Integral domains; characteristic

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2015, 2018)
• Priority tier: T3
• Marks (count): 10 (1), 15 (1)
• Average solve time: ~6 min
• Difficulty mix: easy 2
• Section: A | Dominant type: proof / verification

Why This Chapter Matters

Integral domain questions appear in two distinct flavours: testing whether a given set is an integral domain or field (2015, 15 marks — three sub-parts) and proving a structural theorem about polynomial rings over a domain (2018, 10 marks). The test question requires checking ring axioms one-by-one for each set, which is fast if you have a clear checklist. The polynomial-ring question hinges on the single fact that degree is additive over a domain, which reduces the entire proof to two lines of algebra. Together they cover ring structure from the axiomatic ground up through polynomial rings.

Minimum Theory

Integral domain. A commutative ring $R$ with unity $1\ne0$ is an integral domain if it has no zero divisors: $ab=0\Rightarrow a=0$ or $b=0$. A field is an integral domain in which every non-zero element has a multiplicative inverse. The chain of containments is: fields $\subset$ integral domains $\subset$ commutative rings.

Degree in a polynomial domain. If $R$ is an integral domain and $f,g\in R[x]$ are non-zero, then $\deg(fg)=\deg f+\deg g$ (the leading coefficients multiply to a non-zero element since $R$ has no zero divisors). In particular $R[x]$ is also an integral domain.

Units in $R[x]$. A unit in $R[x]$ is a polynomial $f$ with a multiplicative inverse $g$ in $R[x]$: $fg=1$. Degree additivity immediately forces $\deg f=\deg g=0$, so $f$ and $g$ are constants; then $fg=1$ means their constant terms multiply to $1$ in $R$, i.e. $f\in R$ and $f$ is a unit of $R$.

Question Archetypes

integral-domain-test (1 question(s); 2015)

Recognition Cues — “Do the following sets form integral domains?”; “If so, state if they are fields”; three sets are given; must check ring axioms and zero-divisor property for each.

Solution Template

For each set $S$:

1. Check closure under addition and multiplication.
2. Check additive identity ($0\in S$?) and additive inverses.
3. Check multiplicative identity ($1\in S$?).
4. Check commutativity (often inherited from $\mathbb Z$ or $\mathbb R$).
5. If all ring axioms pass: check zero-divisors (domain?) and inverses (field?).
6. Conclude.

Worked Example

2015 Paper 2, 2015-P2-Q4a (15 marks)

Do the following sets form integral domains with respect to ordinary addition and multiplication? If so, state if they are fields: (i) $\{b\sqrt 2 : b\in\mathbb Q\}$; (ii) the set of even integers; (iii) the set of positive integers.

Part (i) — $S=\{b\sqrt 2 : b\in\mathbb Q\}$.

Closure under addition: $(b_1\sqrt 2)+(b_2\sqrt 2)=(b_1+b_2)\sqrt 2\in S$. $\checkmark$

Closure under multiplication: $(b_1\sqrt 2)(b_2\sqrt 2)=2b_1b_2$. This is rational, not of the form $b\sqrt 2$ unless $b_1b_2=0$. So $S$ is not closed under multiplication.

Conclusion (i): Not an integral domain (fails closure under multiplication; not even a ring).

Part (ii) — $S=2\mathbb Z=\{\ldots,-4,-2,0,2,4,\ldots\}$.

Closure under $+$ and $\times$: $\checkmark$ (sum and product of even integers are even). Additive identity $0\in S$. $\checkmark$ Additive inverses: $-2k\in S$ for $2k\in S$. $\checkmark$ No zero divisors (inherits from $\mathbb Z$). $\checkmark$

But: no multiplicative identity — $1\notin S$, and no even number $e$ satisfies $e\cdot 2=2$ (would need $e=1$, which is odd).

Conclusion (ii): Not an integral domain (lacks multiplicative identity; $2\mathbb Z$ is a rng, not a ring with unity).

Part (iii) — $S=\mathbb Z^+=\{1,2,3,\ldots\}$.

No additive identity: $0\notin S$. No additive inverses: $-1\notin S$. So $(S,+)$ is not even a group.

Conclusion (iii): Not an integral domain (not a ring — fails additive identity and additive inverses).

