Split (1)

train · 76.7k rows

Q stringlengths 4 3.96k	A stringlengths 1 3k	Result stringclasses 4 values
Theorem 50. If $ \alpha ,\beta $ are algebraic numbers over $ k $, then the same is true for $ \alpha + \beta ,\alpha - \beta ,{\alpha \beta } $, and, if $ \beta \neq 0 $, for $ \alpha /\beta $ .	If $ {\alpha }_{1},\ldots ,{\alpha }_{n} $ are the conjugates of $ \alpha $ and $ {\beta }_{1},\ldots ,{\beta }_{m} $ are those of $ \beta $ with respect to $ k $, then the elementary symmetric functions of $ \alpha $, as well as those of $ \beta $, are numbers in $ k $ . The product\n\n\[ \nH\left( x\r...	Yes
Theorem 51. If $ \omega $ is a root of a polynomial\n\n\[ \varphi \left( x\right) = {x}^{m} + \alpha {x}^{m - 1} + \beta {x}^{m - 2} + \cdots + \lambda \]\n\nwhose coefficients are algebraic numbers over $ k $, then $ \omega $ is also an algebraic number over $ k $ .	Suppose $ {\alpha }_{i} $ runs through the conjugates of $ \alpha ,{\beta }_{k} $ runs through the conjugates of $ \beta $ etc. Then by the theorem on symmetric functions, the polynomial\n\n\[ F\left( x\right) = \mathop{\prod }\limits_{{i, k,\ldots, s}}\left( {{x}^{m} + {\alpha }_{i}{x}^{m - 1} + {\beta }_{k}{x}^...	Yes
Theorem 53. Every number in $ K\left( \theta \right) $ is obtained exactly once in the form\n\n\[ \alpha = {c}_{0} + {c}_{1}\theta + {c}_{2}{\theta }^{2} + \cdots + {c}_{n - 1}{\theta }^{n - 1} \]\n\n(34)\n\nwhere the $ {c}_{0},\ldots ,{c}_{n - 1} $ run through all numbers of the ground field $ k $ .	To prove this suppose $ \alpha = P\left( \theta \right) /Q\left( \theta \right), Q\left( \theta \right) \neq 0 $ . Then $ Q\left( x\right) $ does not have the root $ \theta $ in common with the function $ f\left( x\right) $ belonging to $ \theta $ which is irreducible over $ k $ ; hence by Theorem 49, \( Q\...	Yes
Theorem 54. Every number $ g\left( \theta \right) $ of the field $ K\left( \theta \right) $ is likewise an algebraic number over $ k $ of degree at most $ n $ . The relative conjugates of a number $ \alpha = g\left( \theta \right) $ are the distinct numbers among the numbers \( g\left( {\theta }_{i}\right) \l...	For if $ {\theta }_{1},\ldots ,{\theta }_{n} $ are the conjugates of $ \theta $ with respect to $ k $, we form the product\n\n\[ F\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {x - g\left( {\theta }_{i}\right) }\right) \]\n\nThe coefficients of this polynomial are integral rational combinations of ...	Yes
Theorem 55. Each rational equation $ R\left( {\alpha ,\beta ,\gamma ,\ldots }\right) = 0 $ between numbers $ \alpha ,\beta ,\gamma ,\ldots $ in $ K\left( \theta \right) $ with coefficients in $ k $ remains true if $ \alpha ,\beta ,\gamma ,\ldots $ are replaced by the conjugates with the same index.	As a rational function of $ \alpha ,\beta ,\gamma ,\ldots, R $ is identical to the quotient of two integeral rational expressions $ P $ and $ Q $\n\n\[ R\left( {\alpha ,\beta ,\gamma ,\ldots }\right) = \frac{P\left( {\alpha ,\beta ,\gamma ,\ldots }\right) }{Q\left( {\alpha ,\beta ,\gamma ,\ldots }\right) } \]\n\n...	Yes
