Let F be the ring of continuous functions from $R$ to $R$. Let $A =\{f \in F \; | \; f(0)=0\}$ Show that $A$ is maximal in R.

Question

Let F be the ring of continuous functions from $R$ to $R$. Let $A =\{f \in F \; | \; f(0)=0\}$ Show that $A$ is maximal in R.

Please check my attempt below:- (This question has been posted already but I haven't studied homomorphism as of now, I just need to know if my attempt is fine)

First of all, it can be shown that $A=<x>$ ,i.e, it is the Principle Ideal generated by $x \in F$ and hence it is an Ideal of $F$

To show it is maximal, I am showing $\frac{F }{<x>}$ is field

Since $F$ is a commutative ring with unity $= 1$ $\rightarrow \frac{F}{<x>}$ is commutative ring with unity $= 1 + <x>$

Consider, $\; x\in <x>$

$\rightarrow x + <x> = <x> \; \rightarrow x = 0 \;$ in $\frac{F }{<x>}$

$\frac{F}{<x>} = \{ f \; \;+\; <x> \; \; | \; \; f \in F \} = \{\; k \; + \; <x> \; | \; k \in R\}$ (where $R$ is the set of real functions)

Consider non zero element $k + A \; \in \frac{F}{<x>} \; \rightarrow k\neq 0$ $\; \; \exists \frac1k \; \in R \; \;$ such that $(k+<x>)(\frac1k + <x>) = 1 + <x> = (\frac1k + <x>)(k+<x>)$

Hence $\frac{F}{<x>} $ is field and $<x> $ is maximal.

I am not very sure if this is correct because my textbook has shown $A$ as maximal directly without showing the factor ring is field.

Answer 1

Actually $A$ is not the ideal generated by $x$, in the algebraic sense. It is however the closed ideal generated by $x$ where $F$ is endowed with a suitable topology, but I guess this is not really what you want.

When you take a nonzero element $k+A\in F/A$, $k$ is a function, not a number, so you can't simply take $1/k\in\mathbb{R}$.

But your approach can work: notice that $A=\ker T$ where $T:F\to\mathbb{R}$ is the epimorphism $T(f)=f(0)$ and use the first isomorphism theorem to conclude that $F/A\simeq \mathbb{R}$ is a field.