MEMOIRS

72 4. COMMUTATIVITY AND MULTIPLICITY FREENESS(where on the right hand side the multiplication and the power of x Lare taken inthe orbit algebra sense), and Π(L, σ x ) embeds naturally.□Corollary 4.3.4. Let X be homogeneous. Assume that the function field iscommutative. Then X is multiplicity free. Moreover, there is some unirational pointx such that the orbit algebra Π(L, σ x ) is commutative graded factorial.Proof. Take any point x and form the orbit algebra R =Π(L, σ x )withrespectto the (not necessarily efficient) associated tubular shift σ x . By the precedingproposition R is commutative. Moreover, by Serre’s theorem 2.1.2, R is a homogeneouscoordinate algebra for X, and hence “classical” algebraic geometry showsthat X is multiplicity free (see for example [55]). In particular, there exists someunirational point x and the assertion follows since σ x is exhaustive.□The orbit algebras R =Π(L, σ x ), where σ x is exhaustive and where R is commutative,are described (in case char k ≠ 2) in Theorem 4.2.1. From this we get theexplicit form of the function fields as in Theorem 4.3.1. This explicit description followsagain from the next theorem which provides also the proof for the implication(3)⇒(1) in Theorem 4.3.1.Theorem 4.3.5. Let X be homogeneous. Assume that for all rational pointsx ∈ X we have e(x) =1. Then for each rational point x the orbit algebra Π(L, σ x )is commutative.Moreover, if ε = 1, then there is a finite field extension K/k such thatΠ(L, σ x ) ≃ K[X, Y ], whereX and Y are central variables of degree one. If ε =2and char(k) ≠2, then there is a finite field extension K/k such that Π(L, σ x ) ≃K[X, Y, Z]/(−aX 2 − bY 2 + abZ 2 ),witha, b ∈ K ∗ such that −aX 2 − bY 2 + abZ 2 isan anisotropic quadratic form over K.Proof. There is a rational point x. By assumption x is unirational. Thus theassociated tubular shift σ x is efficient and the orbit algebra R =Π(L, σ x )gradedfactorial. We have R = R 0 〈R 1 〉 and [R 1 : R 0 ]=ε+1. Moreover, [R n : R 0 ]=εn+1.(Note that by Corollary 0.6.2 in case ε = 1 the underlying tame bimodule M isnon-simple.)Fact: Let u be a non-zero element of R, homogeneous of degree one. Then uis prime.Namely, u is irreducible, hence by 1.3.3 a divisor of some prime element π,which because of deg(u) = 1 is associated to a rational, hence multiplicity freepoint. It follows from 1.6.5 that u equals π up to some unit and hence is primeitself.It follows that there are rational points y (and z) such that the prime elementsπ x , π y (and π z ) are linearly independent over R 0 and R = R 0 〈π x ,π y 〉 (in case ε =1)or R = R 0 〈π x ,π y ,π z 〉 (in case ε = 2). Moreover, by 1.7.1 the prime π x is central.We have to show(a) R 0 is commutative;(b) π y (and π z ) commutes with each element from R 0 ;(c) π y π z = π z π y (in case ε =2).