Proceedings of the 44th Symposium on Ring Theory and ...

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Pro<strong>of</strong>. If A embeds in ̂R, <strong>the</strong>n A admits a generating character, by Lemmas 10 and 11. Conversely, let ϱ be a generating character <strong>of</strong> A. We use ϱ to define f : A → ̂R, as follows. For a ∈ A, define f(a) ∈ ̂R by f(a)(r) = ϱ(ra), r ∈ R. It is easy to check that f(a) is indeed in ̂R, i.e., that f(a) is a character <strong>of</strong> R. It is also easy to verify that f is a left R-module homomorphism from A to ̂R. If a ∈ ker f, <strong>the</strong>n ϱ(ra) = 0 for all r ∈ R. Thus <strong>the</strong> left R-submodule Ra ⊂ ker ϱ. Because ϱ is a generating character, we conclude that Ra = 0. Thus a = 0, and f is injective. □ When A = R, Proposition 12 is consistent with <strong>the</strong> definition <strong>of</strong> a generating character <strong>of</strong> a ring. Indeed, if R embeds into ̂R, <strong>the</strong>n R and ̂R are isomorphic as one-sided modules, because <strong>the</strong>y have <strong>the</strong> same number <strong>of</strong> elements. Theorem 13 (†). Let R = M m (F q ) be <strong>the</strong> ring <strong>of</strong> m × m matrices over a finite field F q . Let A = M m×k (F q ) be <strong>the</strong> left R-module <strong>of</strong> all m × k matrices over F q . Then A admits a left generating character if and only if m ≥ k. Pro<strong>of</strong>. If m ≥ k, <strong>the</strong>n, by appending m − k columns <strong>of</strong> zeros, A can be embedded inside R as a left ideal. By Example 3 and Lemma 10, A admits a generating character. Conversely, suppose m < k. We will show that no character <strong>of</strong> A is a generating character <strong>of</strong> A. To that end, let ϖ be any character <strong>of</strong> A. By Example 3, ϖ has <strong>the</strong> form ϖ Q for some k × m matrix Q over F q . Because k > m, <strong>the</strong> rows <strong>of</strong> Q are linearly dependent over F q . Let P be any nonzero matrix over F q <strong>of</strong> size m × k such that P Q = 0. Such a P exists because <strong>the</strong> rows <strong>of</strong> Q are linearly dependent: use <strong>the</strong> coefficients <strong>of</strong> a nonzero dependency relation as <strong>the</strong> entries for a row <strong>of</strong> P . We claim that <strong>the</strong> nonzero left submodule <strong>of</strong> A generated by P is contained in ker ϖ Q . Indeed, for any B ∈ R, ϖ Q (BP ) = ϑ q (tr(Q(BP ))) = ϑ q (tr((BP )Q)) = ϑ q (tr(B(P Q))) = 0, using P Q = 0 and <strong>the</strong> well-known property tr(BC) = tr(CB) <strong>of</strong> <strong>the</strong> matrix trace. Thus, no character <strong>of</strong> A is a generating character. □ Proposition 14 (†). Suppose A is a finite left R-module. Then A admits a left generating character if and only if soc(A) admits a left generating character. Pro<strong>of</strong>. If A admits a generating character, <strong>the</strong>n so does soc(A), by Lemma 10. Conversely, suppose soc(A) admits a generating character ϑ. Utilizing <strong>the</strong> short exact sequence (2.4), let ϱ be any extension <strong>of</strong> ϑ to a character <strong>of</strong> A . We claim that ϱ is a generating character <strong>of</strong> A. To that end, suppose B is a submodule <strong>of</strong> A such that B ⊂ ker ϱ. Then soc(B) ⊂ soc(A) ∩ ker ϱ = soc(A) ∩ ker ϑ, because ϱ is an extension <strong>of</strong> ϑ. But ϑ is a generating character <strong>of</strong> soc(A), so soc(B) = 0. Since B is a finite module, we conclude that B = 0. Thus ϱ is a generating character <strong>of</strong> A. □ Corollary 15 (†). Let A be a finite left R-module. Suppose soc(A) admits a left generating character ϑ. Then any extension <strong>of</strong> ϑ to a character <strong>of</strong> A is a left generating character <strong>of</strong> A. We now (finally) turn to <strong>the</strong> pro<strong>of</strong> <strong>of</strong> Theorem 5. (†) Pro<strong>of</strong> <strong>of</strong> Theorem 5. Statements (2) and (3) are equivalent by Proposition 8. We next show that (3) implies (1). –229–
By Example 3, <strong>the</strong> right R-module (R/ rad(R)) R equals <strong>the</strong> character module <strong>of</strong> <strong>the</strong> left R-module R (R/ rad(R)). By Proposition 9 applied to <strong>the</strong> left R-module A = R R, we have ( R (R/ rad(R)))̂ ∼ = soc( ̂R R ) ∼ = soc(R R ), because ̂R is assumed to be right free. We thus have an isomorphism (R/ rad(R)) R ∼ = soc(RR ) <strong>of</strong> right R-modules. One can ei<strong>the</strong>r repeat <strong>the</strong> argument for a left isomorphism (using (2)) or appeal to <strong>the</strong> <strong>the</strong>orem <strong>of</strong> Honold [15, Theorem 2] mentioned after Definition 1. Now assume (1). Referring to (2.1), we see that R being Frobenius implies that soc(R) is a sum <strong>of</strong> matrix modules <strong>of</strong> <strong>the</strong> form M mi (F qi ). By Theorem 13 and summing, soc(R) admits a left generating character. By Propositions 7 and 14, R itself admits a left generating character. Thus (2) holds. □ 2.4. Frobenius Algebras. In this subsection I want to point out <strong>the</strong> similarity between a general (not necessarily finite) Frobenius algebra and a finite Frobenius ring. I thank Pr<strong>of</strong>essor Yun Fan for suggesting this short exposition. Definition 16. A finite-dimensional algebra A over a field F is a Frobenius algebra if <strong>the</strong>re exists a linear functional λ : A → F such that ker λ contains no nonzero left ideals <strong>of</strong> A. It is apparent that <strong>the</strong> structure functional λ plays a role for a Frobenius algebra comparable to that played by a left generating character ϱ <strong>of</strong> a finite Frobenius ring. As one might expect, <strong>the</strong> connection between λ and ϱ is even stronger when one considers a finite Frobenius algebra. Recall that every finite field F q admits a