Theorem 56. A number $ \alpha $ in $ K\left( \theta \right) $ belongs to the ground field $ k $ if and only if it is equal to its $ n $ conjugates. A number $ \alpha $ in $ K\left( \theta \right) $ has degree $ n $ with respect to $ k $ if and only if it is distinct from all its conjugates. The latter c...	The first two statements follow immediately from Theorem 54 and the definition which follows it. Moreover, if $ \alpha $ in $ K\left( \theta \right) $ is to generate the field $ K\left( \theta \right) $ , thus if $ K\left( \theta \right) = K\left( \alpha \right) $ is to hold, then the degree of $ \alpha $ mus...	Yes
Theorem 57. In order that the numbers\n\n\[ \n{\omega }^{\left( i\right) } = \mathop{\sum }\limits_{{k = 1}}^{n}{c}_{ik}{\theta }^{k - 1}\;\left( {{c}_{ik}\text{ numbers in }k}\right) \n\]\n\n(35)\n\nform a fundamental system of $ K\left( \theta \right) $, it is necessary and sufficient that the determinant \( \begin...	Obviously we need only investigate when the numbers $ 1,\theta ,\ldots ,{\theta }^{n - 1} $ can be represented in terms of the $ {\omega }^{\left( i\right) } $ as\n\n\[ \n{\theta }^{p - 1} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{pi}{\omega }^{\left( i\right) }\;\left( {p = 1,\ldots, n}\right) \left( {a}_{pi}\righ...	Yes
Theorem 58. The n numbers $ {\omega }^{\left( 1\right) },\ldots ,{\omega }^{\left( n\right) } $ in $ K\left( \theta \right) $ form a fundamental system if and only if there is no linear relation\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i}{\omega }^{\left( i\right) } = 0 \]\n\n(37)\n\nwith coefficients in \( k ...	The $ n $ numbers $ {\omega }^{\left( i\right) } $ of this type are said to be linearly independent. For, in the notation used above it would follow from (37) that\n\n\[ 0 = \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i}\mathop{\sum }\limits_{{k = 1}}^{n}{c}_{ik}{\theta }^{k - 1} \]\n\nand as before, if the \( {u}_{i} ...	Yes
Theorem 60. If $ \alpha $ satisfies any equation at all with integral coefficients whose leading coefficient is equal to 1, then $ \alpha $ is an integer.	Let $ \varphi \left( x\right) = {x}^{N} + {a}_{1}{x}^{N - 1} + \cdots + {a}_{N} $ with rational integral $ a $’s and $ \varphi \left( \alpha \right) = 0 $. Moreover let\n\n\[ f\left( x\right) = {c}_{0}{x}^{n} + {c}_{1}{x}^{n - 1} + \cdots + {c}_{n} \]\n\nbe the irreducible polynomial in $ k\left( 1\right) $ whi...	Yes
Theorem 61. The sum, difference, and product of two integers is again an integer. Hence every rational integral function (polynomial) of integers with rational integral coefficients is again an integer.	For if $ {\alpha }_{1},\ldots ,{\alpha }_{n} $ are the conjugates of a number $ \alpha $ and if $ {\beta }_{1},\ldots ,{\beta }_{m} $ are the conjugates of a number $ \beta $, then\n\n\[ F\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\mathop{\prod }\limits_{{k = 1}}^{m}\left( {x - \left( {{\alpha }_{i} ...	No
Theorem 63. Every algebraic number $ \alpha $ can be transformed into an integer by multiplication by a suitable nonzero rational number.	To prove this assume that\n\n\[ \n{c}_{0}{x}^{n} + {c}_{1}{x}^{n - 1} + \cdots + {c}_{n - 1}x + {c}_{n} = 0 \n\]\n\nis an equation for $ \alpha $ with rational integral coefficients and $ {c}_{0} \neq 0 $ . Then by multiplication by $ {c}_{0}^{n - 1} $ we obtain an integer equation for $ y = {c}_{0}x $ with lea...	Yes
Theorem 65. Every ideal a can be written in the form $ \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) $ with the $ \alpha $ suitably chosen integers in $ K $ . Moreover, we may even take $ r \leq n $ .	The numbers of an ideal $ \mathfrak{a} $ which is not $ \left( 0\right) $ (the case $ \mathfrak{a} = \left( 0\right) $ is trivial) obviously form an infinite Abelian group, under composition by addition, which is a subgroup of the group of all integers in $ K $ . Consequently by Theorem 34 the ideal \( \mathfra...	Yes
