Ring homomorphisms; quotient rings

At a Glance

• Frequency: 3 sub-parts across 3 of 14 years (2013, 2015, 2025)
• Priority tier: T3
• Marks (count): 15 (3)
• Average solve time: ~8 min
• Difficulty mix: medium 2, easy 1
• Section: A | Dominant type: proof

Why This Chapter Matters

All three UPSC ring homomorphism questions are 15-mark proofs, making this the single highest-yield ring theory topic by total marks. They share a common engine: exhibit an evaluation homomorphism, identify its kernel as the target ideal, apply FIT to get the quotient ring, then classify it as a field (maximal ideal) or integral domain (prime ideal) or neither. Once you have this pipeline in muscle memory, all three archetypes — and any future variant — are solved with the same four-step argument.

Minimum Theory

Ring homomorphism. A map $\phi:R\to R'$ is a ring homomorphism if $\phi(a+b)=\phi(a)+\phi(b)$ and $\phi(ab)=\phi(a)\phi(b)$ for all $a,b\in R$. Many definitions also require $\phi(1_R)=1_{R'}$ (unital); UPSC questions sometimes ask you to prove this from surjectivity.

Ideal and quotient ring. A subset $I\subseteq R$ is an ideal if $(I,+)$ is a subgroup of $(R,+)$ and $rI\subseteq I$, $Ir\subseteq I$ for all $r\in R$. The quotient ring $R/I=\{r+I:r\in R\}$ is a ring under $(r+I)+(s+I)=(r+s)+I$ and $(r+I)(s+I)=rs+I$.

Key characterizations (commutative ring with unity). (i) $I$ is maximal $\iff R/I$ is a field. (ii) $I$ is prime $\iff R/I$ is an integral domain.

FIT for rings. If $\phi:R\to R'$ is a surjective ring homomorphism, then $R/\ker\phi\cong R'$.

Question Archetypes

ring-hom-property (1 question(s); 2015)

Recognition Cues — “Prove $\phi(1)$ is the unit element of $R'$”; or “prove a surjective ring homomorphism preserves [property]”; the word onto or surjective always appears — it is the load-bearing hypothesis.

Solution Template

1. Let $r'\in R'$ be arbitrary.
2. By surjectivity, choose $r\in R$ with $\phi(r)=r'$.
3. Compute $\phi(1)\cdot r' = \phi(1)\cdot\phi(r) = \phi(1\cdot r) = \phi(r) = r'$.
4. Compute $r'\cdot\phi(1) = \phi(r)\cdot\phi(1) = \phi(r\cdot 1) = \phi(r) = r'$.
5. Since $\phi(1)\cdot r'=r'=r'\cdot\phi(1)$ for all $r'\in R'$, $\phi(1)=1_{R'}$.

Worked Example

2015 Paper 2, 2015-P2-Q2a (15 marks)

If $R$ is a ring with unit element $1$ and $\phi$ is a homomorphism of $R$ onto $R'$, prove that $\phi(1)$ is the unit element of $R'$.

Let $r'\in R'$ be arbitrary. By surjectivity, choose $r\in R$ with $\phi(r)=r'$. Then:

$\phi(1)\cdot r' = \phi(1)\cdot\phi(r) = \phi(1\cdot r) = \phi(r) = r'.$

$r'\cdot\phi(1) = \phi(r)\cdot\phi(1) = \phi(r\cdot 1) = \phi(r) = r'.$

Since $\phi(1)$ acts as a two-sided identity on every element of $R'$, and identities are unique, $\phi(1)=1_{R'}$.

$\boxed{\phi(1)\text{ is the unit element of }R'.}\;\blacksquare$

Why surjectivity is essential. Without it, $\phi(1)$ is only an identity for the subring $\phi(R)\subseteq R'$. Counterexample: $\phi:\mathbb Z\to\mathbb Z\times\mathbb Z$, $\phi(n)=(n,0)$, is a non-surjective homomorphism with $\phi(1)=(1,0)\ne(1,1)=1_{\mathbb Z\times\mathbb Z}$.

Common Traps

• Do not use $\phi(1)=1_{R'}$ as the definition of a ring homomorphism — the question asks you to prove it.
• Check both sides: $\phi(1)\cdot r'=r'$ and $r'\cdot\phi(1)=r'$. For non-commutative $R'$, one side alone is not enough.
• Surjectivity is used exactly once: to find a preimage of the arbitrary $r'$.

maximal-ideal-quotient (1 question(s); 2013)

Recognition Cues — an ideal defined by a vanishing condition at a point, e.g. $M=\{f\in C[0,1]:f(c)=0\}$; asked whether $M$ is a maximal ideal; the ring is typically a function ring.

Solution Template

1. Define the evaluation homomorphism $\phi:R\to\mathbb R$ (or $\mathbb F$) by $\phi(f)=f(c)$.
2. Show $\phi$ is a ring homomorphism (additive and multiplicative).
3. Show $\phi$ is surjective.
4. Show $\ker\phi = M$ (elements vanishing at $c$).
5. Apply FIT: $R/M\cong\mathbb R$. Since $\mathbb R$ is a field, $M$ is a maximal ideal.

Worked Example

2013 Paper 2, 2013-P2-Q3b (15 marks)

Let $R^C$ be the ring of all real-valued continuous functions on $[0,1]$ under pointwise addition and multiplication. Let $M=\{f\in R^C : f(1/2)=0\}$. Is $M$ a maximal ideal? Justify.

Step 1 — Evaluation homomorphism. Define $\phi:R^C\to\mathbb R$ by $\phi(f)=f(1/2)$.

Additive: $\phi(f+g)=(f+g)(1/2)=f(1/2)+g(1/2)=\phi(f)+\phi(g)$. $\checkmark$

Multiplicative: $\phi(fg)=(fg)(1/2)=f(1/2)g(1/2)=\phi(f)\phi(g)$. $\checkmark$

Sends $1$ to $1$: $\phi(1)=1$ (constant function). $\checkmark$

Step 2 — Surjectivity. For any $c\in\mathbb R$, the constant function $f\equiv c$ satisfies $\phi(f)=c$. So $\operatorname{Im}\phi=\mathbb R$.

Step 3 — Kernel. $\ker\phi=\{f:f(1/2)=0\}=M$.

Step 4 — FIT. $R^C/M = R^C/\ker\phi \cong \mathbb R$.

Step 5 — Maximality. $\mathbb R$ is a field. In a commutative ring with unity, $M$ is maximal $\iff R^C/M$ is a field. Hence $M$ is maximal.

$\boxed{M\text{ is a maximal ideal of }R^C,\ \text{with }R^C/M\cong\mathbb R.}\;\blacksquare$

Common Traps

• Must verify all three parts of “homomorphism”: additivity, multiplicativity, and $\phi(1)=1$. Omitting multiplicativity loses marks.
• The characterisation used is: $I$ maximal $\iff R/I$ a field (in a commutative ring with unity). State it explicitly.
• $M$ is also a prime ideal (since $\mathbb R$ is also an integral domain), but the question asks for maximality — the field criterion suffices.

prime-maximal-via-hom (1 question(s); 2025)

Recognition Cues — evaluation homomorphism on a polynomial ring ($\mathbb Z[x]$, $\mathbb Q[x]$); asked to prove a principal ideal $\langle f\rangle$ is prime but not maximal; the quotient is a domain but not a field.

Solution Template

1. Define the evaluation homomorphism $\phi:R[x]\to R$ by $\phi(f)=f(a)$.
2. Verify it is a (unital) ring homomorphism.
3. Show $\phi$ is surjective.
4. Identify $\ker\phi = \langle x-a\rangle$ (polynomials with root $a$).
5. FIT: $R[x]/\langle x-a\rangle\cong R$.
6. If $R$ is an integral domain but not a field: $\langle x-a\rangle$ is prime (but not maximal).

Worked Example

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

Examine whether $\phi:\mathbb Z[x]\to\mathbb Z$ defined by $\phi(f(x))=f(0)$ is a homomorphism. Deduce that $\langle x\rangle$ is a prime ideal in $\mathbb Z[x]$, but not a maximal ideal.

Step 1 — Homomorphism. For $f,g\in\mathbb Z[x]$:

$\phi(f+g)=(f+g)(0)=f(0)+g(0)=\phi(f)+\phi(g),$

$\phi(fg)=(fg)(0)=f(0)g(0)=\phi(f)\phi(g),\quad\phi(1)=1.$

So $\phi$ is a (unital) ring homomorphism.

Step 2 — Surjectivity. For any $n\in\mathbb Z$, the constant polynomial $f(x)=n$ gives $\phi(f)=n$. So $\operatorname{Im}\phi=\mathbb Z$.

Step 3 — Kernel. $\ker\phi=\{f\in\mathbb Z[x]:f(0)=0\}=\{f:\text{constant term }=0\}=\{xg(x):g\in\mathbb Z[x]\}=\langle x\rangle$.

Step 4 — FIT. $\mathbb Z[x]/\langle x\rangle\cong\mathbb Z$.

Step 5 — $\langle x\rangle$ is prime. $\mathbb Z[x]/\langle x\rangle\cong\mathbb Z$ is an integral domain (no zero divisors). Hence $\langle x\rangle$ is a prime ideal.

Step 6 — $\langle x\rangle$ is not maximal. $\mathbb Z$ is not a field ($2$ has no inverse in $\mathbb Z$). So $\mathbb Z[x]/\langle x\rangle\not\cong$ a field, and $\langle x\rangle$ is not maximal. (Witness: $\langle x\rangle\subsetneq\langle 2,x\rangle\subsetneq\mathbb Z[x]$.)

$\boxed{\langle x\rangle\text{ is prime but not maximal in }\mathbb Z[x].}\;\blacksquare$

Common Traps

• Identifying the kernel exactly as $\langle x\rangle$ (polynomials with zero constant term = divisible by $x$) is the technical heart; state this clearly.
• Confusing the two characterisations: prime $\iff R/I$ is an integral domain; maximal $\iff R/I$ is a field. $\mathbb Z$ is a domain but not a field.
• For non-maximality, state a concrete witness: the ideal $\langle 2,x\rangle$ strictly contains $\langle x\rangle$ and is strictly smaller than $\mathbb Z[x]$.

Marks-Aware Writing

All three questions are 15 marks. The common structure is: homomorphism verified (3 marks), surjectivity (2 marks), kernel identified (3 marks), FIT stated and applied (3 marks), ideal classified (4 marks). For prime/maximal classification, you must state the relevant theorem (“prime $\iff$ domain, maximal $\iff$ field”) and apply it — a bare assertion without the theorem earns at most half marks.

Practice Set

• 2023-P2-Q3a (15 m) — — Hint: use an evaluation or natural projection homomorphism; identify kernel and quotient.
• 2016-P2-Q1a (10 m) — — Hint: verify ring axioms; check whether the ideal is maximal via the quotient.