proceedings of the
american mathematical
society
Volume 87, Number 3, March 1983
THE CENTER OF THE QUOTIENT DIVISION RING
OF THE UNIVERSAL ENVELOPE OF A LIE ALGEBRA
ALFONS I. OOMS
Abstract. Let L be a finite dimensional Lie algebra over a field k of characteristic
zero, D(L) the quotient division ring of U(L). We compare the center Z(D(L))
with Z( D(H)) where H is an ideal of L of codimension one.
Let k be a field of characteristic zero. We first recall the following result due to
Bernât [1;3, p. 134-135].
Lemma 1. Let A be an associative algebra over k with a unit element, B a subalgebra
containing \, x E A an element symmetric (i.e. [x, B] C B) and transcendental with
respect to B such that A = B[x\. Hence every element a E A can be considered as a
polynomial in x with coefficients in B. We denote by deg a the degree (in x) of a. We
assume further that A and B have quotient division rings K and L (so that L is the
division subring of K generated by B ) and that A is a set of derivations of A leaving
stable B such that deg(£x) < 1 for all E E A. We extend the elements of A to
derivations of K and denote by KA the set of elements of K which are annihilated by all
£ G A. Then the following hold:
(1) x is symmetric and transcendental with respect to L and if a E A then degB a =
deg¿ a. LD A — B and the algebra L[x] has a quotient division ring which is K.
(2) K^ is the quotient division ring of L[x]A.
(3) Suppose KA ¥= £A (i.e. L[x]A t- L) and let z be an element of L[x]A of minimal
degree > 0. Then z is transcendental over LA, L[x]A = LA[z] and KA = LA(z).
Corollary.
Let us assume in addition
to the hypotheses of the lemma that
deg(£x) = 0 for all £ G A. Then z is of the form ex + d with c E LA, c ¥= 0, and
d E L.
Proof.
cn_xx"~]
Choose uEL[x]x\L.
We can find c, G L such that
+ ••• +cxx + c0 withc„ ¥" Oand/i s* 1. Let £ G A; then
£« = (Ecn)xn
+ cnEx"
+(Ecn_x)x"-]
+ c„_xExn~]
u = cnx" +
+ ■■■ +£c0.
On the other hand, for each m > 1 we have that
m
Exm = 2 xm~'(Ex)x'~]
i= i
- m(Ex)xm~
' + terms of lower degree.
Received by the editors June 1, 1982.
1980 Mathematics Subject Classification. Primary 17B35; Secondary I6A08.
Key words and phrases. Finite dimensional
ring.
Lie algebra, universal enveloping algebra, quotient division
*'J|983 American
Mathematical
Society
OO02-9939/82/OOOO-O87O/$OI50
394
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395
THE UNIVERSAL ENVELOPE OF A LIE ALGEBRA
Hence,
(*)
Eu = (Ecn)x"
+ (ncnEx
+ Ecn_x)x"
""' 4- terms of lower degree.
This is a polynomial with coefficients in L since £c, G L and Ex E B C L. Now,
u G L[x]A implies that £w = 0 for all £ G A. Taking into account that x is
transcendental
with respect to L, we obtain Ecn = 0 and ncnEx + £c„ _, = 0 so that
E(ncnx + c„_|) = 0 for all £ G A. Consequently. nc„x + c„ , G L[.v]A with cH G
LA, c„ ¥= 0. Therefore deg z = 1. Finally, it follows easily that r = c.v + ¿/ for some
c E Là,c^0,d
E L.
Remark. If we assume furthermore that A2, # ßA (i.e. /1A (¿ Z? and a fortiori
A"A# LA), then a similar argument shows that we can find an element : = bxx + b2
G/1A with 6, G 5a, bx ^0, and/>2 G £ such that L[x]A = LA[r] and KA = LA(r).
Next, let I be a finite dimensional Lie algebra over A, U(L) its universal
enveloping algebra and D(L) the quotient division ring of U(L). We denote by
Z(U(L)) (resp. Z(D(L)))
the center of U(L) (resp. D(L)). We now recall the
following result due to Dixmier [2, p. 339].
Lemma 2. Let H be an ideal of L of codimension one. If Z(U(L))
Z(U(H))CZ(U(L))and
<£.U(H ) then
Z(U(H)) = Z(U(L)) D U(H).
It follows that we have either Z(U(L)) C Z(U(H')) or Z(U( H )) C Z(U(L)).
We are now in a position to prove the following
Theorem.
Let H be an ideal of L of codimension one, x G L \ H. Then we have:
\.(a)JfZ(D(L)) C Z(D(H))thenZ(U(L)) C Z(U(H)).
(b)IfZ(D(H)) C Z(D(L)) then Z(U(H)) C Z(U(L)).
2. (a) //Z(U(L))C Z(U(H)) then Z(D(L))C Z(D(H)).
