Euclidean domains

At a Glance

• Frequency: 6 sub-parts across 6 of 14 years (2013, 2016, 2017, 2019, 2025, 2026)
• Priority tier: T2
• Marks (count): 10 (2), 15 (2), 20 (1)
• Average solve time: ~11 min
• Difficulty mix: medium 4, easy 1
• Section: A | Dominant type: proof

Why This Chapter Matters

This atom has appeared in 6 of the last 14 years — always a proof, usually in Section A — and the questions cluster tightly around three ideas: the division algorithm (for $F[X]$ or $\mathbb{Z}[i]$), the implication chain Euclidean $\Rightarrow$ PID $\Rightarrow$ UFD, and the link between irreducibility and maximality of ideals. Once you understand why the norm $N(a+bi)=a^2+b^2$ is multiplicative and how rounding controls the remainder, every variant reduces to a two-page structural argument. The irreducibility-via-norm technique (2025 Q) is the same norm argument in miniature and takes under five minutes.

Minimum Theory

Euclidean domain. An integral domain $R$ is a Euclidean domain (ED) if there exists a function $d:R\setminus\{0\}\to\mathbb{Z}_{\ge0}$ (the Euclidean norm or valuation) such that for every $a\in R$ and $b\in R\setminus\{0\}$, there exist $q,r\in R$ with $a = qb + r, \quad r=0 \text{ or } d(r) < d(b).$ The key examples are: $\mathbb{Z}$ with $d(n)=|n|$; $F[X]$ (for a field $F$) with $d(f)=\deg f$; and the Gaussian integers $\mathbb{Z}[i]$ with $d(a+bi)=a^2+b^2$.

The implication chain. Every Euclidean domain is a PID, and every PID is a UFD: $\text{ED} \;\Rightarrow\; \text{PID} \;\Rightarrow\; \text{UFD} \;\Rightarrow\; \text{Integral Domain.}$ Each implication is strict. $\mathbb{Z}[X]$ is a UFD but not a PID (the ideal $(2,X)$ is not principal). The ring $\mathbb{Z}\!\left[\tfrac{1+\sqrt{-19}}{2}\right]$ is a PID but not Euclidean (Dedekind–Hasse counterexample). In a PID: irreducible $\Leftrightarrow$ prime $\Leftrightarrow$ the generated ideal is maximal.

Prime ideals versus maximal ideals. $P$ is prime if $P\ne R$ and $ab\in P\Rightarrow a\in P$ or $b\in P$; $M$ is maximal if $M\ne R$ and no ideal sits strictly between $M$ and $R$. Maximal always implies prime. The converse fails in general ($\mathbb{Z}[X]$ has $(X)$ prime but $(X)\subsetneq(2,X)\subsetneq\mathbb{Z}[X]$) but holds in a PID for non-zero primes. The word non-zero is load-bearing: $(0)$ is prime in every domain (that is the definition of a domain) yet is maximal only if $R$ is a field — in $\mathbb{Z}$, $(0)\subsetneq(2)\subsetneq\mathbb{Z}$. Equivalently, a PID that is not a field has Krull dimension 1: its prime chains are exactly $(0)\subsetneq(p)$.

$\mathbb{Z}[i]$ in detail. The norm $N(a+bi)=a^2+b^2$ is multiplicative: $N(\alpha\beta)=N(\alpha)N(\beta)$. Units are elements of norm 1: $\{\pm1,\pm i\}$. To divide $\alpha$ by $\beta\ne0$, write $\alpha/\beta = x+iy$ in $\mathbb{Q}[i]$, then round each of $x$ and $y$ to the nearest integer $m$ and $n$. Set $q=m+ni$ and $r=\alpha-q\beta$. The rounding error on each component is at most $\tfrac{1}{2}$, so $N(r) = N\!\left(\tfrac{\alpha}{\beta}-q\right)\cdot N(\beta) \le \left(\tfrac{1}{4}+\tfrac{1}{4}\right)N(\beta) = \tfrac{1}{2}N(\beta) < N(\beta).$ This is the Euclidean property for $\mathbb{Z}[i]$.

