← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q3d — Step-by-Step Solution

10 marks · Section A

Euclidean domains · Algebra · asked 5× in 13 yrs · Read the full method →

Question

Let $a$ be an irreducible element of the Euclidean ring $R$, then prove that $R/(a)$ is a field.

Technique

Show $(a)$ is maximal using the PID structure of a Euclidean ring and irreducibility; then apply ”$R/M$ field $\iff$ $M$ maximal”. The Bézout/gcd route gives inverses directly.

Solution

Throughout, $R$ is a Euclidean ring (commutative ring with unity, an integral domain, equipped with a Euclidean valuation $d:R\setminus\{0\}\to\mathbb Z_{\ge0}$). Recall a Euclidean ring is a principal ideal domain (PID), and in a PID irreducible $\iff$ prime, and the key fact: the ideals containing $(a)$ correspond to divisors of $a$.

Step 1 — $(a)$ is a maximal ideal

We show the ideal $(a)$ is maximal. Suppose $J$ is an ideal with

Since $R$ is a PID, $J=(b)$ for some $b\in R$. From $(a)\subseteq(b)$ we get $b\mid a$, say $a=bc$ for some $c\in R$.

Because $a$ is irreducible, in the factorization $a=bc$ one of $b,c$ must be a unit:

• If $b$ is a unit, then $(b)=R$, so $J=R$.
• If $c$ is a unit, then $b=ac^{-1}\in(a)$, so $(b)\subseteq(a)$; combined with $(a)\subseteq(b)$ this gives $J=(b)=(a)$.

Hence the only ideals between $(a)$ and $R$ are $(a)$ and $R$ themselves; i.e. $(a)$ is maximal. (Note $(a)\ne R$ since $a$ is irreducible, hence a non-unit, so $1\notin(a)$.)

Step 2 — Maximal ideal $\Rightarrow$ quotient is a field

Standard theorem: for a commutative ring $R$ with unity, the quotient $R/M$ is a field iff $M$ is a maximal ideal. Since $(a)$ is maximal,

Answer