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The class number formula

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The class number formula

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Content THE CLASS NUMBER FORMULA by Christopher D. Berry A Thesis Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulllment of the Requirements for the Degree MASTER OF ARTS (MATHEMATICS) May 2012 Copyright 2012 Christopher D. Berry In memory of Cecil E. Jordan ii Acknowledgments I cannot thank Professor Sheldon Kamienny enough for his tireless committment to me and to seeing this thesis completed. Without his guidance and encouragement I'm not sure I would have made it to the end of this road. iii Table of Contents Dedication ii Acknowledgments iii Abstract v The Class Number Formula 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Number Fields and their Properties . . . . . . . . . . . . . . . . . . . . . . . . 2 Norm and Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Discriminants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Lattices and Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Regulators and the Fundamental Domain . . . . . . . . . . . . . . . . . . . . . 5 Dedekind Zeta Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 The Class Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Finiteness of the Class Number . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 The Class Number Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Proof of the Class Number Formula . . . . . . . . . . . . . . . . . . . . . . . . 12 Bibliography 16 iv Abstract The analytic class number formula equates the residue of a dedekind zeta function at 1 with various properties of a related number eld. These properties are dened, their correlation explained, and then they are related to the formula. The class number, a measure of a number eld's failure to be a principal ideal domain, is dened and analyzed. Finally, the formula is explicitly stated and proven. v The Class Number Formula Introduction The Class Number Formula was rst proven by Dirichlet in 1839, but for classes of quadratic forms rather than ideals. Since then, it has been expanded and used in a wide variety of circumstances, from being a test for whether or not a ring is a principal ideal domain, to calculations of the properties of number elds. Its use comes up in the study of everything from ring theory, to elliptic curves, to residues of zeta functions in analysis. Herein, an explanation and proof. Preliminaries Any complex numberz withz n = 1 forn2Z is a root of unity, sometimes called annth root of unity. When discussing complex functions, it is often helpful to understand the behavior of a function near a pole. The value of the function goes to innity as the argument approaches the pole. For simple poles we can calculate a nite residue as follows, which provides insight into the function's behavior. Let f(z) be a function with a pole at s. If there exists a holomorphic function g(z) such thatf(z) =g(z)=(zs) on the domain of f, then s is a simple pole of f, and lim z!s (zs)f(z) is the residue of f at s. 1 Number Fields and their Properties A number eld is a nite eld extension of the rational numbersQ. Namely, it is itself a eld, which containsQ as a subeld and can therefore be viewed as a nite-dimensional Q-vector space. Since every number eld is a nite extension of Q, it is an algebraic extension ofQ. And as such, every element of a given number eld is also algebraic. Thus every element is an algebraic number, or a zero of a polynomial with coecients in Q (and in fact the zero of a monic polynomial, found by dividing the original polynomial by the leading coecient). And since all of the elements of a given number eld are algebraic, we know that the number eld is itself a subeld ofQ, the algebraic closure ofQ. Those elements satisfying monic polynomials with coecients in Z, the ring of inte- gers, are called algebraic integers. All rational integers are algebraic integers, and the algebraic integers in a particular number eld are closed under addition and multipli- cation, thereby forming a ring called the ring of integers of the number eld (for the number eld K, the ring of integers is denotedO K ). This ring is a Dedekind domain. One property of a number eld