Spectrum of Ring of Smooth Functions on $\mathbb{R}^n$ When we define smooth manifold, we starting with topological space $M$ which localy homeomorphic to $\mathbb{R}^n$ and setting up sheaf $\mathscr{F}(M)$ of functions on it which localy isomorphic to $\mathscr{F}(\mathbb{R}^n)$.
Let's $M = \mathbb{R}^n$ and $R = \mathscr{F}(\mathbb{R}^n)$ is a ring of smooth functions on $M$.
Is it true that Spec($R$) with Zariski topology homeomorphic to $\mathbb{R}^n$?
 A: There is a canonical embedding $\psi: \mathbb R^n \hookrightarrow \text{Spec}(R)$ which carries a point $p \in \mathbb R^n$ to the maximal ideal consisting of functions $f: \mathbb R^n \to \mathbb R$ with $f(p)=0$. To see that this is a maximal ideal, consider the ring homomorphism $\phi: R \to \mathbb R$ given by evaluation at $p$, $\phi(f) = f(p)$.
By the smooth version of Urysohn's lemma, two disjoint closed sets in $C_0, C_1 \subset \mathbb R^n$ can be separated by a function $f$ so that $f(C_0) = \{0\}$ and $f(C_1)=\{1\}$; the closed subset of $\text{Spec}(R)$ defined by the ideal $fR$, intersected with the image of $\psi$, is precisely $\psi(f^{-1}(0))$. Together with the obvious fact that smooth functions are continuous, this is enough to see that the embedding $\psi$ is indeed a homeomorphism onto its image.
Note that $\text{Spec}(R)$, or even, for that matter, the maximal ideals $\text{mSpec}(R)$, are much bigger than the image of $\psi$. For example, there is a maximal ideal containing the ideal of all functions with compact support. Since the ideal of functions with compact support is not contained in $\psi(p)$ for any $p$, a maximal ideal containing it cannot be equal to $\psi(p)$ for any $p$.
A: A trivial way you can tell this is false is that Spec of any ring is (quasi)compact, but $\mathbb{R}^n$ is not (if $n>0$).
A bit less trivially, this is still false if you replace $\mathbb{R}^n$ with a compact manifold (of dimension $>0$).  This follows from the following theorem.

Theorem:  Let $R$ be a commutative ring.  If $\operatorname{Spec}(R)$ is Hausdorff, it is totally disconnected.

Proof: Suppose $\operatorname{Spec}(R)$ is Hausdorff and let $p,q\in\operatorname{Spec}(R)$ be distinct.  Then there are disjoint open sets $U,V\subset\operatorname{Spec}(R)$ such that $p\in U$ and $q\in V$.  We may further choose $U$ and $V$ to be distinguished open sets, so there exist $f,g\in R$ such that $U=\operatorname{Spec}(R_f)$ and $V=\operatorname{Spec}(R_g)$.  But this implies $U$ and $V$ are compact, and hence closed since $\operatorname{Spec}(R)$ is Hausdorff.  Thus $U$ and $V$ are clopen sets separating $p$ and $q$.
However, it is true that if $M$ is a compact manifold, then $M\cong\operatorname{MaxSpec}(C^\infty(M))$.  To prove this, first note that for each $x\in M$, the ideal $m_x$ of functions vanishing at $x$ is a maximal ideal.  If $I\subseteq C^\infty(M)$ is any ideal not contained in $m_x$ for any $x$, then for each $x\in M$ we can choose an element $f_x\in I$ which does not vanish at $x$.  Replacing $f_x$ by $f_x\bar{f_x}$, we may assume that $f_x\geq 0$ everywhere.  The sets $U_x=\{y:f_x(y)\neq 0\}$ are an open cover of $M$, so there are finitely many $x_1,\dots, x_n$ such that $M=\bigcup_{i=1}^n U_{x_i}$.  The function $f=\sum_{i=1}^n f_{x_i}$ is then an element of $I$ which vanishes nowhere, and hence is a unit.  Thus $I$ is all of $C^\infty(M)$.  This shows that every proper ideal is contained in some $m_x$, and hence every maximal ideal is of the form $m_x$.  Now note that using bump functions, it is easy to show that the bijection $m_x\mapsto x$ is a continuous map $\operatorname{MaxSpec}(C^\infty(M))\to M$.  Since $\operatorname{MaxSpec}(C^\infty(M))$ is compact and $M$ is Hausdorff, this map is thus a homeomorphism.
