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Let $\phi : C([0,1]) \rightarrow \mathbb{R}$ be a surjective ring homomorphism. How would I prove that $\phi$ is the evaluation map $\phi(f) = f(x)$ for some $x \in [0,1]$?

I'm not even sure this conclusion is true. All I've done is started a proof by contradiction, assuming there is no such x, played around with quantifiers to determine that I need to show that $\forall x\in[0,1], \exists f\in C([0,1])$ such that $\phi(f) \neq f(x)$ leads to a contradiction. How do I go about finishing this?

user113857
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  • Quite possibly. My biggest fear is that the conclusion isn't even true. I just don't yet have enough intuition with abstract algebra to know the right way to go with these proofs. – user113857 Dec 06 '13 at 11:23
  • Hmmm... I shall attempt it. – user113857 Dec 06 '13 at 11:27
  • Pardon the poor latex skills...I literally just started using it tonight so I could type here.

    Well if $g \in ker(ev_x)$ then $g(x) = 0$. Why does that imply that $g \in ker(\phi)$ ? Knowing $\phi$ is a homomorphism only assures that the zero function is in the kernel; we don't know that anything else is in it, do we?

     – user113857 Dec 06 '13 at 11:37
  • Turns out my latex skills are better than I thought they would be. – user113857 Dec 06 '13 at 11:37
  • Nah, I actually just learned tonight as well.

    If $g \in ker(\varphi)$ then $\varphi(g)$ equals zero. You're right that the fact that zero function is in $ker\varphi$ is immediately obvious, but not much else. I'm still working on the part after that, but g in $ker(ev_a)$ implies g has a zero on $[0,1]$ (or $\mathbb{R}$ - which is it?). I'm thinking we'll need to use maximal ideals in some way.

     – Lost Dec 06 '13 at 11:44
  • Yea... $ker(\phi)$ is definitely a maximal ideal of $C([0,1])$. So if, say, $ker(ev_a)$ is an ideal of $C([0,1])$, then it either equals $ker(\phi)$ or is a proper subset of it (or is the whole set C, which would be interesting). I think. I'm not really sure about any of this. – user113857 Dec 06 '13 at 11:50
  • Actually, notice that $ev_a$ is a surjective homomorphism from $C([0,1])$ to $\mathbb{R}$ which tells us that the kernel is a maximal ideal as well. I also don't think $ker(ev_a)$ could equal the whole set since it is surjective. – Lost Dec 06 '13 at 11:52
  • Right. Is it possible for a ring to have two maximal non-identical ideals? I think it is. – user113857 Dec 06 '13 at 11:54
  • It sounds reasonable, yes. Those ideals would have to be disjoint, of course.

    I think that what we need to do is to get a better intuitive understanding of what $ker(\varphi)$ is compared to $ker(ev_a)$.

     – Lost Dec 06 '13 at 11:56
  • Take $p \in ker(\phi)$ and $e \in ker(ev_a)$. Then $\phi(pe) = \phi(p)\phi(e) = 0\phi(e) = 0$. So $pe \in ker(\phi)$. Similarly, using $ev_a$, we see that $pe \in ker(ev_a)$. So THEY ARE NOT DISJOINT. Does this help us? – user113857 Dec 06 '13 at 12:01
  • Well, I did assume neither kernel is empty. That's a safe assumption, isn't it? – user113857 Dec 06 '13 at 12:04
  • Yes, because both maps are surjective. They have to be nonempty, fortunately.

    As for your other comment, that's true. I wonder if that does give us what we need. However, what if the only element in $ker(\varphi)$ is the zero function? I'm not sure if maximal ideals include trivial ones.

