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Let $R$ be the ring of continuous real valued functions on $[0,1]$. I'm trying to prove that every homomorphism $R \mapsto \mathbb{R}$ is of the form

\begin{equation} \phi_t : R \to \mathbb{R} \quad \text{given by} \quad \phi_t(f)= f(t), \end{equation}

This is an exercise from Basic Algebra Nathan Jacobson, section 2.7 exercise 14. And it comes with a hint. Which is the following

Hint: If $\phi \neq \phi_t$ there is an $f_t \in R$ such that $\phi(f_t) \neq f_t(t)$. Then $g_t = f_t - \phi(f_t)1 \in R$ and $g_t(t) \neq 0$ but $\phi(g_t) = 0$. Show that there exist a finite number of $t_i$ such that $g(x) = \sum g_{t_i}^2(x) \neq 0$ for all x. Then $g^{-1} \in R$ but $\phi(g) = 0$.

I was able to show that $\phi(g_t) = 0$, but i don't know how to proceed with the other part of the hint, i think that i can say that the sets $\{ x: g_t(x) \neq 0 \}$ are a finite cover for $[0,1]$, and that for each set i have $t$ in them, so i have finite $t$ such that $g(x) = \sum g_{t_i}^2(x) \neq 0$. (Am i right? or i am missing some details).

Thank you in advance.

Olivier Roche
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Alek Murt
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  • What do you mean by unique? – Robert Shore Nov 26 '19 at 16:37
  • The only homomorphism from R to $\mathbb{R}$ it's of the form $\phi_t(f) = f(t)$. – Alek Murt Nov 26 '19 at 16:40
  • Something similar was discussed at https://math.stackexchange.com/questions/159801/characterizing-all-ring-homomorphisms-c0-1-to-mathbbr?rq=1 . Maybe you should try to find something there. – FMont Nov 26 '19 at 16:41
  • I'm trying to follow the same idea from that post, that's why i try to make the question around the compactness of the set and why we have finite $t$. – Alek Murt Nov 26 '19 at 16:44
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    Ah. You're not trying to prove the homomorphism is unique. You're trying to prove that all possible ring homomorphisms are of the form $\phi_t$. – Robert Shore Nov 26 '19 at 16:50
  • @RobertShore yup, that's what it is – Alek Murt Nov 26 '19 at 16:53
  • The set ${ x: g_t(x) \neq 0 }$ is a subset of $\mathbb{R}$. It can't be a cover since a cover is a subset of the topology of $\mathbb{R}$. – Olivier Roche Nov 26 '19 at 19:08

1 Answers1

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The hint has so much implicit quantifiers, explicit them and everything becomes clear :

Assume for contradiction that the morphism $\phi : R \mapsto \mathbb{R}$ is not of ,the form $\phi_t$, ie $\boxed{\forall t \in [0;1], \, \phi \neq \phi_t}$.

For each $t \in [0;1]$, since $g_t(t)\neq 0$, there's an open neighborhood $V_t$ of $t$ such that $\forall x \in V_t, \, g_t(x)\neq 0$.

$\{V_t\}_{t \in [0;1]}$ forms an open cover of the compact set $[0;1]$, hence there is a finite subset, say $\{t_1;\dots;t_n\}$, of the index set $[0;1]$ such that $\{V_{t_1};\dots;V_{t_n}\}$ covers $[0;1]$.

Now take $g(x) := \sum_{i=1}^n g_{t_i}(x)$, etc.

Olivier Roche
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