UPSC Maths 2025 Paper 2 Q4a — Solution

15 marks · Section A

Question

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

Technique

Recognise $\phi$ as the evaluation homomorphism $\text{ev}_0$ , apply the First Isomorphism Theorem, then use the criteria $R/I$ is an integral domain $\iff I$ prime, and $R/I$ is a field $\iff I$ maximal.

Solution

Step 1 — $\phi$ is a ring homomorphism. For $f,g\in\mathbb{Z}[x]$ , $\phi(f)=f(0)$ is the constant term. Evaluation at $0$ respects addition and multiplication: $\phi(f+g) = (f+g)(0) = f(0)+g(0) = \phi(f)+\phi(g),$ $\phi(fg) = (fg)(0) = f(0)\,g(0) = \phi(f)\,\phi(g),$ and $\phi(1) = 1$ . Hence $\phi$ is a unital ring homomorphism.

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

Step 3 — Kernel of $\phi$ . $\ker\phi = \{f\in\mathbb{Z}[x] : f(0)=0\} = \{f : \text{constant term} = 0\} = \langle x\rangle.$ Indeed $f(0)=0$ iff $f$ has no constant term iff $x \mid f$ in $\mathbb{Z}[x]$ .

Step 4 — First Isomorphism Theorem. Since $\phi$ is a surjective homomorphism with kernel $\langle x\rangle$ , $\mathbb{Z}[x]/\langle x\rangle \;\cong\; \mathbb{Z}.$

Step 5 — $\langle x\rangle$ is prime. A quotient $R/I$ is an integral domain iff $I$ is prime. Here $\mathbb{Z}[x]/\langle x\rangle \cong \mathbb{Z}$ , and $\mathbb{Z}$ is an integral domain. Therefore $\langle x\rangle$ is a prime ideal.

Step 6 — $\langle x\rangle$ is not maximal. A quotient $R/I$ is a field iff $I$ is maximal. Here $\mathbb{Z}[x]/\langle x\rangle \cong \mathbb{Z}$ , and $\mathbb{Z}$ is not a field (e.g. $2$ has no inverse in $\mathbb{Z}$ ). Therefore $\langle x\rangle$ is not maximal.

(Concretely, $\langle x\rangle$ is properly contained in the larger proper ideal $\langle 2, x\rangle$ — the polynomials with even constant term — which strictly contains $\langle x\rangle$ and is strictly contained in $\mathbb{Z}[x]$ , witnessing non-maximality. Here $\mathbb{Z}[x]/\langle2,x\rangle\cong\mathbb{F}_2$ , a field, so $\langle 2,x\rangle$ is maximal.)

Answer

$\phi:\mathbb{Z}[x]\to\mathbb{Z},\ f\mapsto f(0)$ is a surjective ring homomorphism with $\ker\phi=\langle x\rangle$ .
By the 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 $\Rightarrow \langle x\rangle$ is not maximal.

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