paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0007121 | We will construct a pseudotensor functor MATH from MATH to MATH. On objects MATH we let MATH. In order to define it on polylinear maps, we need to find out how MATH can be recovered from MATH and the action of MATH. Denote by MATH the embedding MATH, and let MATH be the composition MATH . Explicitly, MATH is given by t... |
math/0007121 | Can be found in CITE. |
math/0007121 | By the construction of the filtrations it is evident that we can assume MATH. Then MATH, and MATH acts on it by linear differential operators. The rest of the proof is clear. |
math/0007121 | For a proof of REF see CITE. In order to prove REF , notice that if MATH is a derivation of MATH, then its action on MATH is given by MATH for all MATH, where the MATH form a topological basis of MATH and the MATH are continuous derivations of MATH. Subtracting MATH from MATH, we get a derivation MATH acting trivially ... |
math/0007121 | REF follows from REF. The skew-commutativity of the bracket REF follows immediately from that of MATH. The proof of the NAME identity is straightforward by using REF. Let us check for example that the associativity of MATH is equivalent to that of MATH; the case of the NAME identity is similar. We will use the notation... |
math/0007121 | We have MATH for every MATH. Since the right action of MATH on MATH is surjective (see REF ), it follows that MATH for any MATH. |
math/0007121 | Let us choose bases of MATH and MATH, and let us fix expressions of elements from the first basis as MATH-linear combinations of elements from the second basis. Denote by MATH the highest degree of the coefficients of all these expressions. Using REF, we see that MATH for all MATH. Repeating the same reasoning after sw... |
math/0007121 | CASE: The linear-compactness follows from REF , since REF a filtration by finite-codimensional subspaces with trivial intersection and MATH is complete with respect to this filtration. The continuity of the MATH-action follows from REF: MATH . CASE: It only remains to check that the NAME bracket of MATH is continuous. ... |
math/0007121 | Follows immediately from REF . |
math/0007121 | First of all, notice that we can assume MATH is torsion-free, since by REF , MATH where MATH is the torsion submodule of MATH. The proof of the proposition is then based on REF and the following two lemmas. The map MATH induced by the inclusion MATH constructed in REF is uniformly continuous, that is, for every MATH we... |
math/0007121 | It is well known that MATH is trivial for MATH and MATH-dimensional for MATH; see for example, CITE. |
math/0007121 | The proof is by induction on the number of MATH not contained in MATH, the basis of induction being trivial. Consider MATH, with all MATH. We can write MATH so that MATH is a linear combination of NAME - NAME - NAME basis elements of MATH not containing MATH in their expression, and MATH. Then: MATH . Since the third s... |
math/0007121 | First of all, it is easy to check that the elements REF indeed belong to MATH. REF is easy, and the computation of the pseudobrackets is straightforward using REF, reformulated as MATH . Now let us consider an element MATH, MATH. We will prove that MATH can be expressed as MATH-linear combination of the above elements ... |
math/0007121 | Straightforward computation, using REF . |
math/0007121 | Similar to that of REF. If MATH, there is nothing to prove. Let MATH be elements that complement MATH to a basis of MATH. If MATH is not in MATH we take it to be one of the MATH's. Write out REF as MATH . Now, if MATH involves some MATH's, there is no way to cancel out the terms MATH. This proves that MATH. Similarly, ... |
math/0007121 | Let us write MATH and MATH (summation over repeated indices). Then REF is equivalent to MATH where ``cyclic" means summing over cyclic permutations of the indices MATH. Multiply this equation by MATH and sum over MATH. Using that MATH, we get MATH where now the cyclic permutations are over MATH. This is exactly REF . C... |
