# A surjective ring homomorphism $\phi : C([0,1]) \rightarrow \mathbb{R}$ is evaluation at a point [duplicate]

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?

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## marked as duplicate by YACP, Najib Idrissi, Davide Giraudo, egreg, TMMDec 6 '13 at 19:53

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 6 '13 at 11:23
Hmmm... I shall attempt it. –  user113857 Dec 6 '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 6 '13 at 11:37
Turns out my latex skills are better than I thought they would be. –  user113857 Dec 6 '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 6 '13 at 11:44

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.

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!

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Would all of this really be expected in a first course on modern algebra requiring no prior knowledge of topology? –  user113857 Dec 6 '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? –  user38268 Dec 6 '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 6 '13 at 12:16
He certainly wants to challenge us. –  user113857 Dec 6 '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 6 '13 at 12:20