Ring of continuous real-valued functions

[Mr Davis 97]

Homework Statement

Let $R$ be the ring of all continuous real-valued functions $f : [0,1] \to \mathbb{R}$ with pointwise addition and pointwise multiplication of functions as its two operations. Let $c \in [0,1]$ and denote $M_c = \{f\in R : f(c) = 0\}$.

a) Show that any $f\in R$ such that $f(c) \not = 0$ for all $c\in [0,1]$ is a unit in $R$.

b) Show that $M_c$ is a maximal ideal of $R$.

Homework Equations

The Attempt at a Solution

I need some pointers for this.

For a), it's clear that I need to produce some inverse function $g$ such that $f \circ g = id$ and $g\circ f = id$. But I'm not seeing why the condition that $f(c) \not = 0$ for all $c\in [0,1]$ guarantees that $f$ has an inverse.

For b), I think I might have figured this out.
For each fixed $c\in [0,1]$ the map $E_c: R \to \mathbb{R}$ such that $E_c(f) = f(c)$ (called evaluation at $c$) is a ring homomorphism because the operations in $R$ are pointwise addition and multiplication of functions. The kernel of $E_c$ is given by $M_c = \{f\in R : f(c) = 0\}$. Also, $E_c$ is surjective: given any $a\in \mathbb{R}$ the constant function $f(x)=a$ maps to $a$ under evaluation at $c$. Thus $R/M_c \cong \mathbb{R}$. Since $\mathbb{R}$ is a field, $M_c$ is a maximal ideal.

 

[WWGD]

Homework Statement

Let $R$ be the ring of all continuous real-valued functions $f : [0,1] \to \mathbb{R}$ with pointwise addition and pointwise multiplication of functions as its two operations. Let $c \in [0,1]$ and denote $M_c = \{f\in R : f(c) = 0\}$.

a) Show that any $f\in R$ such that $f(c) \not = 0$ for all $c\in [0,1]$ is a unit in $R$.

b) Show that $M_c$ is a maximal ideal of $R$.

Homework Equations

The Attempt at a Solution

I need some pointers for this.

For a), it's clear that I need to produce some inverse function $g$ such that $f \circ g = id$ and $g\circ f = id$. But I'm not seeing why the condition that $f(c) \not = 0$ for all $c\in [0,1]$ guarantees that $f$ has an inverse.

I assume they mean a unit under the product they defined, not in terms of invertibility of the function .

 

[Mr Davis 97]

I assume they mean a unit uner the product they defined, not in terms of invertibility of the function .

Oh, right. Well since $f(x) \in \mathbb{R}$ and is never $0$, then define $g(x) = \frac{1}{f(x)}$. Does this work?

Also, note that I added my attempt for part b).

 

[member 587159]

Oh, right. Well since $f(x) \in \mathbb{R}$ and is never $0$, then define $g(x) = \frac{1}{f(x)}$. Does this work?

Also, note that I added my attempt for part b).

Both (b) and the corrected attempt for (1) are correct. Good job!

 

[WWGD]

Oh, right. Well since $f(x) \in \mathbb{R}$ and is never $0$, then define $g(x) = \frac{1}{f(x)}$. Does this work?

Also, note that I added my attempt for part b).

Yes, good work. I think it is also true that all maximal ifeals are of this form.

 

[member 587159]

Yes, good work. I think it is also true that all maximal ifeals are of this form.

Out of curiosity, why do you think all maximal ideals are of this form? I don't see this immediately.

 

[WWGD]

Out of curiosity, why do you think all maximal ideals are of this form? I don't see this immediately.

Just remem

Out of curiosity, why do you think all maximal ideals are of this form? I don't see this immediately.

I just remember having seen the result; no particularly deep insight.

 

[haruspex]

The solution to part b seems a little over the top.
Suppose we try to extend the ideal by inclusion of g for which g(c)=a>0 (say).
By completeness under addition, the extended ideal must include all continuous functions f such that f(c) is an integer multiple of a.
Multiplying by the constant functions from $\mathbb{R}$ yields the whole of $\mathbb{R}$.
Doesn't that work?

 

[fresh_42]

I don't understand:

... such that f(c) is an integer multiple of a.

Assume $M_c$ is not maximal. Then there is an ideal $M\subseteq R$ with $M_c \subsetneq M$. So we have $g(c)=a \neq 0$ for some function $g \in M - M_c\,$, and thus $((ra^{-1})g)(c)=r \in M$ for any $r \in \mathbb{R}$. Next we get for an arbitrary function $h\in R$ with $h(c)=s$ that $h = (h-(sa^{-1})g) + (sa^{-1})g$ where $(h-(sa^{-1})g)(c)=0$, i.e. $(h-(sa^{-1})g)\in M_c$, and $(sa^{-1})g \in M$. Thus by additive closure $R \subseteq M_c+M =M$.

Sorry, I had to write it out in order to see whether this is "shorter". It's more basic than what @Mr Davis 97 has written, but I think not shorter. With respect to part a), his solution is probably what the author had in mind, esp. to establish the automatisms: prime $\leftrightarrow$ integral domain, maximal $\leftrightarrow$ field. In the end we have also shown, that $M$ contains a unit.