     – Lost Dec 06 '13 at 12:11

1 Answers1

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The problem is very tricky and actually requires something topological. I will now sketch the argument. The idea is the following. Let $X= [0,1]$, we have a mapping $\Psi:X \mapsto \operatorname{MaxSpec} (X)$ given by sending a point $x \in X$ to the maximal ideal $\mathfrak{m}_x$ that is the kernel of the evaluation homomorphism at $x$. The evaluation homomorphism is surjective (why?) and since $\Bbb{R}$ is a field the kernel has to be a maximal ideal.

What your homework assignment is asking you to show:

The map $\Psi$ defined above is surjective.

Once you do this, you will have shown that every maximal ideal of $C[0,1]$ is equal to the kernel of some evaluation homomorphism (at some point $x$) which is (almost) enough to complete your problem. The following steps lead to proving that $\Psi$ is surjective :

  1. Given a maximal ideal $\mathfrak{m}$, consider $V(\mathfrak{m}) = \{f \in C[0,1] : f(x) = 0 \hspace{2mm} \forall f\in \mathfrak{m}\}$. Show that if $V(\mathfrak{m})$ contains some point $x$ (i.e. is non-empty) then this is enough to show $\mathfrak{m} = \mathfrak{m}_x$.

  2. If $V(\mathfrak{m}) = \emptyset$, then argue using compactness of $X$ that we can find a function $f \in \mathfrak{m}$ that does not vanish at any point $x \in X$. Why is this a contradiction?

However, you're not done yet! You've only shown that the kernel of your map $\phi$ is equal to the kernel of some evaluation homomorphism $\phi_x$. Why is $\phi = \phi_x$? To show that $\phi = \phi_x$ recall we can factor $\phi$ as $\pi \circ \overline{\phi}$ and $\phi_x$ as $\pi \circ \overline{\phi_x}$ where the bars indicate induced maps on the quotient. The $\pi$'s here are exactly the same because you know their kernels are equal! So to show $\phi = \phi_x$, it is enough to show that the induced map on the quotients (which is an automorphism of $\Bbb{R}$) is the same.

You now have to prove: Any ring automorphism $f$ of $\Bbb{R}$ is the identity map. Do this in the following steps.

  1. Show that $f$ has to fix the rationals pointwise (easy).

  2. Show that $f$ has to be order preserving, i.e. if $x <y $ then $f(x) < f(y)$. Since $f$ is a ring homomorphism, it is enough to show that if $x > 0$ then $f(x) > 0$.

  3. Now show $f$ is the identity map! Hint: If $f(x)$ for some real number $x$ is not equal to $x$ (say it is less than), choose a rational number $q$ such that $f(x) < q < x$. Recall $f^{-1}$ fixes the rationals, so what contradiction comes now?

  4. Conclude the result for your homework problem!

  • Would all of this really be expected in a first course on modern algebra requiring no prior knowledge of topology? – user113857 Dec 06 '13 at 12:12
  • @user113857 I honestly don't know. I am just giving you my honest thoughts about how to solve a problem. Perhaps your lecturer wanted to give you people a challenge? –  Dec 06 '13 at 12:13
  • Thanks for the help. This may look intimidating, but this actually looks similar to what the professor talked about (not in class though). In particular, 2 tells us that if there exists such a function then $ev_a$ is not surjective, a contradiction. – Lost Dec 06 '13 at 12:16
  • He certainly wants to challenge us. – user113857 Dec 06 '13 at 12:16
  • Very tricky indeed, but after further digestion, this is still an application of topics we covered - the universal mapping theorem of quotient rings for instance. – Lost Dec 06 '13 at 12:20
  • @Lost See the edits I made. You're not done yet! Has my answer helped? –  Dec 06 '13 at 12:23
  • @user113857 See the edit. –  Dec 06 '13 at 12:25
  • I haven't worked through it just yet, but I'm about to.

    However, yes, it helps - the major issue is that I didn't know what had to be proved. Thanks again!

     – Lost Dec 06 '13 at 12:30
  • @Lost Just link your lecturer to this thread, he can contact me if he wishes. –  Dec 06 '13 at 12:32