math/0007121 | It remains to show that, conversely, any vector field that preserves the form MATH is equal to MATH for some MATH. Indeed, let MATH be such that MATH. Since MATH and MATH, this is equivalent to MATH which implies MATH for some MATH. In other words, MATH for any MATH. Using MATH, we get MATH for MATH. This, together wit... |
math/0007121 | The proof is very similar to that of REF . There we showed that MATH is equivalent to REF, and the same argument applies here. Similarly, REF is equivalent to REF . Now if MATH, then MATH implies MATH, which together with REF leads to MATH. |
math/0007121 | Given MATH, above we have defined the MATH-form MATH such that MATH, MATH and MATH. Since MATH is nondegenerate, we have MATH. Conversely, starting with a contact MATH-form MATH, we can define MATH and MATH satisfying REF - REF. |
math/0007121 | It remains to show that, conversely, any vector field from MATH is equal to MATH for some MATH. Indeed, let MATH be such that MATH for some MATH. Let us write MATH for some MATH. Then MATH and MATH, while MATH. Therefore MATH, which implies MATH, MATH, and MATH. |
math/0007121 | CASE: We have seen in REF that MATH. Let MATH be one of the pseudoalgebras MATH, MATH or MATH and MATH be its natural embedding in MATH. We have shown in REF that in this case the image of MATH in MATH under MATH is MATH, MATH or MATH, respectively. REF below implies that MATH is a central extension of its image in MAT... |
math/0007121 | We define MATH in terms of its NAME coefficients - the MATH-products MATH. We have already seen, when we discussed associativity, that REF is equivalent to the following equation (compare REF): MATH . This can be inverted to find (compare REF): MATH . The MATH-sesqui-linearity properties of MATH with respect to MATH an... |
math/0007121 | REF is an immediate consequence of REF . Indeed, the only thing that remains to be checked is the associativity of MATH, and it follows from REF: MATH . To prove REF , we associate with each MATH the MATH-pseudolinear map MATH given by MATH. Then MATH which shows that MATH. |
math/0007121 | Since MATH, we can identify MATH with MATH so that its action on MATH is given by REF. To prove REF, we use REF and the explicit definition of associativity from REF. Due to MATH-bilinearity, we can assume that MATH. Then: MATH . On the other hand, we have: MATH . Now REF follows from the uniqueness from REF . |
math/0007121 | Since MATH is a MATH-submodule of MATH, after replacing MATH with MATH, we can assume that MATH. By definition, MATH. Since, by REF , MATH, we have MATH. On the other hand, since MATH is finite over MATH and MATH satisfies REF, there exists a finite-dimensional subspace MATH of MATH such that MATH. It follows that MATH... |
math/0007121 | REF follows from REF : if MATH with linearly independent MATH, then MATH can be zero only if all MATH, which implies MATH. Similarly, REF follows from REF, since we can write MATH uniquely in the form MATH. |
math/0007121 | REF are obvious. CASE: By NAME 's REF, MATH with MATH. If MATH is finite dimensional, any continuos homomorphism MATH must contain some MATH in its kernel. Let MATH but MATH. Then, by REF , MATH so that for each MATH, MATH for some MATH. This implies MATH, since MATH, proving REF . Similarly, REF follows from the fact ... |
math/0007121 | This can be verified by straightforward but rather tedious computations. To illustrate them, let us check REF. By definition, we have: MATH while MATH . Hence REF is a consequence of the following identity: MATH which can be checked by pairing both sides with MATH. Indeed, MATH . A more conceptual proof can be given by... |
math/0007121 | To construct the inverse of MATH, identify MATH with MATH and consider MATH . Here, as before, MATH, MATH are dual bases in MATH and MATH, and MATH for MATH. This is well defined, that is, the sum is finite, because MATH for all but a finite number of MATH and MATH. Using identity REF, it is easy to see that MATH is MA... |
