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This is to expand further on something I wrote in the comments. Martin wrote:

Here are some necessary conditions: [...] Are they sufficient?

I think no since a polynomial algebra $\mathbb{R}[x_1,\ldots,x_n]$ satisfies all these conditions but is not isomorphic to the algebra of smooth functions on a manifold. A short explanation for this is that smooth functions satisfy Borel's lemma while polynomial algebras don't. Here's a more detailed explanation:

Let $m_p$ the maximal ideal corresponding to a point $p\in Spec_{\mathbb{R}}A$. Then for polynomial algebras as well as algebras of the form $C^\infty(M)$, the $m_p$-completion $\hat{A}=\varprojlim \hat{A}=\lim_{r \geq 1} A/m_p^r$ is (after fixing local coordinates) a formal power series algebra $R[[x_1,\ldots,x_n]]$. The natural map $A\to \hat A$ may be interpreted as associating to a function its Taylor expansion at $p$. One version of the lemma of Borel says that for smooth function algebras this map is surjective. In other words: for every power series I give you (even non convergent) you can find a smooth function which has it as its Taylor series. Obviously this does not hold for polynomial algebras. So this gives you another necessary condition.

In the comments I said that polynomial algebras are finitely generated while algebras $C^\infty(M)$ are not. You asked me how to see this. I don't know if there is a simpler proof, but I would apply the same lemma of Borel: finitely generated algebras have a countable basis as vector spaces. But formal power series have no countable basis, and since the map $A\to \hat A$ is surjective also $A$ cannot have a countable basis. (Obviously if you just wanted to know that polynomial algebras are not isomorphic to smooth function algebras you didn't need this anymore).

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This is to expand further on something I wrote in the comments. Martin wrote:

Here are some necessary conditions: [...] Are they sufficient?

I think no since a polynomial algebra $\mathbb{R}[x_1,\ldots,x_n]$ satisfies all these conditions but is not isomorphic to the algebra of smooth functions on a manifold. A short explanation for this is that smooth functions satisfy Borel's lemma while polynomial algebras don't. Here's a more detailed explanation:

Let $m_p$ the maximal ideal corresponding to a point $p\in Spec_{\mathbb{R}}A$. Then for polynomial algebras as well as algebras of the form $C^\infty(M)$, the $m_p$-completion $\hat{A}=\varprojlim A/m_p^r$ is (after fixing local coordinates) a formal power series algebra $R[[x_1,\ldots,x_n]]$. The natural map $A\to \hat A$ may be interpreted as associating to a function its Taylor expansion at $p$. One version of the lemma of Borel says that for smooth function algebras this map is surjective. In other words: for every power series I give you (even non convergent) you can find a smooth function which has it as its Taylor series. Obviously this does not hold for polynomial algebras. So this gives you another necessary condition.

In the comments I said that polynomial algebras are finitely generated while algebras $C^\infty(M)$ are not. You asked me how to see this. I don't know if there is a simpler proof, but I would apply the same lemma of Borel: finitely generated algebras have a countable basis as vector spaces. But formal power series have no countable basis, and since the map $A\to \hat A$ is surjective also $A$ cannot have a countable basis. (Obviously if you just wanted to know that polynomial algebras are not isomorphic to smooth function algebras you didn't need this anymore).