Subrings and ideals

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2015, 2022, 2023, 2024)
• Priority tier: T3
• Marks (count): 10 (2), 15 (2)
• Average solve time: ~8 min
• Difficulty mix: easy 2, medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Ideal questions in Section A reward knowledge of three templates: (a) give an example of a ring with a subring having a different identity; (b) show a set is an ideal, identify it as a principal ideal, and analyse the quotient ring; (c) use the First Isomorphism Theorem for rings to prove an ideal is prime (quotient is an integral domain) but not maximal (quotient is not a field). All three appear in the 10–15 mark range and are algorithmic once you know which template applies.

Minimum Theory

Ideal. A non-empty subset $I$ of a ring $R$ is a (two-sided) ideal if: (1) $a-b\in I$ for all $a,b\in I$; (2) $ra,ar\in I$ for all $r\in R$, $a\in I$.

Principal ideal. The ideal generated by a single element $a\in R$: $\langle a\rangle=\{ra : r\in R\}$ (in a commutative ring).

Quotient ring. If $I$ is an ideal of $R$, the quotient $R/I$ is a ring with cosets as elements.

Prime vs maximal ideal.

• $I$ is prime $\iff$ $R/I$ is an integral domain.
• $I$ is maximal $\iff$ $R/I$ is a field.
• Every maximal ideal is prime; the converse fails (e.g. $\langle x\rangle$ in $\mathbb{Z}[x]$).

First Isomorphism Theorem (rings). If $\phi:R\to S$ is a surjective ring homomorphism with $\ker\phi=I$, then $R/I\cong S$.

Idempotent subring trick. In $\mathbb{Z}_n$, an element $e$ with $e^2=e$ (idempotent, $e\ne0,1$) generates a subring $\{0,e,2e,\ldots\}$ with identity $e\ne1_R$. This gives the standard example of a subring with a different identity.

Question Archetypes

ring-example (1 question(s); 2015)

Worked Example

2015 Paper 2, 2015-P2-Q1b (10 marks)

Give an example of a ring having identity but a subring of this having a different identity.

Take $R=\mathbb{Z}_6$ (identity $1$) and $S=\{0,2,4\}\subset R$.

$S$ is a subring: $4+4=8\equiv2\in S$; $2\cdot4=8\equiv2\in S$; $4\cdot4=16\equiv4\in S$, etc. Closure under $+$ and $\cdot$ holds.

Identity of $S$: $4\cdot0=0$, $4\cdot2=8\equiv2$, $4\cdot4=16\equiv4$ — so $4$ acts as the multiplicative identity of $S$.

Since $4\ne1=1_R$, $R$ and $S$ have different identities. $\blacksquare$

Recognition pattern for other examples: in $\mathbb{Z}_n$ for composite $n$, any idempotent $e\ne0,1$ (satisfying $e^2\equiv e\pmod n$) generates a subring $eR$ with identity $e$. In $\mathbb{Z}_6$: $4^2=16\equiv4$ ✓ and $3^2=9\equiv3$ ✓ (so $\{0,3\}$ also works with identity $3$).

Common Traps

• Don’t confuse “subring” conventions. Some authors require a subring to contain the parent ring’s identity; UPSC uses the broader definition (closed under $+,\cdot$, with its own identity that may differ).
• Check idempotent: $e^2=e$. Without this, $e$ need not act as an identity on the generated subring.

ideal-quotient (1 question(s); 2022)

Worked Example

2022 Paper 2, 2022-P2-Q4a (15 marks)

Let $S=\{f\in\mathbb{R}[x]:f(0)=f(1)=0\}$. Prove $S$ is an ideal of $\mathbb{R}[x]$. Is $\mathbb{R}[x]/S$ an integral domain?

$S$ is an ideal: $0\in S$; if $f,g\in S$ then $(f-g)(0)=(f-g)(1)=0$, so $f-g\in S$; if $f\in S$ and $h\in\mathbb{R}[x]$, then $(hf)(0)=h(0)\cdot0=0$ and $(hf)(1)=h(1)\cdot0=0$, so $hf\in S$. ✓

Identify $S$: $f(0)=f(1)=0$ iff both $x$ and $(x-1)$ divide $f$. Since $\gcd(x,x-1)=1$, this means $x(x-1)\mid f$. So $S=\langle x(x-1)\rangle=\langle x^2-x\rangle$.