Theorem 68. If $ \mathfrak{a}\mathfrak{b} = \mathfrak{a}\mathfrak{c} $, then if $ \mathfrak{a} \neq 0,\mathfrak{b} = \mathfrak{c} $ .	To see this we determine an ideal $ m $ such that $ {am} = \left( \delta \right) $ is a principal ideal. Then\n\n\[ \n{amb} = {amc},\;\left( \alpha \right) b = \left( \alpha \right) c.\n\]\n\nThe latter equation asserts that $ \alpha $ times every number from $ \mathfrak{b} $ is of the form $ \alpha $ times a...	No
Theorem 69. An ideal $ \mathfrak{c} = \left( {{\gamma }_{1},\ldots ,{\gamma }_{r}}\right) $ is a divisor of $ \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right) $ if and only if every number of $ \mathfrak{a} $ belongs to $ \mathfrak{c} $ .	If $ \mathfrak{c} \mid \mathfrak{a} $, then there is $ \mathrm{a}\mathfrak{b} = \left( {{\beta }_{1},\ldots ,{\beta }_{p}}\right) $ for which $ \mathfrak{b} \neq \left( 0\right) $ and\n\n\[ \left( {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right) = \left( {\beta ,\ldots ,{\beta }_{p}}\right) \cdot \left( {{\gamma }_{1...	Yes
Theorem 70. For every two ideals $ \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) ,\mathfrak{b} = \left( {{\beta }_{1},\ldots ,{\beta }_{s}}\right) $ which are not both $ = \left( 0\right) $, there is a uniquely determined greatest common divisor \( \mathfrak{d} = \left( {\mathfrak{a},\mathfrak{...	We show that $ \mathfrak{d} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r},{\beta }_{1},\ldots ,{\beta }_{s}}\right) $ has the stated properties of divisibility. Since every sum \	No
Theorem 71. If $ \mathfrak{p} $ is a prime ideal and $ \mathfrak{p} \mid \mathfrak{a}\mathfrak{b} $, then $ \mathfrak{p} $ divides either $ \mathfrak{a} $ or $ \mathfrak{b} $ or both.	For if $ \mathfrak{p} $ does not divide the factor $ \mathfrak{b} $, then\n\n\[ \left( {p, b}\right) = \left( 1\right) \]\n\nsince, as a prime ideal, $ \mathfrak{p} $ has no factors except (1) and $ \mathfrak{p} $ . It follows from (41) that\n\n\[ \mathfrak{a} = \mathfrak{a}\left( 1\right) = \mathfrak{a}\left( ...	Yes
Theorem 73. There are infinitely many prime ideals in each field.	Each rational prime $ p $ defines an ideal $ \left( p\right) $, and moreover if $ p $ and $ q $ are distinct positive primes, then $ \left( {p, q}\right) = 1 $ in the sense of our ideal theory, since the number 1 occurs in the form $ {px} + {qy} $ in $ \left( {p, q}\right) $ . Consequently, the same prime...	Yes
Theorem 74. If $ \mathfrak{a} $ and $ \mathfrak{b} $ are ideals distinct from (0), then there is always a number $ \omega $ for which\n\n\[ \left( {\omega ,\mathfrak{a}\mathfrak{b}}\right) = \mathfrak{a}. \]\n\nThis $ \omega $ then obviously has a decomposition $ \omega = \mathfrak{{ac}} $ where \( \left( {\m...	For a proof, let $ {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{r} $ be all the distinct prime ideals which divide $ \mathfrak{a}\mathfrak{b} $, and let $ \mathfrak{a} = {\mathfrak{p}}_{1}^{{a}_{1}}\cdots {\mathfrak{p}}_{r}^{{a}_{r}}\left( {{a}_{i} \geq 0}\right) $ . We define the $ r $ ideals \( {\mathfrak{d}}_{...	Yes
