I would like to compile all questions I have encountered with Ideals in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions and ask if there are any gaps.
Question is to see if :
the ideal $\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=0\}$ is maximal ideal in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions.
What I have done so far is :
consider $\eta :\mathcal{C}[0,1]\rightarrow \mathbb{R}$ where $f\rightarrow f(0)$
$Ker (\eta) = \{ f\in \mathcal{C}[0,1] : f(0)=0\}$
I could see that for each $r\in \mathbb{R}$, I would set $f$ such that $f(x)=x+r$
So, $\eta$ is surjective.
So, I would get $\mathcal{C}[0,1]/\mathcal{I}\cong \mathbb{R}$
As $\mathbb{R}$ is a field so ia the quotient $\mathcal{C}[0,1]/\mathcal{I}$ which means that $\mathcal{I}$ is an ideal in $\mathcal{C}[0,1]$.
Now Another Question on same idea :
Question is to see if :
the ideal $\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=f(1)=0\}$ is maximal ideal in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions.
I thought of using same ideas as above but i am unable to choose perfect $\eta$
I believe that this is not even a prime ideal..
I set $f(x)=x$ and $g(x)=x-1$ and consider $f(x)g(x)=(x)(x-1)$ then,
$(fg)(0)=f(0)g(0)=0.(-1)=0$
$(fg)(1)=f(1)g(1)=1.0=0$
So, $fg\in \mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=f(1)=0\}$ but the neither $f$ nor $g$ is in $\mathcal{I}$.
I have two questions:
I see that $0$ does not play any role in the question $\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=0\}$ I mean I would have same case if i replace $0$ by any real number.
So, I would like to say that $$\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(r)=0\}$$ is a maximal ideal in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions.
And one more thing I would claim is that
If the collection $\mathcal{I}$ is such that all elements of $\mathcal{I}$ has more than $1$ common zero then $\mathcal{I}$ is not a maximal ideal and not even a prime ideal..
I would like to learn more about this kind of ideals in $\mathcal{C}[0,1]$.
I would be thankful if some one can assure that what have done is sufficient to conlcude what i have concluded and I would be thankful if some one can suggest some material to read regarding this.
Thank you
Best Answer
What you say is correct and is true in more generality. You should be interested by the following result:
Here are some hints to prove this theorem:
1) Determine the invertibles of $A$ (easy).
2) For any $x$ in $K$, show that the set of functions of $A$ vanishing in $x$ is a maximal ideal of $A$ (easy).
3) Conversely, let $I$ be a maximal ideal of $A$. Suppose by contradiction that for any $x$ in $K$, you can find a function $f_x$ in $A$ which does not vanish at $x$. By continuity, there is a neighborhood $U_x$ of $x$ such that $f_x$ does not vanish on $U_x$.
4) Using compactness, construct a function of $A$ that vanishes nowhere and conclude (clever). Hint: in the real numbers, a sum $\sum \lambda_i^2$ vanishes if, and only if, all $\lambda_i$ vanish.
Remark: adapt the proof, for functions with values in $\mathbb{C}$.
Edit: concerning the question of primality of ideals of functions vanishing at more than one point, you can prove this as follows:
Let $I_C$ be the set of functions of $A$ vanishing one some closed set $C$ of $[0,1]$. Write $C = X \cup Y$ where $X$ and $Y$ are closed proper subsets of $[0,1]$ in $C$ (this is always possible if $C$ has at least two points x < y: take for $X$ the set of points $t \in A$ with $t \leq \frac{x+y}{2}$ and for $Y$ the set of points $t \in A$ with $t \geq \frac{x+y}{2}$).
Let $f$ and $g$ be functions in $A$ vanishing exactly on $X$ and $Y$. For instance, you can take the distance functions to $X$ and $Y$. Then $f.g$ is in $I_C$ but neither $f$ nor $g$ are in $I_C$.
This can certainly be generalized for a more general topological space, but maybe the compactness condition is not sufficient.