paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9912224 | Note that MATH is the unitary corresponding to MATH. It follows that MATH together with MATH generates MATH. Hence MATH is a weak generator. For any fixed MATH, we have MATH since the set MATH is free from MATH with amalgamation over MATH (see REF ). On the other hand, MATH by REF . Hence MATH, and thus MATH. Hence MAT... |
math/9912224 | We have that MATH, since MATH is a weak generator. But MATH, since MATH are free with amalgamation over MATH. Since each MATH implements a free action of the integers, MATH, so that MATH. Thus MATH. |
math/9912225 | This claim hold trivially for MATH. Suppose that it holds for MATH, we prove it for MATH. Let MATH and MATH. By induction the minimum energy contained in the given springs is MATH where we have for convenience let MATH denote MATH and MATH denote MATH. This energy is minimized when MATH at which point the energy takes ... |
math/9912226 | CASE: We have : MATH where we used the definition of MATH and REF . The second identity is similar. CASE: Since MATH we can compute : MATH using REF , definition of the source counital subalgebra, and the identity MATH that follows from REF . Observe that MATH is a separability idempotent CITE of MATH. CASE: Using REF ... |
math/9912226 | First, we need to check that MATH is well defined. For all MATH and MATH we have : MATH where we used definition of the target counital subalgebra, REF , and that MATH for all MATH. Next, we verify that MATH for all MATH. For all MATH we have : MATH using the identity MATH and REF . The following computation shows that... |
math/9912226 | We need to check that MATH . For all MATH, and MATH we compute MATH where we used REF and the properties of the element MATH. Also, for every MATH we have : MATH where we used that MATH commutes with the right multiplication by elements from MATH and identities from REF . |
math/9912226 | Follows from REF . |
math/9912226 | We know that MATH, where MATH is the trivial MATH-module algebra, therefore applying REF to MATH we see that MATH is a projective generating MATH-module such that MATH. Therefore, MATH and MATH are NAME equivalent. Since MATH is always semisimple (as a separable algebra), MATH is semisimple. |
math/9912228 | REF implies: MATH . The first part of the statement in the theorem follows directly from REF . If we define MATH then REF is a direct consequence of the equality REF . |
math/9912228 | We only have to prove the existence of the representation of the residues of MATH as stated in the theorem. For MATH and MATH let MATH be the NAME distribution given by the sum of the smooth densities MATH each defined on the connected component MATH of the fixed point set MATH as described in REF and by REF . We have ... |
math/9912228 | Let us fix a MATH invariant metric on MATH. This can be done by averaging any metric over the compact group MATH. Let MATH be a point in the quotient space and MATH such MATH. Let MATH be the isotropy group at MATH. Then MATH acts on the tangent space MATH and keeps invariant the tangent space MATH to the MATH-orbit of... |
math/9912228 | Let us fix a MATH invariant Riemannian metric on the base space and a MATH invariant linear connection MATH in the vector bundle. For a fixed MATH with MATH, consider, as described in the proof of REF , the isotropy group MATH and the direct sum decomposition of the tangent space at MATH as MATH modules MATH. Denote by... |
math/9912228 | It is obvious that any MATH invariant section MATH defines a section MATH by MATH. Let MATH and MATH be two MATH invariant sections in MATH which induce equal sections MATH. If MATH then there exists MATH such that MATH. Nevertheless, MATH. Then there exists MATH such that MATH. But MATH so MATH. So MATH. We reached a ... |
math/9912228 | Let us fix a metric MATH on the base space MATH. Let MATH be the orthogonal group, MATH. We will describe the orthonormal frame bundle MATH. We will show that MATH is a manifold and that the group MATH acts on it with finite isotropy groups. The MATH vector bundle we need to construct will be essentially the pull-back ... |
math/9912228 | The multiplication operator by MATH is a continuous map on MATH endowed with the NAME norm MATH, being a differential operator of order zero. The the restriction to the MATH invariant smooth functions remains a continuous operator. |
math/9912228 | It is sufficient to prove that any point MATH has a neighborhood MATH such that MATH is a submanifold in MATH. Let MATH be an linear orbifold chart at MATH, MATH. Observe that MATH. Then for any MATH we have MATH. So MATH. The fixed point set MATH is a submanifold given by linear equations on which MATH acts trivially.... |
