← 2026 Paper 2

UPSC 2026 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

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

Question

Show that every non-zero prime ideal in a Euclidean Domain is maximal.

Technique

A Euclidean domain gives you exactly one tool — the division algorithm — and the standard way to cash it in is the minimal-norm argument: in any non-zero ideal, an element of least norm generates the ideal. That converts “Euclidean domain” into “PID”, after which the result is a three-line argument on generators: if $(p)\subseteq(a)$ with $(p)$ prime, write $p=ab$ and let primality of $(p)$ decide which of $a,b$ lies in $(p)$ — one case gives $(a)=(p)$, the other forces $a$ to be a unit, so $(a)=R$. The recognisable trigger: “prime $\Rightarrow$ maximal” in a Euclidean/principal-ideal setting is never a quotient-ring computation; it is “no ideal can sit strictly in between”, tested on generators.

Solution

Step 0 — Definitions and conventions, stated because the marks depend on them.

Throughout, $R$ is a Euclidean domain: a commutative integral domain with $1\neq 0$ together with a function

such that for all $a,b\in R$ with $b\neq 0$ there exist $q,r\in R$ with

(Some texts add the axiom $d(a)\le d(ab)$ for $a,b\neq0$. It is not needed anywhere below — only the division property (E) is used. Saying so is worth a line.)

An ideal $P\subseteq R$ is prime if $P\neq R$ and $ab\in P\Rightarrow a\in P$ or $b\in P$. An ideal $M\subseteq R$ is maximal if $M\neq R$ and there is no ideal $J$ with $M\subsetneq J\subsetneq R$.

Note that "$P\neq R$" is part of both definitions, so for maximality of a prime $P$ only the no-intermediate-ideal condition has to be proved.

Step 1 — Every Euclidean domain is a principal ideal domain.

Let $I\subseteq R$ be an ideal. If $I=\{0\}$ then $I=(0)$ is principal. So assume $I\neq\{0\}$ and consider the set of norms of its non-zero elements,

$S$ is a non-empty set of non-negative integers, so by the well-ordering principle it has a least element. Choose $b\in I$, $b\neq0$, with $d(b)=\min S$.

Claim: $I=(b)$. Since $b\in I$ and $I$ is an ideal, $(b)=bR\subseteq I$. Conversely let $a\in I$. By the division property (E) applied to $a$ and $b\ (\neq 0)$ there are $q,r\in R$ with

Now

because $a\in I$, $b\in I$ and $I$ is an ideal. If $r\neq0$ then $d(r)\in S$ with $d(r)<d(b)=\min S$ — impossible. Hence $r=0$ and $a=bq\in(b)$.

Therefore $I\subseteq(b)$, so $I=(b)$: every ideal of $R$ is principal. $\qquad\blacksquare$

Step 2 — A non-zero prime ideal admits no ideal strictly between it and $R$.

Let $P$ be a non-zero prime ideal of $R$. By Step 1, $P=(p)$ for some $p\in R$, and $p\neq0$ since $P\neq\{0\}$.

Let $J$ be any ideal with

By Step 1 again, $J=(a)$ for some $a\in R$. Since $p\in P\subseteq J=(a)$, there exists $b\in R$ with

Now $ab=p\in P$ and $P$ is prime, so

Case 1: $a\in P$. Then $(a)\subseteq P$ (as $P$ is an ideal containing $a$), while $P\subseteq(a)=J$ by hypothesis. Hence

Case 2: $b\in P=(p)$. Then $b=pc$ for some $c\in R$, and substituting into $p=ab$,

Since $p\neq0$ and $R$ is an integral domain (no zero divisors — this is where the “domain” hypothesis is spent), we may cancel $p$:

So $a$ is a unit of $R$, and therefore

Thus every ideal $J$ with $P\subseteq J\subseteq R$ satisfies $J=P$ or $J=R$: there is no ideal strictly between $P$ and $R$.

Step 3 — Conclude.

$P$ is prime, so by definition $P\neq R$. Combined with Step 2, $P$ is a maximal ideal of $R$.

Hence every non-zero prime ideal in a Euclidean domain is maximal. $\qquad\blacksquare$

Step 4 — Why the hypothesis “non-zero” cannot be dropped.

The zero ideal $(0)$ of a domain $R$ is always prime, because $ab=0\Rightarrow a=0$ or $b=0$ is exactly the definition of a domain. But $(0)$ is maximal iff $R$ is a field. So in the Euclidean domain $\mathbb{Z}$,

exhibits $(0)$ as a prime ideal that is not maximal. The word “non-zero” in the question is therefore load-bearing, not decorative.

(Consistency check: the theorem is not violated, since $(0)$ is excluded. And in a field — which is trivially Euclidean with $d\equiv0$ — the only ideals are $(0)$ and $R$, so $(0)$ is maximal and there is no non-zero proper ideal at all; the theorem is vacuously true there.)

Step 5 — Two remarks for depth.

(i) Where “Euclidean” was actually used. Only in Step 1, and only through (E). The result therefore holds verbatim in any PID, and the true statement is: in a PID, every non-zero prime ideal is maximal — equivalently, a PID that is not a field has Krull dimension $1$: its chains of prime ideals are exactly $(0)\subsetneq(p)$.

(ii) The hypothesis is not removable in a general domain. $\mathbb{Z}[x]$ is a UFD but not a PID. There $(x)$ is a non-zero prime ideal, since

is an integral domain; but $\mathbb{Z}$ is not a field, so $(x)$ is not maximal — indeed

So the theorem genuinely needs principality, i.e. it is the division algorithm doing the work.

(iii) Quotient-ring formulation (equivalent one-line restatement). Using $P$ prime $\iff R/P$ a domain and $M$ maximal $\iff R/M$ a field, the theorem says: for a non-zero ideal $I$ of a Euclidean domain, $R/I$ is an integral domain if and only if it is a field.

Answer