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The Circle Problem on Surfaces of Variable Negative Curvature - Pollicott, Sharp (1997) (Correct)
....proof of Theorem 1. Note that for Re(s) h, we may write #(s) as a Stieltjes integral with respect to N(t) e st dN(t) Since we have that #(s) has an extension as an analytic function to a neighbourhood of the line Re(s) h, apart from a simple pole at s = h, the Ikehara Tauberian theorem [2] tells us that N(t) #, where C 0 is the residue of #(s) at s = h. 4. Some generalizations In this final section we discuss some generalizations of Theorem 1. First we give a higher dimensional version of the result. Observe that the only point at which we use the fact that M is a ....
S. Lang, Algebraic Number Theory, Addison-Wesley, New York, 1970.
Fermat's Last Theorem - Darmon, Diamond, Taylor (2000) (1 citation) (Correct)
....one has a well de ned conjugacy class [c] in G consisting of those elements that arise as complex conjugation for some embedding , We will denote by G1 the subgroup f1; cg for one such element c. We have the following fundamental results concerning the structure of G (see for instance [La1]) 51 Theorem 2.1 If F= is a nite extension then F is only rami ed at nitely many primes (those dividing the discriminant of F= Theorem 2.2 (Hermite Minkowski) If S is a nite set of primes and if d 2 0 then there are only nitely many extensions F= of degree d which are unrami ed outside ....
S. Lang, Algebraic Number Theory, Addison-Wesley, Reading, MA 1970.
The Circle Problem on Surfaces of Variable Negative Curvature - Pollicott, Sharp (1997) (Correct)
.... that for Re(s) h, we may write j(s) as a Stieltjes integral with respect to N(t) j(s) Z 1 0 e Gammast dN(t) Since we have that j(s) has an extension as an analytic function to a neighbourhood of the line Re(s) h, apart from a simple pole at s = h, the Ikehara Tauberian theorem [2] tells us that N(t) Ce ht , as t 1, where C 0 is the residue of j(s) at s = h. 4. Some generalizations In this final section we discuss some generalizations of Theorem 1. First we give a higher dimensional version of the result. Observe that the only point at which we use the fact that M ....
S. Lang, Algebraic Number Theory, Addison-Wesley, New York, 1970.
On the rank of Picard groups of modular varieties attached to.. - Bruinier (2001) (Correct)
.... as a Dirichlet series: A r (N) 1 2 i X n2Z f0g 1 n G(n; N) r = 1 X n 1 1 n = G(n; N) r ) Using the fact G(n; N) aG(n=a; N=a) for aj(n; N ) we nd A r (N) N r 1 X ajN X m 1 (m;a) 1 1 m a 1 r = G(m; a) r ) If we insert the explicit formula for G(m; a) cf. [La] chapter 4.3) we obtain by a lengthy but straightforward calculation A r (N) N r 1 X ajN a 1 r=2 L( a ; 1) 8 : 0; if a 1; 2 (mod 4) 4 (r) if a 1 (mod 4) 2 (r 1) 2 8 (r) if a 0 (mod 4) 14 Here L( a ; s) denotes the Dirichlet series associated to the ....
S. Lang, Algebraic Number Theory, Addison-Wesley (1970).
On Diophantine definability and decidability in large subrings.. - Shlapentokh (2001) (1 citation) (Correct)
....Q, for every x # E, ord Q H(x) # 0. Proof. Let # and Q be as in the statement of the lemma. Then powers of # constitute a local integral basis of M over E with respect to Q. Thus the factorization of the minimal polynomial of # modulo Q corresponds to the factorization of Q in M . See [9], Proposition 25, page 27] Let x # E and assume x has a negative order at Q. Then H(x) has a negative order at Q. On the other hand, suppose x is integral at Q and H(x) has positive order at Q. Then H(T ) has a root modulo Q and thus a linear factor modulo Q. This implies Q has a factor of ....
S. Lang, Algebraic Number Theory, Addison Wesley, Reading, MA, 1970
Diophantine Undecidability of Function Fields of Characteristic .. - Shlapentokh (2001) (Correct)
....N [T ] the monic irreducible polynomial of a generator of N l over N . Since the constant extension is separable we can conclude the following. The constant field of N l M contains N l . Factorization of f(T ) modulo Q corresponds to the factorization of Q in the extension N l M M. See [16], Proposition 25, page 27. Modulo Q, f(T ) will have a linear factor. Therefore, Q will have a factor of relative degree 1 in MN l . Let Q l be this factor. Then the residue field of Q l is the same as the residue field of Q, i.e. the residue field is N l . Thus, the constant field of MN l is N l ....
....residue field is N l . Thus, the constant field of MN l is N l and Q l is of degree 1. 3. Let f(T ) # N [T ] be the monic irreducible polynomial of a generator of M over M . Then f(T)does not factor modulo Q. Therefore, since this constant extension is separable, by Proposition 25, page 27 of [16] again, Q has only one factor of relative degree deg(f)in M . Thus, the residue field of Q is of degree deg(f) over N . This residue field also contains the constant field of M which is of degree at least deg(f) over N . Thus, the residue field of Q must be equal to the constant field of M . ....
S. Lang. Algebraic Number Theory. Addison Wesley, Reading, MA, 1970.
Automorphisms of even unimodular lattices and unramified.. - Gross, McMullen (2001) (Correct)
....(An equivalent condition is that S( 1)S(1) 1) n ; see Proposition 3.3. We will show: Proposition 3.1 If the polynomial S(x) is unrami ed, then the eld extension K=k is also unrami ed (at all nite primes) Fields and traces. We start with some algebraic preliminaries; see e.g. FT] or [La] for more background. Let OK K be the ring of integers in a number eld K=Q . The trace form on K is de ned by h ; i = Tr K Q ( For any Z module M K generated by a basis of K over Q , we de ne the dual module by M = f 2 K : h ; i 2 Z for all 2 Mg = Hom(M;Z) If I K is a ....
S. Lang. Algebraic Number Theory. Addison-Wesley, 1970.
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Lang, S 1996 Algebraic Number theory
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