is that it can be embedded into R, the real num- bers, or into C, the complex numbers, in a specic number of dierent ways. A real embedding of a number eld K is an injective eld homomorphism f : K! R and a complex embedding is an injective eld homomorphism g :K!C. Since both are eld homomorphisms, they x all of the elements of Q. The number of embeddings is equal to the dimension of K over Q as a vector space (and is therefore always nite). It is customary to denote this number as r 1 + 2r 2 wherer 1 is the number of real embeddings and 2r 2 is the number of complex embeddings. Complex embeddings come in pairs by composing with the complex conjugation operation, and as such, r 2 is always an integer. So we have: [K :Q] =r 1 + 2r 2 with r 1 ;r 2 2Z 2 With these embeddings in mind, we can construct two important mappings. The rst, denoted by', is constructed by ordering the embeddings ofK intoR asf 1 ; 2 ;:::; r 1 g and then choosing a representative embedding from each pair of complex conjugates r 1 +1 ; r 1 +2 ;:::; r 1 +r 2 so that by the usual association ofC withR 2 , we get: ' :K!R n by x7! ( 1 (x); 2 (x);:::; r 1 +r 2 (x)) The second, denoted by , uses the same ordering of embeddings, and is given by: :K!R r 1 +r 2 x7! (logj 1 (x)j; logj 2 (x)j;:::; logj r 1 (x)j; 2 logj r 1 +1 (x)j;:::; 2 logj r 1 +r 2 (x)j) Finally, dene a function :R n !R r 1 +r 2 such that =. Namely, (x 1 ;x 2 ;;x n ) = (logjx 1 j;:::; logjx r 1 j; log x r 1 +1 2 +x r 1 +2 2 ;:::; log x n1 2 +x n 2 ) Notice that for a complex embedding i , 2 logj i (x)j = logj i (x)j 2 = log < ( i (x)) 2 += ( i (x)) 2 Norm and Trace For a number eld K, we dene the norm and trace of an element based on how mul- tiplication by that element aects the eld. For any element k2 K, we notice that multiplication by k is a Q-linear transformation of K. As such, we can constuct a matrix for this tranformation relative to a basis ofK as aQ-module. The normN(k) of k is the determinant of this matrix, and the trace Tr(k) of k is the trace of this matrix, as dened in linear algebra. From linear algebra, we know that the trace and norm are 3 independent of the basis we choose forK. We can view Tr(k) as a sum of the embeddings of k, so that: Tr(k) = n X i=1 i (k) Additionally, we can view N(k) as a product of the embeddings of k, so that: N(k) = n Y i=1 i (k) The ordering of the embeddings is clearly not relevant in either formulation. Discriminants Let K be a number eld of degree n (namely [K : Q] = n) with ring of integersO K , and letfe 1 ;e 2 ;:::;e n g be a basis forO K as a Z-module. Letf 1 ; 2 ;:::; n g be the collection of real and complex embeddings ofK, and dene a matrixA byA = [a ij ] where a ij = i (e j ). We dene the discriminant D K of K to be the square of the determinant of the matrix A, so D K =det(A) 2 . Since we are taking the determinant squared, the properties of the determinant tell us that the discriminant remains unchanged if any of the rows or columns of A are switched. As such, the ordering of the basisfe j g and the ordering of the embeddings f i g do not aect the discriminant. Looking back at our earlier denition of the discriminant, we can use the trace to see that the discriminant is independent of our choice of basis. Since P n k=1 k (e i e j ) = Tr(e i e j ), we get that: D K = det(A) 2 = det(A) det(A) = det(A) det(A t ) = det(AA t ) = det([ n X k=1 k (e i ) k (e j )]) = det([ n X k=1 k (e i e j )] = det([Tr(e i e j )]) 4 And since we then see the discriminant calculated purely in terms of the trace, which does not depend on our choice of basis, and without reference to the embeddings, we know it is a value which is independent of both and can therefore be considered an essential property of the number eld itself. Similarly, we can calculate the discriminant D(I) of an idealI in the ring of integers O K by beginning with a Z-basis for I instead of for all ofO K . As with D K , D(I) is a non-zero integer. Lattices and Volume Dene a latticeL of rankm inR n to be an additive subgroupfc 1 v 1 ++c m v m ; c i 2Zg for some linearly independent setfv i g2R n . A lattice has full rank if m = n. We can compute the volume of the fundamental parallelepiped of a lattice by taking the absolute value of the determinant given by a matrix whose rows are thefv i g. One example would be to see