math/0007121 | It remains to show that a non-torsion element MATH does not lie in the kernel of MATH. Consider the map MATH constructed in REF . Then MATH. The map MATH induced by MATH maps the NAME coefficient MATH of MATH to the corresponding NAME coefficient MATH of a nonzero element in the free MATH-module MATH. Now, it is clear ... |
math/0007121 | MATH means that MATH, hence MATH is torsion. |
math/0007121 | To show the equivalence of REF , it is enough to tensor the exact sequence MATH with MATH, getting the exact sequence MATH, and to apply REF . Assume that REF holds but MATH is not torsion. Then it contains a nonzero element MATH which generates a free MATH-module. Then MATH has growth MATH, which is a contradiction. |
math/0007121 | Since MATH is torsionless, MATH is injective by REF . Hence, for MATH, MATH, one has MATH iff MATH for all MATH. By REF, this is equivalent to MATH. Hence, for MATH, MATH for all MATH iff MATH lies in the center of MATH. Now MATH is in the kernel of MATH iff MATH for all MATH and MATH, which means that MATH is central ... |
math/0007121 | Let MATH be in the kernel of MATH; then by the previous lemma MATH is contained in the center of MATH. The latter is finite dimensional by assumption, so the kernel MATH of MATH is of finite codimension in MATH. This implies that MATH is open in MATH, and it contains MATH for some MATH. Let MATH but MATH; then by REF ,... |
math/0007121 | The only thing that remains to be checked is the locality property for MATH. It follows from that of MATH, since in this case MATH is injective. |
math/0007121 | The kernel of MATH acts trivially on MATH and hence on MATH. |
math/0007121 | Assume that MATH belongs to MATH; then all MATH, MATH. This implies that all NAME coefficients MATH of MATH are zero, hence MATH. |
math/0007121 | Follows easily from REF. Say that MATH lies in the kernel of MATH. Since MATH is a homomorphism of NAME MATH-differential algebras, its kernel is a MATH-stable ideal of MATH. Then by REF, MATH for all MATH, MATH, because in the right-hand side all elements MATH lie in MATH. This means that MATH for all MATH, hence MATH... |
math/0007121 | By REF, MATH is a current NAME algebra over MATH, where MATH is either a finite-dimensional simple NAME algebra MATH (for MATH), or one of the NAME algebras of vector fields MATH, MATH, MATH or MATH (MATH). In particular, we know that MATH. By REF , a sufficiently high power of any nonzero element MATH maps any given o... |
math/0007121 | As was already remarked, MATH is a subalgebra of MATH. Since MATH is a free MATH-module, we have MATH. For MATH, write MATH with MATH and linearly independent MATH. Then for any MATH we have (compare REF): MATH . This is zero for any MATH iff MATH for all MATH, which means MATH. |
math/0007121 | For MATH, both sides of REF are trivial. For MATH, we will proceed by induction on MATH. First of all, note that the compatibility or solvability of the systems REF or REF does not change when we apply an automorphism of MATH. The same is true when we make an elementary transformation: multiply one equation by a functi... |
math/0007121 | Let us choose a basis MATH of MATH and write MATH REF where MATH and MATH, MATH. Note that MATH and MATH is a NAME algebra homomorphism. We can choose the basis MATH in such a way that MATH for MATH, and MATH for MATH. Then MATH is a basis of MATH. Moreover, note that MATH is given by MATH. Let MATH be another transiti... |
math/0007121 | It is enough to show that the reconstruction functor MATH gives the same result on the topological MATH-modules MATH and MATH. In order to do so, we must show that every MATH-linear continuous homomorphism of MATH to MATH can be obtained from a unique MATH-linear continuous homomorphism of MATH to itself by composing w... |
math/0007121 | REF is standard. REF follows from REF and the fact that MATH. CASE: If MATH has an abelian ideal MATH, then the preimage of MATH under the natural projection MATH must be solvable and strictly bigger than MATH, which is a contradiction. |
math/0007121 | REF has already been proved, when MATH is finite, in REF . In the general case, it can be deduced from REF . Let MATH, then MATH for any MATH. Hence the action of MATH on MATH is trivial, and by REF, MATH for any MATH. In order to prove REF , notice that MATH. Then build a NAME MATH-pseudoalgebra structure on MATH by l... |
