Ideals in ring of continuous functions $\mathcal{C}[0,1]$ ... NBHM- Algebra

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

Solution 1:

What you say is correct and is true in more generality. You should be interested by the following result:

Let $K$ be a topological compact space. Then, the maximal ideals of $A := \mathcal{C}(K, \mathbb{R})$ are exactly the ideals of functions vanishing at a fixed point of $K$.

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.