math/9912228 | We will give the construction of the density MATH on a linear orbifold chart MATH. Let MATH be the MATH . NAME density on MATH that represents MATH. For MATH let MATH. Then MATH is a submanifold of MATH. We have MATH and MATH. We will need the following lemma: The restriction of the projection map MATH is a covering ma... |
math/9912228 | If we define MATH by MATH then MATH, but MATH will not be, in general, a pseudodifferential operator, unless MATH is smoothing and then the smooth kernel MATH of MATH is defined uniquely by MATH. We will define MATH using a partition of unity of MATH with smooth functions MATH subordinated to a cover of MATH with coord... |
math/9912228 | For each MATH such that MATH choose MATH with MATH. Because the orbibundles are isomorphic, there exists a group isomorphism MATH between MATH - the corresponding isotropy groups of MATH in MATH and one can find neighborhoods MATH of MATH in MATH which are MATH invariant and a MATH-equivariant bundle diffeomorphism MAT... |
math/9912228 | It is sufficient to prove that if MATH is a classical pseudodifferential operator that induces the zero operator on MATH, then MATH is smoothing. Let MATH be a point such that its projection MATH onto MATH is a smooth point. We will show that the total symbol of the operator MATH on a neighborhood of MATH is smoothing ... |
math/9912228 | As described in REF one can find an open cover MATH of the orbifold MATH, a family of finite groups MATH , MATH, and a vector space MATH such that MATH is isomorphic to the associated vector orbibundle of MATH orbits of the MATH vector bundle MATH and MATH can be covered with a family of open sets MATH which are MATH e... |
math/9912228 | Let MATH and denote by MATH the orthogonal group MATH. Let MATH be the MATH vector bundle, with MATH, such that the vector orbibundle of orbits is canonically isomorphic to MATH, as in REF . Let us fix a MATH bi-invariant metric MATH on MATH so that MATH and a MATH bi-invariant positive selfadjoint pseudodifferential o... |
math/9912228 | The proof is analogous to the proof of REF We construct the operator MATH acting on sections of the MATH vector bundle MATH with the help of a MATH bi-invariant elliptic selfadjoint pseudodifferential operator MATH acting on MATH. The principal symbol of MATH is equal to MATH with MATH the principal symbol of MATH and ... |
math/9912228 | Let MATH be a fixed finite atlas of linear orbibundle charts . Using a partition of unity we can decompose the complex powers of the operator MATH as MATH where MATH are pseudodifferential operators of order MATH with support inside MATH and MATH is a smoothing operator. We can arrange for MATH to take smooth sections ... |
math/9912228 | For each MATH and MATH, the density MATH can be obtained as follows: let MATH and MATH be a linear vector orbibundle chart centered at MATH. Observe that MATH and the map MATH is a local diffeomorphism. REF gives us NAME densities MATH on MATH whose integrals are equal to the residues of the zeta function and can be co... |
math/9912228 | Let MATH be a MATH-invariant open neighborhood of MATH in MATH such that MATH for MATH. Let MATH be the linear representation of MATH on the tangent space at MATH generated by the action of MATH on MATH. Choose a MATH invariant metric on MATH (this can always be done by averaging a metric over the finite group MATH). T... |
math/9912228 | Let MATH be a finite cover of MATH with open sets as in REF so that the restrictions of the MATH bundle MATH to the subsets MATH are isomorphic to MATH trivial vector bundles for some MATH. Then MATH is a finite open cover with MATH invariant open sets. Consider a partition of unity MATH subordinated to the open cover ... |
math/9912228 | We have MATH so for MATH the integral defining MATH is absolutely convergent so one can change the order of integration in REF. Consider the NAME expansion of MATH near MATH . We have MATH the double integral being the composition of the NAME transform in MATH and the inverse NAME transform in MATH evaluated at MATH. B... |
math/9912238 | Without loss of generality, MATH is Legendrian and MATH. Given MATH with MATH, let MATH denote the affine MATH-line through MATH and MATH; let MATH denote the contact plane at MATH. Thus MATH is a (round) circle for MATH, and MATH reduces to the singleton MATH. Let MATH denote the (euclidean) diameter of MATH. Because ... |
math/9912238 | If MATH is not connected, then MATH is a split link and so not fibered. If MATH is connected, then (by REF ) there is a small perturbation MATH of MATH such that MATH and MATH are ambient isotopic and MATH is in general position; of course MATH is also connected, so its closure MATH is a divide. An inspection of NAME '... |
