MEMOIRS

1.3. PRIME IDEALS AS ANNIHILATORS 35Hence the zero morphism 0 : S e −→ L(d + n) satisfiesr x = p(n) ◦ 0. By theHomotopy-Lemma [45, Lemma B.1] (applied to this special situation) there is ans ∈ Hom(L(d),L(n)) such that r = sπ. Hence P ′ = Rπ follows.Note that End(S e ) ≃ M e (D) (whereD = End(S) is a skew field) is a prime ringand σ induces an automorphism of this ring. Using the formula (s∗r) x = σ m (s x )◦r x(where m is the degree of r) it is sufficient to show that for some n ≥ 0themapR n −→ End(S e ),r↦→ r xis surjective. But this follows from Lemma 1.2.2.(3) Let e =1. Ifa, b ∈ R are homogeneous such that ab ∈ P then (ab) x =0.Since End(S) isaskewfield,a x =0orb x = 0, hence a ∈ P or b ∈ P and P iscompletely prime. For the converse, if e>1, then there are non-zero matrices A,B ∈ M e (End S) such that A · B = 0. By the proof of (2) there are homogeneous a,b ∈ R such that b x = B and a x = σ −m (A) (wherem is the degree of b). It followsthat ab ∈ P , but a ∉ P and b ∉ P . Hence P is not completely prime.□1.3. Prime ideals as annihilatorsIn this section we give another description of the homogeneous prime idealswhich occur in Theorem 1.2.3. We assume R =Π(L, σ), where σ is efficient. Asusual, we set F (n) =σ n (F ) for all F ∈H.1.3.1 (Fibre map). Let S be simple, concentrated in x, lete = e(x). For anf ∈ Hom(L, L ′ ), where L ′ is some line bundle, we have the following commutativediagram with universal exact sequenceswith fibre map f x .0 LfπL(d)f ′ S e f x00 L ′ π ′ L ′ (d) S e 0,1.3.2 (1-irreducible maps). Let f be a (non-zero) morphism between line bundles.Then f is called 1-irreducible, if whenever f = gh with morphisms g andh between line bundles, then g or h is an isomorphism. The following facts areobvious:(1) Each non-zero map between line bundles has a factorization into 1-irreduciblemaps.(2) A morphism between line bundles is 1-irreducible if and only if its cokernelis a simple object.(3) Each simple object is cokernel of a 1-irreducible map. Moreover, one of theline bundles can be chosen arbitrarily.(4) If u : L −→ L(n) is a 1-irreducible map, then it is an irreducible element inR. The converse does not hold in general, in orbit case III.The following lemma is a fundamental statement on 1-irreducible maps.Lemma 1.3.3. Let S be simple, concentrated in x, letπ = π x and e = e(x),and let0 −→ L −→ uL ′ −→ S −→ 0