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


*

*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 $\mathfrak{m} = \mathfrak{m}_x$.

*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.




*

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

*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 $f(x) > 0$.

*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?

*Conclude the result for your homework problem!