$\mathbb{R}[x]/S$ is NOT an integral domain: the generator $x^2-x=x(x-1)$ is reducible, so the quotient has zero divisors: $\bar{x}\cdot\overline{(x-1)}=\overline{x(x-1)}=\bar{0}$ but $\bar{x}\ne\bar{0}$ and $\overline{(x-1)}\ne\bar{0}$. ($\mathbb{R}[x]/(x^2-x)\cong\mathbb{R}\times\mathbb{R}$ by CRT.) $\blacksquare$

Common Traps

• Closure under right-multiplication by $R$. An ideal requires $hf\in I$ for all $h\in R$, not just for $h\in I$. Write the proof for an arbitrary $h\in\mathbb{R}[x]$.
• Identifying the generator. Both $x$ and $x-1$ must divide $f$ because the conditions are $f(0)=0$ AND $f(1)=0$ simultaneously. Don’t stop at ”$x\mid f$.“

prime-maximal-ideal (1 question(s); 2025)

Worked Example

2025 Paper 2, 2025-P2-Q4a (15 marks)

Show $\phi:\mathbb{Z}[x]\to\mathbb{Z}$, $\phi(f)=f(0)$, is a homomorphism. Deduce $\langle x\rangle$ is prime but not maximal in $\mathbb{Z}[x]$.

Homomorphism: $\phi(f+g)=(f+g)(0)=f(0)+g(0)=\phi(f)+\phi(g)$ and $\phi(fg)=f(0)g(0)=\phi(f)\phi(g)$, and $\phi(1)=1$. ✓

Surjectivity: the constant polynomial $n$ maps to $n\in\mathbb{Z}$.

Kernel: $f\in\ker\phi\iff f(0)=0\iff x\mid f\iff f\in\langle x\rangle$.

First Isomorphism Theorem: $\mathbb{Z}[x]/\langle x\rangle\cong\mathbb{Z}$.

• $\mathbb{Z}$ is an integral domain $\Rightarrow$ $\langle x\rangle$ is prime.
• $\mathbb{Z}$ is not a field ($2$ has no inverse) $\Rightarrow$ $\langle x\rangle$ is not maximal. $\blacksquare$

(Concrete witness of non-maximality: $\langle x\rangle\subsetneq\langle2,x\rangle\subsetneq\mathbb{Z}[x]$, where $\langle2,x\rangle=\{f:f(0)\text{ even}\}$.)

Common Traps

• Surjectivity must be verified. Without confirming $\phi$ is onto, the First Isomorphism Theorem only gives $\mathbb{Z}[x]/\langle x\rangle\cong\operatorname{Im}\phi$, not $\cong\mathbb{Z}$.
• “Not maximal” needs a witness or the field criterion. Saying ”$\mathbb{Z}$ is not a field” is the clean argument; alternatively, exhibit the chain $\langle x\rangle\subsetneq\langle2,x\rangle\subsetneq\mathbb{Z}[x]$.
• Prime $\ne$ maximal in general. In $\mathbb{Z}$ (a PID), every nonzero prime is maximal. In $\mathbb{Z}[x]$ (not a PID by Krull dimension 2), there are prime ideals that are not maximal — $\langle x\rangle$ is the canonical example.

Marks-Aware Writing

For a 10-mark example question: give the ring, identify the subring, construct the multiplication table of the subring (or verify closure by explicit computation), and identify the identity. Three lines of explicit arithmetic plus the conclusion.

For a 15-mark ideal+quotient proof: three parts — (a) ideal axioms verified explicitly, (b) the ideal identified as principal with the generator named, (c) the quotient analysis with zero divisors exhibited or the CRT structure stated. Skipping any of the three parts costs 4–5 marks.

Practice Set

• 2023-P2-Q1b (10 m) — — Hint: Bezout’s theorem; $\gcd(4,6)=2$, so $\langle4\rangle+\langle6\rangle=\langle2\rangle$ in $\mathbb{Z}$.