that '(O K ) is in fact a lattice of full rank in R n and that by using the same Z-basis we used to calculate the discriminant of K to create a matrix ' K given by [ i (x j )], ' K looks remarkably similar to the matrix A above. In fact, simple matrix manipulations show that the determinant of the matrix of '(O K ) is equal to (2i) r 2 times det(A). From this we see that the volume, given byjdet(' K )j, is 2 r2 p D K . A similar result can be computed for the discriminant of an idealI inO K , by taking a Z-basis ofI instead ofO K . The volume of the lattice for'(I) is the normN(I) = [O K :I] times the volume of the lattice for '(O K ), we get that the volume is 2 r2 N(I) p D K . Regulators and the Fundamental Domain Take a number eld K, with ring of integersO K , and look at the group of units ofO K . Dirichlet's Unit Theorem states that the group of units is nitely generated with rank 5 r =r 1 +r 2 1. One way to see this is to recall that units have norm with absolute value equal to 1, so that by our denition of the norm, any given unit u inO K has: 1 =jN(u)j = n Y i=1 j i (u)j = r 1 Y i=1 j i (u)j r 1 +r 2 Y i=r 1 +1 i (u) i (u) = r 1 Y i=1 j i (u)j r 1 +r 2 Y i=r 1 +1 j i (u)j 2 Taking the natural logarithm of this equation gives us a sum of the coordinates of (u) on the right and 0 on the left, telling us that all the units ofO K map to a hyperplane in R r 1 +r 2 under , and as such have rank at most r 1 +r 2 1. The remainder of the proof spells out the exact nature of the group of units as a product of a cyclic group of roots of unity andZ r 1 +r 2 1 by virtue of the fact that there arer 1 +r 2 1 linearly independent units, ensuring that the group of units will have rank r 1 +r 2 1 and not less than that. As such, choose a set of fundamental units fu 1 ;u 2 ;:::;u r g, which are gen- erators of the group of units of O K modulo roots of unity. For each u i , let f 1 (u i ); 2 (u i );:::; j (u i );:::; r+1 (u i )g be the images of u i under embeddings into R or C as dened before, and let the d j be the degree of each embedding j (so d j = 1 for the real embeddings and d j = 2 for the complex embeddings). From this we can construct an r by r + 1 matrix with entries: [d j logj j (u i )j]. And since all units have norm with absolute value of 1, and the sum of the entries in a row of this matrix is the log of the norm, which is 0. If we take the determinant of the matrix formed by deleting a column, this value will be independent of which column we choose (up to sign). The absolute value of this determinant is the regulator of the number eld, denotedR K . The regulator is also independent of our choice of generators. Dene = (1;:::; 1; 2;:::; 2) with 1 in the rst r 1 coordinates and 2 in the next r 2 coordinates. Now since we saw that the embeddings of eachu i sum to 0, we know that the (u i ) span anR r -dimensional hyperplane ofR r+1 . Since lies outside this hyperplane, we see thatf; (u 1 ); (u 2 );::: (u r )g form a basis for R r+1 =R r 1 +r 2 . Now we dene a fundamental domain X K inR n as the set of x2R n such that: 1. N(x)6= 0 6 2. (x) =c 0 + P r 1 +r 2 1 i=1 c i (u i ) with 0ci< 1 for i> 0 3. 0 arg(x 1 )< 2 w K forr 1 = 0 andx 1 > 0 forr 1 > 0, wherex 1 is the rst coordinate of x inR n =R r 1 C r 2 and w K is the number of roots of unity inO K From this we can see that the fundamental domain is a cone in R n . And in fact, we can relate the volume of a portion of this cone to the regulator R K . To do so, we look at the portion we call Y K , namelyfx2X K ;N(x) 1g. Let us start by taking Y j = e 2ij w K Y K for 0 j < w K . Since e 2ij w K is a unit and multiplication by a unit is volume preserving, we get that each Y j has the same volume as Y K and thus Y K =[ w K 1 j=0 Y j has w K times the volume of Y K . If we look at this domain inR n =R r 1 C r 2 , we can translate it into polar coordinates by settingx i = (p i ;s i ) wherep i 0 for 1ir 1 +r 2 . As such, ther 1 real embeddings have s i =1 and the r 2 complex embeddings after that have 0 s i < 2. In this notation, we see that N(x) 1 becomes: 0< r 1 +r 2 Y i=1 p d i i 1 Let P describe the points (p 1 ;p 2 ;:::;p r 1 +r 2 ) which meet the conditions given above (note that those conditions do not restrict thefs i g). As the jacobian of our transforma- tion to polar coordinates can be determined