math/9912240 | We begin by introducing some notation and recalling some basic material from CITE and CITE. Let MATH be the trivial bundle over MATH with fiber MATH. Impose the MATH pre-Hilbert space structure on the fibers and consider the smooth subbundle MATH whose fiber above MATH is MATH . The bundle MATH over MATH is MATH-equiva... |
math/9912240 | Since the argument is nearly identical to the proof of REF , we only explain the modifications one needs to make. The tangent space to the gauge group MATH at the identity is given by the space of REF completed in the MATH norm. Therefore, the tangent space to the subgroup MATH of based gauge transformations is the sub... |
math/9912240 | We first show that REF is independent of the choices of MATH and MATH. The argument of REF shows that REF depends only on the gauge orbits MATH and not on their gauge representatives. That argument also shows that REF is independent of the choice of MATH. So, it suffices to show that REF is independent of the choice of... |
math/9912241 | Assume that MATH. Then MATH is an algebraic fiber space and MATH is an elliptic curve. Let MATH be a general fiber of MATH. Then MATH is a smooth projective surface, so MATH is ample REF . Hence MATH for MATH by NAME Vanishing Theorem (compare REF ) and so MATH for all MATH. Then applying REF , we know that MATH is a M... |
math/9912241 | Note that MATH is an algebraic fiber space in the case MATH when we regard MATH as a morphism from MATH to MATH. Assume MATH. Denote a terminalization of MATH by MATH (see REF ). Let MATH be a general fiber of MATH. Then MATH is smooth and connected, so MATH is an irreducible reduced NAME surface. Therefore MATH is amp... |
math/9912241 | Assume that MATH. Then we know that MATH because of REF and MATH is an elliptic curve. Let MATH be a MATH-factorialization. Then there exists a sequence MATH such that for MATH each morphism MATH is constructed by REF and MATH has at most MATH-factorial terminal singulalities (if MATH, MATH). Put MATH. Since MATH is no... |
math/9912241 | It is enough to consider the case that MATH is a terminal REF-fold by terminalization. Since a general hyperplane section of a terminal REF-fold is smooth, we can use the proof of CITE. |
math/9912241 | Assume that MATH. Write MATH. Now that the NAME morphism MATH is an algebraic fiber space REF , a general fiber of MATH is MATH. Using REF , we know MATH. Let MATH be a MATH-factorial terminalization as in the proof of REF . Since MATH is not nef, there exists an extremal contraction MATH and we know MATH because REF s... |
math/9912241 | Taking a MATH-factorial terminalization, we may assume that MATH has at most MATH-factorial terminal singularities. By NAME theory, MATH is birational to a MATH-factorial terminal REF-fold MATH such that MATH has a NAME fiber space structure MATH, that is, MATH is an extremal contraction with MATH. If MATH, then MATH i... |
math/9912241 | Assume that MATH is infinite. Then there exists an infinite tower of normal finite étale NAME covers MATH . We know MATH by MATH, hence MATH for a sufficiently large MATH. Then MATH being strictly nef contradicts REF . |
math/9912241 | Because MATH for a general hyperplane section MATH by NAME REF, page REF, MATH is finite. Assume that MATH below. Write MATH. Let MATH be a terminalization and set MATH. Since MATH is nef and not big, MATH by REF . If MATH, then MATH is ample by REF and so it contradicts MATH. Hence NAME Theorem (compare CITE) says tha... |
math/9912241 | Let MATH be a sufficiently ample general hyperplane section and MATH. Consider the exact sequence MATH . Therefore MATH. Let MATH be a terminalization. If MATH, then MATH by REF . Furthermore by MATH, MATH . So MATH. This implies MATH is ample by REF . On the other hand, MATH is ample when MATH REF . This completes the... |
math/9912241 | Note that MATH by REF . Let MATH be the locus of cDV points of MATH. If MATH is finite, combining REF , we know that MATH is ample because MATH. If MATH is infinite, then there exists an infinite tower of normal finite NAME covers MATH such that MATH is étale where MATH is the morphism from MATH to MATH and MATH. Let M... |
math/9912241 | Assume that MATH. Then we know readily that MATH. Let MATH be a resolution of singularities. Because the NAME morphism of MATH is an algebraic fiber space CITE, so is the one of MATH, MATH and MATH. Consider the case that the equality holds. Then we can regard terminalization of MATH as a resolution of singularities MA... |
math/9912242 | The distribution of MATH has a support which remains in a fixed compact set, and it converges weakly towards that of MATH. REF follows from this and the fact that the function MATH is a limit of a decreasing sequence of continuous functions. If MATH is large enough, then the union of the supports of the distributions o... |