Theorem 76. The number of residue classes $ {\;\operatorname{mod}\;\mathfrak{a}} $ is finite. If the number of residue classes is denoted by $ N\left( \mathfrak{a}\right) $ and if $ {\alpha }_{1},\ldots ,{\alpha }_{n} $ is a basis for $ \mathfrak{a} $, then \( N\left( \mathfrak{a}\right) = \left\| {\Delta \left(...	The numbers of a form a subgroup of the group $ \mathfrak{G} $ of all integers of the field. The different cosets in $ \mathfrak{G} $ determined by a obviously form the different residue classes mod a. Hence the number of distinct residue classes mod a is the index of $ \mathfrak{a} $ in $ \mathfrak{G} $ . This...	Yes
Theorem 78. For given $ \alpha $ and $ \beta $ the congruence\n\n\[ \n{\alpha \xi } \equiv \beta \left( {\;\operatorname{mod}\;a}\right)\n\]\n\ncan be solved by an integer $ \xi $ in $ K $ if and only if $ \left( {\alpha ,\mathfrak{a}}\right) \mid \beta $ . If $ \left( {\alpha ,\mathfrak{a}}\right) = 1 $, t...	If we assume $ \left( {\alpha ,\mathfrak{a}}\right) = 1 $ to begin with, and let $ \xi $ run through a system of $ N\left( \mathfrak{a}\right) $ numbers which are incongruent $ {\;\operatorname{mod}\;\mathfrak{a}} $, then $ {\alpha \xi } $ runs through all the residue classes \( {\;\operatorname{mod}\;\mathfr...	Yes
Theorem 79. For two ideals $ \mathfrak{a} $ and $ \mathfrak{b} $, we always have\n\n\[ N\left( \mathfrak{{ab}}\right) = N\left( \mathfrak{a}\right) \cdot N\left( \mathfrak{b}\right) \]	Let $ \alpha $ be a number divisible by $ \mathfrak{a} $ such that $ \left( {\alpha ,\mathfrak{a}\mathfrak{b}}\right) = \mathfrak{a} $ . If we let $ {\xi }_{i} $ $ \left( {i = 1,2,\ldots, N\left( \mathrm{\;b}\right) }\right) $ run through a complete system of residues mod $ \mathrm{b} $ and let \( {\eta }_{...	Yes
Theorem 80. If $ \left( {\mathfrak{a},\mathfrak{b}}\right) = 1 $, then $ \varphi \left( {\mathfrak{a}\mathfrak{b}}\right) = \varphi \left( \mathfrak{a}\right) \cdot \varphi \left( \mathfrak{b}\right) $ and in general\n\n\[ \varphi \left( \mathfrak{a}\right) = N\left( \mathfrak{a}\right) \mathop{\prod }\limits_{{\ma...	To see this let $ \alpha $ be chosen so that $ \left( {\alpha ,{ab}}\right) = \alpha $ and $ \beta $ so that $ \left( {\beta ,{ab}}\right) = b $ . Then if $ \xi $ runs through a complete system of residues $ {\;\operatorname{mod}\;b} $ and $ \eta $ runs through such a system mod a in \( {\alpha \xi } + {\...	Yes
Theorem 81. The norm of a prime ideal $ \mathfrak{p} $ is a power of a certain rational prime $ p, N\left( \mathfrak{p}\right) = {p}^{f}.f $ is called the degree of $ \mathfrak{p} $ . Every ideal $ \left( p\right) $, where $ p $ is a rational prime, can be decomposed into at most $ n $ factors.	For each prime ideal $ \mathfrak{p} $ divides certain rational numbers and consequently also certain rational primes $ p $ . Suppose that $ \mathfrak{p} \mid p, p = \mathfrak{{pa}} $ . Then $ N\left( p\right) = $ $ N\left( \mathfrak{p}\right) \cdot N\left( \mathfrak{a}\right) $ and consequently the rational i...	Yes
Theorem 83. The group of residue classes $ {\;\operatorname{mod}\;\mathfrak{p}} $ is an Abelian group $ \mathfrak{G}\left( \mathfrak{p}\right) $ of order $ N\left( \mathfrak{p}\right) = {p}^{f} $ under composition by addition and the number of basis elements is equal to the degree $ f $ of the prime ideal \( \m...	For since $ \mathfrak{p} \mid p $, the number of residue classes whose elements $ \alpha $ satisfy the congruence\n\n\[ \n{p\alpha } \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\right)\n\]\n\nis equal to the number of all residue classes, thus $ {p}^{f} $. Consequently, by Theorem $ {27}, f $ is equal to...	Yes
