Let $k$ denote a field of characteristic $0$ (assume algebraically closed for convenience). Define $J=k\langle x,y|[x,y]=y^{2}\rangle$. This noncommutative algebra (which can be viewed as a derivation ring over a commutative polynomial ring) is often referred to as the "Jordan plane" in the literature.


It can be see that $J\ncong \mathcal{U}(\mathfrak{g})$ (as algebras) for any Lie algebra $\mathfrak{g}$ (since $J/\langle [J,J]\rangle $ is not semiprime). My question is as follows: does there exist a Lie algebra $\mathfrak{g}$ and ideal $I\subseteq \mathcal{U}(\mathfrak{g})$ such that $J\cong \mathcal{U}(\mathfrak{g})/I$ (as algebras)? I guess that the answer should be "no", but I'm not sure how to go about proving it. I also assume this should be well known, given the depth of literature/ knowledge about the Jordan plane.


EDIT: I'm looking for a finite dimensional Lie algebra $\mathfrak{g}$ with these properties.



The Jordan Plane and Enveloping Algebras
Original answer without the finite-dimensionality requirement:


Yes, trivially so. For *every* algebra $A$, one can consider the underlying Lie-algebra $\mathfrak{a}$ of $A$ and gets a surjection $U(\mathfrak{a}) \twoheadrightarrow A$ because of the universal mapping property of $U(\mathfrak{a})$.




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New answer:


An algebra-homomorphism $f:U(\mathfrak{g})\to J$ is surjective iff $f(\mathfrak{g})\subseteq J$ is an algebra-generating-set of $J$.


It is easy to see that $J$ is a $\mathbb{N}$-graded algebra with $\deg(x)=\deg(y)=1$ and $\{y^a x^b \mid a,b\in\mathbb{N}\}$ is a homogenuous $k$-basis of $J$. This proves that if $f$ is surjective, then there must be elements $\hat{x},\hat{y}\in\mathfrak{g}$ with $f(\hat{x})=x+\text{higher degree terms}$ and $f(\hat{y})=y+\text{higher degree terms}$.


Now observe that $x\cdot y^a x^b = y^a x^{b+1} + a y^{a+1} x^b$ holds in $J$ so that $[x,y^a] = a y^{a+1}$. This shows
$$[\underbrace{x,[x,\ldots,[x}\_{a}, y]]] = a! y^{a+1}$$
$$\implies [\underbrace{f(\hat{x}),[f(\hat{x}),\ldots,[f(\hat{x})}\_{a}, f(\hat{y})]]] \equiv a! f(\hat{y})^{a+1} \mod\text{terms of degree}>a+1$$
Since the right hand sides are linearly independent by degree reasons, this shows that the elements $[\underbrace{\hat{x},[\hat{x},\ldots,[\hat{x}}\_{a},\hat{y}]]]$ of $\mathfrak{g}$ must also be linearly independent so that $\mathfrak{g}$ cannot be finite-dimensional.