 

[WWGD]

I don't understand:

Assume $M_c$ is not maximal. Then there is an ideal $M\subseteq R$ with $M_c \subsetneq M$. So we have $g(c)=a \neq 0$ for some function $g \in M - M_c\,$, and thus $((ra^{-1})g)(c)=r \in M$ for any $r \in \mathbb{R}$. Next we get for an arbitrary function $h\in R$ with $h(c)=s$ that $h = (h-(sa^{-1})g) + (sa^{-1})g$ where $(h-(sa^{-1})g)(c)=0$, i.e. $(h-(sa^{-1})g)\in M_c$, and $(sa^{-1})g \in M$. Thus by additive closure $R \subseteq M_c+M =M$.

Sorry, I had to write it out in order to see whether this is "shorter". It's more basic than what @Mr Davis 97 has written, but I think not shorter. With respect to part a), his solution is probably what the author had in mind, esp. to establish the automatisms: prime $\leftrightarrow$ integral domain, maximal $\leftrightarrow$ field. In the end we have also shown, that $M$ contains a unit.[/QUOTEMaybe we can use: if $g(c)=a\neq 0$ then $a^{-1}g(c)= a^{-1}a=1...$. And we just saw 1 is the unit.

 

[fresh_42]

Maybe we can use: if $g(c)=a\neq 0$ then $a^{-1}g(c)=a^{-1}a=1...$. And we just saw 1 is the unit.

Is it? We do not have an $h$ such that for all $f$ we get $(h(x)\cdot g(x))\cdot f(x)=f(x)\,,$ we only have $[g]\cdot [a^{-2}g] = [1]$, if I'm right, so $R/M_c$ is a field, which was what the OP had anyway.

 

[WWGD]

Is it? We do not have an $h$ such that for all $f$ we get $(h(x)\cdot g(x))\cdot f(x)=f(x)\,,$ we only have $[g]\cdot [a^{-2}g] = [1]$, if I'm right, so $R/M_c$ is a field, which was what the OP had anyway.

But,isn'tit the case that if we have [1] in our ideal, then the ideal is equal to the full ring? If [1] is in I , then r[1] is also in I for any r inR.

 

[fresh_42]

But,isn'tit the case that if we have [1] in our ideal, then the ideal is equal to the full ring? If [1] is in I , then r[1] is also in I for any r inR.

Yes, so we have shown that $R/M_c$ is a field. Just wanted to say, that we have still no $1\in M \supsetneq M_c$ and so did the same as the OP.

 

[mathwonk]

Suppose M is an ideal with no common zero. Then for every point c of [0,1] there is a function fc in M that does not vanish at c. Now use compactness to cover [0,1] by a finite number of open intervals Jc1,...Jcn where some function fci in M fails to vanish anywhere on Jci. Then try to cook up a function in M that does not vanish anywhere on [0,1], and conclude that M contains a unit.

 

[haruspex]

I don't understand:

Assume $M_c$ is not maximal. Then there is an ideal $M\subseteq R$ with $M_c \subsetneq M$. So we have $g(c)=a \neq 0$ for some function $g \in M - M_c\,$, and thus $((ra^{-1})g)(c)=r \in M$ for any $r \in \mathbb{R}$. Next we get for an arbitrary function $h\in R$ with $h(c)=s$ that $h = (h-(sa^{-1})g) + (sa^{-1})g$ where $(h-(sa^{-1})g)(c)=0$, i.e. $(h-(sa^{-1})g)\in M_c$, and $(sa^{-1})g \in M$. Thus by additive closure $R \subseteq M_c+M =M$.

Sorry, I had to write it out in order to see whether this is "shorter". It's more basic than what @Mr Davis 97 has written, but I think not shorter. With respect to part a), his solution is probably what the author had in mind, esp. to establish the automatisms: prime $\leftrightarrow$ integral domain, maximal $\leftrightarrow$ field. In the end we have also shown, that $M$ contains a unit.

Ok. I suppose I prefer the more basic proof because it gives me more of a feel for what is going on.

Wrt whether there are any other maximal ideals of this ring..

Suppose I is an ideal that does not have this property. So for each point c there is some member f that is nonzero at c.
Since f is continuous, it is nonzero in a half open interval [c,d_f). Let [c,d) be the union of these over all f such that f(c) is nonzero.
There must be a member g such that g(d) is nonzero. Likewise, it is nonzero in some interval (d', d].
Choose x in the intersection of (d',d) and (c,d). We have g nonzero in [x ,d] and some f nonzero on [c,x]. So h=f²+g² is nonzero on [c,d].
If d<1, h must also be nonzero for some interval beyond d, contradicting the definition of d. Thus, there must be a member which is nonzero on [0,1].

From there it is easy to see that the ideal must be the whole ring.

Interesting that this proof depends on the continuity, which had not been necessary for the posted problem.

Edit... got interrupted while typing all that, and thus beaten by mathwonk!