1. ## Maximal Ideal

Let $R$ be the ring of all real valued continuous functions on the closed unit interval. If $M$ is a maximal ideal of $R$, prove that there exists a real number $\gamma, \ 0 \leq \gamma \leq 1$ such that $M=M_{\gamma}=\{ f(x) \in R \ | \ f(\gamma)=0\}$

2. Originally Posted by Chandru1
Let $R$ be the ring of all real valued continuous functions on the closed unit interval. If $M$ is a maximal ideal of $R$, prove that there exists a real number $\gamma, \ 0 \leq \gamma \leq 1$ such that $M=M_{\gamma}=\{ f(x) \in R \ | \ f(\gamma)=0\}$
First, notice that each of the sets $M_{\gamma}$ is a maximal ideal in $R$. In fact, if 1 denotes the constant function 1 then for any $f\in R$ we can write $f = (f-f(\gamma)1) + f(\gamma)1$. Since $f-f(\gamma)1\in M_{\gamma}$, it follows that $M_{\gamma}$ (which is trivially an ideal) has codimension 1 in $R$ and is therefore maximal.

Now let $M$ be a maximal ideal of $R$. If there exists $\gamma\in[0,1]$ such that each element of $M$ vanishes at $\gamma$, then $M$ is contained in (and hence by maximality is equal to) $M_{\gamma}$, and we are done.

So suppose that there is no such $\gamma$. Then for each $t\in[0,1]$ there exists $f_t\in M$ such that $f_t(t)\ne0$. The set $\{s\in[0,1]:f_t(s)\ne0\}$ is an open neighbourhood of t. These neighbourhoods cover the unit interval, so by compactness there is a finite subcover, given by elements $f_{t_n}$ of $M$. Then $\textstyle\sum f_{t_n}^2$ is an element of $M$ that is strictly positive throughout the interval and hence invertible. But an ideal containing an invertible element must by the whole ring. So $M$ is not a proper ideal.

3. ## Not Clear

Hi--

I am not able to understand what do u mean by Codimension. By the way the proof is excellent, shows how algebra and analysis are closely related.

" Algebra is properly speaking the analysis of equations".

4. Originally Posted by Chandru1
I am not able to understand what do u mean by Codimension.
To say that a subspace M of a vector space R has codimension 1 means that the subspace generated by adding just one additional element to M is the whole space. (Equivalently, the quotient space R/M has dimension 1.) So if M has codimension 1 in R then M is a maximal proper subspace of R. In the particular case where R is a ring and M is an ideal, this obviously implies that M is a maximal ideal.