math/9912242 | The function MATH is harmonic in MATH, while MATH is subharmonic there, consequently MATH is a nonnegative superharmonic function on MATH. Since this function attains the value MATH by MATH, it is identically MATH by the minimum principle, therefore MATH is harmonic on MATH, and thus the support of MATH is included in ... |
math/9912242 | Let MATH be a symmetry free with MATH, then by CITE MATH and MATH are MATH-free, thus MATH is distributed as MATH. Now using the fact that multiplying with a free NAME unitary MATH does not change the MATH-distribution of MATH, we can replace the latter according to MATH, and get the following equalities of MATH-distri... |
math/9912245 | By induction on the dimension of simplices, it clearly suffices to show the following. If MATH consists of all simplices MATH such that MATH for each strict subset MATH, then MATH is a direct summand of MATH. As MATH is projective the canonical epimorphism MATH admits a section MATH . The definition of MATH yields for ... |
math/9912245 | As said above, MATH is an exact sequence of projective MATH - modules and so splits, whence the induced map MATH is surjective. The first claim follows immediately and the second one is obtained replacing skeleton by degree filtration. For the final assertion, note that the hypothesis yields a natural epimorphism of DG... |
math/9912245 | Consider on each simplex MATH the canonical flabby resolution, say, MATH. As this resolution is functorial, each component of the complex MATH is naturally again a DG MATH - module and the resulting system MATH is a resolution by DG MATH - modules. Now cut this resolution on the simplex MATH at the place MATH to obtain... |
math/9912245 | REF can be verified locally and there CITE applies. To prove REF , we may assume by the first part of REF that MATH. In this case the spectral sequence associated to the skeleton filtration on MATH converges and so it is sufficient to show the claim in case that MATH with a projective DG MATH - module MATH. In view of ... |
math/9912245 | It follows from REF that any quasiisomorphism between projective DG MATH - modules that locally vanish above is represented by an actual morphism MATH of DG MATH - modules. Such a morphism is a homotopy equivalence if the identity on the mapping cone MATH is homotopic to REF. With MATH also MATH is projective, thus MAT... |
math/9912245 | If MATH is a projective approximation, then MATH is necessarily bounded above with coherent cohomology as this holds for MATH and MATH is a quasiisomorphism. Conversely, it suffices by REF to show the existence of a projective approximation when MATH is furthermore MATH-acyclic. But if MATH is MATH-acyclic with coheren... |
math/9912245 | Let MATH and MATH be projective approximations as DG MATH - modules. By REF , MATH is a quasiisomorphism and so MATH . As MATH and MATH is a projective DG MATH - module that locally vanishes above, REF follows from REF. To obtain REF , let MATH be as before and MATH a MATH - acyclic resolution that is a morphism of DG ... |
math/9912245 | As mentioned above, the natural morphism of functors MATH is an isomorphism on the level of modules. It induces an isomorphism MATH of the corresponding derived functors on MATH, whence the functor MATH is still fully faithful on MATH. To prove the projection formula, observe first that the natural map MATH is an isomo... |
math/9912245 | According to our general assumption on DG algebras, the MATH - module MATH is in MATH, and so REF guarantee a projective approximation of MATH in form of a morphism of DG MATH - modules from a graded free DG MATH - module, say MATH. The induced morphism MATH of DG algebras induces a surjection in cohomology and the str... |
math/9912245 | In view of REF, the quasiisomorphism REF for MATH shows that MATH and the term on the right is isomorphic to MATH. According to REF, the term on the left is isomorphic to MATH, and so the first isomorphism follows. The second one follows from the fact that MATH is a projective MATH - module. The final assertion is a co... |
math/9912245 | As the direct sum of a family of connections is again a connection, we may restrict by REF to the case that MATH, where MATH is a projective MATH - module generated in a single degree, say MATH. If MATH is graded free, then MATH with MATH a finite dimensional vector space over MATH, and the collection of maps MATH defi... |