Question Archetypes

euclidean-domain (1 question; 2013)

Show $\mathbb{Z}[i]$ is a Euclidean domain, hence a PID and a UFD

Recognition Cues

The question names $\mathbb{Z}[i]$ (or a similar ring with a known norm) and asks you to verify one or more properties in the chain Euclidean $\Rightarrow$ PID $\Rightarrow$ UFD. All three may be asked together (as in 2013). The tell is that a specific norm function is either given or obviously available (sum of squares, polynomial degree).

Solution Template

1. Define the norm. State $N(a+bi)=a^2+b^2$ and note it is a non-negative integer with $N(\alpha)=0$ iff $\alpha=0$.
2. Establish multiplicativity (briefly): $N(\alpha\beta)=N(\alpha)N(\beta)$ — this follows from $|z_1 z_2|^2=|z_1|^2|z_2|^2$.
3. Prove division with remainder. For $\alpha,\beta\ne0$: write $\alpha/\beta = x+iy\in\mathbb{Q}[i]$, round to $q=m+ni$, set $r=\alpha-q\beta$. Bound $N(r/\beta)=(x-m)^2+(y-n)^2\le\tfrac{1}{4}+\tfrac{1}{4}=\tfrac{1}{2}<1$, so $N(r)<N(\beta)$. Conclude ED.
4. PID from ED. Let $I$ be any ideal. If $I=\{0\}$, then $I=(0)$. Otherwise pick $b\in I$ with $N(b)$ minimal; for any $a\in I$ write $a=qb+r$ with $N(r)<N(b)$ — since $r=a-qb\in I$, minimality forces $r=0$, so $a\in(b)$. Hence $I=(b)$.
5. UFD from PID. In a PID every irreducible is prime (standard), and the ascending chain condition holds (bounded norm), so unique factorisation follows.

Worked Example

2013 Paper 2, 2013-P2-Q3a (15 marks)

State whether the Gaussian integers $\mathbb{Z}[i]$ form a Euclidean domain, a principal ideal domain, and a unique factorisation domain. Justify your answer in each case.

Euclidean domain. Let $\alpha,\beta\in\mathbb{Z}[i]$, $\beta\ne0$. Compute $\alpha/\beta = x+iy\in\mathbb{Q}[i]$. Choose $m,n\in\mathbb{Z}$ with $|x-m|\le\tfrac{1}{2}$ and $|y-n|\le\tfrac{1}{2}$. Set $q=m+ni$ and $r=\alpha-q\beta$. Then: $N(r) = N(\alpha - q\beta) = N\!\left(\tfrac{\alpha}{\beta}-q\right)\cdot N(\beta) \le \left(\tfrac{1}{4}+\tfrac{1}{4}\right)N(\beta) = \tfrac{N(\beta)}{2} < N(\beta).$ So $\alpha = q\beta + r$ with $r=0$ or $N(r)<N(\beta)$. Since $N(a+bi)=a^2+b^2\ge 0$ and $N(\alpha)=0\Leftrightarrow\alpha=0$, this makes $\mathbb{Z}[i]$ a Euclidean domain with norm $N$.

PID. Let $I\subseteq\mathbb{Z}[i]$ be any ideal. If $I=\{0\}$ then $I=(0)$. Otherwise choose $b\in I$ with $N(b)$ minimal (possible since $N$ takes non-negative integer values). For any $a\in I$, apply the Euclidean property: $a=qb+r$ with $N(r)<N(b)$. Since $r=a-qb\in I$ and $N(r)<N(b)$, minimality forces $r=0$, so $a=qb\in(b)$. Thus $I=(b)$ and $\mathbb{Z}[i]$ is a PID.

UFD. Every PID is a UFD by a standard theorem: in a PID every irreducible element is prime (since the ideal $(p)$ is maximal, hence prime), and the Noetherian property (which holds because ideals are generated by elements of finite norm) prevents infinite ascending chains. Unique factorisation into irreducibles then follows from the prime property plus Noetherian induction. Hence $\mathbb{Z}[i]$ is a UFD.

