$f(x)$ such that $f (0)$ is not $0$ but $f(1) = 0$. So $(f(x) + I)$ is an element. But how would this element have an inverse

Question

The ideal $I = \{f \mid f (0) = 0\}$ in the ring $C [0, 1]$ of all continuous real valued functions on the interval $[0, 1]$ is a maximal ideal..But I can not understand how do it's elements look like.

There must be one element $g(x)$ such that $g (0)$ is not $0$ but $g(1) = 0$. So $(g(x) + I)$ is an element. But how would this element have an inverse?(We know that the $(C[0,1] / I)$ is a field as I is Maximal and $C[0,1]$ is commutative Ring with Unity.)

Answer 1

The unity in the ring $C[0,1]/I$ is $(1+I)$, meaning the set $1+f(x)$ where $f(0)=0$.

You may notice that $$g(x) \in (1+I) \iff g(x) = 1 + f(x) : f(x)\in I\iff g(x) -1\in I$$ The last condition, in particular, means that $$ g(x) -1\in I \iff g(0) -1 = 0 \iff g(0) =1 $$ so $g(x)$ is a unit function if and only if $g(0)=1$.

Given now any $h(x)$ such that $h(0) = c \ne 0$, it belongs to the class $(h(x)+I)$, and you're looking for a function $h'(x)$ such that $$(h(x)+I)*(h'(x)+I) = (h(x)h'(x) + I)= (1+I)$$ that is $$ h(x)h'(x) \in (1+I). $$ Considering what we have shown, we're looking for a function $h'(x)$ such that $h(0)h'(0)=1$. Can you conclude?