math/9912245 | That MATH is a homomorphism of right MATH - modules is easily verified by explicit calculation. Moreover, MATH whence MATH is a homomorphism of DG modules, thus a cycle. If MATH are connections, then MATH is MATH - linear and so MATH, which means that the cycles MATH are cohomologous. |
math/9912245 | If MATH, the classes in question are equal to MATH itself. If MATH, the map MATH of degree MATH is MATH-linear. As MATH is equal to MATH the claim follows. |
math/9912245 | To compare the NAME classes formed with DG algebra resolutions MATH and MATH of MATH over MATH, note first that by REF there is a free MATH algebra that constitutes a resolvent of MATH. Thus we may suppose that MATH is a free MATH-algebra. In this REF gives the independence from the choice of DG algebra resolution. The... |
math/9912245 | Consider as in REF a resolvent of MATH over MATH and choose projective approximations MATH and MATH. There is a morphism MATH lifting the given morphism MATH and the assertion follows now easily from REF. |
math/9912245 | By the preceding result and by REF the diagram commutes. |
math/9912245 | Let MATH be a quasiisomorphism into a MATH-acyclic module, MATH a projective approximation, and let MATH be a connection. Using the product rule, the composed map can be extended to a connection MATH on MATH. Hence under the natural map MATH the NAME class MATH maps onto MATH. Since the module on the left is isomorphic... |
math/9912245 | This is an immediate consequence of REF and the fact that MATH. |
math/9912245 | Let MATH be a resolvent of MATH as in REF. There are projective resolutions MATH, MATH, MATH of MATH, MATH, MATH, respectively that fit into a commutative diagram Consider connections MATH, MATH on MATH, MATH, respectively, and equip MATH with the connection MATH. Using the isomorphisms MATH can be computed as follows.... |
math/9912245 | We need to show for MATH that MATH . As the trace map is compatible with taking cup products, the diagram commutes, where as before MATH. Therefore MATH . As MATH is a derivation of degree MATH on MATH and MATH, we obtain that MATH . For homogeneous endomorphisms MATH the trace satisfies MATH, whence taking traces yiel... |
math/9912245 | In the algebraic case this is shown in CITE. In the analytic case we can proceed as follows. Let MATH be a resolvent of MATH over MATH as in REF and let MATH be a projective resolution of MATH as a MATH - module. A MATH-extension MATH of MATH gives rise to an extension MATH of MATH by MATH. Since MATH is smooth over MA... |
math/9912245 | For the convenience of the reader we include the simple argument. We may assume that MATH is a complex of finite free MATH - modules with MATH for MATH. By REF is a perfect complex and so for MATH the complex MATH has the property that MATH is exact with MATH a free MATH - module. Using induction on MATH and the long e... |
math/9912245 | Let MATH be a resolvent for MATH over MATH as in REF and choose a quasiisomorphism MATH into a MATH-acyclic complex of MATH - modules, where MATH and MATH denote the simplicial sheaves on MATH associated to MATH and MATH, respectively. We choose an algebra resolution MATH of the composition MATH so that MATH is a grade... |
math/9912245 | For a manifold, MATH, and according to NAME, CITE, see also CITE, the fundamental class of MATH is given by MATH whence the result follows from REF. |
math/9912245 | Let MATH be a resolvent of MATH and choose a free graded DG MATH - algebra MATH that provides a DG algebra resolution MATH of the composition MATH. By REF, the induced map MATH is a quasiisomorphism and so MATH constitutes a resolvent for MATH over MATH. Thus MATH and moreover MATH where MATH is the NAME functor as in ... |
math/9912245 | For suitable representative of the cotangent complexes involved there is a commutative diagram of exact sequences of complexes of MATH - modules by REF and our assumption that MATH is injective on MATH. In MATH this gives rise to a commutative diagram MATH . As well, the commutative diagram of exact sequences of MATH -... |
math/9912245 | Let MATH be the class corresponding to the extension MATH, given by a morphism MATH in the derived category. The result REF induces a commutative diagram where MATH is the NAME connection. The relative partial NAME character MATH is the image of the absolute partial NAME character MATH in MATH. Using REF and the inject... |
math/9912245 | Let MATH be the MATH infinitesimal neighbourhood of MATH in MATH so that MATH is an extension of MATH by MATH, where MATH is the maximal ideal. The map MATH is injective and applying REF repeatedly we see that MATH can be lifted to a deformation MATH on MATH for all MATH. Let MATH be a versal deformation of MATH which ... |