generating character ϑ q , by Example 2. Theorem 17 (†). Let R be a Frobenius algebra over a finite field F q , with structure functional λ : R → F q . Then R is a finite Frobenius ring with left generating character ϱ = ϑ q ◦ λ. Conversely, suppose R is a finite-dimensional algebra over a finite field F q and that R is a Frobenius ring with generating character ϱ. Then R is a Frobenius algebra, and <strong>the</strong>re exists a structure functional λ : R → F q such that ϱ = ϑ q ◦ λ. Pro<strong>of</strong>. Both R ∗ := Hom Fq (R, F q ) and ̂R = Hom Z (R, Q/Z) are (R, R)-bimodules satisfying |R ∗ | = | ̂R| = |R|. A generating character ϑ q <strong>of</strong> F q induces a bimodule homomorphism f : R ∗ → ̂R via λ ↦→ ϑ q ◦ λ. We claim that f is injective. To that end, suppose λ ∈ ker f. Then ϑ q ◦λ = 0, so that λ(R) ⊂ ker ϑ q . Note that λ(R) is an F q -vector subspace contained in ker ϑ q ⊂ F q . Because ϑ q is a generating character <strong>of</strong> F q , λ(R) = 0, by Proposition 7. Thus λ = 0, and f is injective. Because |R ∗ | = | ̂R|, f is in fact a bimodule isomorphism. We next claim that <strong>the</strong> structure functionals in R ∗ correspond under f to <strong>the</strong> generating characters in ̂R. That is, if ϖ = f(λ), where λ ∈ R ∗ and ϖ ∈ ̂R, <strong>the</strong>n λ satisfies <strong>the</strong> condition that ker λ contains no nonzero left ideals <strong>of</strong> R if and only if ϖ is a generating character <strong>of</strong> R (i.e., ker ϖ contains no nonzero left ideals <strong>of</strong> R). Suppose ϖ is a generating character <strong>of</strong> R, and suppose that I is a left ideal <strong>of</strong> R with I ⊂ ker λ. Since ϖ = ϑ q ◦ λ, we also have I ⊂ ker ϖ. Because ϖ is a generating character, Proposition 7 implies I = 0, as desired. Conversely, suppose λ satisfies <strong>the</strong> condition that ker λ contains no nonzero left ideals <strong>of</strong> R, and suppose that I is a left ideal <strong>of</strong> R with I ⊂ ker ϖ. Then λ(I) is an F q -linear –230–
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Proceedings <stron
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Organizing Committee of</st
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Polycyclic codes and sequential cod
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Preface The 44th <
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8:40-9:30 Dan ZachariaSyracuse Univ
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The 44th S
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Tuesday September 27 8:40-9:30 Atsu
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Conversely, let (T , F) be a stable
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Then T = gen(X) and X is Ext-projec
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DIMENSIONS OF DERIVED CATEGORIES TA
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Notation 4. (1) Let A be an abelian
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of finitely genera
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is the map given b
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(2) Let R be a commutative ring whi
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QUIVER PRESENTATIONS OF GROTHENDIEC
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Proposition 3. Let C be a category,
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Example 11. Let Q be the</s
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and Gr(X ′ ) is given by
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particular, the de
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Proposition 1 and 2 leave us with d
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4. Examples Example 6. Let M = M(A
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EXAMPLE OF CATEGORIFICATION OF A CL
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The aim of <strong
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X ∈ mod Π Q S 1 S 2 S 3 1 ❁
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The previous proposition has a dual
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Computing inductively all t
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Example 23. We have ⎛ ⎞ ⎛ ⎞
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classification of
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quiver Q over K, and let D b (H) <s
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Then Q ∈ Q 17 . Suppose char K
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Moreover the Hochs
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DERIVED AUTOEQUIVALENCES AND BRAID
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3. Periodic Twists We now describe
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We now specialise to the</s
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and τ − n := Ext n Λ(DΛ, −)
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where r v ij := a i a j −a j a i
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ON A DEGENERATION PROBLEM FOR COHEN
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We make several othe</stron
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Let A be a commutative Gorenstein r
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3. Extended orders In the</
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WEAK GORENSTEIN DIMENSION FOR MODUL
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1.2. Introduction. The notion <stro
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e 2 A z 1 −→ X → 0 in mod-A,
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5. Gorenstein algebras In this sect
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ON Ω-PERFECT MODULES AND SEQUENCES
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when R is a Nakayama algebra <stron
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says that components of</st
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then we obtain a c