math/0007121 | REF is immediate from REF . Assume that REF holds but MATH is not simple. Then MATH has a nontrivial central regular ideal. If MATH is central in MATH for every MATH, then by REF MATH for every MATH, MATH. When MATH is either finite or free, MATH for all MATH if and only if MATH (compare REF ). Therefore MATH for all M... |
math/0007121 | It is easy to check that MATH satisfies the assumptions of REF (see REF). |
math/0007121 | Let MATH be non-central. Assume that MATH for all MATH. Note that by REF , we have: MATH for MATH. This implies: MATH, which gives MATH for any MATH, MATH. Then we can use REF to show that MATH is central in MATH, which is a contradiction. Therefore, there is some MATH such that MATH has a nonzero NAME coefficient, tha... |
math/0007121 | First of all, we observe that MATH is a linearly compact NAME algebra with respect to the topology defined in REF, see REF . Consider the extended annihilation algebra MATH, obtained by letting MATH act on MATH according to its MATH-module structure. MATH is a linearly compact NAME algebra possessing a fundamental suba... |
math/0007121 | The only place in the proof of REF where we used the simplicity of MATH was where we deduced that any nonzero regular ideal of MATH must equal the whole MATH. This argument is modified as follows. Let MATH be a nonabelian regular ideal of MATH contained in MATH. Then the minimality of MATH implies that MATH and MATH. T... |
math/0007121 | By REF, the annihilation algebra MATH of MATH is an irreducible central extension of a current NAME algebra MATH, where MATH is a simple linearly compact NAME algebra of growth MATH. We have surjective maps MATH where MATH is the universal central extension of MATH. By REF, MATH is either finite dimensional (when MATH)... |
math/0007121 | By NAME 's Lemma, it is enough to show that MATH for any collection of finite ideals MATH of MATH such that MATH for MATH, where MATH is a totally ordered index set. Assume that MATH. Then there is a chain of ideals MATH of the same rank whose intersection is zero. Fix some MATH. Then for any MATH, MATH, the module MAT... |
math/0007121 | By REF , the annihilation algebra MATH. By REF , the latter is isomorphic to MATH, since MATH. Then MATH (see REF). Now by REF , MATH. |
math/0007121 | Let MATH be a current pseudoalgebra over MATH, and MATH be one of the primitive pseudoalgebras of vector fields, where MATH is a NAME subalgebra of MATH. Then, by REF, the annihilation algebra MATH is a current NAME algebra over MATH: MATH, and MATH is isomorphic to MATH, MATH, MATH or MATH, where MATH, MATH. As in the... |
math/0007121 | Consider the set MATH of all minimal nonzero ideals of MATH. This set is nonempty by REF , and finite because MATH is a NAME MATH-module. The adjoint action of MATH on MATH gives a homomorphism of NAME pseudoalgebras MATH, compare REF . We claim that the direct sum of these homomorphisms is an injective map. Indeed, le... |
math/0007121 | Let us write MATH where MATH and MATH is a basis of MATH. Denote by MATH (respectively MATH) the maximal degree of the MATH (respectively MATH). We have (compare REF): MATH . Assume that MATH. Notice that only the third summand contains coefficients from MATH of degree MATH, hence it must be zero modulo MATH. Since the... |
math/0007121 | Assume that MATH is not a direct sum of pseudoalgebras of vector fields. If MATH is not semisimple, then MATH contains nonzero abelian elements. If MATH is semisimple, REF implies that MATH contains a subalgebra of the form MATH with MATH, and therefore contains nonzero abelian elements (for example MATH for any MATH).... |
math/0007121 | By REF , a pseudoalgebra MATH of vector fields does not contain nonzero abelian elements, and hence is semisimple. Then, by REF, MATH is a direct sum of finite simple NAME pseudoalgebras and of pseudoalgebras of the form MATH. In fact, there is only one summand, as MATH for any two nonzero elements MATH. Furthermore, M... |
