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.
Best Answer
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.