math/9912245 | For REF observe that the map MATH is surjective due to the versality of MATH. In turn, by REF below, the functor MATH is right exact as MATH is smooth over a subspace of MATH. Thus, MATH is right exact in MATH as well. To show REF , let MATH, MATH be as in REF . By versality, there is a morphism MATH that lies over som... |
math/9912245 | The implications REF are obvious. To show REF , note first that MATH as MATH is complete. Hence REF implies that MATH is surjective. This means that there are derivations MATH and elements MATH such that MATH, where MATH. By REF and NAME, see CITE, there is then an isomorphism MATH, where MATH is a MATH - subalgebra of... |
math/9912245 | As was already observed in the proof of REF, we have MATH, and, as MATH is formally semiuniversal, the map MATH is bijective. Hence we can write MATH with MATH. By definition, MATH, and so the dimension of MATH is just the minimal number of generators of MATH. In the extended NAME sequence MATH the module MATH vanishes... |
math/9912245 | Note that MATH as MATH injects into MATH. By REF, the vector space MATH is dual to MATH, where MATH is as in the proof of REF and MATH is the integral closure of MATH in MATH. Now the result follows from REF. |
math/9912245 | The injective hull MATH of MATH can be obtained as a limit MATH with MATH and each MATH finite artinian. As MATH is the zero map, there is an index MATH so that MATH is already zero. Restricting the map MATH to MATH yields a commutative diagram and REF follows. In order to derive REF , note that the map MATH embeds MAT... |
math/9912245 | Indeed, the functor MATH is left exact on MATH and satisfies the assumptions of REF. |
math/9912245 | Let MATH be a deformation of MATH over an artinian germ MATH, whence MATH is a MATH - module that is flat over MATH and restricts to MATH on MATH. By REF, for every coherent module MATH on MATH there is a semiregularity map MATH and by REF, this map is compatible with base change. According to CITE, the functor MATH is... |
math/9912245 | Let MATH be a deformation of MATH over an artinian germ MATH, whence MATH is a MATH - module that is flat over MATH and induces MATH on MATH. The functor MATH is exact. As the semiregularity map is compatible with base change MATH, see REF, the result follows as before from REF. |
math/9912245 | To prove this statement, we use the existence of tangent functors MATH for holomorphic mappings as constructed in CITE, see also CITE. These functors fit into an exact sequence MATH see CITE. The group MATH is canonically isomorphic to the set of all isomorphism classes of extensions of MATH by MATH. Such extension is ... |
math/9912245 | Let MATH be a resolvent for MATH over MATH and let MATH be a quasiisomorphism into a MATH-acyclic complex MATH of MATH - modules as in REF. We need to construct a map MATH . Let MATH be the topological preimage of the sheaf MATH and let MATH be the associated simplicial sheaf of rings on MATH. As the topological restri... |
math/9912245 | For every coherent MATH - module MATH the natural map MATH is a quasiisomorphism and so the claim follows from REF. |
math/9912245 | This is a local calculation, so we may suppose MATH . NAME and NAME defined by REF so that the NAME complex MATH is a MATH-resolution of the sheaf MATH. More explicitly, set MATH with the NAME differential given by MATH. Note that MATH is the free graded algebra over MATH with generators MATH, MATH, in (cohomological) ... |
math/9912245 | As MATH is a sheaf of flat MATH - modules, the complex MATH is quasiisomorphic to MATH. Therefore, the canonical projection MATH induces a quasiisomorphism MATH . In a first step, we construct a section of this projection, MATH as follows. The map MATH, where MATH is as above, realizes MATH as a subcomplex of MATH. We ... |
math/9912245 | For REF , let MATH be a curvilinear extension of MATH by MATH and let MATH be a morphism of MATH - algebras lifting the given map MATH, so that MATH corresponds under MATH to the extension MATH. By the valuative criterion of integral dependence mentioned above, MATH is in the kernel of MATH, whence MATH. Thus MATH. To ... |
math/9912245 | Replacing MATH by MATH we may assume that MATH. Choose elements MATH that form a basis of the MATH-vector space MATH and consider the natural ring homomorphism MATH given by MATH. In a first step we prove that this map is finite. In fact, the elements MATH generate the ring MATH as an algebra over MATH, and if MATH is ... |
math/9912245 | In the absolute case, where MATH, this is a result due to NAME and NAME, see CITE. Alternatively, it follows from the chain of REF where we have applied REF. To deduce the general case, set MATH. The spectral sequence MATH yields MATH . Hence the result follows from the chain of REF . |
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