Geometrie algebrique Algebraic Geometry
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Geometrie algebrique/Algebraic Geometry Un lemme de descente Arnaud BEAUVILLE et Yves LASZLO Resume – Soient A un anneau, f un element simplifiable de A , A? le separe complete de A pour la topologie (f)-adique. Nous prouvons que la donnee d'un fibre vectoriel sur Spec(A) equivaut a celle d'un fibre sur l'ouvert f 6= 0 de Spec(A) et sur Spec(A?) , et d'un isomorphisme de leurs images reciproques sur l'ouvert f 6= 0 de Spec(A?) . A descent lemma Abstract – Let A be a ring, f a nonzero divisor in A , A? the completion of A for the (f)-adic topology. We prove that the data of a vector bundle on Spec(A) is equivalent to the data of a vector bundle on the open subset f 6= 0 of Spec(A) and on Spec(A?) , together with an isomorphism of their pull back to the open subset f 6= 0 of Spec(A?) . Abridged English Version: Let A be a ring, f a nonzero divisor in A , A? the completion of A for the (f)-adic topology. We denote by Af and A?f the ring of fractions A[1/f ] and A?[1/f ] .
Voir
- isomorphisme
- a?
- a? a??a
- donnee de descente sur f?g
- gf ?
- mf ??
- homomorphisme fidele
- assertion
- a?f
G´eom´etriealg´ebrique/Algebraic Geometry
Unlemmededescente Arnaud BEAUVILLEet Yves LASZLO
Re´sume´A un anneau,– Soient f,afiulbdeAeeme´le´nilpmistnblAsee´par´e compl´et´edeApourlatopologie(fdanouqleovsnrpuoefibr´d’unn´eee.quusNo)di-a vectorielsurSpec(A)´equivauta`celled’unfibr´esurl’ouvertf6= 0 de Spec(A) et sur Spec(buqorpiceo’lrusserseueledr´esagimdtu’inosomprihmstuverA),ef6de Spec(A= 0 b) .
A descent lemma Abstracta ring,– Let A be fAA , a nonzero divisor in bA forthe completion of the (fSpec(A) is equivalent)-adic topology. We prove that the data of a vector bundle on to the data of a vector bundle on the open subsetf6on Spec(of Spec(A) and = 0 bA) , together with an isomorphism of their pull back to the open subsetf6of Spec(= 0 bA) .
b Abridged English Version:Let A be a ring,fA thea nonzero divisor in A , b completion of A for the (f)-adic topology. We denote by Afand Afthe ring of b fractions A[1/f] and A[1/fA-module we can associate an ATo any ] . f-module b b b b F = Mf, an A-module G = A⊗AM and an Af-isomorphismϕ: A⊗AF→Gf. Conversely, can we recover an A-module from the data (F,G, ϕ) ? This turns out to be true under the assumption that M isf-regularthat the homothety, i.e. fM is injective: THEOREM.−Suppose given: – anAf-moduleF; b – anf-regularA-moduleG; b b – anAf-linear isomorphismϕ: A⊗AF−→Gf. Then there exists af-regularA-moduleMand isomorphismsα: Mf−→F, b β: A⊗AM−→Gsuch thatϕis the composition of −1 β 1⊗αf b b A⊗AF−−−−→A⊗AMf−−−−→Gf.
The triple(M, α, β)is uniquely determined up to a unique isomorphism. IfFandGare finitely generated(resp.flat,resp.projective and finitely generated), thenMhas the same property. Though this looks like a classical descent problem, it does not seem to follow directly from Grothendieck’s faithfully flat descent theory: if A is not noetherian, b the map A→A is not flat.
b Proof: Put B = Af×A , and letρ: A→The homomor-B be the canonical map. phismρisfaithfulA-module M the B-module, which means that for each nonzero
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