Theorem 86. If $ \mathfrak{p} $ is a prime ideal, and if for two polynomials $ P $ and $ Q $ the product\n\n\[ P\left( {{x}_{1},\ldots ,{x}_{m}}\right) \cdot Q\left( {{x}_{1},\ldots ,{x}_{m}}\right) \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) ,\]\n\nthen at least one of the polynomials is \( \equiv...	The theorem is true for polynomials of 0 variables, that is, for constants. We show that it is correct in general by passing from $ m $ to $ m + 1 $ . Assume it is already proven for all polynomials with $ m $ or fewer variables. Each polynomial of $ m + 1 $ variables can be put into the form\n\n\[ P\left( {{x}...	Yes
Theorem 88. For each ideal $ \mathfrak{a} $ of a Galois field the principal ideal $ \left( {N\left( \mathfrak{a}\right) }\right) = $ $ {\mathfrak{a}}^{\left( 1\right) }{\mathfrak{a}}^{\left( 2\right) }\cdots {\mathfrak{a}}^{\left( n\right) } $ (compare with Theorem 107).	For the proof, we form the polynomial $ P\left( x\right) = {\alpha }_{1}x + {\alpha }_{2}{x}^{2} + \cdots + {\alpha }_{r}{x}^{r} $ from a new variable $ x $ and $ \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) $ where the GCD of the coefficients $ = \mathfrak{a} $ . The product of the conjug...	Yes
Theorem 89. Let $ p $ be a rational prime which does not divide $ d $ . Then $ p $ splits in the field $ K\left( \sqrt{d}\right) $ into two distinct prime ideals $ \mathfrak{p},{\mathfrak{p}}^{\prime } $ provided the congruence\n\n\[ \n{x}^{2} \equiv d\left( {\;\operatorname{mod}\;{4p}}\right) \n\]\n\ncan be ...	If the prime $ p $ which does not divide $ d $ splits in $ K\left( \sqrt{d}\right) $, then $ p $ can only split into prime factors $ \mathfrak{p},{\mathfrak{p}}^{\prime } $ which are of degree 1 . By Theorem 85, each integer in $ K $ is congruent $ {\;\operatorname{mod}\;\mathfrak{p}} $ to a rational numb...	Yes
Theorem 91. If the prime $ p $ does not divide $ D \cdot m $, then $ p $ is not divisible by the square of a prime ideal in $ K\left( \zeta \right) $ .	For if $ {\mathfrak{p}}^{2} \mid p $, then let us choose a number $ \omega $ which is divisible by $ \mathfrak{p} $ but not by $ {\mathfrak{p}}^{2} $ . It follows from the lemma that\n\n\[ \n{\omega }^{pf} \equiv \omega \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{2}}\right)\n\]\n\nSince $ {p}^{f} \geq 2 $, ...	Yes
Theorem 92. If the prime $ p $ does not divide $ D \cdot m $, and if $ f $ is the smallest positive exponent such that $ {p}^{f} \equiv 1\\left( {\\operatorname{mod}m}\\right) $, then $ p $ splits into exactly $ e = h/f $ distinct prime factors in $ K\\left( \\zeta \\right) $. Each factor has degree \( f ...	Let $ \\mathfrak{p} $ be a prime factor of $ p $ of degree $ {f}_{1} $. Then, by (43), for each integer $ \\omega $ in $ K\\left( \\zeta \\right) $, \[ {\\omega }^{{p}^{{f}_{1}}} \equiv \\omega \\left( {\\operatorname{mod}\\mathfrak{p}}\\right) \] (46) and this congruence holds for each integer $ \\omega $ ...	Yes