to bep r 1 +1 p r 1 +2 p r 1 +r 2 , and thefs i g can be eliminated, we get: V Y = 2 r 1 (2) r 2 w K Z Z P p r 1 +1 p r 1 +2 p r 1 +r 2 dp 1 dp 2 dp r 1 +r 2 Now we set t 0 = Q r1+r2 i=1 p d i i and dene 0 t i < 1 for i = 1; 2;:::;r so that our conditions on x become: logp d i i = d i N r 1 +r 2 Y j=1 p d j j + r X k=1 t k logj i (u k )j = d i t 0 N + r X k=1 t k logj i (u k )j 7 As such, we note that a change of variables again from the fp i g to the ft i g has a jacobian given by the determinant of an r 1 + r 2 by r 1 + r 2 matrix with columns [ p i nt 0 p i logj i (u 1 )j ::: p i logj i (u r )j]. Dividing out the fp i g from each column, multiplying out the nt 0 from the rst row, and multiplying the last r 2 rows by 2 gives p 1 p 2 p r 1 +r 2 nt 0 2 r 2 times the determinant of a new matrix with columns [d i d i logj i (u 1 )j ::: d i logj i (u r )j] so that adding all of the columns to the rst makes the rst column [n 0 ::: 0] revealing the determinant of the matrix to be nR K . From the denition of t 0 , we get that the jacobian of this transformation is R K 2 r 2p r 1 +1 p r 1 +r 2 . Finally, we apply this transformation to our integral, noting that the cube T dened by theft i g has volume 1 and see: V Y = 2 r 1 (2) r 2 w K Z Z T p r 1 +1 p r 1 +r 2 R K 2 r 2 p r 1 +1 p r 1 +r 2 dt 0 dt 1 dt r = 2 r 1 r 2 R K w K Dedekind Zeta Functions The Riemann Zeta Function Q (s) is dened as P 1 n=1 1 n s , continued analytically to the entire complex plane. The sum converges absolutely for s with real part greater than 1. The function has a simple pole at s = 1. Additionally, the Riemann Zeta Function satises a functional equation, given as: Q (s) = 2 s s1 sin s 2 (1s) Q (1s) where (s) is the analytic continuation of (s) = (s 1)! denited on positive integers and (s) = R 1 0 e t t s1 dt dened on complex numbers with positive real part. Since the factors in the functional equation are meromorphic functions on the complex plane, it provides a way to calculate values of the Riemann Zeta Function for a value s based on the function's value at 1s. This equation also demonstrates how the Riemann Zeta Function has zeros, called trivial zeros, at negative even integers due to its sin component. 8 The same cannot be said of positive even integers, since the component has poles at these values. Number theorists have shown a strong link between the behavior of the Riemann Zeta Function and the distribution of prime numbers amongst the integers. They have shown that all nontrivial zeros of the function lie in the so called critical strip of complex numbers with real value between 0 and 1. The Riemann Hypothesis states that all of these nontrivial zeros have real part equal to 1 2 , and remains an open question. Dedekind zeta functions generalize this concept to number elds. So that for a number eld K, K (s) is dened as the analytic continuation of P I 1 (N(I)) s , where the sum runs over all nonzero ideals ofO K . From this vantage point, the Riemann Zeta Function is simply the special case whereK =Q. Similar to the Riemann Zeta Function, general dedekind zeta functions converge for s with real part greater than 1 and have a simple pole at s = 1. Dedekind zeta functions are also given by a functional equation: f K (s) =jD K j s 2 h ( s 2 ) s 2 i r 1 2(2) s (s) r 2 K (s) f K (s) =f K (1s) The Riemann Hypothesis has an analogue for dedekind zeta functions, known as the Extended Riemann Hypothesis, which states that any nontrivial zeros of a dedekind zeta function will have real part equal to 1 2 . This hypothesis is important because the zeros of a dedekind zeta function have been connected to the distribution of prime ideals within the number eld's ring of integers. Like the Riemann Hypothesis, this is also an open question. The Class Number Let K be a number eld and letO K be its ring of integers. A fractional ideal I is an O K -module contained in K for which there exists a nonzero element c2O K such that cIO K . Every fractional ideal is nitely generated as anO K -module. 