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UPSC 2026 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Integral domains; characteristic · Algebra · asked 3× in 14 yrs · Read the full method →

Question

Let $n$ be a positive integer. Show that the following statements are equivalent:

(i) $\mathbb{Z}/n\mathbb{Z}$ is an integral domain.

(ii) $\mathbb{Z}/n\mathbb{Z}$ is a field.

(iii) $n$ is prime.

Technique

Three statements, so six implications in principle — but a cycle $(\mathrm{i})\Rightarrow(\mathrm{iii})\Rightarrow(\mathrm{ii})\Rightarrow(\mathrm{i})$ proves all six by transitivity and costs less than half the writing. Announce the cycle, then run it: the composite factorisation $n=ab$ manufactures the zero divisor, Bézout manufactures the inverse, and “field $\Rightarrow$ domain” is the one-line generic implication. The whole question turns on one axiom most candidates never write down — an integral domain and a field are both required to satisfy $1\neq 0$ — because $n=1$ is a live case here and is only excluded by that axiom.

Solution

Step 0 — Conventions, stated because the marks depend on them.

All rings below are commutative with identity. Recall the two definitions in the precise form:

• A ring $R$ is an integral domain if $R$ is commutative, has an identity $1_R$ with $1_R\neq 0_R$, and has no zero divisors: for $a,b\in R$, $ab=0\Rightarrow a=0$ or $b=0$.
• A ring $R$ is a field if $R$ is commutative, has an identity $1_R$ with $1_R\neq 0_R$, and every $a\neq 0$ in $R$ has a multiplicative inverse.

The clause $1_R\neq 0_R$ (equivalently, $R$ is not the zero ring) is part of both definitions. It is the clause that does the work at $n=1$; Step 6 is devoted to it.

An integer $p$ is prime if $p\ge 2$ and its only positive divisors are $1$ and $p$. In particular $1$ is not prime, and no prime is $1$.

Step 1 — The ring $\mathbb{Z}/n\mathbb{Z}$, and two facts we shall use repeatedly.

For $n\ge 1$, $n\mathbb{Z}$ is an ideal of $\mathbb{Z}$ and the quotient

is a commutative ring with identity $\bar 1$ under $\bar a+\bar b=\overline{a+b}$, $\bar a\,\bar b=\overline{ab}$ (these are well defined precisely because $n\mathbb{Z}$ is an ideal). It has exactly $n$ elements.

Fact A. $\bar a=\bar 0 \iff n\mid a$. In particular, if $0<a<n$ then $\bar a\neq\bar 0$.

Fact B. $\bar 1=\bar 0\iff n\mid 1\iff n=1$ (as $n\ge 1$). Hence

Step 2 — Strategy: prove a cycle, not six implications.

The three statements are equivalent as soon as we prove the single cycle

because then any one of them implies any other by following the arrows round (e.g. $(\mathrm{ii})\Rightarrow(\mathrm{iii})$ is $(\mathrm{ii})\Rightarrow(\mathrm{i})\Rightarrow(\mathrm{iii})$). We prove exactly these three implications.

Step 3 — $(\mathrm{i})\Rightarrow(\mathrm{iii})$: if $\mathbb{Z}/n\mathbb{Z}$ is an integral domain then $n$ is prime.

Assume $\mathbb{Z}/n\mathbb{Z}$ is an integral domain.

First, $n\ge 2$. By the definition of an integral domain, $\bar 1\neq \bar 0$; by Fact B this forces $n\ge 2$. (This is where $n=1$ is killed. Do not skip it — see Step 6.)

Next, $n$ has no non-trivial factorisation. Suppose, for contradiction, that $n$ is not prime. Since $n\ge2$ and $n$ is not prime, $n$ has a positive divisor other than $1$ and $n$; so we may write

By Fact A, $0<a<n$ gives $\bar a\neq\bar 0$ and $0<b<n$ gives $\bar b\neq\bar 0$. But

so $\bar a$ and $\bar b$ are zero divisors in $\mathbb{Z}/n\mathbb{Z}$ — contradicting the hypothesis that it is an integral domain.

Hence $n$ is prime. $\qquad\blacksquare$

Step 4 — $(\mathrm{iii})\Rightarrow(\mathrm{ii})$: if $n$ is prime then $\mathbb{Z}/n\mathbb{Z}$ is a field.

Let $n=p$ be prime, so $p\ge2$.

The ring is non-trivial and commutative with identity. By Step 1 it is commutative with identity $\bar1$, and by Fact B, $p\ge2$ gives $\bar 1\neq\bar 0$.

