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Let $M$ be a Hausdorff, paracompact, finite-dimensional real manifold. How can I show that any algebra homomorphism from $C^{\infty}_{\mathbb{R}}(M)$ to $\mathbb{R}$ is an evaluation $\varphi_p$ i.e. $f\mapsto f(p)$ for some $p\in M.$

I can easily see that the converse of this is true, but have no idea how start with the given statement.

Bumblebee
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  • I guess what you should use that the product of functions with disjoint support is zero. Thus the after applying the homomorphism on one of the two is zero. Then throw somehow a partition of unity in (which will be mapped to 1) :) only happy guessing, I did not check anything – Severin Schraven Oct 25 '19 at 20:30
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    If $M=\mathbb{R}$, then there is a maximal ideal $I$ of $C^{\infty}{\mathbb{R}}(M)$ that contains the smooth functions with compact support. The quotient $C^{\infty}{\mathbb{R}}(M)\to C^{\infty}_{\mathbb{R}}(M)/I$ gives you a homomorphism that isn't an evaluation at a point of $\mathbb{R}$. – conditionalMethod Oct 25 '19 at 20:35
  • @conditionalMethod: Is $C^{\infty}_{\mathbb{R}}(\mathbb{R})/I\subset \mathbb{R}$? – Bumblebee Oct 25 '19 at 20:37
  • The quotient by a maximal ideal gives you a field. – conditionalMethod Oct 25 '19 at 20:39
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    But how do you know the field in question is the reals? – Severin Schraven Oct 25 '19 at 20:42
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    https://math.stackexchange.com/questions/1577744/if-m-is-compact-every-maximal-ideal-in-f-arises-in-this-way-as-a-point-of here is the compact case – Severin Schraven Oct 25 '19 at 21:49
  • @SeverinSchraven: Thank you for helping me. – Bumblebee Oct 25 '19 at 21:55
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    See https://math.stackexchange.com/questions/1577985/smooth-manifold-m-is-completely-determined-by-the-ring-f in the case that $M$ is second-countable. Without that assumption, the result is not necessarily true. – Eric Wofsey Oct 25 '19 at 21:59
  • In particular, if $M$ is discrete and of size equal to a measurable cardinal, then the limit with respect to a complete nonprincipal ultrafilter is a homomorphism to $\mathbb{R}$ which is not evaluation at a point. – Eric Wofsey Oct 25 '19 at 22:04
  • @EricWofsey: I am tring to a different proof for this :) For such an algebra homomorphism $\varphi,$ what can you say about the codimension of $\text{ker}(\varphi)$ without referring to this result? – Bumblebee Oct 28 '19 at 22:27
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    Well by definition $\ker(\varphi)$ has codimension $1$, since the quotient has dimension $1$. – Eric Wofsey Oct 28 '19 at 22:30

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