It is proved that for any complex semisimple Lie algebra \(\mathfrak g\) the Krull dimension (in the sense of Gabriel and Rentschler) of its universal enveloping algebra is equal to the dimension of the Borel subalgebra of \(\mathfrak g\).
It is easy to see that \(\text{Kdim\,}{\mathbf U}({\mathfrak g})\geq\dim{\mathfrak g}\). The opposite inequality is the hard part of the proof.
The conjecture was previously proved by S. P. Smith for \({\mathfrak g}=\text{sl}(2,\mathbb{C})\) [J. Algebra 71, 189-194 (1981; Zbl 0468.17003)].
Let \(G\) be a simply connected semisimple algebraic group with the Lie algebra \(\mathfrak g\), \(U\) be a maximal unipotent subgroup of \(G\) and set \(X=G/U\). It was shown previously by the author [Lect. Notes Math. 924, 173-183 (1982; Zbl 0478.17003)] that the fact in question would follow from the inequality \(\text{Kdim\,}D(X)\leq\dim X\), where \(D(X)\) is the ring of globally defined differential operators in the sense of Grothendieck. In the same paper the desired inequality was established when \(\mathfrak g\) is a direct sum of copies of \(\text{sl}(2,\mathbb{C})\) and in another paper of the author [J. Algebra 102, 39-59 (1986; Zbl 0595.17006)] when \({\mathfrak g}=\text{sl}(3,\mathbb{C})\). These were the only cases known.
In a recent paper of R. Bezrukavnikov, A. Braverman and L. Positselskii [J. Inst. Math. Jussieu 1, No. 4, 543-557 (2002; Zbl 1044.16020)] among other things it was proved that \(D(X)\) is a Noetherian ring. In this paper it is explained how this result implies that \(\text{Kdim\,}D(X)\leq\dim X\) and consequently \(\text{Kdim\,}\mathbf U({\mathfrak g})\geq\dim{\mathfrak g}\).