Every non-zero element is invertible. Let $\bar a\neq\bar 0$, i.e. (Fact A) $p\nmid a$. Consider $d=\gcd(a,p)$. Since $d\mid p$ and $p$ is prime, $d=1$ or $d=p$; and $d=p$ would give $p\mid a$, which is false. Hence

By Bézout’s identity there exist integers $u,v$ with

Reducing modulo $p$ (i.e. applying the ring homomorphism $\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$, and using $\bar p=\bar 0$):

So $\bar u$ is a multiplicative inverse of $\bar a$. As $\bar a\neq\bar0$ was arbitrary, every non-zero element of $\mathbb{Z}/p\mathbb{Z}$ is a unit.

Therefore $\mathbb{Z}/p\mathbb{Z}$ is a field. $\qquad\blacksquare$

(Bézout is the route to write, because it is constructive — it names the inverse $\bar u$. An alternative, equally valid, is the pigeonhole argument: for $\bar a\neq \bar 0$ the map $\varphi:\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}$, $\varphi(\bar x)=\bar a\bar x$, is injective — $\bar a\bar x=\bar a\bar y\Rightarrow p\mid a(x-y)\Rightarrow p\mid x-y$ by Euclid’s lemma — hence surjective on the finite set $\mathbb{Z}/p\mathbb{Z}$, so $\bar 1$ is attained.)

Step 5 — $(\mathrm{ii})\Rightarrow(\mathrm{i})$: every field is an integral domain.

This implication uses nothing about $\mathbb{Z}/n\mathbb{Z}$; it is the general fact.

Let $F$ be a field. By definition $F$ is commutative with $1\neq 0$, so only the absence of zero divisors needs checking. Suppose $ab=0$ in $F$ with $a\neq 0$. Since $F$ is a field, $a^{-1}$ exists, and

So $ab=0$ forces $a=0$ or $b=0$: $F$ has no zero divisors, and $F$ is an integral domain.

Applying this with $F=\mathbb{Z}/n\mathbb{Z}$ closes the cycle. $\qquad\blacksquare$

Step 6 — The case $n=1$, addressed explicitly (this is where the question bites).

The hypothesis is that $n$ is a positive integer, so $n=1$ is admitted and must be dealt with. Here

Now note carefully:

• The condition ”$ab=0\Rightarrow a=0$ or $b=0$” holds in $\{\bar0\}$ vacuously — there is no non-zero element at all, so there is no zero divisor to find. A candidate who defines an integral domain merely as “a commutative ring with no zero divisors” therefore concludes that $\mathbb{Z}/1\mathbb{Z}$ is an integral domain, while $1$ is not prime — and the stated equivalence collapses at $n=1$.
• It does not collapse, because the definition also demands $1\neq 0$, which fails here. So $\mathbb{Z}/1\mathbb{Z}$ is not an integral domain and not a field.
• And $1$ is not prime.

Hence at $n=1$ all three statements (i), (ii), (iii) are false, so they are (trivially but genuinely) equivalent there. The general argument of Steps 3–5 already covers this: the only place $n=1$ could have leaked through is Step 3, and it was excluded there by the $\bar1\neq\bar0$ axiom, which is exactly why that half-line was written out.

Step 7 — Conclusion.

The cycle $(\mathrm{i})\Rightarrow(\mathrm{iii})\Rightarrow(\mathrm{ii})\Rightarrow(\mathrm{i})$ is established for every positive integer $n$, hence

Step 8 — Two remarks that show the structure behind the result.

(a) The ideal-theoretic proof (one line each). For a commutative ring $R$ with identity and an ideal $I$:

In $R=\mathbb{Z}$ the ideals are exactly $n\mathbb{Z}$, $n\ge0$, and for $n\ge1$

the middle-to-right step being the standard fact that in a PID every non-zero prime ideal is maximal. Substituting $I=n\mathbb{Z}$ gives (i) $\iff$ (iii) $\iff$ (ii) at once. The elementary proof above is still the one to present — it needs no quotient-ring machinery — but quoting this in two lines at the end is cheap and reads as mastery.

(b) Why the hypothesis ”$n$ positive” is not decoration. The ring $\mathbb{Z}/0\mathbb{Z}\cong\mathbb{Z}$ is an integral domain but is not a field, and $0$ is not prime. So at $n=0$ statement (i) is true while (ii) and (iii) are false: the three statements are not equivalent. The restriction $n\ge1$ in the question is doing real work. (In the ideal language: $\{0\}$ is a prime ideal of $\mathbb{Z}$ that is not maximal — the single exception in the PID fact quoted in (a).)

(c) The trichotomy behind (i) $\Rightarrow$ (ii). For $n\ge2$, $\bar a$ is a unit of $\mathbb{Z}/n\mathbb{Z}$ iff $\gcd(a,n)=1$, and a non-zero $\bar a$ is a zero divisor iff $\gcd(a,n)>1$; so in $\mathbb{Z}/n\mathbb{Z}$ every non-zero element is either a unit or a zero divisor, with no third possibility. This is the concrete face of the general theorem every finite integral domain is a field, which supplies $(\mathrm{i})\Rightarrow(\mathrm{ii})$ directly if one prefers.

Answer