Common Traps

• Forgetting to bound $N(r)$, not $N(r/\beta)$. The rounding argument gives $N(r/\beta)\le\tfrac{1}{2}$, so $N(r)=N(r/\beta)\cdot N(\beta)\le\tfrac{1}{2}N(\beta)<N(\beta)$. Write this as a product of norms — the step is one line but must appear explicitly.
• Claiming “PID therefore UFD” without justification. The UPSC mark scheme expects the intermediate fact: every irreducible in a PID is prime, and the ascending chain condition holds. A bare citation loses marks; one sentence of explanation suffices.
• Units. The units of $\mathbb{Z}[i]$ are exactly $\{\pm1,\pm i\}$ (the elements with $N=1$). Confusing units with primes invalidates the UFD argument.

ideal-irreducible (1 question; 2016)

Show the principal ideal $(f)$ in $K[X]$ is maximal if and only if $f$ is irreducible over the field $K$

Recognition Cues

The question mentions the ideal generated by a polynomial in $K[X]$ (where $K$ is a field) and asks for a maximality–irreducibility equivalence, or equivalently asks to show $K[X]/(f)$ is a field iff $f$ is irreducible. The biconditional structure calls for two separate proofs ($\Rightarrow$ and $\Leftarrow$).

Solution Template

1. Setup. Note $K[X]$ is a PID (proved by the division algorithm since $K$ is a field). The ideals containing $(f)$ are exactly the $(g)$ with $g\mid f$.
2. (Irreducible $\Rightarrow$ Maximal). Suppose $f$ is irreducible and $(f)\subseteq J=(g)\subseteq K[X]$ for some ideal $J$. Then $g\mid f$. Since $f$ is irreducible, $g$ is either a unit or an associate of $f$. If $g$ is a unit, $J=K[X]$. If $g\sim f$, then $J=(f)$. So $(f)$ is maximal.
3. (Maximal $\Rightarrow$ Irreducible). Suppose $(f)$ is maximal. If $f=gh$ with $g,h$ non-units, then $(f)\subsetneq(g)\subsetneq K[X]$, contradicting maximality. So $f$ is irreducible.
4. Corollary (often asked): $K[X]/(f)$ is a field $\Leftrightarrow$ $(f)$ is maximal $\Leftrightarrow$ $f$ is irreducible.

Worked Example

2016 Paper 2, 2016-P2-Q1a (10 marks)

Show that the ideal $(f)$ in $K[X]$ is a maximal ideal if and only if $f$ is an irreducible polynomial over the field $K$.

First note: $K[X]$ is a PID because $K$ is a field and the division algorithm holds in $K[X]$ (leading coefficient of $g$ is invertible in $K$). In a PID every ideal is principal, so every ideal containing $(f)$ has the form $(g)$ with $g\mid f$.

($\Leftarrow$) Suppose $f$ is irreducible. Let $(f)\subseteq (g)\subseteq K[X]$. Then $g\mid f$. Since $f$ is irreducible, the only divisors of $f$ are units and associates. If $g$ is a unit then $(g)=K[X]$; if $g$ is an associate of $f$ then $(g)=(f)$. Hence there is no ideal strictly between $(f)$ and $K[X]$, so $(f)$ is maximal.

($\Rightarrow$) Suppose $(f)$ is maximal. Assume for contradiction that $f=gh$ with neither $g$ nor $h$ a unit. Then $\deg g<\deg f$ and $\deg h<\deg f$, so $f\mid g$ is false, meaning $(f)\subsetneq (g)$. Also $g$ is a non-unit non-zero polynomial, so $(g)\subsetneq K[X]$. This gives a proper chain $(f)\subsetneq(g)\subsetneq K[X]$, contradicting maximality. Hence $f$ is irreducible.

Consequence. $(f)$ maximal $\Rightarrow K[X]/(f)$ is a field (a standard ring-theory fact: $R/I$ is a field iff $I$ is maximal).