Theorem 93. Each ideal $ \mathfrak{g} $ is the range of values of a linear form\n\n\[{\xi }_{1}{\rho }_{1} + \cdots + {\xi }_{r}{\rho }_{r}\]\n\nwhere $ {\rho }_{1},\ldots ,{\rho }_{r} $ are certain integers or fractions in $ \mathfrak{g} $, while the $ {\xi }_{i} $ run through all integers in $ K $ . We writ...	Let $ v $ be chosen for $ g $ according to (2). Then all products of $ v $ with the numbers in $ \mathfrak{g} $ obviously form an integral ideal $ \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) $ and then $ \mathrm{g} = \left( {{\alpha }_{1}/v,\ldots ,{\alpha }_{r}/v}\right) $.\n\nIf \( ...	No
Theorem 94. Assume we are given $ n $ linear homogeneous expressions\n\n\[ \n{L}_{p}\left( x\right) = \mathop{\sum }\limits_{{q = 1}}^{n}{a}_{pq}{x}_{q}\;\left( {p = 1,2,\ldots, n}\right) ,\n\]\n\nwith real coefficients $ {a}_{pq} $, whose determinant $ D = \left\| {a}_{pq}\right\| $ is different from zero, as well...	The proof is along the lines of Minkowski's contribution to the geometry of numbers. To begin with we ask: \	No
Theorem 95. Let $ n $ linear forms $ {L}_{p}\left( x\right) = \mathop{\sum }\limits_{{q = 1}}^{n}{a}_{pq}{x}_{q}\left( {p = 1,\ldots, n}\right) $ be given with real or complex coefficients whose determinant $ D \neq 0 $ . Moreover if one of the forms is not real, we assume the complex conjugate of a form also occ...	To prove this we replace the system $ {L}_{p}\left( x\right) $ by that system of real forms $ {L}^{\prime }\left( x\right) $ which arises if the real and imaginary components of the $ {L}_{p}\left( x\right) $ are considered by themselves. We take $ {L}_{p}^{\prime }\left( x\right) = {L}_{p}\left( x\right) $ if ...	Yes
Theorem 96. In each ideal class of $ K $ there is an integral ideal whose norm is $ \leq \left\| \sqrt{d}\right\| $ . Thus the number of ideal classes in $ K $ is finite.	To prove this let $ \mathfrak{a} $ be an integral ideal in the class $ {B}^{-1} $, where $ B $ is an arbitrarily given class. If $ {\alpha }_{1},\ldots ,{\alpha }_{n} $ denotes a basis of $ a $, then, by Theorem 95, there are rational integers $ {x}_{1},\ldots ,{x}_{n} $, not all vanishing, such that\n\n\[ ...	Yes
Theorem 98. For each ideal $ \mathfrak{a} $ in $ K $ there is a number $ A $ which generally does not belong to the field $ K $, such that the numbers of $ \mathfrak{a} $ are identical with those numbers of the field $ K $ which are divisible by $ A $ .	By Theorem $ {97}{\mathfrak{a}}^{h} $ is equal to a principal ideal $ \left( \omega \right) $ . The number $ A = \sqrt[h]{\omega } $ has the asserted property for if $ \alpha $ is a number in $ \mathfrak{a} $, then $ {\alpha }^{h} $ belongs to $ {\mathfrak{a}}^{h} $ and therefore \( {\alpha }^{h}/\omega \...	Yes
Theorem 99. The group $ \mathfrak{W} $ of all roots of unity in $ K $ is finite, and indeed it is a cyclic group of order $ w \geq 2 $ .	Since all roots of unity, including all conjugates, have absolute value 1, the first assertion follows from the lemma. Moreover if $ p $ is a prime dividing the order of $ \mathfrak{W} $, then the number of solutions of $ {x}^{p} = 1 $ is equal to $ {p}^{1} $, and thus, by Theorem 28, the basis number of the gr...	No
Theorem 100. The group $ \mathfrak{E} $ of all units in $ K $ has a finite basis. Furthermore this basis consists of precisely $ r = {r}_{1} + {r}_{2} - 1 $ elements of infinite order, while the remaining basis elements are roots of unity.	Thus this means:\n\nThere are $ r + 1 $ units $ \zeta ,{\eta }_{1},{\eta }_{2},\ldots ,{\eta }_{r} $, where $ \zeta $ is a wth root of unity, such that each unit of the field is obtained exactly once in the form\n\n\[ \varepsilon = {\zeta }^{a}{\eta }_{1}^{{a}_{1}}\cdots {\eta }_{r}^{{a}_{r}} \]\n\nwhere \( {a}_{...	Yes
