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It is part of an exercise in the book Basic Homological Algebra, Chapter 2 Exercise 16.

Suppose $R$ is the subring of $C^\infty(\mathbb{R})$ of all functions with period $2\pi$, and let $I$ be the maximal ideal of $R$ consisting of all functions of $R$ taking $0$ to $0$.

How can I prove that $I$ is generated by $\sin(x)$ and any (one) function in $I$ which take nonzero value at $\pi$?

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    given a fctn in I use the 'any (one) function' to make it also vanish at $\pi$, then divide by sin2012-07-11

1 Answers 1

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I realized that my previous answer did not address the 'any' function part. So here is a second attempt which might be what Mike suggest in the comment:

Call $g(x)$ that function in $I$ with $g(\pi)=c\neq 0$ an let $g_2:=1-g(x)/c$ so that $1=g(x)/c+g_2(x)$. Observe that $g_2$ vanishes at $\pi$.

Now multiplying an $f\in I$ with the above partition of unity you get $f=fg/c+fg_2$ and the second summand vanishes at $0$ and $\pi$. Dividing $fg_2$ by $sin(x)$ gives you a smooth periodic function $h(x)$ defined everywhere (by Hadamards Lemma) so $f=fg/c+h\sin(x)$ which proves the claim.