Let $A$ be a Poisson $\mathbb{C}$-algebra: $A$ has a the structure of a complex commutative algebra and at the same time it carries the structure of a Lie algebra, with Lie bracket $\{\cdot, \cdot\}$. The latter should be a biderivation of the commutative multiplication, as well as the Lie multiplication.


Just as the definition of an $A$-module may be seen as an abstraction of left multiplication of $A$ on itself, we can define a Poisson $A$-module to be a vector space $V$ together with two multiplicative rules $A \times V\rightarrow V$ satisfying compatibility conditions. For more detail see [one of the original papers by Farkas](http://www.tandfonline.com/doi/abs/10.1080/00927870008827025?journalCode=lagb20).


My question is the following:


**Is every simple Poisson module finitely generated as an $A$-module?**


I expect that the answer is `no' although I have been unable to find a counterexample or indeed any reference which considers this question. I am especially interested in the case where $A$ is an affine algebra, ie. finitely generated and reduced. Proposition 1.1 in the paper cited above states that when $A$ is symplectic every Poisson module is induced by a $\mathcal{D}$-module, so this may be a fertile source of counterexamples: is every simple $\mathcal{D}(A)$-module finitely generated over $A$?


Thanks in advance!


**Edit**: Following Ben's response I am now specifically looking for examples where $A$ is a reduced Poisson algebras, ie. no nilpotent elements.



Are simple Poisson $A$-modules finitely generated as $A$-modules?
This is definitely not true; in fact, it's easy to construct counterexamples with $A$ finite dimensional.


For any Lie algebra $\mathfrak{g}$, you can define a Poisson algebra structure on $\mathbb{C}\oplus \mathfrak{g}$ such that $(a\_1+X\_1)(a\_2+X\_2)=a\_1a\_2+a\_1X\_2+a\_2X\_1$ for $a\_i\in \mathbb{C}$ and $X\_i\in \mathfrak{g}$ (you can more fancily define this as $\mathrm{Sym}(\mathfrak{g})/ \mathfrak{g}^2$); the Poisson bracket is defined by $\{a\_1+X\_1,a\_2+X\_2\} =[X\_1,X\_2]$. The simple Poisson modules over this algebra are precisely the simple modules over the Lie algebra (with the "commutative" action of $\mathfrak{g}$ trivial, and the "bracket" action being the representation). It's well known that finite dimensional Lie algebras have lots of simple infinite dimensional modules; for example, consider the Verma module for a generic highest weight over $\mathfrak{sl}\_2$. These aren't finitely generated over $A$, since a finitely generated $A$ module is finite dimensional.


**EDIT:** I hadn't noticed the comment about reducedness in the question, but you can think of these same examples as modules over $\mathrm{Sym}(\mathfrak{g})$; more generally, as argued in [this paper](https://arxiv.org/abs/0708.3928) of David Jordan, these sort of quotients show up as $A/I^2$ whenever you have a maximal ideal $I$ which is Poisson (and every finite dimensional simple is pulled back from this construction). 