Theorem 102. If $ {\alpha }_{1},\ldots ,{\alpha }_{n} $ are basis elements of the ideal $ \mathfrak{a} $, then the $ n $ sequences of numbers $ {\beta }_{1}^{\left( p\right) },\ldots ,{\beta }_{n}^{\left( p\right) }\left( {p = 1,\ldots, n}\right) $, which are defined by (65), are conjugate sequences of numbers ...	Since, moreover,\n\n\[ \n{\Delta }^{2}\left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) = \frac{1}{{\Delta }^{2}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) } = \frac{1}{d{N}^{2}\left( \mathfrak{a}\right) }\n\]\n\nand by (47)\n\n\[ \n{\Delta }^{2}\left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) = {N}^{2}\left( ...	No
Theorem 104. All numbers of the ideal $ \mathfrak{f} = {F}^{\prime }\left( \theta \right) /\mathfrak{d} $ belong to the ring $ R\left( \theta \right) $, and if all numbers of an ideal a belong to the ring $ R\left( \theta \right) $, then $ \mathfrak{a} $ is divisible by $ \mathfrak{f} $ .	If $ \omega \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{f}}\right) $, then $ \alpha = \omega /{F}^{\prime }\left( \theta \right) $ is a number with denominator $ \mathfrak{d} $, and by Lemma (a), $ \alpha {F}^{\prime }\left( \theta \right) $ must be a number of this ring. Hence the first part of our theorem...	Yes
Theorem 107. The ideal $ {N}_{k}\left( \mathfrak{A}\right) $ is an ideal in $ k $ . If $ k $ is the field of rational numbers, then $ {N}_{k}\left( \mathfrak{A}\right) = \left( {N\left( \mathfrak{A}\right) }\right) $ .	To begin let $ \mathfrak{A} = \left( {{A}_{1},\ldots ,{A}_{s}}\right) $ be an integral ideal, where the $ {A}_{i} $ are numbers in $ K $ . Then, by $ \$ {28} $, for any variables $ {u}_{1},\ldots ,{u}_{s} $, the content of the conjugate polynomials\n\n\[ \n{F}^{\left( i\right) }\left( u\right) = {A}_{1}^{\lef...	Yes
Theorem 108. For each prime ideal $ \mathfrak{P} $ of $ K $ there is exactly one prime ideal $ \mathfrak{p} $ in $ k $ which is divisible by $ \mathfrak{P} $ . Then\n\n\[ \n{N}_{k}\left( \mathfrak{P}\right) = {\mathfrak{p}}^{{f}_{1}} \n\]\n\nwhere $ {f}_{1} $ is a natural number $ \leq m.{f}_{1} $ is call...	By Theorem $ {107}{N}_{k}\left( \mathfrak{P}\right) $ is an ideal in $ k $ which, by definition, is divisible by $ \mathfrak{P} $ . If $ {N}_{k}\left( \mathfrak{P}\right) $ is decomposed into its prime factors, then by the fundamental theorem $ \mathfrak{P} $ must divide at least one of these prime ideals in ...	Yes
Theorem 109. If $ N $ denotes the norm in $ K $ and $ n $ denotes the norm in $ k $, then for each ideal $ \mathfrak{A} $ in $ K $\n\n\[ N\left( \mathfrak{A}\right) = n\left( {{N}_{k}\left( \mathfrak{A}\right) }\right) \]	To begin with, this assertion follows immediately for each number $ A $ in $ K $, by (74). By Theorem 107, or also by consideration of the principal ideal $ {\mathfrak{A}}^{h} $, the result is also obtained for each ideal in $ K $ .	Yes
Theorem 110. If the relative degree of the prime ideal $ \mathfrak{P} $ is equal to 1, then each number in $ K $ is congruent modulo $ \mathfrak{P} $ to a number in $ k $ .	By Theorem 108, $ N\left( \mathfrak{P}\right) = n{\left( \mathfrak{p}\right) }^{{f}_{1}} $ ; hence for $ {f}_{1} = 1 $ the number of residue classes $ {\;\operatorname{mod}\;\mathfrak{P}} $ in $ K $ is equal to the number of residue classes $ {\;\operatorname{mod}\;\mathfrak{p}} $ in $ k $ . However if a nu...	Yes
