermat's theorem on sums of squares says this is possible if and only if p ≡ 1 (mod 4) (or p = 2). The norm's multiplicativity is what connects splitting to sums of squares: if p = (a+bi)(a−bi), then N(p) = p² = N(a+bi)·N(a−bi) = (a²+b²)², forcing a²+b² = p. Primes ≡ 3 (mod 4) cannot be expressed as sums of two squares, so they cannot split.

This is the payoff of the Gaussian integer machinery: a 2,000-year-old number theory result (Fermat's two-square theorem) becomes a corollary of unique factorization in ℤ[i]. The mod 4 condition emerges from quadratic reciprocity and properties of the Legendre symbol, but in ℤ[i] its geometric meaning is clear: whether p = a² + b² has a solution is exactly whether p factors as a norm in ℤ[i].