(b) // Z(U(H))CZ(U(L))
then Z( D( H )) E*Z( D( L )) and in this case Z( D( L )) is
a purely transcendental
extension of Z( D( H )) of degree one. Indeed,
there exists an
element z = bxx + b2 E Z(U(L)), transcendental over Z( D(H)), with bx G Z(U( H )),
b] =£ 0, and b2 E U(H) and such that Z(D(L)) is generated as a field by : and
Z(D(H)). Finally, Z(D(H)) = Z(D(L)) n D(H).
Proof.
1. (a) Z(U(L)) C Z(D(L)) C Z(D(H))
C D(H).
Hence Z(U(L))C
D(H)n U(L)= U(H). Consequently Z(U(L)) C Z( U(H)).
(b) Z(U(H)) C Z(D(H)) C Z(D(L)). Therefore Z(U(H)) C Z(D(L)) n U(L)
= Z(U(L)).
2. Put A = U(L), B = U(H) and A = ad L. Then A, B. x and A satisfy the
conditions of Lemma 1. Indeed, U(L) and U(H) have quotient division rings D(L)
and D(H). Since H is an ideal of L, every derivation £ G ad L leaves stable U(H).
This holds in particular for ad x, which means that x is symmetric with respect to
U(H). By the Poincaré-Birkhoff-Witt
Theorem x is also transcendental over U(H)
and U(L) — U(H)[x]. Hence each element of U(L) can be written as a polynomial
in x with coefficients in U(H) (say on the left). Since H is an ideal of L of
codimension one, we have that [L, L] C H and so deg(£r) = 0 for all £ G A =
ad L. We now proceed with the proof.
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396
A. I. OOMS
(a) Suppose Z(U(L)) C Z(U(H)). Take y E Z(U(H))\Z(U(L))
and consider the
derivation £ = ad y in U(L). Clearly, E(U(H)) = 0 and Ex = -ad x(y) E U(H),
Ex ¥= 0. Take
w = uv~{ and
h£ü = H'£t;.
Now suppose
now w G Z(D(L)). We can choose elements w, v E U(L),v¥=0,
with
such that deg v is minimal. From u — wv we obtain Eu = (£k»)ü +
Write v = cnx" + c„_ ,x"~ ' + ■■• +c0 with c, E U(H) and c„ ¥= 0.
£t> ¥= 0. This forces n > 1. Then formula (*) gives us
£u = ( £c„ )x" + ( nc„Ex + £c„_, )jc"-1 + terms of lower degree
= wc„( Ex )x"
' + terms of lower degree,
since Ecl = 0 for all /'. In particular deg(£u) = degu — 1, which contradicts the
minimality of deg v since w = (Eu)(Ev)~]. Therefore Ev = 0, which implies that
n = 0 and so v = c0 G U(H). It follows that £w = 0 and u G U(H). Consequently,
w = uv-[ E D(H) and thus u>G Z(£>(//)). We have shown that Z(D(L)) C
Z(D(H)). Moreover Z(D(L)) ¥= Z(D(H)) since equality would imply Z(U(L)) =
Z(U(H)) by (1).
(b) Suppose Z(U(H))CZ(U(L)).
Then Z(U(L)) £ U(H) and Z(U(H)) =
Z(U(L)) n i/(//). In other*words, Í7(L)A (¿ t/(//) and Z(U(H)) = U(H)A. Then
the corollary and the remark following it assures the existence of an element
z = bxx + b2 G t/(L)A = Z(t/(L)) with ¿>,G L/(//)A = Z(U(H)),bx ¥= 0, and 62 G
¿/(/Y), such that D(//)[jc]a = D(H)A[z\ and D(L)A = D(H)A(z) with z transcen-
dental over D( H )A.Clearly £>(L)A = Z( D(L)), so it remains to show that D(Hf
=
Z(D(H)). For this purpose we take w G Z(D(H)). Then 0 = [w, z] = [w, bx]x +
bx[w, x] + [w, b2] = bx[w, x]. Hence [w, x] — 0 which implies that w G Z(D(L)).
So we have proved that Z(D(H))
Z(D(H))
Corollary.
C Z(D(L)). It follows that
= Z(Z)(L))
n D(/7) = Z)(L)A n £>(//) = D(H)A.
Let H be an ideal of L of codimension one such that Z(U(H))
Z(U(L)). Then we have either Z(D(H))C
¥•
Z(D(L)) or Z(D(L)) C Z(D(H)).
References
1. P. Bernât. Sur le corps enveloppant d 'une algèbre de Lie résoluble. Bull. Soc. Math. France, Mém No.
7(1966). MR 37 #6334
2. J. Dixmier, Sur les représentations
unitaires des groupes de Lie nilpotents. II, Bull. Soc. Math. France
85(1957). 325-388. MR 20 #1928
3._Algebres
enveloppantes. Cahiers Scientifiques, fase. 37. Gauthier-Villars.
Paris. 1974. MR 58
# 16803a
Department
of Mathematics,
Universitair
Centrum Limburg, 3610 Diepenbeek, Belgium
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