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I have read in some differential geometry works that the ring of smooth functions $C^{\infty}(U)$ is a semi-simple ring, for $U\subseteq\mathbb{R}^n$ an open set; right now I can cite a remark immediately preceding Proposition 2.4.1 on Kostant's notes "Graded Manifolds, graded Lie theory and prequantisation".

How can one prove such a thing? Is there anywhere I can find this result for citation purposes?

This question earned me the "Tumbleweed" badge at math.SE; I hope it doesn't happen in here! Here's the link for math.SE: https://math.stackexchange.com/questions/2318033/proof-and-reference-if-possible-of-semisimplicity-of-the-ring-of-smooth-functi

Thanks a lot! Cheers!

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  • $\begingroup$ By semisimple you mean "trivial Jacobson radical", right? (and not what the ring theorists often call "semisimple") $\endgroup$
    – Yemon Choi
    Commented Jun 22, 2017 at 2:47
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    $\begingroup$ If indeed you mean (or Kostant meant) that the Jacobson radical is zero, this is trivial: the functions vanishing at one point form a maximal ideal of $C^{\infty}(U)$, and the intersection of these ideals is clearly $(0)$. $\endgroup$
    – abx
    Commented Jun 22, 2017 at 4:47
  • $\begingroup$ Choi: what ring theorists call "semisimple" is equivalent (as far as I know) to "trivial Jacobson radical". abx: Spot on! You are right, it slipped my mind. Please answer the question (even if it's trivial) so I can mark it. Cheers! $\endgroup$ Commented Jun 22, 2017 at 21:53
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    $\begingroup$ Regarding the previous comment: that used to be my belief, but then I kept finding that algebraists I talked to would interpret "semisimple" to mean "artinian and has trivial Jacobson radical" $\endgroup$
    – Yemon Choi
    Commented Jun 22, 2017 at 22:51
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    $\begingroup$ Nowadays most ring theorist use semiprimitive to mean trivial jacobson radical and semisimple to mean the regular module is semisimple which is equivalent to semiprimitive and Artinian. $\endgroup$ Commented Jun 23, 2017 at 0:49

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