← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Principal Ideal Domains (PID) · Algebra · Read the full method →

Question

Let $R$ be a principal ideal domain. Show that every ideal of a quotient ring of $R$ is a principal ideal, and $R/P$ is a principal ideal domain for a prime ideal $P$ of $R$.

Technique

Correspondence theorem for ideals + surjectivity gives $\pi(aR)=(\bar a)$; prime ideal $\Leftrightarrow$ quotient is a domain.

Solution

Throughout, $R$ is a commutative ring with $1$ which is a principal ideal domain (PID): every ideal of $R$ is of the form $(a)=aR$ for some $a\in R$.

Part 1 — Every ideal of a quotient ring $R/I$ is principal

Let $I$ be any ideal of $R$ and let $\pi:R\to R/I$ be the canonical surjection, $\pi(r)=r+I$.

Correspondence (lattice) theorem. The ideals of $R/I$ correspond bijectively to the ideals of $R$ containing $I$, via $J\mapsto J/I$ (equivalently $\bar J\mapsto \pi^{-1}(\bar J)$).

Let $\bar J$ be an arbitrary ideal of $R/I$. Its preimage $J=\pi^{-1}(\bar J)$ is an ideal of $R$ (preimage of an ideal under a ring homomorphism), and $I=\pi^{-1}(0)\subseteq J$. Since $R$ is a PID,

Now apply $\pi$. Because $\pi$ is surjective, $\pi(J)=\bar J$, and

where $\bar a=a+I$. (More precisely $\pi(aR)=\pi(a)\pi(R)=\pi(a)(R/I)$ since $\pi(R)=R/I$.)

Hence $\bar J$ is the principal ideal generated by $\bar a$. As $\bar J$ was arbitrary:

(Thus $R/I$ is a principal ideal ring; it need not be a domain unless $I$ is prime.)

Part 2 — $R/P$ is a PID for a prime ideal $P$

Recall: in a commutative ring with $1$, an ideal $P$ is prime iff the quotient $R/P$ is an integral domain.

Let $P$ be a prime ideal of $R$. Then:

1. $R/P$ is an integral domain — this is the standard characterization of a prime ideal. (For $\bar a\bar b=0$ in $R/P$ means $ab\in P$, so $a\in P$ or $b\in P$ by primeness, i.e. $\bar a=0$ or $\bar b=0$. Also $R/P\ne 0$ because a prime ideal is proper, $P\ne R$, so $\bar 1\ne 0$.)

2. Every ideal of $R/P$ is principal — by Part 1 with $I=P$.

A commutative integral domain in which every ideal is principal is, by definition, a principal ideal domain. Therefore

Answer