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UPSC 2013 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Euclidean domains · Algebra · asked 6× in 14 yrs · Read the full method →

Question

Let J={a+bi∣a,b∈Z}J=\{a+bi\mid a,b\in\mathbb Z\} be the ring of Gaussian integers (subring of C\mathbb C). Which of the following is JJ: Euclidean domain, principal ideal domain, unique factorization domain? Justify your answer.

Technique

Standard verification of the Euclidean norm N(a+bi)=a2+b2N(a+bi)=a^{2}+b^{2} with the division algorithm via rounding to nearest integer.

Solution

Claim: J=Z[i]J=\mathbb Z[i] is all three — Euclidean domain, principal ideal domain, and unique factorisation domain.

Strategy. Prove JJ is Euclidean (the strongest of the three). The implications Euclidean ⇒\Rightarrow PID ⇒\Rightarrow UFD are standard, so the other two follow.

Step 1 — JJ is a Euclidean domain

Define the norm

N(α)=N(a+bi)=a2+b2=∣α∣2.N(\alpha)=N(a+bi)=a^{2}+b^{2}=|\alpha|^{2}.

This is non-negative integer-valued, with N(αβ)=N(α)N(β)N(\alpha\beta)=N(\alpha)N(\beta) (multiplicativity), and N(α)=0  ⟺  α=0N(\alpha)=0\iff\alpha=0.

Division algorithm: for α∈J,  β∈J∖{0}\alpha\in J,\;\beta\in J\setminus\{0\}, find γ,ρ∈J\gamma,\rho\in J with α=γβ+ρ\alpha=\gamma\beta+\rho and N(ρ)<N(β)N(\rho)<N(\beta).

Proof. Compute the rational quotient α/β=p+qi∈Q(i)\alpha/\beta=p+qi\in\mathbb Q(i) (well-defined since β≠0\beta\ne 0). Choose integers m,nm,n with ∣m−p∣≤1/2|m-p|\le 1/2 and ∣n−q∣≤1/2|n-q|\le 1/2 (round to nearest integer). Set γ=m+ni∈J\gamma=m+ni\in J. Then

ρ=α−γβ=β[(p−m)+(q−n)i].\rho=\alpha-\gamma\beta=\beta\bigl[(p-m)+(q-n)i\bigr].

Take norms:

N(ρ)=N(β)⋅[(p−m)2+(q−n)2]≤N(β)⋅(14+14)=N(β)2<N(β).N(\rho)=N(\beta)\cdot\bigl[(p-m)^{2}+(q-n)^{2}\bigr]\le N(\beta)\cdot(\tfrac{1}{4}+\tfrac{1}{4})=\tfrac{N(\beta)}{2}<N(\beta).

So NN is a Euclidean function on JJ, and JJ is a Euclidean domain.

Step 2 — Euclidean ⇒\Rightarrow PID

Standard theorem. Sketch. For any non-zero ideal I⊂JI\subset J, pick β∈I\beta\in I minimising N(β)N(\beta). For any α∈I\alpha\in I, division gives α=γβ+ρ\alpha=\gamma\beta+\rho with N(ρ)<N(β)N(\rho)<N(\beta). But ρ=α−γβ∈I\rho=\alpha-\gamma\beta\in I, and by minimality of N(β)N(\beta), ρ=0\rho=0. So α=γβ∈(β)\alpha=\gamma\beta\in(\beta). Hence I=(β)I=(\beta) — principal.

Step 3 — PID ⇒\Rightarrow UFD

Standard theorem. Sketch. In a PID, every irreducible is prime (since irreducibles generate maximal ideals = prime ideals). Combined with the ascending-chain condition (Noetherian, automatic in PIDs), every non-zero non-unit has a factorisation into irreducibles, and the factorisation is unique up to units and reordering.

Conclusion

Answer

  J=Z[i] is a Euclidean domain, hence a PID, hence a UFD.  \boxed{\;J=\mathbb Z[i]\text{ is a Euclidean domain, hence a PID, hence a UFD.}\;}
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