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Let $R$ be the ring of real-valued continuous functions on $[0,1].$ Note that if $0 \le t \le 1$ then the evaluation map $\psi_t :f \mapsto f(t)$ is a homomorphism of $R$ into $\Bbb R$. Show that any homomorphism $\varphi$ of $R$ into $\Bbb R$ is of this form.

Let $\varphi : R \to \Bbb R$ be a homomorphism such that $\varphi \ne \psi_t$ for any $t \in [0,1]$. This means that for all $t$ there exists $f \in R$ such that $\varphi(f) \ne \psi_t(f)=f(t)$.

I'm trying to show that $\varphi$ cannot be a homomorphism unless it's of the form $\psi_t$, but I don't know how to argue from here. I'm told to consider something called a partition of unity, but I haven't heard of these beforehand and don't really now how to use them here. Is there another way to proceed?

Walker
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  • The kernel of the homomorphism would have to be a maximal ideal. The question about the maximal ideals of $C[0,1]$ is asked very frequently and has many solutions on the site. – rschwieb Aug 25 '22 at 20:11
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    Why would it need to be maximal? I don't think the kernels are maximal ideals neccessarily. @rschwieb – Walker Aug 25 '22 at 20:22
  • when the image is a field, then kernel is maximal, because $R/K\cong \mathbb R$. The ideals of the field correspond with the ideals of the quotient. All two of them. – rschwieb Aug 26 '22 at 01:45
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    That I know, but there is no guarantee that for an arbitary homomorphism $\varphi$ we have that $\text{Im}(\varphi)$ is a field. There is no mention that $\varphi$ should be surjective. @rschwieb – Walker Aug 26 '22 at 07:27
  • there is no guarantee that... Yes, that is true. There is no mention that... No explicit mention, but nevertheless, the evaluation map is surjective. I'm pretty sure "of this form" is meant to be "a ring homomorphism onto $\mathbb R$." – rschwieb Aug 26 '22 at 11:13
  • It's also not that hard to show that any such ring homomorphisms is surjective. – Jacob FG Aug 26 '22 at 11:37
  • I think we're talking about different things here. The evaluation map certainly is surjective so for $\varphi$ a evaluation map $R/\ker(\varphi) \cong \Bbb R$. And since $\Bbb R$ is a field $\ker(\varphi)$ is maximal and thus of the form $M_c = { f \in R \mid f(c) = 0}$ for $c \in [0,1]$. But I'm not talking about evaluation maps. What I'm trying to show is that if I pick any homomorphism (not neccessarily an evaluation map), then I'll arrive at a contradiction if it's not an evaluation map. I understood from your comment that even this kind of homomorphism should give a maximal kernel? – Walker Aug 26 '22 at 12:16
  • @rschwieb I think I need to mention you in order for you to get a notification. – Walker Aug 27 '22 at 09:16
  • @SleepWalker This duplicate seems like a good starting point for that. – rschwieb Aug 27 '22 at 18:01

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