# Thread: characterizing the maximal ideals of a function ring

1. ## characterizing the maximal ideals of a function ring

Hello!!!
So much homework :-\
I need to show that every maximal ideal in the ring R, which is made of all continuous functions [0,1]-->R, is of the form M(c)={f in R| f(c)=0).

I think I have a kinda of a topological proof formed, and of course it can be specifically used here (the proof is for any hausdorff-compact space), but I'm seeking a more algebraic proof, which I'm sure exists since most of us haven't studied topology yet.
I was to show that if M is a maximal ideal in R, then there must exist a point c in [0,1] which satisfies f(c)=0 for all f in M. Therefore, I start by contradically assuming that for all x in [0,1] there exists a function f_x in M such that f_x(x)=0. I cannot reach a contradiction though...

Tomer.

2. Originally Posted by aurora
Hello!!!
So much homework :-\
I need to show that every maximal ideal in the ring R, which is made of all continuous functions [0,1]-->R, is of the form M(c)={f in R| f(c)=0).

I think I have a kinda of a topological proof formed, and of course it can be specifically used here (the proof is for any hausdorff-compact space), but I'm seeking a more algebraic proof, which I'm sure exists since most of us haven't studied topology yet.
I was to show that if M is a maximal ideal in R, then there must exist a point c in [0,1] which satisfies f(c)=0 for all f in M. Therefore, I start by contradically assuming that for all x in [0,1] there exists a function f_x in M such that f_x(x)=0. I cannot reach a contradiction though...

Tomer.

Use the fact that if $\displaystyle M$ is a maximal ideal in the ring $\displaystyle R$, then $\displaystyle R/M$ is a field. This is easy to see. Using $\displaystyle I$ as the ideal in the problem and $\displaystyle f$ in $\displaystyle C$, then $\displaystyle I + f = I$ iff $\displaystyle f$ is in $\displaystyle I$, that is $\displaystyle f(a) = 0$ for some $\displaystyle a$ in$\displaystyle [0,1]$, thus if $\displaystyle I + f$ is a distinct coset, then $\displaystyle f$ is not in $\displaystyle I$, hence $\displaystyle f(x)$ is nonzero for any $\displaystyle x$ in $\displaystyle [0,1]$, hence $\displaystyle 1/f(x)$ is continuous and is, of course, nonzero on $\displaystyle [0,1]$ and $\displaystyle 1/f + I$ is a distinct coset and $\displaystyle (f+I)(1/f+I) = 1 + I$, so that $\displaystyle R/M$ is a field.

3. Originally Posted by aurora
I need to show that every maximal ideal in the ring R, which is made of all continuous functions [0,1]-->R, is of the form M(c)={f in R| f(c)=0).

I think I have a kinda of a topological proof formed, and of course it can be specifically used here (the proof is for any hausdorff-compact space), but I'm seeking a more algebraic proof, which I'm sure exists since most of us haven't studied topology yet.
I was to show that if M is a maximal ideal in R, then there must exist a point c in [0,1] which satisfies f(c)=0 for all f in M. Therefore, I start by contradically assuming that for all x in [0,1] there exists a function f_x in M such that f_x(x)=0. [You mean f_x(x)≠0, of course.] I cannot reach a contradiction though...
That's a good start. But you will have to resort to analysis at some stage because this is a result about a ring of continuous functions (and it's no longer true if you drop the continuity condition).

So for each x in [0,1] you have a function f_x in M such that f_x(x)≠0. Replacing f_x by –f_x if necessary, you can even assume that f_x(x)>0. The set $\displaystyle U_x = \{y:f_x(y)>0\}$ is then an open neighbourhood of x. By the compactness of the unit interval, finitely many of these sets U_x, say $\displaystyle U_{x_1},\ldots,U_{x_n}$ will cover the unit interval. Then the function $\displaystyle f_{x_1}+\ldots+f_{x_n}$ in M will be strictly positive throughout the interval and hence invertible. Hence M is the whole of R. Contradiction!

4. Galois, thanks. I know that R/I is a field, and have even used it in my topological proof.
Opalg, thank you. However, doesn't that count for a topological proof? The whole use of the terms open sets, neighborhoods and coverings is a bit problematic here, since as I've said, most of us shouldn't be familiar with these terms (I specifically have taken topology but others did not).

After I show there exists some point c in [0,1] so that f(c)=0 for all f in M, it's pretty easy to show that M=M_c, though I still needed to use the fact that R/M is a field.

But so far, I have written down a topological proof Isn't there a proof not dealing with topological concepts?

5. Originally Posted by aurora
But so far, I have written down a topological proof Isn't there a proof not dealing with topological concepts?
I don't think so. As I said above, the result is false if you drop the continuity condition. The maximal ideal space of the ring of all functions on the unit interval is a huge space, far bigger than the unit interval.

So you have to bring continuity into the proof somehow, and that implies that you need to use some analysis or topology.

6. Well, I guess I'll sattle with this one then. :-)

I thank you guys very much.