Let $R$ be the ring of continuous functions $f:[-1, 1]\to \mathbb{R}$ and $M$ is the kernel of $R\to \mathbb{R}, \ f\mapsto f(0)$.
One can easily check that $M$ is a maximal ideal of $R$. Therefore $M\in \operatorname{Spec} R$ and there exists localization $A:=R_M$.
Here are the questions:
— Is $A$ isomorphic to the ring of germs of functions?
— May other localizations of $R$ be applied to analytical problems?
Thanks for paying attention.
Let us denote by $B$ the ring of germs at $0$. Our goal is to prove that $A$ is isomorphic to $B$. The first thing we then need to do is to understand the elements of $A$.
We know that every element of $A$ is of the form $\frac{f}{g}$ with $f \in R$ and $g \in R \smallsetminus M$. However, we also need to understand when two such fractions are equal. Recall that by definition $\frac{f_1}{g_1} = \frac{f_2}{g_2}$ if and only if there exists $h \in R \setminus M$ such that $h(f_1g_2 - f_2g_1) = 0$ in $R$. Also recall that the natural map $R \to A$ is given by $f \mapsto \frac{f}{1}$.
Lemma 1. The natural map $R \to A$ is surjective.
Proof. Given $\frac{f}{g} \in A$ we need to show that there exists $k \in R$ and $h \in R \setminus M$ such that $h(kg - f) = 0$. Since $g(0) \neq 0$ and $g$ is continuous, there exists a closed neighbourhood $U$ of $0$ such that $g$ is never zero on $U$. Then $\frac{f}{g}$ is a continuous function on $U$, and hence by the Tietze extension theorem there exists a continuous extension of $\frac{f}{g}$ to all of $[-1, 1]$. Let us denote this extension by $k$. Lastly choose a bump function $h$ with support contained in $U$ and $h(0) \neq 0$. Then $k \in R$, $h \in R \setminus M$ and $h(kg - f) = 0$, as we wanted. $\square$
Lemma 2. Let $f, g \in R$. In $A$ we have $\frac{f}{1} = \frac{g}{1}$ if and only if there exists a neighbourhood $U$ of $0$ such that $f|_U = g|_U$.
Proof. $\Rightarrow$: If $\frac{f}{1} = \frac{g}{1}$, then there exists $h \in R \setminus M$ such that $h(f - g) = 0$. Since $h(0) \neq 0$ and $h$ is continuous, there exists a neighbourhood $U$ of $0$ such that $h$ is never zero on $U$. This implies that $f|_U = g|_U$.
$\Leftarrow$: Conversely, assume $f|_U = g|_U$ for some neighbourhood $U$ of $0$. Choose a bump function $h$ with support contained in $U$ and $h(0) \neq 0$. Then $h \in R \setminus M$ and $h(f - g) = 0$, i.e. $\frac{f}{1} = \frac{g}{1}$. $\square$
We now have everything we need to show that $A$ and $B$ are isomorphic. However, I leave this as an exercise to you. The idea is to map $[f] \in B$ to $\frac{f}{1} \in A$. Lemma 2 then shows you that this map is well-defined and injective, and lemma 1 shows you that it is surjective.