Common Traps

• Not excluding the trivial cases. The ideal $(0)$ is not maximal and $0$ is not irreducible; the ideal $K[X]$ itself is not proper. Assume $f$ is a nonzero non-unit throughout.
• Strictness of the inclusion. To show $(f)\subsetneq(g)$ in the ($\Rightarrow$) direction, you need to verify $g\nmid f$ (since $g$ is not an associate of $f$ — $h$ is also a non-unit). A sentence confirming $g\nmid f$ (i.e., $f/g=h$ is not a unit) is required.
• “Prime and maximal coincide” only in a PID. In a general commutative ring, irreducible $\ne$ prime. Here the PID hypothesis is essential; without establishing $K[X]$ is a PID first, the argument fails.

division-algorithm (1 question; 2017)

Prove the division algorithm in $F[X]$ for a field $F$

Recognition Cues

The question asks you to prove that for $f,g\in F[X]$ with $g\ne0$, there exist unique $q,r\in F[X]$ with $f=qg+r$ and $\deg r < \deg g$ (with the convention $\deg 0 = -\infty$). This is an existence-plus-uniqueness proof; the key tool is induction on $\deg f$.

Solution Template

1. Trivial case. If $\deg f < \deg g$, set $q=0$, $r=f$; the condition $\deg r < \deg g$ is satisfied.
2. Inductive step. Let $\deg f = n \ge \deg g = m$, with leading coefficients $a_n$ and $b_m$. Define $f_1 = f - (a_n/b_m)X^{n-m}g$. Note $b_m$ is invertible in $F$ (field hypothesis). Then $\deg f_1 < n$ (the leading terms cancel). Apply the induction hypothesis to $f_1$ and $g$: $f_1 = q_1 g + r_1$ with $\deg r_1 < \deg g$. Set $q = q_1 + (a_n/b_m)X^{n-m}$, $r=r_1$.
3. Uniqueness. Suppose $f=qg+r=q'g+r'$. Then $(q-q')g = r'-r$. If $q\ne q'$, then $\deg((q-q')g)=\deg(q-q')+\deg g\ge\deg g$, but $\deg(r'-r)<\deg g$ — contradiction. So $q=q'$ and $r=r'$.

Worked Example

2017 Paper 2, 2017-P2-Q2c (20 marks)

Let $F$ be a field and $f(X),\,g(X)\in F[X]$ with $g(X)\ne0$. Prove that there exist unique polynomials $q(X)$ and $r(X)$ in $F[X]$ such that $f(X)=q(X)g(X)+r(X)$ and either $r(X)=0$ or $\deg r(X)<\deg g(X)$.

Existence (by induction on $n=\deg f$).

Base. If $f=0$ or $\deg f < \deg g = m$, take $q=0$, $r=f$. Then $f=0\cdot g+f$ and $\deg r<\deg g$. $\checkmark$

Step. Suppose existence holds for all polynomials of degree $< n$, and let $\deg f=n\ge m$. Write $f=a_n X^n+\cdots$ and $g=b_m X^m+\cdots$. Since $F$ is a field, $b_m$ has an inverse $b_m^{-1}\in F$. Define $f_1 = f - a_n b_m^{-1} X^{n-m} g.$ The coefficient of $X^n$ in $f_1$ is $a_n - a_n b_m^{-1}\cdot b_m = 0$, so $\deg f_1 \le n-1 < n$. By the induction hypothesis, $f_1 = q_1 g + r_1$ with $r_1=0$ or $\deg r_1 < m$. Then $f = a_n b_m^{-1} X^{n-m} g + f_1 = \underbrace{\bigl(a_n b_m^{-1} X^{n-m} + q_1\bigr)}_{q} g + r_1.$ Setting $r=r_1$ gives $f=qg+r$ with $\deg r < \deg g$. $\checkmark$

Uniqueness. Suppose $f=qg+r=q'g+r'$ with $\deg r,\deg r'<\deg g$. Then $(q-q')g=r'-r$. If $q\ne q'$ then $\deg\bigl((q-q')g\bigr) = \deg(q-q')+\deg g \ge \deg g$ (using that $F[X]$ is a domain, so degree is additive). But $\deg(r'-r)\le\max(\deg r',\deg r)<\deg g$ — contradiction. Hence $q=q'$ and therefore $r=r'$. $\square$