math/0007121 | We have already noticed (see the paragraph before REF) that any nonzero ideal MATH of MATH contains MATH. Then MATH is an ideal of MATH, but MATH is simple by REF. |
math/0007121 | Any homomorphism MATH leads to abelian elements in MATH, and therefore is zero (see REF ). Let MATH be a homomorphism from MATH to MATH. Then MATH induces a homomorphism of NAME algebras MATH. By REF, MATH is simple, so MATH where MATH is a primitive MATH-pseudoalgebra of vector fields (MATH and MATH is a subalgebra of... |
math/0007121 | As a MATH-module, MATH is isomorphic to a direct sum of several copies of MATH. Any ideal MATH of MATH is in particular a MATH-module, hence it is spanned over MATH by elements of the form MATH where MATH and MATH is a root vector in MATH. If MATH is such that MATH for some nonzero MATH, then MATH, since MATH is an ide... |
math/0007121 | Define a map MATH by the formula: MATH . Then for MATH, we have: MATH . It is easy to see that MATH is a surjective NAME algebra homomorphism. Any isomorphism MATH induces an isomorphism of NAME algebras MATH. By REF , MATH for some proper ideal MATH of MATH. Recall that MATH is isomorphic as a topological algebra to M... |
math/0007121 | The statement of the proposition is equivalent to saying that all solutions MATH of REF with MATH are either trivial or of the form MATH. It is easy to check that the latter is indeed a solution (compare REF ). Let us choose a basis MATH of MATH, and write MATH and MATH for some MATH, MATH. We will assume throughout th... |
math/0007121 | REF is an immediate consequence of REF. By the above remarks, it remains to prove REF in the cases when MATH is a current pseudoalgebra over either MATH or MATH, where MATH is a subalgebra of MATH. In the former case (MATH, MATH), note that MATH is spanned over MATH by elements MATH for MATH. Then MATH, and by REF we k... |
math/0007121 | CASE: If MATH then all its coefficients in front of MATH are zero for different MATH. Since MATH, it follows that all MATH. REF are clear by REF. REF follows from REF - REF and the fact that MATH is MATH-invariant. (MATH is MATH-invariant because MATH is MATH-equivariant, see REF.) |
math/0007121 | Follows from REF . |
math/0007121 | By REF , MATH is simple iff MATH is not abelian and has no nontrivial MATH-invariant ideals. In particular, MATH is semisimple. Using REF and the fact that MATH is an ideal of MATH if MATH is an ideal, we see that MATH is a direct sum of isomorphic finite simple NAME MATH-pseudoalgebras. |
math/0007121 | It only remains to check REF, which is straightforward. |
math/0007121 | The second statement is obvious by the definitions. Since MATH REF and MATH REF act irreducibly on MATH, we only have to check that the action of MATH on MATH is irreducible for MATH. Using diagonal matrices, we see that it suffices to check that MATH acts irreducibly on MATH. Recall that this action is given by (see R... |
math/0007121 | Let MATH be a left ideal, MATH, and MATH with linearly independent MATH. Then: MATH . Taking MATH, we see that all MATH. In particular, MATH, and hence each element from MATH is a MATH-linear combination of elements of the form MATH. For such MATH, we have MATH. For MATH, MATH, we get that MATH. This proves the first p... |
math/0007121 | Assume that MATH for some MATH, MATH. Let REF be a relation of this form with MATH so that MATH is minimal. We call MATH the degree of the relation REF. Assume that MATH's are linearly independent, so that the degree of REF is positive. We can find MATH, MATH such that MATH. Applying MATH to REF and using REF, we obtai... |
math/0007121 | Fix nonzero elements MATH, MATH, and let MATH. Let MATH be the linear span of MATH; we set MATH for MATH. For MATH, MATH, we have: MATH, and by induction, MATH . In particular, all MATH are MATH-modules. Since MATH is a NAME MATH-module, there exists MATH such that MATH. In particular, MATH for some MATH. Writing REF f... |
