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UPSC 2024 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Subrings and ideals · Algebra · asked 4× in 13 yrs · Read the full method →

Question

Let $S=(x)$ be the ideal of $\mathbb{Z}[x]$ generated by $x$. Show that $S$ is a prime ideal but not a maximal ideal of $\mathbb{Z}[x]$.

Technique

Identify $\mathbb Z[x]/(x)\cong\mathbb Z$ via the evaluation map; use the equivalences: prime $\iff$ quotient is integral domain, maximal $\iff$ quotient is field.

Solution

Key equivalences.

• $S$ is prime $\iff\mathbb Z[x]/S$ is an integral domain.
• $S$ is maximal $\iff\mathbb Z[x]/S$ is a field.

Step 1 — Identify $\mathbb Z[x]/(x)$

The evaluation map $\varepsilon_0:\mathbb Z[x]\to\mathbb Z$, $p(x)\mapsto p(0)$, is a surjective ring homomorphism with $\ker\varepsilon_0=\{p(x):p(0)=0\}=(x)=S$.

By the First Isomorphism Theorem:

Step 2 — $S$ is prime

$\mathbb Z$ is an integral domain (no zero-divisors), so $S$ is a prime ideal.

Step 3 — $S$ is not maximal

$\mathbb Z$ is not a field ($2$ has no multiplicative inverse in $\mathbb Z$), so $S$ is not a maximal ideal.

Explicitly: $(x,2)\supsetneq(x)$ is a strictly larger proper ideal (since $2\notin(x)$ but $1\notin(x,2)$), confirming $(x)$ is not maximal.

Answer