$\boxed{\text{None of the three sets forms an integral domain.}}\;\blacksquare$

Common Traps

• Part (i): the product $(b_1\sqrt 2)(b_2\sqrt 2)=2b_1b_2$ is rational, not irrational. Students often assume “irrational times irrational is irrational” — this is false.
• Part (ii): even integers form a rng (ring without identity). UPSC convention requires a multiplicative identity for “ring” and “integral domain”; $2\mathbb Z$ fails this.
• Part (iii): the most fundamental failure is no additive identity (0 is not a positive integer) and no additive inverses. This means $\mathbb Z^+$ is not even closed under the group structure of $(\mathbb Z,+)$.
• If part (i) had been $\{a+b\sqrt 2 : a,b\in\mathbb Q\}=\mathbb Q(\sqrt 2)$, that is a field. The restriction to elements of the form $b\sqrt 2$ (no rational part) destroys closure under multiplication.

units-in-polynomial-ring (1 question(s); 2018)

Recognition Cues — “Let $R$ be an integral domain with unit; show any unit in $R[x]$ is a unit in $R$”; key words: integral domain, unit, polynomial ring $R[x]$.

Solution Template

1. State the degree additivity lemma: for $R$ a domain, $\deg(fg)=\deg f+\deg g$ (leading coefficients multiply to a non-zero element).
2. Let $f\in R[x]$ be a unit; choose $g\in R[x]$ with $fg=1$.
3. Apply degree: $\deg f+\deg g=\deg(fg)=\deg(1)=0$.
4. Since both degrees are non-negative integers summing to 0, both are 0. So $f=a_0$, $g=b_0$ are constants.
5. $fg=a_0b_0=1$ in $R$, so $a_0$ is a unit of $R$.

Worked Example

2018 Paper 2, 2018-P2-Q1a (10 marks)

Let $R$ be an integral domain with unit element. Show that any unit in $R[x]$ is a unit in $R$.

Step 1 — Degree additivity. For non-zero $f,g\in R[x]$ with leading coefficients $a_m,b_n$: the leading coefficient of $fg$ is $a_mb_n$. Since $R$ has no zero divisors, $a_mb_n\ne0$, so $\deg(fg)=\deg f+\deg g$.

Step 2 — Let $f\in R[x]$ be a unit. There exists $g\in R[x]$ with $fg=1$.

Step 3 — Degree equation. $\deg f+\deg g=\deg(fg)=\deg(1)=0$.

Step 4 — Both degrees are zero. Since $\deg f\ge0$ and $\deg g\ge0$ are non-negative integers summing to 0:

$\deg f=0\quad\text{and}\quad\deg g=0.$

So $f=a_0$ and $g=b_0$ are constant polynomials.

Step 5 — $a_0$ is a unit of $R$. From $fg=1$: $a_0b_0=1$ in $R$. Hence $a_0$ is a unit of $R$.

$\boxed{f(x)\in R[x]\text{ a unit}\;\Longrightarrow\;f\in R\text{ is a unit of }R.}\;\blacksquare$

(Converse: every unit of $R$ is trivially a unit of $R[x]$, so $U(R[x])=U(R)$.)

(Why the domain hypothesis is necessary: in $\mathbb Z_4[x]$, $(1+2x)^2=1+4x+4x^2=1$, so $1+2x$ is a unit of degree 1 — the argument breaks down because $\mathbb Z_4$ has zero divisors.)

Common Traps

• The single load-bearing fact is $\deg(fg)=\deg f+\deg g$, which requires $R$ to be a domain (otherwise the leading coefficients might multiply to zero). State this explicitly and say why.
• After concluding $\deg f=0$, you must still extract $a_0b_0=1$ from $fg=1$ to conclude $a_0$ is a unit of $R$ — don’t stop at “constant polynomial.”
• Provide the $\mathbb Z_4[x]$ counterexample or at least mention that without the domain hypothesis the result fails.

Marks-Aware Writing

For the 15-mark domain-test question: address each of the three sets separately with clear sub-headings. For each, identify the first axiom that fails — wasting time checking all axioms when the first fails loses time. State the conclusion explicitly for each set. The summary box and three clear conclusions account for the last 3 marks.

For the 10-mark units-in-$R[x]$ proof: the argument has five steps (degree lemma, unit equation, degree equation, both degrees zero, unit of $R$). Skipping the degree lemma and just asserting "$\deg f=0$" earns at most 5 marks. Include the $\mathbb Z_4$ counterexample to show the domain hypothesis is essential — this is worth 1–2 marks of “completeness.”

Practice Set

• 2025-P2-Q4a (15 m) — — Hint: use the evaluation homomorphism $f\mapsto f(0)$ and check $\mathbb Z[x]/\langle x\rangle\cong\mathbb Z$ is a domain but not a field.
• 2024-P2-Q4a (15 m) — — Hint: test the ring axioms and zero-divisor property for the specific ring given.