math/0007121 | Using REF , we can assume that MATH. The proof will be by induction on the length of the derived series of MATH. First consider the case when MATH is abelian. By a NAME 's Lemma argument, it is enough to find an eigenvector when MATH is abelian generated by one element MATH. We may assume that MATH; then by REF all MAT... |
math/0007121 | Assume that MATH is not semisimple, that is, it has a nonzero abelian ideal MATH. Then, by REF, MATH has an eigenvector in MATH. If MATH is the corresponding eigenspace in MATH, then, by REF , MATH is a MATH-submodule of MATH. The irreducibility of MATH implies that MATH. Now, by REF , MATH is a free MATH-module, since... |
math/0007121 | First of all, note that the property that any element MATH of MATH acts nilpotently on MATH remains valid when we replace MATH by any quotient of MATH by an ideal. In particular, MATH will have that property. However, MATH is semisimple, and from the classification of finite semisimple NAME pseudoalgebras we see that t... |
math/0007121 | In order to prove the statement, it is enough to show that all MATH-module extensions between generalized weight modules relative to distinct weights are trivial. The strategy is to consider first the case when MATH is the NAME pseudoalgebra generated by one element MATH. Then in the general case, we show that the gene... |
math/0007121 | The proof is an application of REF , using the following fact. Let MATH be a vector space and MATH REF be subspaces of MATH such that all MATH are finite dimensional, then MATH is finite dimensional. It is enough to show this for MATH, in which case it follows from the isomorphism MATH . |
math/0007121 | Any MATH is contained in some MATH, which is finite dimensional by REF , and MATH-invariant because MATH. |
math/0007121 | CASE: Let MATH be an extension of MATH-modules, which is split over MATH. Choose a splitting MATH as MATH-modules. The fact that MATH and MATH are homomorphisms of MATH-modules implies (MATH, MATH, MATH): MATH . It is clear that MATH and MATH is MATH-linear; in other words, MATH. For MATH, MATH, we have (compare REF): ... |
math/0007121 | Since the adjoint action of MATH on MATH is trivial, we obtain that MATH maps to zero in the exact sequence REF. Therefore we have exact sequences MATH which lead to isomorphisms REF follows from REF . |
math/0007121 | Let MATH be a central extension of MATH with a pseudobracket MATH where MATH for MATH. The skew-symmetry of MATH is equivalent to REF. The NAME identity is equivalent to NAME identity for MATH together with the following cocycle condition for MATH (compare REF ): MATH . With the usual notation MATH, MATH, etc., we have... |
math/0007121 | CASE: The NAME pseudoalgebra MATH is free of rank one, with MATH, MATH, hence we can use REF . In this case MATH, and REF becomes MATH for MATH. We choose a basis MATH of MATH such that MATH, and express MATH in a NAME - NAME - NAME basis as MATH, MATH (see REF). Then the above equation becomes: MATH . Comparing terms ... |
math/0007121 | MATH is free of rank one and MATH, hence REF becomes MATH, MATH. This is satisfied only by multiples of MATH if MATH and by all elements of MATH otherwise. Since MATH is abelian, then MATH and trivial cocycles are multiples of MATH. |
math/0007121 | MATH is free of rank one, and MATH where MATH is a basis of MATH with the only nonzero commutation relations MATH, MATH (see REF). It is immediate to check that MATH for all MATH. Moreover, the element MATH from REF equals MATH. Then, if MATH with MATH, MATH, REF becomes: MATH . All solutions MATH of this equation are ... |
math/0007121 | By REF , MATH is spanned over MATH by elements MATH satisfying the relations MATH and MATH . The pseudobrackets are (see REF): MATH . Trivial cocycles MATH are determined by the identity (see REF): MATH where MATH, which gives: MATH . Let MATH be a cocycle for MATH representing a central extension. Write MATH for short... |
math/0007121 | Let MATH be a cocycle for MATH. We will write MATH for MATH. Then NAME identity leads to the equation MATH . This immediately implies: MATH. Since MATH this shows that MATH for all MATH. We can now solve the homogeneous REF degree by degree. Solutions of degree zero are cocycles of the NAME algebra MATH, hence they are... |
