UPSC 2023 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Subrings and ideals · Algebra · asked 4× in 13 yrs · Read the full method →

Question

Express the ideal $4\mathbb{Z}+6\mathbb{Z}$ as a principal ideal in the integral domain $\mathbb{Z}$ .

Technique

Bézout’s identity in $\mathbb{Z}$ ; $m\mathbb{Z}+n\mathbb{Z}=\gcd(m,n)\mathbb{Z}$ .

Solution

Strategy. $\mathbb{Z}$ is a PID; the sum of two principal ideals equals the ideal generated by the gcd of their generators.

Step 1 — Definition of the sum.

By definition,

4\mathbb{Z}+6\mathbb{Z}=\{4a+6b:a,b\in\mathbb{Z}\}.

Step 2 — Containment in $2\mathbb{Z}$ .

For any $a,b\in\mathbb{Z}$ , $4a+6b=2(2a+3b)\in 2\mathbb{Z}$ , so $4\mathbb{Z}+6\mathbb{Z}\subseteq 2\mathbb{Z}$ .

Step 3 — Reverse containment.

By Bézout, $\gcd(4,6)=2$ and indeed $2=(-1)\cdot 4+1\cdot 6\in 4\mathbb{Z}+6\mathbb{Z}$ . Since an ideal is closed under multiplication by ring elements,

2\mathbb{Z}=\{2k:k\in\mathbb{Z}\}\subseteq 4\mathbb{Z}+6\mathbb{Z}.

Step 4 — Conclude.

Answer

\boxed{\;4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}=\langle 2\rangle.\;}