Common Traps

• Field hypothesis is used exactly once. The invertibility of $b_m$ is the only step that requires $F$ to be a field. Make this explicit; examiners look for it.
• Degree convention for zero. When $f=0$ or $f_1=0$ arises during induction, use $\deg 0 = -\infty$ so the base case applies cleanly. Treating $\deg 0$ as $0$ creates a spurious edge case.
• Uniqueness needs the domain property. The step $\deg((q-q')g)=\deg(q-q')+\deg g$ uses the fact that $F[X]$ is an integral domain (no zero divisors). If you only assumed $F$ is a ring, this step fails.

quotient-field-irreducible (1 question; 2019)

Show $R/(a)$ is a field when $a$ is irreducible in a Euclidean ring $R$

Recognition Cues

The question gives a Euclidean ring $R$ (or Euclidean domain), an irreducible element $a\in R$, and asks to show the quotient ring $R/(a)$ is a field. The Euclidean hypothesis is load-bearing: it guarantees a division algorithm and hence a gcd theory (Bézout’s identity).

Solution Template

1. Show $(a)\ne R$. Since $a$ is irreducible, it is a non-unit, so $1\notin(a)$, meaning $(a)\subsetneq R$. Hence $R/(a)$ is a non-trivial ring.
2. Show every nonzero coset is invertible. Take $\bar{x}=x+(a)\ne\bar{0}$, so $a\nmid x$. Since $a$ is irreducible and $a\nmid x$, we have $\gcd(a,x)=1$ (an irreducible has only unit and associate divisors; $x$ is not a multiple of $a$, so $\gcd=1$).
3. Bézout. In a Euclidean domain, $\gcd(a,x)=1$ means there exist $u,v\in R$ with $ua+vx=1$.
4. Conclude. In $R/(a)$: $\bar{u}\cdot\bar{a}+\bar{v}\cdot\bar{x}=\bar{1}$, and $\bar{a}=\bar{0}$, so $\bar{v}\cdot\bar{x}=\bar{1}$. Thus $\bar{x}$ has inverse $\bar{v}$. Since $\bar{x}$ was arbitrary, $R/(a)$ is a field.

Worked Example

2019 Paper 2, 2019-P2-Q3d (10 marks)

Let $R$ be a Euclidean ring and $a\in R$ an irreducible element. Prove that $R/(a)$ is a field.

Step 1: $(a)$ is a proper ideal. Since $a$ is irreducible, $a$ is not a unit. Hence $1\notin (a)$, so $(a)\ne R$, and $R/(a)$ is a non-trivial quotient ring.

Step 2: Every non-zero element of $R/(a)$ is a unit. Let $\bar{x}=x+(a)\ne\bar{0}$, so $a\nmid x$. Consider $d=\gcd(a,x)$ in $R$ (which exists since $R$ is Euclidean, hence a PID). Since $d\mid a$ and $a$ is irreducible, $d$ is either a unit or an associate of $a$. If $d\sim a$ then $a\mid x$, contradicting $a\nmid x$. So $d$ is a unit, i.e., $\gcd(a,x)=1$ (up to units).

Step 3: Bézout. Since $R$ is a Euclidean domain and $\gcd(a,x)=1$, the ideal $(a,x)=R$ (in a PID, $(a,x)=(\gcd(a,x))=(1)=R$). Hence there exist $u,v\in R$ with $ua + vx = 1.$

Step 4: Invertibility. Reduce modulo $(a)$: $\bar{u}\cdot\bar{0}+\bar{v}\cdot\bar{x}=\bar{1}$, giving $\bar{v}\cdot\bar{x}=\bar{1}$. So $\bar{x}$ has multiplicative inverse $\bar{v}$ in $R/(a)$.