math/0007121 | REF follow from REF , and REF from a direct application of REF . For any other simple pseudoalgebra MATH, the strategy is to construct a continuous family of pseudoalgebras MATH, indexed by MATH endowed with the NAME topology, that are all isomorphic to MATH when MATH, and whose fiber at MATH is one of the pseudoalgebr... |
math/0007127 | Since MATH and MATH are discrete, and MATH is totally disconnected, there exists a compact, open subgroup MATH of MATH, such that MATH. Let MATH and MATH, so MATH and MATH are finite-index, open subgroups of MATH, and define MATH by MATH for MATH and MATH. |
math/0007127 | Let MATH and MATH be finite-index subgroups of MATH, such that MATH is an isomorphism from MATH to MATH. Then MATH induces isomorphisms MATH . By identifying each of MATH and MATH with MATH in the natural way (and noting that MATH), we may think of MATH and MATH as MATH-subspaces of MATH. By replacing MATH and MATH wit... |
math/0007127 | By assumption, there exists a lattice MATH such that MATH. Since MATH, and MATH is abelian, we see that MATH. Therefore MATH, so, by the choice of MATH, we have MATH. By assumption, MATH is a lattice in MATH, so MATH is closed CITE, hence discrete. Thus, there is an open compact subgroup MATH, such that MATH. Let MATH,... |
math/0007127 | We may assume MATH. (Otherwise, we must have MATH, which means MATH is abelian, so REF applies.) From REF , we may assume there exist CASE: a standard automorphism MATH of MATH, and CASE: a homomorphism MATH, such that MATH, for all MATH. From REF , we may assume that there is a finite-index subgroup MATH of MATH, such... |
math/0007127 | Because MATH and MATH have finite codimension in MATH and MATH, respectively, we see that MATH has finite codimension in MATH. Thus, in proving REF, we may assume that MATH. CASE: This follows from our proof of REF below. CASE: There are some nonzero MATH, such that MATH. Let MATH. We claim that MATH. Otherwise, we hav... |
math/0007127 | CASE: Choose MATH. Then REF implies the desired conclusion. CASE: From REF, we have MATH and MATH, so MATH. |
math/0007127 | Assume, for the moment, that MATH. For MATH, define MATH iff MATH. For nonzero MATH, we see, from Notation REF, that MATH has finite codimension in MATH if and only if MATH has finite codimension in MATH. Therefore, REF implies that MATH iff MATH. The equivalence classes are precisely the sets of the form MATH, for som... |
math/0007127 | Choose MATH, such that MATH and MATH; let MATH. For MATH, we have MATH so MATH contains MATH. Because MATH and MATH contain codimension-MATH subspaces of MATH and MATH, respectively, this implies that MATH contains a codimension-MATH subspace of MATH. Because both MATH and MATH belong to MATH, the desired conclusion fo... |
math/0007127 | Choose MATH as in REF , and let MATH and MATH. It suffices to show MATH and MATH . Let MATH be the irreducible factors of MATH. Then we may write MATH where MATH. From the NAME Remainder Theorem, we know that the natural ring homomorphism from MATH to MATH is an isomorphism. Thus, we may work in each factor MATH, and a... |
math/0007127 | For any MATH, we have MATH . Then the proof is completed by induction on MATH. |
math/0007127 | Let MATH be the codimension of MATH. Choose a power MATH of MATH so large that MATH is MATH-separable. Then REF implies MATH. Choose MATH, such that MATH and MATH. We have MATH . So MATH has small codimension in MATH. Therefore MATH must have small codimension in MATH, so MATH must be small, as desired. |
math/0007127 | Apply REF to the map MATH of REF . |
math/0007127 | Let MATH be the codimension of MATH in MATH, and choose MATH so large that, for every MATH, the subspace MATH contains elements of degree MATH whose leading coefficients span MATH. For any element of MATH of degree MATH, we show that there is a MATH-separable element of MATH of degree MATH with the same leading coeffic... |
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