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Proposition 16. Let C s be a stable
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3.1. ZD ∞ -components. We assume
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Theorem 26. Let R be a local artini
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where each f i is an irreducible ep
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QUANTUM UNIPOTENT SUBGROUP AND DUAL
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{F i } i∈I and q-Serre relations
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Theorem 8. (1)Let U − q (w) → U
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E-mail address: ykimura@kurims.kyot
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Indeed, (1, 2, 3, 4) {1,2} −−
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By comparing this the</stro
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(a) (b) Proposition 17. Let G be a
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POLYCYCLIC CODES AND SEQUENTIAL COD
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⎛ ⎞ 0 0 c 0 1 c t D c = ⎜ 1
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References [1] D. Boucher and P. So
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2. Dimension of tr
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APR TILTING MODULES AND QUIVER MUTA
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(1) Q ′ is a quiver obtained from
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1 arrows of ˜µ L
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[11] S. Fomin, A. Zelevinsky, Clust
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Theorem. Assume r is a common prime
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References [1] Apostol, T. M., The
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e(n, s) n = 6 7 8 s = 1 36 63 120 2
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Now we will show that the</
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is equal to ( n 2s−1 ); hence we
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THE NOETHERIAN PROPERTIES OF THE RI
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Proposition 4 (Paper III, Propositi
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HOCHSCHILD COHOMOLOGY OF QUIVER ALG
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(4) In [10], Green, Snashall and So
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4. Quiver algebras defined by two c
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[6] L. Evens, The cohomology ring <
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Then there is an i
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• ( ˜F , ˜m) = ({ (1, 3), (2, 3
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Corollary 7 ([16, Theorem 3.4]). Th
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We have the direct
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We say a CW complex is regular, if
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SHARP BOUNDS FOR HILBERT COEFFICIEN
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Let us briefly note how this paper
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whence the set Λ
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Proof of</
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We have Q·H j m(A/(a 1 , a 2 , ·
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Let B = A/U(a d−1 ). We t
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• quiver Γ = (I, Ω) I Ω
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B = (B τ ) relations ρ 1 , · ·
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Definition 1. double quiver ˜Γ p
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◦ side Γ = (I, Ω) cycle (ac
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Definition 6. n × n A(Γ Dyn ) =
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(2) P Lie theory
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Page 195 and 196: P (Γ)-module i-th radical B →
Page 197 and 198: B ∈ Λ(d) ε ∗ i (B) := dim C
Page 199 and 200: I-graded vector space V (d) (B τ )
Page 201 and 202: A ∗(i) l (a) = ∑n+1 t=l+1 (a i,
Page 203 and 204: wt(a) = (−d 1 , −d 2 , −d 3 )
Page 205 and 206: Ω Q Ω ( B Ω ; wt, ε i , ϕ
Page 207 and 208: “Λ(d) G(d)-” Definition 25.
Page 209 and 210: (( ( ) ( ) 0 s 0 0 0 , , , s) ( u 0
Page 211 and 212: MATRIX FACTORIZATIONS, ORBIFOLD CUR
Page 213 and 214: ( ∂f Jac(f t t ) := C[x 1 , . . .
Page 215 and 216: Definition 14. CM L f (R f ) Ob(C
Page 217 and 218: 4. Dolgachev Proposition 24 ([1]
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Page 221 and 222: (f, G f ) Dolgachev A Gf f Berg
Page 223 and 224: ON A GENERALIZATION OF COSTABLE TOR
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Page 231 and 232: GRADED FROBENIUS ALGEBRAS AND QUANT
Page 233 and 234: Definition 2. A graded algebra A is
Page 235 and 236: In this case, ν ∈ Aut k E induce
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Page 239 and 240: The topics covered are ring-<strong
Page 241 and 242: (2.3) soc( R R) ∼ = k⊕ s i T i
Page 243: principal ideal Rr ⊂ ker ϖ. Thus
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Page 259 and 260: ψ : ε(A) → Â. In this way, C
Page 261 and 262: REALIZING STABLE CATEGORIES AS DERI
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Page 267 and 268: Next we consider the</stron
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Page 271 and 272: ALGEBRAIC STRATIFICATIONS OF DERIVE
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Page 283 and 284: INTRODUCTION TO REPRESENTATION THEO
Page 285 and 286: is exact. Since E n ∼ = En+r , Mi
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P = P ′ t by t is a projective R
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SUBCATEGORIES OF EXTENSION MODULES
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Lemma 6. Let S 1 and S 2 be Serre s
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In the first row <
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