Since every nonzero element of $R/(a)$ is invertible, $R/(a)$ is a field. $\square$

Common Traps

• Not establishing $(a)\ne R$ first. The quotient ring $R/R=\{0\}$ is not a field; the argument requires $1\notin(a)$, which follows from $a$ being a non-unit (i.e., irreducible). State this explicitly at the start.
• Euclidean hypothesis is essential. Bézout’s identity holds in any PID (and hence in any ED). If $R$ were only a UFD, $\gcd$ exists but Bézout need not hold as a ring identity, so the proof breaks. The Euclidean/PID structure is precisely the hypothesis that makes $\gcd(a,x)=ua+vx$ possible.
• "$\gcd=1$" needs a sentence of justification. Don’t just assert $\gcd(a,x)=1$; show that $d\mid a$ with $a$ irreducible forces $d$ to be a unit (since $x$ is not a multiple of $a$).

irreducibility-via-norm (1 question; 2025)

Show that a specific element is irreducible in $\mathbb{Z}[i]$ using the norm

Recognition Cues

The question names an element $\alpha\in\mathbb{Z}[i]$ (a rational integer like $3$, or a Gaussian integer like $1+i$) and asks to show it is irreducible. The method: compute $N(\alpha)$ and note that any factorisation $\alpha=\beta\gamma$ gives $N(\alpha)=N(\beta)N(\gamma)$; then show the only factorisations of $N(\alpha)$ force one factor to have norm 1 (i.e., to be a unit).

Solution Template

1. Compute $N(\alpha)$. For $\alpha\in\mathbb{Z}$, $N(\alpha)=\alpha^2$.
2. Norm factorisation. If $\alpha=\beta\gamma$ then $N(\alpha)=N(\beta)N(\gamma)$; list all ways to write $N(\alpha)$ as a product of two positive integers.
3. Eliminate non-trivial splits. For each split $\{d_1,d_2\}$ of $N(\alpha)$: if $d_1=1$ or $d_2=1$, one factor is a unit — this is the irreducible conclusion. Otherwise, show no Gaussian integer of that norm exists (sum of two squares argument) or arrive at a contradiction.
4. Conclude. Since every factorisation forces one factor to be a unit, $\alpha$ is irreducible in $\mathbb{Z}[i]$.

Worked Example

2025 Paper 2, 2025-P2-Q2b (15 marks)

Show that $3$ is an irreducible element in the integral domain $\mathbb{Z}[i]$.

Setup. In $\mathbb{Z}[i]$, the norm is $N(a+bi)=a^2+b^2$, and $N$ is multiplicative. The units are elements of norm 1: $\{\pm1,\pm i\}$. Suppose $3=\beta\gamma$ for some $\beta,\gamma\in\mathbb{Z}[i]$. Then $N(3) = N(\beta)\,N(\gamma) \implies 9 = N(\beta)\,N(\gamma).$

Cases. The factorisation $9 = d_1 d_2$ with $d_1,d_2\in\mathbb{Z}_{>0}$ has three forms: $\{1,9\}$ or $\{3,3\}$ or $\{9,1\}$.

Case $\{1,9\}$ or $\{9,1\}$. $N(\beta)=1$ means $\beta$ is a unit, so $3$ is irreducible in this case. $\checkmark$

Case $\{3,3\}$. Suppose $N(\beta)=a^2+b^2=3$ for some $a,b\in\mathbb{Z}$. Check: $a^2+b^2=3$ has no integer solution (since $3\equiv 3\pmod{4}$, and any sum of two squares satisfies $a^2+b^2\not\equiv 3\pmod{4}$ — because squares are $\equiv 0$ or $1\pmod{4}$, so sums of two squares are $\equiv 0,1,$ or $2\pmod{4}$). Hence no Gaussian integer has norm $3$, and this case is impossible.

Conclusion. In every factorisation $3=\beta\gamma$, one factor has norm 1 and is therefore a unit. Hence $3$ is irreducible in $\mathbb{Z}[i]$. $\square$

Remark. Alternatively: since $3\equiv 3\pmod{4}$, the prime $3$ is inert in $\mathbb{Z}[i]$ (by the two-squares characterisation), meaning it remains a Gaussian prime. But the norm argument above is self-contained and does not require this theorem.

