UPSC Maths 2025 Paper 2 Q2b — Solution

15 marks · Section A

Question

Show that 3 is an irreducible element in the integral domain $\mathbb{Z}[i]$ .

Technique

Use the multiplicative norm $N(a+bi) = a^2+b^2$ : an element is irreducible if it is a non-unit whose only factorizations have a unit factor. Since no Gaussian integer has norm $3$ , any factorization of $3$ is trivial.

Solution

Recall $\mathbb{Z}[i] = \{a+bi : a,b\in\mathbb{Z}\}$ is an integral domain (in fact a Euclidean domain). Define the norm $N(a+bi) = (a+bi)(a-bi) = a^2 + b^2 \in \mathbb{Z}_{\ge 0}.$

Two key properties:

Multiplicativity: $N(\alpha\beta) = N(\alpha)\,N(\beta)$ for all $\alpha,\beta\in\mathbb{Z}[i]$ (since $|\alpha\beta|^2 = |\alpha|^2|\beta|^2$ ).
Units: $\alpha\in\mathbb{Z}[i]$ is a unit $\iff N(\alpha) = 1$ . (If $\alpha\beta=1$ then $N(\alpha)N(\beta)=1$ , forcing $N(\alpha)=1$ ; conversely $a^2+b^2=1$ gives $\alpha\in\{\pm1,\pm i\}$ , all units.)

Step 1 — $3$ is a non-unit. $N(3) = 3^2+0^2 = 9 \ne 1$ , so $3$ is not a unit.

Step 2 — Suppose $3 = \alpha\beta$ with $\alpha,\beta\in\mathbb{Z}[i]$ . Taking norms, $N(\alpha)\,N(\beta) = N(3) = 9.$ Since $N(\alpha),N(\beta)$ are non-negative integers, the possibilities are $\{N(\alpha),N(\beta)\} = \{1,9\}\quad\text{or}\quad \{3,3\}.$

Step 3 — Rule out $N(\alpha)=N(\beta)=3$ . This would require a Gaussian integer with $a^2+b^2 = 3$ . But there is no integer solution to $a^2+b^2=3$ (indeed $3\equiv 3\pmod 4$ , and a prime $\equiv 3\pmod 4$ is never a sum of two squares). Hence $N(\alpha)=3$ is impossible.

Step 4 — Conclusion. The only possibility is $\{N(\alpha),N(\beta)\}=\{1,9\}$ , i.e. one of $\alpha,\beta$ has norm $1$ and is therefore a unit. Thus every factorization $3=\alpha\beta$ has a unit factor.

Since $3$ is a non-zero non-unit whose only factorizations are trivial, $3$ is irreducible in $\mathbb{Z}[i]$ . $\qquad\blacksquare$

(Since $\mathbb{Z}[i]$ is a PID, irreducible $\iff$ prime, so $3$ is also a Gaussian prime — consistent with the fact that rational primes $p\equiv 3\pmod 4$ remain prime in $\mathbb{Z}[i]$ .)

Answer

$\boxed{\,3 \text{ is irreducible in } \mathbb{Z}[i].\,}$ Reason: $N(3)=9$ , and any factorization $3=\alpha\beta$ forces $N(\alpha)N(\beta)=9$ ; the case $N(\alpha)=N(\beta)=3$ is impossible (no Gaussian integer has norm $3$ ), so one factor has norm $1$ and is a unit.