9 Next we endow the set of fractional ideals with a multiplication, given as follows. If I and J are fractional ideals, then their product is given by: IJ =a 1 b 1 +a 2 b 2 + +a n b n with a i 2I; b i 2J; n2Z >0 Now sinceO K is a dedekind domain, every fractional ideal is invertible, and we see that the fractional ideals form a group, F(O K ), withO K as the identity element. Any fractional ideal which is generated by a single element is called a principal fractional ideal. It is easy to see that the inverse of a principal fractional ideal is also a principal fractional idea, turning the set of principal fractional ideals into a group, P(O K ), a subgroup of F(O K ). By taking the quotient of these groups, we get the Ideal Class Group Cl(O K ): Cl(O K ) =F(O K )=P(O K ) The order of this group is called the class number ofO K , denoted h K . It is easy to see from this denition that this number measures the failure of a ring of integers to be a principal ideal domain. IfO K is a principle ideal domain, every fractional ideal is principal, and therefore F(O K ) = P(O K ), which implies that Cl(O K ) is trivial and therefore h K = 1. Finiteness of the Class Number To show that the class number is nite, we will show that there are only nitely many classes of ideals in the class group following the path laid out by Dedekind. To do this will require two main steps. First, that every class contains an ideal of bounded norm. And second, that there are only nitely many ideals of bounded norm. It follows from those two statements that there can only be nitely many classes and therefore the class number must be a nite integer. 10 To see that each class contains an ideal of bounded norm, we start by selecting an integral basis for our number eldK, such asfe 1 ;e 2 ;:::;e n g. Then any algebraic integer k2 K will be a Z-linear combination of the e i as k = c 1 e 1 +c 2 e 2 + +c n e n . For a xed k, dene c k to be an upper bound onjc i j for 1in. Next for each embedding j of K, we look at where it sends the basis, and set b j = P n i=1 j (e i ). Dene b to be an upper bound onjb j j for 1jn. Now if we take the norm of k, we get: jN(k)j = n Y j=1 j j (k)j = n Y j=1 n X i=1 c i j (e i ) n Y j=1 c k n X i=1 j (e i ) = n Y j=1 c k jb j j n Y j=1 c k b = (c k b) n So thatjN(k)j (c k b) n . Since the value of b is dependent only on the basis we use, b n is a nite positive real number which is constant for any algebraic interger we take the norm of. Now chose any class of ideals,C, in the ideal class group and letI be an ideal not in C such thatm n N(I)< (m + 1) n for some positive integer m. If we take all (m + 1) n of the algebraic integers formed as linear combinations of the basisfe 1 ;:::;e n g with integer coecients from 0 to m, the Pigeonhole Principle tells us that at least two must be in the coset ofI. Because of this, their dierence d must be inI, and the coecients of d as a linear combination of basis elements will each have absolute value less than or equal to m. By our demonstration above, d has norm N(d) (bm) n . Since d2 I, we know thathdiI inO K . As such, there must be an ideal I 0 2C such that II 0 =hdi. From this we see that: N(I)N(I 0 ) =N(hdi) =jN(d)j (bm) n b n N(I) So that dividing out by N(I), we're left with N(I 0 ) b n which we know to be a xed positive real number. ThusC contains an ideal of bounded norm, and sinceC was chosen arbitrarily, all the classes in Cl(O K ) have an ideal with norm bounded by b n . 11 Now to see that there are nitely many ideals of a given norm, x a positive integer m. Any ideal I with norm m will have m 2 I, and as suchhmi I. By unique factorization,hmi is the product of nitely many prime ideals, and therefore I must be a product of some combination, but not necessarily all, of those same prime ideals. But there are only nitely many dierent combinations that can be made up of those prime ideals, so there can only be nitely many unique ideals I with norm m. And so there can only be nitely many classes in Cl(O K ), so h K must be nite. The Class Number Formula The Class Number Formula can be stated as follows: For a number eld K, with ring of integersO K , r 1 real embeddings, 2r 2 complex embeddings, class number h K , regulator R K , discriminant D K , and w K the number of roots of unity in K, the dedekind zeta function K (s) converges absolutely for Re(s)> 1 and extends to a meromorphic function dened on all ofC with a simple pole at s = 1, with residue given by: lim s!1 (s 1) K (s) = 2 r 1 (2) r 2 h K R K w K p jD K j Proof of the Class Number Formula We start by looking at the individual classes of ideals within the ideal class group. Choose any such class and call it C. We shall then dene a related zeta function, K;C (s), to be the analytic continuation of P I2C 1 (N(I)) s , where the sum runs only over ideals in C, rather than over all the ideals ofO K . From this we see that if lim s!1 (s 1) K;C (s) = 2 r 1 (2) r 2 R K w K p jD K j then the theorem is proved, as summing the zeta functions for each class gives us the original zeta function and the formula holds. 