Homological dimensions of rings of smooth functions What is the global dimension of the algebra $C^\infty\mathbb R$ of smooth functions $\mathbb R\to\mathbb R$? What is the global dimension of the algebra $(C^\infty\mathbb R)_0$ of germs of smooth functions at $0\in\mathbb R$?
Answers concerning other kinds of homological dimension or other manifolds are welcome; but this question is not about topological homology theories, just pure algebra.
 A: Let $R$ be the ring of germs of smooth functions $f:\mathbb R^d\to\mathbb R$ at 0; I claim that $R$ has global dimension $d$.
In particular, for $M$ a $d$-dimensional manifold (without boundary), the ring of germs of smooth functions $f:M\to\mathbb R$ at $x\in M$ has global dimension $d$, and the ring of smooth functions on $M$ is of dimension at least $d$. I conjecture that it is precisely $d$ in the case of compact manifolds, but to be honest I do not have a clue how to prove it. I would not dare conjecture something for non-compact manifolds because not all maximal ideals are of the form $\{f:f(x)=0\}$ for some $x\in M$.
We will need the following fact from analysis: the ideal $\langle x_1,\ldots,x_k\rangle$ is precisely the ideal of functions vanishing at all points $(0,\ldots,0,x_{k+1},\ldots,x_d)$. This can be deduced from the fact that any smooth function $f:\mathbb R^d\to\mathbb R$ can be written
$$ f:x\mapsto f(0)+x_1g_1(x) + x_2g(x) + \ldots + x_dg(x) $$
with
$$ g_k:x\mapsto\int_0^1(\partial_kf)(0,\ldots,0,ux_k,x_{k+1}\ldots,x_d)\mathrm du. $$
In particular each $g_k$ is smooth and is identically zero whenever $f$ vanishes over $\{x_1=\cdots=x_{k-1}=0\}$.
Proof of $\operatorname{gl dim}R=d$. The ring $R$ is local, with maximal ideal $\mathfrak m$ the collection of functions vanishing at 0, hence its global dimension is the projective dimension of $R/\mathfrak m$ (Corollary 19.5 in [E]). By the above fact, $\mathfrak m=\langle x_1,\ldots,x_d\rangle$, and because of this characterisation again it is not hard to see that $x_1,\ldots,x_d$ is an $R$-regular sequence. In particular, the Kozsul complex
$$0\longrightarrow\Lambda^dR^d\longrightarrow\cdots\longrightarrow\Lambda^2R^d\longrightarrow\underbrace{\Lambda^1R^d}_{\simeq R^d}\longrightarrow\underbrace{\Lambda^0R^d}_{\simeq R}\longrightarrow R/\mathfrak m\longrightarrow0$$
sending $(g_1,\ldots,g_d)$ to $x_1g_1+\cdots+x_dg_d$ in degree 1 is a free resolution of $R/\mathfrak m$, so the quotient has projective dimension at most $d$. Moreover, upon tensoring the resolution by $R/\mathfrak m$ itself, all differentials are sent to zero, hence $\operatorname{Tor}^R_k(R/\mathfrak m,R/\mathfrak m)=(\Lambda^kR^d)\otimes(R/\mathfrak m)\simeq\Lambda^k(R/\mathfrak m)^d$, which is non-zero for $k=d$. This means $R/\mathfrak m$ has projective dimension at least $d$, i.e. precisely $d$, and indeed we find the global dimension of $R$ to be $d$.
[E] D. Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry (1995).
