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Jul 29, 2014 at 15:57 vote accept Allen Knutson
Jul 29, 2014 at 14:29 comment added Brent Pym Nicola's answer below certainly gives the desired counterexample. I just wanted to mention that if $I$ is already radical then there is a simple obstruction to equality of $I$ and $\langle I\cap Z \rangle$. Namely, the Poisson bracket descends to a Lie algebra structure on $I/I^2$ (the "co-normal Lie algebra"). If $I$ is generated by elements of $Z$ then this Lie bracket will be abelian, but most conormal algebras are not. For example, for the maximal ideal $\mathfrak{m}$ defining the origin in the dual of a Lie algebra $\mathfrak{g}$, we have $\mathfrak{m}/\mathfrak{m}^2\cong\mathfrak{g}$
Jul 29, 2014 at 7:55 answer added Nicola Ciccoli timeline score: 5
Jul 29, 2014 at 6:38 answer added Peter Michor timeline score: 0
Jul 29, 2014 at 0:17 history asked Allen Knutson CC BY-SA 3.0