Let $R$ be the ring all continous functions from $[0,1]$ to $\mathbb{R}$. Let $I$ be the ideal generated by $f(x)$ where $f(x)=x^2+1$ .Then
$(a) I$ is maximal ideal of $R$
$(b) I$ is prime ideal of $R$.
$(c) \frac RI $ is finite
$(d)char\frac RI$ is a prime number.
My attempt:-
For $(a)$
$I$ is not a maximal ideal since every maximal ideal of $R$ is of the form $\{f\in R: f(c)=0\}$, for some $c\in \mathbb{R}$, and the given polynomial doesn't vanish anywhere.
But I have one question here: $R$ is a ring with identity ,namely $f(x)=1, \forall x\in [0,1]$, so every proper ideal is contained in a maximal ideal.
But how can a non-vanishing function be in the set of functions vanishing at a particular point?[or if the question is simply put , in which maximal ideal is $I$ contained?]
For $(c)$
We have $\{ax+b+\langle x^2+1\rangle : a,b \in \mathbb{R}\}\subset\frac RI$ thus $\frac RI$ is not finite.
I have no idea about other options
Am I thinking properly? Please help me clear my doubts and answer rest of the parts.
Thanks for your valuable time.
You actually have $I = R$ since the generator is invertible with inverse $x \mapsto \dfrac{1}{x^2+1}$.
(a) Your observation that it not a maximal ideal was correct. However, in this case, the reason is that it is not even a proper ideal!
This is precisely why you struggled with the last part of the doubt. As you noted, every proper ideal is part of a maximal ideal.
(c) We clearly know that $R/I$ is finite now. The error in your reasoning is that you were probably thinking that you are in the polynomial ring.
In that case, $$ax + b + \langle x^2 + 1\rangle \neq a'x + b' + \langle x^2 + 1\rangle$$ iff $(a, b) \neq (a', b')$.
However, here you don't have that anymore.
(b) and (d) should be answered now based on the knowledge that $I = R$.
But just for completeness -
(b) $I$ is not prime since it is not proper.
(d) ${\rm char}\ R/I = 1$, not a prime.