# Is the universal enveloping algebra of a finite-dimensional Lie algebra (left) noetherian?

The universal enveloping algebra of a Lie algebra $\mathfrak{g}$ is a flat deformation of $S(\mathfrak{g})$, so these algebras should be similar in many ways. Does at least this general similarity hold?

• The universal enveloping algebra of a finite dimensional Lie algebra is a so-called G-alegbra, hence is left and right Noetherian (see e.g. singular.uni-kl.de/Manual/3-1-5/sing_510.htm). Note that this includes quantized enveloping algebra as well. – Adrien Jan 10 '13 at 10:47
• Adrien, thank you very much for the reference. – Oleg Jan 10 '13 at 11:36

Yes, if a filtered ring $R$ has the property that its associated graded ring is Noetherian, then $R$ is Noetherian. Universal enveloping algebras have a PBW filtration such that the associated graded algebra is $S(\mathfrak{g})$. This is proved in Noncommutative Noetherian Rings by McConnell, Robson, Small - see sections 1.6 and 1.7.
• I don't have a reference, but I can just explain you how I see it. $U(g)$ is the quotient of the tensor algebra $T(g)$ modulo the relations $x\otimes y−y\otimes x−[x,y]$ for $x,y\in g$. Let $k$ be the base field, $t$ be a variable and $B$ be the quotient of the algebra $T(g)\otimes_k k[t]$ modulo the relations $x\otimes y−y\otimes x−t[x,y]$. Then $B$ is (I hope --- I haven't checked) a free $k[t]$-module with the usual PBW basis and an algebra over $k[t]$, the fiber of $B$ over the point $t=0$ is $S(g)$, and the fiber over $t−a$ for $a\in k^*$ is $U(g)$. – Oleg Jan 10 '13 at 13:34