12 In this context, let us choose an ideal J = 2 C such that any fractional ideal I is in C if and only if JI = (j) for some j2K with j2J. From this, our denition of K;C becomes: K;C (s) = X I2C 1 (N(I)) s = X (j);j2J N(J) s N((j)) s =N(J) s X (j);j2J 1 N((j)) s Now we want to translate this sum from an algebraic one to a geometric one, in order to be able to calculate it. Using the fact that the norm in K is preserved under ', we get that: K;C (s) =N(J) s X x2'(J)\X K 1 N(x) s Dene V J to be the volume of the fundamental parallelepiped of the lattice '(J). Let V Y be the volume of Y K , namely the fundamental domain X K intersected with fx2 R n ;N(x) 1g. We know that for a positive real number p, the volume of the lattice 1 p '(J) will be V J p n , so we can relate V J and V Y by noticing that: V Y = lim p!1 V J p n f 1 p '(J)\Y K g Pulling the constant V J out of the limit and dividing by it yields: V Y V J = lim p!1 f 1 p '(J)\Y K g p n And by noticing that the number of points inf 1 p '(J)\Y K g is equal to the number of points inf'(J)\fx2R n ;N(x)p n gg for a positive realp, and dropping the exponent n because p was chosen arbitrarily, we get: V Y V J = lim p!1 jf'(J)\fx2R n ;N(x)pggj p 13 Now because the number of points of '(J)\X K with bounded norm is nite, the norms of those points form a discrete set, and as such, we can order the points as f'(J)\X K g =fx 1 ;x 2 ;:::g whereN(x i )N(x i+1 ) for alli. This condition will ensure that the sequence ofN(x i ) will be nondecreasing. From this we see that for some > 0: jfx2'(J)\X K ;N(x)N(x k )gj<kjfx2'(J)\X K :N(x)N(x k )gj Dividing through by N(x k ) yields: jfx2'(J)\X K ;N(x)N(x k )gj N(x k ) N(x k ) N(x k ) < k N(x k ) jfx2'(J)\X K ;N(x)N(x k )gj N(x k ) So that by taking limits, noticing that lim k!1 N(x k ) N(x k ) = 1, and substituting in using our earlier equality, we get: V Y V J = lim k!1 k N(x k ) From this limit, we know that for suciently large k, V Y V J k < 1 N(x k ) < V Y V J + k Taking the exponential for s> 1 and summing over suciently large k gives: ( V Y V J ) s 1 X m 1 k s < 1 X m 1 N(x k ) s < ( V Y V J +) s 1 X m 1 k s We know that the sums P 1 m 1 k s converge for s > 1 by the integral test. Since we dened the sequencef 1 N(x k ) g to be nonincreasing (likef 1 k g), the sum P 1 m 1 N(x k ) s also 14 converges by the integral test. Now multiplying by (s 1) and taking limits as s goes to 1 from the positive side, we get: ( V Y V J ) lim s!1 + (s 1) 1 X k=1 1 k s < lim s!1 + (s 1) 1 X k=1 1 N(x k ) s < ( V Y V J +) lim s!1 + (s 1) 1 X k=1 1 k s Now we know that lim s!1 + (s 1) P m k=0 1 k s and lim s!1 + (s 1) P m k=0 1 N(x k ) s are both equal to 0 wherem is the positive integer below which thek aren't suciently large for our initial inequality. As such, we take the well-known fact that lim s!1 + (s 1) P 1 k=1 1 k s = 1 as a residue of the Riemann Zeta Function, and get: V Y V J < lim s!1 + (s 1) 1 X k=0 1 N(x k ) s < V Y V J + Namely: V Y V J = lim s!1 + (s 1) 1 X k=0 1 N(x k ) s = lim s!1 + (s 1) K;C (s) But we already calculated V Y and V J , which gives: lim s!1 + (s 1) K;C (s) = N(I)( 2 r 1 r 2 R K w K ) 2 r 2 N(I) p jD K j = 2 r 1 (2) r 2 R K w K p jD K j Summing over all classes C proves the formula. 15 Bibliography [Helenius, 2006] Helenius, F. (2006). Finiteness of the class group a la Dedekind. Available at http://userwww.service.emory.edu/cmagnan/ACEStalks/dedekind.pdf. [Lang, 1994] Lang, S. (1994). Algebraic Number Theory. Springer, second edition. [Milne, 2009] Milne, J. S. (2009). Algebraic number theory (v3.02). Available at http://www.jmilne.org/math/. [Osserman, 2005] Osserman, B. (2005). Lecture notes. Available at http://www.math.ucdavis.edu/osserman/classes/254A/. [Sivek, 2005] Sivek, G. (2005). The analytic class number formula. Available at http://modular.math.washington.edu/129-05/nal papers/. 16

