Hello Everyone, I have this homework problem, I'm going to share what i have so far, not sure if Im in the right path. First, I have: $$f \sim g \, \Leftrightarrow \,x_0 \in \mathbb{R^n}, \exists \,U \ni x_0\,\mid f|_U = g|_U$$
(A) Reflexivity: $f \sim f \Rightarrow$ for $x_0 \in \mathbb{R^n}, \exists \,U \ni x_0\,\mid f|_U = f|_U$
(B)Symmetry: If $f \sim g$ then $g \sim f$ $\Rightarrow$ for $x_0 \in \mathbb{R^n}, \exists \,U \ni x_0\,\mid f|_U = g|_U$ and $g|_U = f|_U$
(C)Transitivity: If $f \sim g$ and $g \sim h$, then $f \sim h \Rightarrow $ for $ x_0 \in \mathbb{R^n}, \exists \,U,V \ni x_0\,\mid f|_U = g|_U$ and $\,\,\,\,\,\,\,\,g|_V = h|_V,$ therefore $f|_{U \cap V} = h|_{U \cap V}$, so $f \sim h$
Now, the equivalence class are defined as follows: $$[f]= \{\varphi \in C_{x_0}\ | \varphi \sim f\} \Rightarrow \varphi|_U = f|_U $$
$$[g]= \{\pi \in C_{x_0}\ | \pi \sim g\} \Rightarrow \pi|_V = g|_V$$
They are welldefined, lets suppose that $[f]=[\phi]$ and $[g]=[\psi]$, then we say that:$[f]+[g] = [\phi]+[\psi] $ and $[f][g]=[\phi][\psi] $. Now by definition,for $x_0 \in \mathbb{R^n}, \exists \,U \ni x_0\, ,f|_U = \phi|_U$ and $\exists \, V \ni x_0\ ,g|_V = \psi|_V $, this implies that:$$(f+g)|_{U \cap V} = (\phi+\psi)|_{U \cap V} \Rightarrow [f + g] = [\phi + \psi],$$ $$(f g)|_{U \cap V} = (\phi \psi)|_{U \cap V} \Rightarrow [f g] = [\phi \psi]$$
So they are well defined.On the other hand, I have: $$\phi:C_{x_0} \longrightarrow \mathbb{R}, \phi[f]=f(x_0) $$
So the kernel of $\phi$ is $M_{x_0}$, so by the first isomorphism theorem, we have that: $$ C_{x_0}/M_{x_0} \cong \mathbb{R} $$Now, $\mathbb{R}$ is a field, therefore $C_{x_0}/M_{x_0} $ is a field $\Rightarrow$ $M_{x_0}$ is maximal. To prove that $M_{x_0}$ is the unique maximal, first we notice that $M_{x_0}$ contains no units, so if $f \notin M_{x_0}.$ Then $f(x_0) \neq 0 \Rightarrow \exists \, U \ni x_0$ such that $f(y) \neq 0, \forall y \in U \Rightarrow 1/f$ exists in a neighborhood of $x_0, \, \Rightarrow 1/f$ is a unit in $C_{x_0} \Rightarrow C_{x_0}$ is a $\mathbf{local \,ring}$, so $M_{x_0}$ is the unique maximal by the characterization of $\mathbf{local\, rings}$
Now to prove that $M_{x_0}$ is finitely generated, We have: $$ h(1) - h(0) = \int^1_0 h'(s)\,ds$$ where $f(x_0)=0 $ in $M_{x_0}$ and $h(s) = f(s(x-x_0) +x_0)$. On the other hand, $$h'(s) = \sum\limits_{i=1}^n \frac{\partial}{\partial x_i} \frac{\partial x_i}{\partial s}, $$ but $\frac{\partial x_i}{\partial s} = (x_i - x_0^{i}),$ this is true because for $s \in \mathbb{R}$ and notice that $h:\mathbb{R} \rightarrow \mathbb{R}$, therefore $(s(x-x_0) +x_0) \in \mathbb{R} \Rightarrow (s(x_i-x_0^{i}) +x_0) \in \mathbb{R^n} $. Now, $h(1) = f(x)$ and $ h(0) = f(x_0)=0$. Hence: $$f(x) = \sum\limits_{i=1}^n (x_i-x_0^{i})\cdot(\int^1_0 \frac{\partial f}{\partial x_i}\,ds),$$ Now, since the integral $\int^1_0 \frac{\partial f}{\partial x_i}\,ds$ depends on the variable $s$, we thus have that the generators of $M_{x_0}$ are $(x_i-x_0^{i})$. In other words, $$M_{x_0} = \langle x_1-x_0^{1},x_2-x_0^{2}, \cdots, x_n-x_0^{n}\rangle$$
Now im not sure about a few thing:
1) this is the right proof for the equivalence relation ${\bf Done}$
2) I don't understand how to proof that they are well defined ${\bf Done}$
3) $C_{x_0}/M_{x_0} $ is a field, then $M_{x_0} $ is maximal, this prove that is the unique one. ${\bf Done}$
4) Is the procedure in $M_{x_0}$ is finitely generated clear or I need some other stuff, my professor told me that the functions in the exercise are smooth too
Suppose $[f] = [\phi], [g] = [\psi].$ Then we need to say that $[f] + [g] = [\phi] + [\psi].$ By definiton, $\exists \,U \ni x_0\,, f|_U = \phi|_U$ and $\exists \,V \ni x_0\,, g|_V = \psi|_V.$ So $(f + g)|_{U \cap V} = (\phi + \psi)|_{U \cap V} \Rightarrow [f + g] = [\phi + \psi].$ This proves that the addition is well defined. Similarly, you can prove that the multiplication is well defined.
${\bf Fact:}$ Let $A$ be a commutative ring with unity. Suppose that the set of all non-units of $A$ forms an ideal $\mathfrak{m}.$ Then $A$ has a unique maximal ideal, and $\mathfrak{m}$ is the maximal ideal.
In this case, all the elements of $M_{x_0}$ are non-unit. Let $f \notin M_{x_0}.$ Then $f(x_0) \neq 0 \Rightarrow \exists $ open nbd $U$ of $x_0$ such that $f(y) \neq 0, \forall y \in U \Rightarrow 1/f$ exists in a nbd of $x_0$ and clearly it is a unit in $C_{x_0}.$ Thus by the above fact $M_{x_0}$ is the unique maximal ideal of $C_{x_0}.$