Common Traps

• Not proving norm multiplicativity. If the question asks you to “show 3 is irreducible,” you must invoke $N(\beta\gamma)=N(\beta)N(\gamma)$ and state this fact (one line of proof or a brief justification is expected). Simply writing "$N(3)=9$" without explanation of why norms multiply loses marks.
• Missing the modular arithmetic step. The conclusion that $a^2+b^2=3$ has no integer solution is not obvious by inspection. The clean proof is: squares are $\equiv 0$ or $1\pmod{4}$, so $a^2+b^2\equiv 0,1,$ or $2\pmod{4}$; but $3\equiv 3\pmod{4}$. Examiners expect this argument.
• Confusing irreducible and prime. In $\mathbb{Z}[i]$, irreducible and prime coincide (because $\mathbb{Z}[i]$ is a UFD). The question asks only for irreducibility; don’t conflate the two without stating the equivalence.

ed-prime-implies-maximal (1 question(s); 2026)

This archetype is the composition of two you have already met, and the reason it stands on its own is that neither half alone answers the question. The ED $\to$ PID half is step 4 of euclidean-domain; the “no ideal strictly between, tested on generators” half is the ($\Leftarrow$) direction of ideal-irreducible, with prime ideal in place of irreducible element. A script that supplies only one of the two loses roughly half the marks.

Recognition Cues

• “Show that every non-zero prime ideal in a Euclidean domain is maximal.”
• The ring is given as a Euclidean domain (not as a PID) — that word tells you which half must be proved rather than quoted.
• The hypothesis “non-zero” is stated explicitly, which is a hint that its necessity should be discussed.

Solution Template

1. State the definitions of ED, prime and maximal. Note that "$\ne R$" is part of both prime and maximal, so only the no-intermediate-ideal condition has to be proved.
2. Prove ED $\Rightarrow$ PID by the minimum-norm argument: in a non-zero ideal $I$ choose $b\ne0$ with $d(b)$ least (well-ordering on $\mathbb{Z}_{\ge0}$); for $a\in I$ divide, $a=bq+r$, note $r=a-bq\in I$, and minimality forces $r=0$. Hence $I=(b)$.
3. Take any $J$ with $P\subseteq J\subseteq R$. Write $P=(p)$, $J=(a)$; since $p\in(a)$, $p=ab$ for some $b$.
4. Use primality on $ab=p\in P$ and split: $a\in P$ gives $J=(a)=P$; $b\in P$ gives $b=pc$, so $p=a(pc)$ and cancelling $p$ (valid — $p\ne0$ and $R$ is a domain) gives $ac=1$, so $a$ is a unit and $J=R$.
5. Conclude and then show the hypothesis is needed: $(0)$ is prime in any domain but not maximal unless $R$ is a field.

Worked Example

2026 Paper 2, 2026-P2-Q1b (10 marks)

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

Source: analysis/solutions/2026-P2-Q1b.md

Step 1 — ED $\Rightarrow$ PID. Let $I\subseteq R$ be an ideal; if $I=\{0\}$ it is $(0)$. Otherwise $S=\{d(x):x\in I,\,x\ne0\}$ is a non-empty set of non-negative integers, so by well-ordering it has a least element; choose $b\in I\setminus\{0\}$ with $d(b)=\min S$. Clearly $(b)\subseteq I$. For $a\in I$, the division property gives $a=bq+r$ with $r=0$ or $d(r)<d(b)$; but $r=a-bq\in I$, so $r\ne0$ would put $d(r)\in S$ below the minimum. Hence $r=0$ and $a\in(b)$, so $I=(b)$.

