Degree of Parabolic Quantum Groups - Dipartimento di Matematica ...

3. Twisted polynomial algebras 33
Proof. We follow De Concini and Kaç (cf. [DCP93] theorem 6.5, page 59 or
[DCK90]). Let
z = bx r + lower term
b ∈ A be an element in the center of R and B be an algebra with R ⊂ B ⊂
z −1 R. We must show that B = R. Let us then take any element u ∈ B and
let y := zu ∈ R.
We develop y := ax s + lower term, a ∈ A, we need to show that u ∈ R
by induction on s. In order to do this we first want to prove that b divides
a. Using hypotheses 6 and 4 we deduce that there is a monomial v ∈ R such
that yv = z(uv) has the form
ax m + lower term
with x m an almost central monomial.
Next write (yv) t = z t−1 z(uv) t and remark that z(uv) t ∈ R. Now
Furthermore, we claim that
(yv) t = a t x tm + lower term.
z h = λhb h x hr + lower term,
for all h, λh ∈ k ∗ . This is easily proved, since z is central, by induction,
remarking that
We deduce that
z h = z h−1 (bx r + lower term) = bz h−1 x r + lower term
a t x tm + lower term = (yv) t
= z t−1 z(uv) t
(3.5a)
(3.5b)
= (λt−1b t−1 x (t−1)r + lower term)z(uv) t (3.5c)
This relation implies that for all t, b t−1 divides a t in A, in other words
(a/b) t ∈ b −1 A. Since A is integrally closed (hypothesis 5), we deduce that b
divides a in A as requested.
Next we claim that r divides s. In fact from y t = z t−1 ku t as for the
identity 3.5, we deduce that in the monoid N k the vector (t − 1)r divides ts
for all t, so r divides underlines and we can find an element w in R so that
x r = x s w + lower term.
We can finish our argument by induction. Assume by contradiction that
there is an element u ∈ B and not in R we may choose it in such a way that
the degree of y = zu ∈ R is minimal. By the previous argument we know
that a = bf, f ∈ A. Then fzw = zfw has the same leading term as y and
u − fw ∈ B. By induction u − fw ∈ R which gives us a contradiction.
Proposition 3.5.4. kH[x1, . . .,xn] is a maximal order.
Proof. All the hypotheses of theorem 3.5.3 are satisfied.