MEMOIRS

42 1. GRADED FACTORIALITYOne can summarize the preceding results by saying that the non-zero normalelements modulo units form a UF-monoid in the sense of [16, Ch. 3], see also [47].Moreover, as in [47, Prop. 2.2] it follows that each non-zero homogeneous idealcontains a normal element. (R is said to be (graded) conformal.)Obviously, each homogeneous element r ∈ R, r ≠ 0, is a product of irreducibleelements. If a normal element is irreducible it is prime.1.6.5 (Ringtheoretic meaning of e(x) andf(x)). Let u be a homogeneous elementof R with cokernel S. Obviously, if S is simple, then u is irreducible. Theconverse also holds in orbit cases I and II, that is, here irreducible elements and 1-irreducible maps are the same concept. Hence, with the notations of Theorem 1.2.3,in orbit cases I and II the prime element π = π x is a product of e = e(x) irreducibleelements: π = u 1 u 2 ...u e . Moreover, each u i is of degree f = f(x) =deg(S). □In orbit case III we get a slightly different result, since then there are irreducibleelements (even of degree one in R) with cokernel of length two; they may have twodifferent points as support.Proposition 1.6.6. With the same notations as in Theorem 1.2.3, assumeorbit case III. Then the following holds for π = π x .(1) If f =deg(S) is even, then π = u 1 u 2 ...u e , where all u i ∈ R are irreducibleof degree f/2 with cokernels isomorphic to S.(2) If f =deg(S) is odd, then e is even and π = u 1 u 2 ...u t ,wheret = e/2 andall u i ∈ R are irreducible of degree f with cokernels isomorphic to S 2 .Proof. A chain of projections S e ↠ S e−1 ↠ ... ↠ S yields a factorizationπ = v 1 v 2 ...v e with v i : L (e−i) −→ L (e−i+1) ,whereL (i) are line bundles such thatL (0) = L and L (e) = L(d). Moreover, the cokernels of the v i are isomorphic to S,hence deg L (i+1) =degL (i) + f. Iff is even, all L (i) lie in the same orbit, whereasin case f is odd, precisely the L (i) ,wherei is even, lie in the same orbit as L. So,inthe first case, all u i := v i are in R and irreducible. In the second case, the elementsu i = v 2i−1 v 2i are in R and irreducible.□Note that in orbit case III there are also irreducible elements in R with indecomposablecokernel S x(2) of length two.The following is a modification of Lemma 1.3.3 so that there is some left andright symmetry:Lemma 1.6.7. Let S be simple, concentrated in x, andletu ∈ R be irreduciblesuch that0 −→ L −→ uL(f) −→ S −→ 0is exact. Then there is a morphism v ′ ∈ R d−f such that π x = uv ′ .Proof. Let π = π x = ab be a product of homogeneous elements. Then thecokernel of a (or b, resp.) is of the form S f for some 0 ≤ f ≤ d. Then, byuniversality, there is a b ′ such that π = b ′ a, and since b ′ π = πb, wegetb ′ = γ(b),where γ : R −→ R is the automorphism such that πr = γ(r)π for each r ∈ R. Nowapply Lemma 1.3.3 to γ −1 (u).□Remark 1.6.8. There is also a version for irreducibles with cokernel S 2 .Thesameargumentshows: