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math/0011090 | It follows from REF and the fact that MATH. |
math/0011090 | The fact that MATH is closed follows easily from MATH REF . Now, we observe: MATH . We have a surjective map: MATH induced by inclusion, and MATH . Finally, using REF we compute: MATH . |
math/0011090 | Let MATH and MATH be given, with MATH. Using the definition of MATH, we compute: MATH . |
math/0011090 | From REF it follows that MATH and MATH are MATH-orthogonal; the conclusion follows from REF and the observation that the kernel of MATH in MATH is contained in MATH. |
math/0011090 | The nondegeneracy assumption of MATH on MATH means that it is represented by an injective operator on MATH. Using the compact inclusion of MATH in MATH, it is easily seen that REF defines a bilinear form which is represented by a compact perturbation of an isomorphism of MATH. Using the additivity of the NAME index of operators it is easily proven that MATH is represented by a NAME operator of index zero in MATH; hence it follows that MATH is indeed represented by an isomorphism of MATH. Using REF , we have inclusions: MATH . Since MATH is an isomorphism, the dimension of the third member in REF equals the dimension of the annihilator of MATH in MATH. The dimension of this annihilator coincides with the codimension of MATH in MATH; by REF , it follows that the inclusions in REF are equalities, which concludes the proof. |
math/0011090 | It is essentially identical to the proof of CITE. |
math/0011090 | Let us denote by MATH the symmetric bilinear form on MATH given by MATH; we regard MATH as a linear map from MATH to MATH by identifying MATH. By the definition of the chart MATH, we have: MATH therefore, we obtain: MATH . It is an easy observation that, since MATH is not a focal instant, we have: MATH where the direct sum is MATH-orthogonal, and so: MATH . The conclusion follows from the fact that, using REF, it is easily seen that the isomorphism MATH carries the restriction of MATH to the bilinear form MATH. |
math/0011090 | By standard regularity arguments (see CITE), REF shows that MATH and MATH are MATH in MATH, which obviously implies that MATH is MATH in MATH. Similarly, REF shows that MATH is of class MATH on MATH; from REF we deduce that MATH is surjective for MATH. The regularity of the family MATH follows then from REF . As to the regularity of the family MATH, we have to show that MATH is surjective for MATH. For MATH it follows directly from the definition of MATH in REF. For MATH, the surjectivity follows from REF . |
math/0011090 | Let MATH be fixed. Using REF we conclude that MATH is nondegenerate on MATH, and therefore MATH is nondegenerate on MATH. Keeping in mind the result of REF , the conclusion follows by applying REF . |
math/0011090 | By REF , the term on the left hand side of REF does not depend on the choice of MATH and MATH. In order to facilitate the computation, we make a suitable choice of the curves MATH and MATH, as follows. Write MATH and MATH, with MATH; by REF , the maps MATH and MATH are solutions of the reduced symplectic system REF, hence they define maps of class MATH on the entire interval MATH. We set MATH again by REF , MATH and MATH are in MATH for all MATH, and so the maps MATH and MATH defined by: MATH are in MATH. Obviously, the maps MATH and MATH are of class MATH, and therefore they define MATH-valued MATH-maps. Once the choice of MATH and MATH is made, we compute as follows: MATH using that MATH and that MATH, MATH do not depend on MATH. |
math/0011090 | Recall that the kernel of MATH is given in REF . For MATH, we have MATH, and MATH is negative definite in MATH. From REF it follows that MATH is negative semi-definite. To conclude the proof we have to show that, if MATH and MATH, then MATH. This follows easily from REF . |
math/0011090 | See CITE. |
math/0011090 | Obviously, for MATH, MATH. The conclusion follows from REF . |
math/0011090 | The proof will be done for NAME - NAME systems (see Subsection REF) with coefficients of class MATH; the geometrical version of the theorem is an immediate corollary. We first consider the case that there are only a finite number of focal instants for the NAME - NAME system REF and that MATH is not a focal instant for the reduced symplectic system REF. Observe that, by REF , the number of focal instants for the reduced symplectic system is finite. By REF , the function MATH is piecewise constant, with jumps at the focal instants of either the NAME - NAME system or the reduced symplectic system. By REF , MATH for MATH sufficiently close to MATH. Let MATH be a focal instant for either the NAME - NAME or the reduced symplectic system; we compute the jump of MATH at MATH. Choose a Lagrangian MATH of MATH which is complementary to both MATH and MATH; such Lagrangian always exists (see for instance CITE). Consider the extended index form MATH defined in REF corresponding to the choice of the bilinear form MATH, where MATH. With such a choice, we have that MATH is nondegenerate on MATH for MATH near MATH (see REF ). Using REF , for MATH sufficiently close to MATH we have: MATH . Using REF , the jump of the function MATH as MATH passes through MATH equals the dimension of MATH, which by REF is equal to the multiplicity of MATH as a focal instant for the reduced symplectic system. The sum of these multiplicities as MATH varies in MATH equals MATH by REF . By REF , the jump of the function MATH as MATH passes through MATH is equal to MATH for MATH sufficiently small. Since MATH is additive by concatenation, the sum of these jumps equals minus the NAME index of the NAME - NAME system. This concludes the proof for the case of a NAME - NAME system REF whose focal instants are isolated and such that MATH is not focal for the reduced symplectic system REF. Consider now the more general case of a NAME - NAME system for which MATH is not focal for the associated reduced symplectic system. Let MATH be a sequence of real analytic curves of MATH-symmetric linear endomorphisms of MATH that converges uniformly to MATH on MATH. Let MATH be the index form of the corresponding NAME - NAME problem and denote by MATH the associated space defined as in REF. Then, MATH converges to MATH in the operator norm topology. Since MATH is nondegenerate (see REF ) and it is represented by a compact perturbation of a negative isomorphism of MATH REF , it follows that, for MATH sufficiently large, MATH REF . Moreover, since MATH is nondegenerate, MATH is represented by a compact perturbation of a positive isomorphism of MATH and MATH converges to MATH, by REF we have MATH for MATH sufficiently large. The conclusion in the case that MATH is not focal for the reduced symplectic system follows from the stability of the NAME index by uniformly small perturbations of the coefficient MATH in REF (see CITE). In the general case that MATH may be focal for the reduced symplectic system, the conclusion follows from the fact that the functions MATH and MATH are left-continuous at MATH. The left-continuity of MATH follows from REF ; the left-continuity of MATH follows from REF and from REF . |
math/0011090 | It follows directly from REF and the proof of REF . |
math/0011091 | There exists a hyperplane MATH in MATH such that MATH if and only if there exist MATH, not all zero, such that MATH for all but finitely many MATH. The last condition is equivalent to MATH, as a non-zero rational function has only finitely many zeros (compare REF); now the result follows. |
math/0011091 | Set MATH and let MATH. Then MATH if and only if MATH; that is, MATH, where MATH . This shows REF . Now MATH is MATH-isomorphic to a MATH-subspace of MATH. Since MATH (see REF), REF follows. |
math/0011091 | Since MATH, MATH for each MATH and each MATH so that MATH. On other hand, as MATH is base-point-free, for each MATH there exists MATH such that MATH; now the result follows. |
math/0011091 | The result follows from the facts that MATH and that MATH, where MATH, MATH being the matrix defining MATH and MATH. |
math/0011091 | Let MATH be a local parameter at MATH. CASE: The proof follows from the equivalences MATH . CASE: The implication REF is trivial. CASE: Let MATH. Then MATH which belong to MATH by REF . Thus, once again by REF we conclude that MATH. CASE: Let MATH be a hyperplane in MATH such that MATH. Denote by MATH the rational function on MATH, obtained by dividing the equation of MATH by the one of MATH. Then we obtain a rational function on MATH, namely MATH (that is, the pull-back of MATH by MATH). The function MATH is regular on MATH and hence MATH is regular on MATH. Moreover, by the election of MATH, we have that MATH and therefore from REF we conclude that MATH. From the definition of MATH we even conclude that MATH. Now suppose that MATH is non-singular; let MATH be a local parameter at MATH and set MATH (the ramification index at MATH). By considering MATH as a function on MATH we have MATH, and by the product formula we also have MATH . Now take MATH such that every point in MATH is non-singular (this is possible because MATH has a finite number of singular points and so we can apply NAME 's theorem). Then from the above equation, MATH . It turns out that MATH (compare CITE), and the result follows. |
math/0011091 | Since MATH, there exists MATH and the result follows. |
math/0011091 | Since MATH and MATH REF follows. REF implies REF . |
math/0011091 | It follows from REF and MATH. |
math/0011091 | We have to show that MATH for each MATH. Suppose that MATH for some MATH. Then MATH and hence MATH. This implies MATH. |
math/0011091 | If MATH were not separable, then MATH, MATH by CITE. Then MATH would be a non-gap at MATH, a contradiction. |
math/0011091 | We have that MATH is the genus of the underlying curve, say MATH. For a canonical divisor MATH on MATH, we observe that MATH by the NAME theorem. Let MATH. From REF , there exists MATH such that MATH for MATH. Then MATH and being the map MATH injective, the result follows. |
math/0011091 | We apply the case MATH in REF . An easy computations shows that MATH. Then MATH and so MATH cannot be NAME. |
math/0011091 | The statements are consequence of REF and the following property of binomial numbers: if MATH, MATH are the MATH-adic expansion of MATH, then MATH. |
math/0011091 | Write MATH. The coefficient of MATH in MATH can be read off from MATH, and the claim follows. |
math/0011091 | If there exists a non-trivial relation MATH with MATH, then we would have MATH for MATH and so MATH, a contradiction. |
math/0011091 | Let MATH, MATH, be the local expansion of MATH at MATH. Set MATH. Then MATH and the result follows. |
math/0011091 | CASE: From REF and its proof, the vectors MATH are MATH-linearly independent and MATH . Let MATH. For MATH, we have vectors of type MATH with MATH zeros and where MATH denotes an element of MATH. Since the last MATH entries of the vector MATH are zeroes, REF follows. CASE: From REF , MATH so that MATH. |
math/0011091 | CASE: Suppose that MATH. Then MATH with some MATH, because MATH. Then we replace the row MATH by MATH in MATH so that MATH, a contradiction to the minimality of MATH. CASE: As in REF we have that MATH if and only if MATH (or equivalently MATH for MATH) for any MATH. Then each MATH is a MATH-power by REF , and so MATH divides MATH by REF ; that is, MATH divides MATH. CASE: Clearly MATH are MATH-linearly independent. Let MATH. Since MATH, there exists a relation of type MATH with MATH. We claim that MATH for MATH. Indeed, suppose that MATH for some MATH. Then by replacing MATH by MATH in MATH we would have that MATH, a contradiction to the minimality of MATH. This finish the proof. |
math/0011091 | From REF , MATH . If MATH, we would have MATH a contradiction. This proves REF . Now REF follows from REF . |
math/0011091 | CASE: It follows from MATH. Note that this result does not depend on the minimality of MATH. CASE: By the product rule (compare REF), we have MATH . Then MATH . By REF we can factor out MATH in each row of MATH, and REF follows. CASE: The proof is similar to REF but here we use the chain REF instead of the product rule. We have MATH where MATH and MATH. Hence MATH and again by REF we can factor out MATH in each row of MATH. |
math/0011091 | (Following CITE) By means of suitable central projections CITE one can assume that MATH. Let MATH be the linear series on MATH in REF , and MATH the MATH-orders. By REF , MATH are the MATH-orders. Then, for each MATH, MATH by REF , and the result follows. |
math/0011091 | We can assume that MATH is a MATH-base. Now there exist MATH such that MATH and the result follows. |
math/0011091 | Set MATH. From the proof of REF with MATH we have a local expansion of type MATH with MATH and the result follows. |
math/0011091 | Let MATH be such that MATH. We apply REF to a point MATH; that is, such that MATH for each MATH. Let MATH. Then the vectors MATH, MATH, are MATH-linearly independent and the result follows. |
math/0011091 | CASE: Set MATH. We have MATH by hypothesis. Then MATH by REF ; that is, MATH for each MATH. CASE: Follows from REF . |
math/0011091 | From REF we can assume that MATH is a MATH-base. Let MATH be the hyperplane corresponding to MATH, where MATH are homogeneous coordinates of MATH. Let MATH be a hyperplane. Then MATH if and only if MATH, since MATH for each MATH. Thus MATH that is, it has dimension MATH. In addition, it is generated by the vectors MATH by the proof of Scholium REF . |
math/0011091 | We have MATH by REF . This proves REF . REF follows from REF is trivial. |
math/0011091 | Let MATH be the minimum positive integer such that MATH. Then MATH and MATH so that MATH. Now we have MATH by REF , where MATH are the gaps at non-Weierstrass points. Since MATH we have a contradiction and the result follows. |
math/0011091 | CITE Suppose MATH. Then for MATH, MATH is a canonical divisor. In particular, MATH for MATH. We consider the isogeny MATH on the Jacobian variety MATH associated to MATH, and the natural map MATH, MATH. Note that MATH if and only MATH since MATH. Then MATH says that there are infinitely points in MATH belonging to the kernel of MATH, a contradiction since this kernel is finite CITE. |
math/0011091 | Let MATH be the smallest integer such that MATH is a MATH-linear combination of MATH. Since MATH is a MATH-base of MATH, then MATH and the result follows. |
math/0011091 | Similar to the proofs of REF . |
math/0011091 | From REF , there exists MATH such that MATH. Hence from REF , MATH for MATH and MATH for MATH. Since MATH are MATH-l.i, from REF follows that MATH for MATH; thus MATH for MATH. The same argument yields MATH; in fact, MATH by the definition of MATH in the proof of REF . Suppose that MATH. Then by REF the vectors MATH would be linearly dependent over MATH so that MATH. This is a contradiction because MATH are MATH-linearly independent. A similar argument shows that MATH if MATH. |
math/0011091 | REF follows from REF taking into consideration that MATH. REF follow as in REF . |
math/0011091 | Let MATH. For MATH, we have MATH by REF . So MATH are the first MATH orders of the morphism MATH, where MATH are as in REF . Then the resul follows from the MATH-adic criterion REF . |
math/0011091 | Set MATH and let MATH, where MATH is a separating variable of MATH. Let MATH be the orders of MATH. Then MATH for each MATH by REF , hypothesis and REF . Then, as MATH, MATH, and the result follows from REF . |
math/0011091 | Note that MATH. Then REF (respectively, REF ) follows from REF with MATH (respectively, from the proof of REF with MATH, and REF ). |
math/0011091 | REF follows from REF with MATH. CASE: If there exists MATH such that MATH, then MATH for each MATH by REF . CASE: Suppose that there exists MATH. Then MATH for each MATH and hence MATH by REF ; that is, MATH for each MATH, a contradiction. |
math/0011091 | Let MATH be the MATH-Frobenius divisor of MATH. Then MATH for each MATH by REF , and so MATH. |
math/0011091 | MATH is MATH-maximal if and only if MATH. By the NAME, this is the case if and only if MATH for each MATH and the equivalences follow. Now we show the formulae on the number of rational points. Let MATH. Then MATH for each MATH so that MATH for MATH; that is, MATH. If MATH, MATH and follows the formula for such MATH's. Finally, if MATH, MATH and the proof is complete. |
math/0011091 | We have MATH. Then from the lemma above, MATH and the result follows. |
math/0011091 | MATH is maximal with respect to NAME 's bound if and only if MATH if and only if MATH. Now, as we can assume MATH by REF so that MATH, the result follows. |
math/0011091 | As in the proof of REF we use MATH. We have MATH and MATH so that MATH; hence MATH and the result follows. |
math/0011091 | It is easy to see that NAME 's bound is not effective to bound MATH; in this case one uses the so-called ``explicit formula" REF: (following CITE) Let MATH, MATH, and write MATH this equation can we rewritten as MATH where MATH. Now suppose that MATH are given real numbers. Then from the above equation we obtain: MATH where MATH and MATH. Note that MATH whenever MATH and MATH. CASE: Here we choose MATH, MATH, MATH. Then MATH. Then from REF we have MATH so that MATH, and hence MATH is optimal. Moreover, as MATH we must have MATH by REF so that MATH. Then MATH and the result on MATH follows. CASE: Here we use MATH, MATH, MATH, MATH, MATH. Then MATH. Then from REF MATH so that MATH. Moreover, MATH whenever MATH. Hence MATH or MATH so that MATH where MATH is the number of MATH's such that MATH. To compute MATH we use the facts that MATH and MATH. We have MATH and hence that MATH. |
math/0011091 | CASE: It follows immediately from REF . REF (Following CITE) Let MATH. From REF , there exists a morphism MATH with MATH. Let MATH be the number of MATH-rational points of MATH which are unramified for MATH. Let MATH be the separable part of MATH. We have that MATH (here MATH is the separable degree of MATH) and from the NAME applied to MATH we find that MATH so that MATH. Thus MATH by hypothesis, and hence there exists MATH, MATH such that MATH with MATH. Let MATH be such that MATH (compare REF ). Then MATH and REF follows. |
math/0011091 | REF follows by REF Let MATH be as in REF , MATH a morphism associated to MATH, MATH such that MATH and MATH. Then MATH, MATH, divides MATH and the result follows. |
math/0011091 | CASE: Fix MATH, and let MATH such that MATH. From REF we have MATH . We claim that MATH; otherwise from MATH we would have MATH, a contradiction. This shows REF . CASE: Applying MATH to REF we have MATH so that MATH and REF follows. |
math/0011091 | REF implies REF since there are infinitely many points MATH such that MATH for MATH. To see REF we take MATH such that MATH. Then MATH by REF . If MATH were classical then MATH so that MATH, a contradiction. |
math/0011091 | CASE: We have MATH for any MATH by REF ; thus MATH by REF . Therefore MATH by REF , and so MATH by REF , where MATH is a morphism associated to MATH. It follows that MATH so that MATH. CASE: By REF MATH for each MATH. Since MATH (compare REF ), the result follows. |
math/0011091 | Let MATH. By REF MATH where MATH. Therefore MATH by REF . Now if MATH, then MATH by REF and the result follows. |
math/0011091 | REF implies REF because the genus of MATH is MATH. Assume REF and suppose that MATH. Then NAME 's genus bound REF applied to MATH would yield MATH, a contradiction. Finally let MATH. By REF MATH for any MATH and hence we can assume that MATH by REF ; in this case, as MATH, MATH and MATH. Let MATH (respectively, MATH) denote the MATH-orders (respectively, MATH-orders) of MATH. Then MATH by REF . Let MATH such that MATH and MATH. We have that-MATH is a separating variable REF and therefore MATH . There exists MATH such that MATH . To proof this we have to show that MATH for MATH by REF . We apply MATH to MATH: MATH and so MATH holds for MATH. Suppose that MATH is true for MATH, MATH. We apply MATH to MATH and using the inductive hypothesis and REF we find that MATH. It turns out that MATH since MATH, and the claim follows. MATH for MATH . From MATH. Let MATH be a local parameter at MATH. Then MATH since MATH by the chain REF . We have that MATH (see REF ) and that MATH for MATH. Therefore MATH; that is, MATH; that is, MATH by REF . It follows that MATH for MATH and so the claim. We conclude that MATH with MATH; moreover MATH since MATH. Then MATH with MATH and MATH gives a relation of type MATH . Finally we have that MATH and with MATH, MATH, we have that REF holds; that is, MATH is MATH-isomorphic to MATH. |
math/0011091 | Suppose that MATH. Then MATH and MATH. By REF MATH, and thus the MATH-Frobenius orders of MATH would be MATH, and MATH. Now from REF MATH for MATH so that MATH. From the identities MATH and MATH we would have MATH . Now, as MATH for MATH by REF , MATH and hence MATH for MATH. In particular, MATH and by the MATH-adic criterion REF we would have MATH for MATH. Then MATH. Now from NAME 's genus bound REF MATH that is, MATH, a contradiction. |
math/0011091 | Since we already observed that MATH for MATH, it is enough to show that there exists MATH such that MATH. Suppose that MATH for any MATH. Then by REF MATH so that MATH because MATH, MATH and MATH. Then from REF we would have MATH; that is, MATH, a contradiction by REF . |
math/0011091 | If MATH, then MATH and it must be a power of two by the MATH-adic criterion REF : that is, REF implies REF . Suppose now that MATH. Then from REF there exists a point MATH such that MATH; thus MATH by REF . In particular MATH. This is a contradiction as follows immediately from the claim below. MATH . In fact, MATH is a complete system of residues module MATH, where MATH . Moreover, for each MATH, MATH and MATH. Hence MATH can be computed by summing up the coefficients of MATH from the above list (see for example, CITE); that is, MATH . |
math/0011091 | CASE: By the identity MATH and REF we can assume MATH. Now the case MATH of REF implies MATH and the result follows since MATH for MATH. CASE: From REF , MATH and MATH. Then the numbers MATH are also non-gaps at MATH. Therefore, by the symmetry of MATH, MATH are gaps at MATH and the proof follows. CASE: Set MATH. We have MATH for MATH by the product rule applied to REF . Then, MATH for MATH, because the matrices MATH have rank two (see REF. Consequently, as MATH is a power of two by REF ), from REF , MATH for some MATH. Finally, from the proof of REF we have that MATH is a local parameter at MATH if MATH. Then, by the election of the MATH's, MATH has no pole but in MATH, and from REF , MATH. |
math/0011091 | We know that MATH. We claim that MATH otherwise we would have MATH, MATH, MATH, MATH, and hence MATH (with MATH being as in REF ). Therefore, after some MATH-linear transformations, the case MATH of REF reads MATH . Now the function MATH satisfies MATH and we find that MATH is a non-gap at MATH (compare CITE). This contradiction eliminates the possibility MATH. Let MATH and MATH. By REF MATH, and since MATH, by REF we have MATH . In particular, MATH. On the other hand, by REF we must have MATH and so, by REF we find that MATH; that is, MATH. Finally we show that MATH. MATH implies MATH. Since MATH (compare REF ), by REF , we have MATH. Therefore MATH so that MATH. |
math/0011091 | CASE: Let MATH be associated to MATH, let MATH be the NAME morphism (relative to MATH) on the dual plane of MATH, and suppose that MATH is not defined over MATH. Then, since the envelope is defined over MATH and MATH is MATH-rational, MATH would belong to two different components of the envelope, namely MATH and MATH. This is a contradiction because the point is non-singular. CASE: This follows from REF . |
math/0011091 | That a set of MATH lines no three concurrent satisfies the bound is trivial. Let MATH be the equation of MATH, let MATH be the factorization of MATH in MATH, and let MATH be the curve given by MATH. For simplicity we assume MATH even, say MATH. Setting MATH, MATH and MATH we have MATH. The singular points of MATH arise from the singular points of each component and from the points in MATH, MATH. Recall that an irreducible plane curve of degree MATH has at most MATH singular points, and that MATH, MATH (NAME 's Theorem). So MATH . |
math/0011091 | CASE: Let MATH be the equation of MATH over MATH. By REF , MATH admits a factorization in MATH of type MATH with MATH and MATH. Let MATH be the plane curve given by MATH . Then MATH satisfies the hypothesis of REF and it has even degree by REF . From REF and NAME 's theorem, for each line MATH (in the dual plane) corresponding to a point MATH, we have MATH where MATH, and so at least MATH points corresponding to unisecants of MATH belong to MATH. Since MATH (see REF ) and MATH, then MATH and from REF we have that at least one of the unisecant points in MATH, says MATH, is non-singular. Suppose that MATH passes through MATH. The point MATH is clearly MATH-rational and MATH is not a point of the curve of REF: otherwise MATH (see REF ). Then, MATH and so MATH is the tangent of MATH at MATH. Therefore MATH is not an inflexion point of MATH, and the proof of REF is complete. |
math/0011091 | REF follows from the proof of REF while REF from REF . |
math/0011091 | Let MATH. Suppose that it is non-singular and an inflexion point of MATH. Then, from REF and the definition of MATH, we have that MATH is not the tangent line of MATH at MATH, that is, we have that MATH. Now suppose that MATH is either singular or a non-inflexion point of MATH. Then from REF we have MATH and the result follows from NAME 's theorem applied to MATH and MATH. |
math/0011091 | Let MATH. From REF we can assume that MATH where MATH . Now MATH, where MATH, and there exists MATH such that MATH . Let MATH. From REF , MATH . Suppose that MATH and MATH. Then MATH so that MATH and hence MATH. Then the result follows from MATH. Now suppose that MATH or MATH. Then MATH and hence MATH for some MATH. Then the result follows from MATH and the fact that MATH. |
math/0011091 | From the computations above and REF , MATH . Now MATH and MATH REF . Then MATH. On the other hand, MATH REF and hence MATH. |
math/0011091 | CASE: As we mentioned in REF , MATH. Since MATH for MATH, REF follows from REF . CASE: If existed a NAME classical irreducible envelope of MATH, then from REF we would have MATH so that MATH. |
math/0011091 | We can assume MATH. If MATH is not a power of MATH, by the MATH-adic criterion REF we have MATH and MATH. |
math/0011091 | From the MATH-adic criterion REF , MATH. Then from REF we have MATH and the result follows. |
math/0011091 | From REF follow that MATH and so that MATH. From MATH and MATH we have that MATH and it follows the assertion on MATH. The bound on MATH follows from REF . |
math/0011091 | Let MATH; so MATH as MATH. From REF we have that MATH so that MATH. Let MATH. If MATH (so MATH), then from REF and MATH we would have that MATH which is a contradiction for MATH. Let MATH. Then MATH and MATH and MATH. Thus from REF we have MATH, that is, MATH. This is a contradiction since by REF we must have MATH. This eliminates the possibility MATH. The other cases can be handled in an analogous way. |
math/0011091 | If MATH, then from MATH we would have that MATH and so, from REF , that MATH, a contradiction. |
math/0011091 | Suppose that MATH. We are going to show that MATH. Notice that MATH by REF . Let MATH be a morphism associated to MATH. From REF there exist MATH, not all zero, such that MATH. Set MATH . (This curve is related to the dual curve of MATH since it is easy to see that MATH is the hyperplane tangent at MATH for infinitely many MATH's.) We have CITE CASE: MATH; CASE: MATH for any MATH; CASE: MATH whenever MATH is singular. It follows from REF that MATH since MATH. Now from REF and the hypothesis on MATH there are three possibilities for MATH-orders: CASE: MATH; CASE: MATH; CASE: MATH. We see that points of type REF cannot occur since MATH and MATH. Now from the proof of CITE we have that MATH so that MATH as MATH. It follows from REF that MATH is non-singular; that is, MATH. In particular the MATH-Weierstrass points are of type REF and we have MATH where MATH is the ramification divisor of MATH and MATH is the number of points of type REF . Now we use the following relation between MATH and MATH CITE: CASE: MATH. Since we already notice that MATH it follows that MATH; that is, MATH. Next we show that MATH. For MATH of type REF , the MATH-orders are MATH. Suppose that MATH. Then MATH is the tangent hyperplane MATH at MATH with respect to MATH, where MATH is the tangent line at MATH with respect to MATH. It is easy to see that MATH so that MATH. This implies MATH, a contradiction. Thus MATH. Finally by means of MATH where MATH is the MATH-Frobenius divisor associated to MATH, we find that MATH, and one easily checks that MATH. |
math/0011096 | Let us take the fields MATH and MATH for all MATH. Let us suppose that MATH is not one to one, then MATH. So let MATH be a non-zero element of MATH such that MATH. Let MATH be the higher index such that MATH. If MATH, trivially we have a contradiction. If MATH, let us take MATH and consider the homomorphism MATH . We know that MATH and this kernel is a prime ideal because MATH is an homomorphism between integral domains. We can write MATH, with MATH and MATH doesn't divide to MATH. This forces MATH to have some non-trivial terms in MATH. Let MATH be the minimum such that MATH is one of these terms. By the NAME preparation theorem we have MATH, where MATH is a unit and MATH . Since MATH is a unit, MATH and MATH . This leads to a contradiction because the roots of MATH are in MATH, with MATH, by the NAME theorem. If MATH let us take MATH and consider the homomorphism MATH . As in the previous case we can write MATH, where MATH. So we have MATH, where MATH is a unit and MATH so MATH . But this is again a contradiction by the NAME theorem: since MATH is a prime ideal, we can suppose that MATH is an irreducible element of the ring MATH. In this situation the NAME theorem says that to obtain the coefficients of a root of MATH, like a NAME series in MATH with coefficients in MATH, we have to resolve a finite number of algebraic equations of degree greater than REF in MATH. Inside MATH we can not obtain MATH and, with a finite number of algebraic equations, we can obtain a finite number of powers of MATH but not all. So this proves the lemma. |
math/0011096 | We know that the homomorphism MATH previously defined is one to one. So we can take the valuation MATH, where MATH is the usual order function over MATH in MATH and MATH is the natural extension to the quotient fields. Let MATH be the residue MATH. Hence, to obtain the lemma we have to prove that MATH and MATH is an algebraic extension of MATH. Let us suppose that MATH. Then there must exist MATH such that MATH, so MATH . This means that MATH. On the other side we have MATH so MATH and we have a contradiction. Hence MATH. Let us prove that MATH is an algebraic extension of MATH. We can consider each element of MATH like a sum of forms with respect to the usual degree. If MATH is a form of degree MATH, then MATH, with MATH a polynomial in MATH and a finite number of elements MATH. Let us take MATH such that MATH and MATH. Then MATH, where MATH is a rational fraction in MATH and a finite number of elements MATH. So MATH is algebraic over MATH. Let us consider MATH a polynomial satisfied by MATH, where MATH for all MATH and MATH. Let MATH be the element MATH . Then we have MATH so MATH and MATH. Subsequently, MATH . This proves that MATH is an algebraic element over MATH and, a fortiori, the lemma. |
math/0011096 | We know, after CITE, that the dimension of a rank-one discrete valuation of MATH is minor or equal than MATH. So we have to prove that there exists a transcendental residue in MATH. Let us suppose that MATH for all MATH. Then the value of MATH is zero, so MATH. If this residue lies in MATH then there exists MATH such that MATH . This implies MATH and then MATH . So we have MATH. Then MATH . If the residue of this element lies too in MATH, then there must exist MATH such that MATH. We can repeat this operation. The previous procedure is finite: if it didn't stop we would construct the power series MATH such that the sequence of partial sums has increasing values. Since MATH is a complete field, then this series amounts to zero in contradiction with MATH and MATH being formally independent. So the procedure must stop and there exists a transcendental element over MATH in MATH. |
math/0011096 | Let us consider the one to one (the proof of injectivity parallels that of REF ) homomorphism MATH . We can take the valuation MATH, with MATH the usual order function in MATH and MATH the natural extension to the quotient fields. We know REF that the residue MATH is transcendental over MATH. Trivially the residue MATH for all MATH are transcendental over MATH because MATH are formally independent variables. Any element MATH is algebraic over MATH parallels that of REF . So the dimension of MATH is MATH. |
math/0011097 | Let MATH be an isomorphism taking MATH to MATH. By the universal property of blowing - up, this extends to an isomorphism between the minimal resolutions MATH. In particular the dual graphs of MATH and MATH are isomorphic. We write MATH. As both MATH and MATH there are exactly MATH very long chains of MATH - curves in both MATH and MATH. Due to their length, they can be recognized. To each endpoint MATH of such a very long chain in MATH there is, by the construction of sandwiched singularities, a MATH - curve MATH in MATH and a smooth curve MATH intersecting MATH. Similar for MATH. The isomorphism MATH sends MATH to a MATH, as the resolution graphs of MATH and MATH are isomorphic. By using the isomorphism MATH, we may glue every MATH (and in it MATH) to the curve MATH in MATH. Thus we may assume that the two resolutions are equal: MATH. Without loss of generality, we may assume that both MATH and MATH are generic elements of MATH. By looking at the image under MATH of the divisors of the pull-back of MATH and MATH on MATH (which is the fundamental cycle of MATH restricted to MATH plus a non-compact curve intersecting the first blown up curve), we see that the divisors of MATH and MATH are the divisors of two functions which generate the maximal ideal of MATH. Thus MATH maps MATH to an isomorphic curve, which except for the MATH - curves has the same resolution as MATH. Thus we may and will suppose that the resolutions MATH and MATH are equal. From the construction of the MATH out of the MATH it follows that MATH. There is an algorithm to get, from the resolution of a plane curve singularity, the resolution of every irreducible component. Following this algorithm, we see that in the resolution of any irreducible component MATH of MATH there is a very long chain of MATH - curves. By the remarks above this is, except for the MATH - curves, also the resolution for MATH. By the formula for the intersection number, see, for example, REF , the intersection number between MATH and MATH increases if the chains of MATH curves, and thus MATH and MATH become bigger. This intersection number is equal to both MATH, and also to the vanishing order of MATH on MATH, where MATH is the normalization of MATH. Here the curves MATH and MATH are defined by irreducible MATH and MATH. Thus we may assume that for all MATH the vanishing order of MATH on MATH is at least MATH, for some large MATH, and where MATH is the MATH'th conductor number of MATH, corresponding to the branch MATH. By definition of the conductor, every function which vanishes with order at least MATH on the normalization MATH for all MATH, is an element of the maximal ideal MATH of MATH. It follows that the class of MATH in MATH lies in MATH. Thus MATH, where we now view MATH as an ideal in MATH. By symmetry, MATH is a unit. By taking MATH large, we may, by the finite determinacy theorem, assume that MATH and MATH are right equivalent. In particular, their zero sets MATH and MATH are isomorphic. This is what we had to show. |
math/0011097 | The proposition follows from the exact sequence REF . |
math/0011097 | We have an R.C. deformation of MATH over MATH. In particular we get a one - parameter deformation of MATH over MATH. Let MATH be the base space of a semi-universal deformation of MATH. It is a smooth space of dimension MATH. By semi-universality we get a map of smooth spaces MATH. This map in general is not an immersion. Indeed, CITE showed that the kernel of the map MATH is equal to MATH, the multiplicity minus the number of branches. However, for a general point of MATH we have that for all singularities MATH, so that the map MATH is an immersion at a general point of MATH. Hence the image of MATH at a general point is smooth of dimension MATH. Thus, by openness of versality, it remains to compute for each infinitely near point MATH, the codimension of the stratum MATH of the ordinary MATH - tuple point in the base space of the ordinary MATH - tuple point. This codimension is MATH, being two less than the number of monomials of degree smaller than MATH. |
math/0011097 | Given MATH, take a decorated curve MATH with MATH. Let MATH be the exceptional curves in the minimal resolution of MATH. Let MATH. By NAME 's result REF, the dimension of the NAME component of MATH is equal to MATH . On the other hand, combining REF we get that this dimension is also equal to MATH . Using REF we thus get MATH . It remains to show that MATH. This is easy, as we know that we get the minimal resolution of MATH out of the minimal good resolution of MATH by doing an extra MATH blowing-ups, for each MATH. We thereby introduce a chain of MATH - curves of length MATH, and decrease the selfintersection of the exceptional curve on the minimal resolution of MATH which MATH intersects by one. From MATH, see REF, the result follows. |
math/0011097 | Let an element MATH in MATH whose first section is given by the ideal MATH for MATH. Then MATH has trivial first section, hence is contained in the image of MATH by the lemma. |
math/0011097 | Clearly we have MATH. Conversely, suppose MATH is an element of MATH. Associated to it is a space MATH, which is obtained by blowing up MATH successively along the sections described by MATH. Hence we get an infinitesimal deformation of some resolution of the curve MATH, in particular a deformation of the minimal resolution of MATH, that is an element of MATH. We look at the induced deformation of a small neighborhood of each MATH. More precisely, from the natural composition MATH we get a map MATH. Thus gives the map MATH. The kernel consists of elements of MATH which induce trivial deformations of all small neighborhoods of each MATH. This is equivalent to saying that we get an infinitesimal equisingular deformation of the minimal resolution MATH, that is, an element of MATH. By NAME, CITE we have a natural smooth functor MATH, whose kernel on tangent spaces is given by the tangent space of the functor MATH, which is the functor of all equisingular deformations of MATH for which all sections can be trivialized, see CITE This proves the Theorem. |
math/0011097 | After blowing up in the trivial section, we get, by construction of MATH equisingular deformations of all connected components of the strict transforms of MATH, which then, by REF , map to the zero element of MATH. |
math/0011099 | Note that what happens during the execution of the inner loop of REF is a jeu de taquin forward (backward) slide performed on MATH into the cell MATH, see REF. First we have to show that MATH is well defined. That is, we have to check that after each jeu de taquin forward slide, after the entry MATH in the cell MATH is deleted from MATH, the cells of MATH which contain an entry form a SSYT as stated in the algorithm. This follows, because after either type of replacement in the inner loop the only possible violations of increase along rows and strict increase along columns in MATH can only involve MATH and the entries to its right and below. When the jeu de taquin forward slide is finished, MATH is a bottom-right corner of MATH, hence after deleting the entry in MATH no violations of increase or strict increase can occur. Next we show that MATH indeed produces a reverse SSYT. In fact, we even show that the tabloid defined by the cells of MATH which have been filled already, is a reverse SSYT at every stage of the algorithm. Clearly, every cell of MATH is filled with an entry exactly once. Furthermore, at the time the cell MATH is filled, the cells in MATH to the right and to the bottom of MATH - if they exist - are filled already, otherwise MATH would not be a bottom-right corner of MATH. Because the sequence of entries chosen is monotonically increasing, rows and columns of MATH are decreasing. So it remains to show that the columns of MATH are in fact strictly decreasing. Suppose that MATH and MATH are cells both containing the same minimal entry MATH, and MATH is right of MATH. When the jeu de taquin forward slide in MATH is performed into the cell MATH, the entry MATH describes a path from MATH to the cell where the slide stops, which we will denote by MATH. Similarly, we have a path from MATH to a cell MATH. Now suppose MATH is in the same column as, but below MATH, as depicted in REF . Clearly, in this case the two paths would have to cross and we had the following situation: First, (the star is a placeholder for an entry we do not know) MATH . In this situation, MATH would have to be smaller then MATH. Then, when the jeu de taquin forward slide into the cell MATH is performed, the following situation would arise at the same four cells: MATH . But this cannot happen, because then MATH would have to be strictly smaller than MATH. It can be shown in a very similar manner that MATH indeed produces a SSYT. We leave the details to the reader. Finally, we want to prove that MATH is inverse to MATH . Suppose that in MATH , a jeu de taquin forward slide into the cell MATH containing the entry MATH is performed on MATH. Suppose that the slide stopped in MATH, MATH is set to MATH and the entry in MATH is deleted from MATH. Among the entries of MATH, MATH is maximal, because smallest entries are chosen first in MATH . Furthermore, among those cells of MATH containing the entry MATH, the cell MATH is most left. This follows, because the tabloid defined by the cells of MATH which have been filled already, is a reverse SSYT, and the paths defined by the jeu de taquin slides cannot cross, as we have shown above. It is straightforward to check that in this situation the jeu de taquin backward slide into MATH performed on MATH in MATH stops in the original cell MATH. By induction we find that MATH is inverse to MATH . |
math/0011099 | First of all, we have to prove that REF . MATH terminates. We required that MATH for all cells MATH, which implies that every time when we replace the entry in cell MATH by MATH (see the beginning of the outer loop of the algorithm) we decrease MATH. It is easy to see that this maximum is never increased in the subsequent steps of the algorithm. It is easy to check that after every type of replacement within the modified jeu de taquin slides, the validity of the equation MATH is preserved. So it remains to show that after every modified jeu de taquin slide of MATH , the resulting filling MATH of MATH is in fact a reverse SSYT: We have that MATH is maximal at the very left of MATH, because rows are decreasing in MATH. Therefore, when MATH, as required for the execution of the outer loop of MATH , we have MATH so MATH is non-negative. Furthermore, after either type of replacement during the modified jeu de taquin slide, the only possible violations of decrease along rows or strict decrease along columns can involve only the entry MATH and the entries to the right and below. By induction, MATH must be a reverse SSYT. The second statement of the lemma is shown with an argument similar to that used in the proof of REF . When the jeu de taquin forward slide in MATH is performed into the cell MATH, the entry MATH describes a path from MATH to the cell MATH, where the slide stops. Similarly, we have a path from MATH to MATH. We conclude that, if MATH were strictly to the right of MATH, that these paths would have to cross. (See REF ). Hence we had the following situation: First, (the star is a placeholder for an entry we do not know) MATH . In this situation, MATH would have to be strictly smaller then MATH. Then, when the modified jeu de taquin slide into MATH is performed, the following situation would arise at the same four cells: MATH . But this cannot happen, because then MATH would have to be at least as big as MATH is. The corresponding statement for REF . MATH is shown similarly. |
math/0011099 | It remains to show, that MATH and MATH are inverse to each other. This is pretty obvious considering the lemma: Suppose that the pair MATH is an intermediate result obtained after a modified jeu de taquin slide into the cell MATH. After this, MATH is increased, where MATH is the cell where the slide stopped. Then the entry in MATH must be among the smallest entries of MATH, so that MATH, because the sequence of MATH's in the cells chosen for the modified jeu de taquin slides is monotonically decreasing. If there is more than one cell MATH which contains a minimal entry of MATH and satisfies MATH, the lemma asserts that the right-most cell was the last cell chosen for the modified jeu de taquin slide MATH . Hence it is certain that the right-most cell containing a minimal entry as selected before the modified jeu de taquin slide of MATH is MATH. It is easy to check, that the replacements done in MATH are exactly inverse to those in MATH . For example, suppose the following replacement is performed in REF: MATH . Then we had MATH and, because of strictly decreasing columns, MATH. Therefore, in MATH , this is reversed and we end up with the original situation. Similarly, we can show that MATH is inverse to MATH , too. |
math/0011101 | We check directly that NAME 's genericity conditions are satisfied for such MATH. First one stratifies MATH by taking as strata the intersections of the hyperplanes, one stratum for each element of the intersection lattice. This stratification is ``bonne" in the sense of CITE, and MATH is transverse to the strata, except MATH, since MATH contains no other intersection of hyperplanes. The condition on MATH is required for NAME 's result to ensure that MATH is NAME. See CITE for further discussion. |
math/0011101 | Clearly the number of critical points of the NAME function MATH on MATH is MATH times the number of critical points of the induced NAME function on MATH. The NAME functions on MATH and MATH are not proper, so we must take care. We follow here the argument given by NAME in CITE for the NAME fiber MATH. The same argument works for MATH; one simply projects the sets chosen for MATH via the map MATH. The actual NAME fiber is MATH. We first note that for a sufficiently large MATH, MATH is diffeomorphic to MATH and homotopy equivalent to MATH. Then for MATH sufficiently large with respect to MATH, we let MATH be the closed ball of radius MATH in the first MATH-coordinates and note that MATH has the homotopy type of MATH, where MATH is the set of complex numbers MATH . Further, from CITE, we note that MATH and MATH can be chosen so that for all MATH, MATH is transverse to MATH in MATH. Then it follows that MATH is a proper relative NAME function on a manifold with boundary, and the map on the boundary is a submersion. Therefore MATH has the homotopy type of MATH with MATH-cells attached. As noted, the same argument works in MATH by considering images under the map MATH. |
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