# Maximal Ideal

• Jan 25th 2010, 10:15 PM
Chandru1
Maximal Ideal
Let $\displaystyle R$ be the ring of all real valued continuous functions on the closed unit interval. If $\displaystyle M$ is a maximal ideal of $\displaystyle R$, prove that there exists a real number $\displaystyle \gamma, \ 0 \leq \gamma \leq 1$ such that $\displaystyle M=M_{\gamma}=\{ f(x) \in R \ | \ f(\gamma)=0\}$
• Jan 26th 2010, 12:46 PM
Opalg
Quote:

Originally Posted by Chandru1
Let $\displaystyle R$ be the ring of all real valued continuous functions on the closed unit interval. If $\displaystyle M$ is a maximal ideal of $\displaystyle R$, prove that there exists a real number $\displaystyle \gamma, \ 0 \leq \gamma \leq 1$ such that $\displaystyle M=M_{\gamma}=\{ f(x) \in R \ | \ f(\gamma)=0\}$

First, notice that each of the sets $\displaystyle M_{\gamma}$ is a maximal ideal in $\displaystyle R$. In fact, if 1 denotes the constant function 1 then for any $\displaystyle f\in R$ we can write $\displaystyle f = (f-f(\gamma)1) + f(\gamma)1$. Since $\displaystyle f-f(\gamma)1\in M_{\gamma}$, it follows that $\displaystyle M_{\gamma}$ (which is trivially an ideal) has codimension 1 in $\displaystyle R$ and is therefore maximal.

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

So suppose that there is no such $\displaystyle \gamma$. Then for each $\displaystyle t\in[0,1]$ there exists $\displaystyle f_t\in M$ such that $\displaystyle f_t(t)\ne0$. The set $\displaystyle \{s\in[0,1]:f_t(s)\ne0\}$ is an open neighbourhood of t. These neighbourhoods cover the unit interval, so by compactness there is a finite subcover, given by elements $\displaystyle f_{t_n}$ of $\displaystyle M$. Then $\displaystyle \textstyle\sum f_{t_n}^2$ is an element of $\displaystyle M$ that is strictly positive throughout the interval and hence invertible. But an ideal containing an invertible element must by the whole ring. So $\displaystyle M$ is not a proper ideal.
• Jan 26th 2010, 11:32 PM
Chandru1
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".
• Jan 27th 2010, 12:23 AM
Opalg
Quote:

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.