← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q3b — Step-by-Step Solution

15 marks · Section A

Ring homomorphisms; quotient rings · Algebra · asked 3× in 13 yrs · Read the full method →

Question

Let $R^C=$ ring of all real valued continuous functions on $[0,1]$, under the operations $(f+g)x=f(x)+g(x)$, $(fg)x=f(x)g(x)$. Let $M=\left\{f\in R^C\;\middle|\;f\!\left(\tfrac{1}{2}\right)=0\right\}$. Is $M$ a maximal ideal of $R$? Justify your answer.

Technique

Recognise $M$ as the kernel of an evaluation homomorphism to a field; First Isomorphism Theorem + “maximal iff quotient is a field.”

Solution

Claim. $M$ is a maximal ideal of $R^C$.

Strategy. Show $M$ is the kernel of the evaluation homomorphism $\mathrm{ev}_{1/2}:R^C\to\mathbb R$. Then by the First Isomorphism Theorem, $R^C/M\cong\mathbb R$. Since $\mathbb R$ is a field, $M$ is maximal.

Step 1 — Evaluation at $1/2$ is a ring homomorphism

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

Additivity: $\phi(f+g)=(f+g)(1/2)=f(1/2)+g(1/2)=\phi(f)+\phi(g)$ ✓.

Multiplicativity: $\phi(fg)=(fg)(1/2)=f(1/2)\,g(1/2)=\phi(f)\,\phi(g)$ ✓.

Sends $1$ to $1$: $\phi(1)=1$ (where $1\in R^C$ is the constant function 1) ✓.

So $\phi$ is a ring homomorphism.

Step 2 — $\phi$ is surjective

For any $c\in\mathbb R$, take the constant function $f\equiv c$. Clearly $f\in R^C$ (constants are continuous) and $\phi(f)=c$.

So $\mathrm{Im}(\phi)=\mathbb R$.

Step 3 — $\ker\phi=M$

By definition, $\ker\phi=\{f\in R^C:\phi(f)=0\}=\{f\in R^C:f(1/2)=0\}=M$.

Step 4 — First Isomorphism Theorem and maximality

By the First Isomorphism Theorem,

$\mathbb R$ is a field. An ideal $I\subset R$ in a commutative ring with unity is maximal iff $R/I$ is a field. Hence $M$ is a maximal ideal of $R^C$.

Answer