The ring of real functions. I have two ideals $I_1=\{ f(x)\in R | f(1)=0\}, I_2=\{ f(x)\in R| f(1)=f(2)=0\}$, where $R$ is the ring of real functions with pointwise multiplication and addition.
Are those Ideals maximal or not? if not then find an intermediate ideal: $I\ne I_i, R$ such that: $I_i\subset I\subset R$.
I think the first one is maximal and second one isn't.
For the first one suppose there exists $I\ne I_1,R$ such that $I_1\subset I\subset R$, then there exist $f(x)\in I$ such that $f(1)\ne 0$, but when adding to it $g(x)\in I_1$ we get $f(1)+g(1)\ne 0$, so $f(x)+g(x)\in I$.
Now I am not so sure, I think both aren't maximal; but then how to show this?
Thanks!
 A: Interesting ring, $R=C[0, 1]$
Define, $\phi_{a} : R\to \Bbb{R}$ by
$\phi_{a}(f) =f(a) $ .
Then, $\phi_{a} $ is a ring homomorphism.
$\ker(\phi_{a}) =\{f\in C[0,1] :\phi_{a}(f) =0\}= \{f\in C[0,1] : f(a) =0 \}$
We know that $I$ is an ideal of $R$ iff there exists a ring homomorphism from $R$ to a suitable ring (!) such that the kernel is $I$.
Now, we claim the ideal $I=\{f\in C[0,1] : f(a) =0 \}$ is maximal.
Clearly, $R/I \cong \Bbb{ R}$ [by 1st Isomorphism theorem ]
And since, $\Bbb{R}$ is a field, it follows that $I$ is a maximal ideal of $R$.

$\{f:C[0, 1]: f(a) =f(b) =0\} = \{f:C[0, 1]: f(a)=0\}\cap \{f:C[0, 1]: f(b) =0\}$
(Intersection of ideals is  again an ideal)
The ideal $\{f:C[0, 1]: f(a) =f(b) =0\}$ is not maximal ideal as
$\{f:C[0, 1]: f(a) =f(b) =0\}\subset \{f:C[0, 1]: f(a) =0\}$ is an ideal of $R$ .
A: Since $I_2 \subsetneq I_1 \subsetneq R$, obviously $I_2$ is not maximal.
Suppose that $I$ is an ideal properly containing $I_1$, so that $I_1 \subsetneq I \subseteq R$, and let $g$ be a real function in $I \setminus I_1$. This means $g(1) \neq 0$.
Let $f$ be any real function in $R$. We can scale $g$ to match the value of $f$ at $x=1$:
$$ h(x) = \frac{f(1)}{g(1)} g(x) $$
Since $g \in I$ and $\frac{f(1)}{g(1)}$ is just a constant real number, $h \in I$. Then we can write $f$ as
$$ f = (f-h) + h $$
Since $h(1) = f(1)$, $(f-h)(1) = 0$, and so $f-h \in I_1 \subset I$. Since both $f-h$ and $h$ are in $I$, their sum $f$ is in also $I$. This has shown that every real function $f$ is in $I$, so $I=R$. In other words, $I_1 \subsetneq I \subsetneq R$ is impossible; $I_1$ is maximal.