Abstract (if available)

Abstract The analytic class number formula equates the residue of a dedekind zeta function at 1 with various properties of a related number field. These properties are defined, their correlation explained, and then they are related to the formula. The class number, a measure of a number field's failure to be a principal ideal domain, is defined and analyzed. Finally, the formula is explicitly stated and proven.

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Asset Metadata

Creator Berry, Christopher D. (author)

Core Title The class number formula

School College of Letters, Arts and Sciences

Degree Master of Arts

Degree Program Mathematics

Publication Date 04/05/2012

Defense Date 03/21/2012

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag analytic class number formula,class number,class number formula,number field,OAI-PMH Harvest,principal ideal domain,zeta function

Language English

Contributor Electronically uploaded by the author (provenance)

Advisor Kamienny, Sheldon (committee chair), Guralnick, Robert M. (committee member), Lanski, Charles (committee member)

Creator Email berryc@usc.edu,chris.berry@alumni.usc.edu

Permanent Link (DOI) https://doi.org/10.25549/usctheses-c3-1542

Unique identifier UC11287856

Identifier usctheses-c3-1542 (legacy record id)

Legacy Identifier etd-BerryChris-573-0.pdf

Dmrecord 1542

Document Type Thesis

Rights Berry, Christopher D.

Type texts

Source University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection)

Access Conditions The author retains rights to his/her dissertation, thesis or other graduate work according to U.S. copyright law. Electronic access is being provided by the USC Libraries in agreement with the a...

Repository Name University of Southern California Digital Library

Repository Location USC Digital Library, University of Southern California, University Park Campus MC 2810, 3434 South Grand Avenue, 2nd Floor, Los Angeles, California 90089-2810, USA