Step 2 — No intermediate ideal. Let $P\ne(0)$ be prime; by Step 1, $P=(p)$ with $p\ne0$. Let $P\subseteq J\subseteq R$; again $J=(a)$. Since $p\in P\subseteq(a)$, there is $b$ with $p=ab$. Now $ab=p\in P$ and $P$ is prime, so:

• $a\in P$: then $(a)\subseteq P$, and $P\subseteq(a)$ by hypothesis, so $J=(a)=P$.
• $b\in P=(p)$: then $b=pc$, so $p=a(pc)=p(ac)$, i.e. $p(1-ac)=0$. Since $p\ne0$ and $R$ is an integral domain we may cancel, giving $ac=1$: $a$ is a unit and $J=(a)=R$.

So $J=P$ or $J=R$, and $P\ne R$ because $P$ is prime. Hence $P$ is maximal. $\blacksquare$

Step 3 — Why “non-zero” cannot be dropped. $(0)$ is prime in every integral domain, since $ab=0\Rightarrow a=0$ or $b=0$ is precisely the definition. But $(0)$ is maximal iff $R$ is a field: in $\mathbb{Z}$, $(0)\subsetneq(2)\subsetneq\mathbb{Z}$. The word in the question is load-bearing, not decorative.

Two remarks for depth. “Euclidean” was used only in Step 1, and only through the division property, so the true statement is in a PID, every non-zero prime ideal is maximal. And by the quotient characterisations ($P$ prime $\iff R/P$ a domain; $M$ 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.

Common Traps

• Quoting “ED $\Rightarrow$ PID” as known. The question hands you Euclidean, so the division algorithm is exactly the hypothesis being tested. Citing the implication and doing only Step 2 answers an easier question.
• Ignoring the “non-zero”. Without it the statement is false, and an answer that never mentions why is incomplete even if Steps 1–2 are perfect.
• Concluding $(a)=(p)$ without the case split. Primality gives either $a\in P$ or $b\in P$, and the two branches land on different conclusions ($J=P$ versus $J=R$). Both are needed.
• Confusing prime element with prime ideal. The hypothesis is on the ideal. (In a PID they correspond, but the argument here is run on the ideal and does not need the correspondence.)
• Forgetting where “domain” is spent. The cancellation $p(1-ac)=0\Rightarrow ac=1$ is the only place the no-zero-divisors hypothesis is used. Point at it.

Marks-Aware Writing

For a 10-mark proof (ideal-irreducible or quotient-field): The mark scheme rewards structure. Write the two directions ($\Leftarrow$ and $\Rightarrow$) as clearly labelled separate paragraphs. Each direction needs: one sentence of setup, the key chain of reasoning (3–5 lines), and a boxed or underlined conclusion. Avoid vague phrases like “by standard theory”; write the one or two lines that constitute the argument.

For a 15-mark proof (Z[i] Euclidean, or irreducibility-via-norm): The Euclidean proof must show the rounding bound $N(r)\le N(b)/2$ explicitly — this is the only novel computation. The subsequent PID and UFD steps are brief (3–4 sentences each). For irreducibility-via-norm, the modular arithmetic argument for “no Gaussian integer of norm 3” is the key step and must be written out in full.

For a 20-mark proof (division algorithm): This is the most substantial proof; allocate roughly 8 marks to the existence induction and 4 marks to uniqueness. The inductive step must display the polynomial $f_1$ with its formula and confirm $\deg f_1 < \deg f$. The field hypothesis (invertibility of the leading coefficient) must be explicitly invoked.

General. Box every final conclusion. Number steps in inductions. If a direction is shorter (e.g., the “maximal $\Rightarrow$ irreducible” direction in 2016), still give it its own paragraph — do not bury it in the other direction.

Practice Set

• 2013-P2-Q3a (15 m) — — full ED-PID-UFD chain for $\mathbb{Z}[i]$; practice stating the rounding bound explicitly
• 2016-P2-Q1a (10 m) — — maximality iff irreducibility; write both directions in separate paragraphs
• 2017-P2-Q2c (20 m) — — full induction proof of division algorithm; identify the single use of the field hypothesis
• 2019-P2-Q3d (10 m) — — quotient field; use Bézout; confirm $(a)\ne R$ as first step
• 2025-P2-Q2b (15 m) — — irreducibility of 3 in $\mathbb{Z}[i]$; supply the mod-4 argument for no norm-3 Gaussian integer