Theorem 111. If $ \mathfrak{D} $ and $ \mathfrak{d} $ are the differents of $ K $ and $ k $ respectively, then for the relative differents $ {\mathfrak{D}}_{k} $ the relation\n\n\[ \mathfrak{D} = {\mathfrak{D}}_{k}\mathfrak{d} \]\n\nholds.	Proof. If $ \Delta $ is a number in $ K $ such that $ \Delta {\mathfrak{D}}_{k}\mathfrak{d} $ is integral, then by the definition of $ {\mathfrak{D}}_{k} $,\n\n\[ \mathfrak{d}{S}_{k}\left( {\Delta A}\right) \text{is integral} \]\n\nfor each integer $ A $, since for each number $ \xi $ in $ k $ which is di...	Yes
Theorem 112. The relative different of $ K $ is the GCD of all relative differents of integers of $ K $ with respect to $ k $ .	For the proof of this theorem we must proceed almost exactly as in the proof of Theorem 105. If $ \theta $ is an integer generating the field $ K $, then the relative ring $ {R}_{k}\left( \theta \right) $ is the set of all numbers \[ {\alpha }_{0} + {\alpha }_{1}\theta + \cdots + {\alpha }_{m - 1}{\theta }^{m - 1...	Yes
Theorem 114. If a prime ideal $ \mathfrak{P} $ in $ K $ divides a prime ideal $ \mathfrak{p} $ in $ k $ to a higher power than the first, then $ \mathfrak{P} $ is a factor of the relative different of $ K $ with respect to $ k $ . Thus there can only exist finitely many prime ideals $ \mathfrak{P} $ of ...	For a proof, let the decomposition of $ \mathfrak{p} $ in $ K $ be\n\n\[ \mathfrak{p} = {\mathfrak{P}}^{e}\mathfrak{A},\;\text{ where }\left( {\mathfrak{A},\mathfrak{P}}\right) = 1, e \geq 2. \]\n\nFor each integer $ A $ in $ K $ we now have, by the often used properties of the binomial coefficients \( \left( \...	Yes
Theorem 115. Suppose that $ K $ is identical with all relative conjugate fields with respect to $ k $ (that is, suppose that $ K $ is a relative Galois field). Then the only prime ideals of $ K $ dividing the relative different $ {\mathfrak{D}}_{k} $ of $ K $ are those which divide a prime ideal of $ k $ ...	For the proof of Theorem 115 it suffices, by Theorem 112, to display a number $ A $ in $ K $ whose different is not divisible by this $ \mathfrak{P} $ . We choose $ A $ to be a primitive root $ {\;\operatorname{mod}\;\mathfrak{P}} $ which is divisible by $ \mathfrak{p}{\mathfrak{P}}^{-1} $ . Now, by the abo...	Yes
Theorem 116. If $ \mu $ is a number in $ k $ which is not the lth power of a number in $ k $, then the field $ K\left( {\sqrt[1]{\mu };k}\right) $ has the relative degree $ l $ with respect to $ k $ . The field $ K\left( {\sqrt[1]{\mu };k}\right) $ is identical with its relative-conjugate fields.	The number $ M = \sqrt[1]{\mu } $ (suppose that the value of the root is somehow fixed) satisfies the equation $ {x}^{i} - \mu = 0 $, whose roots are the $ l $ numbers\n\n\[{\zeta }^{a}M\;\left( {a = 0,1,\ldots, l - 1}\right) .\n\]\n\nIn any case all relative conjugates of $ M $ must occur among them. Let these...	Yes
Theorem 117. Only the following possibilities exist for the behavior of a prime ideal $ \mathfrak{p} $ of $ k $ under passage to $ K $ :\n\n$ \mathfrak{p} $ remains a prime ideal in $ K $ ,\n\n$ \mathfrak{p} $ becomes the lth power of a prime ideal in $ k $ ,\n\n$ \mathfrak{p} $ becomes the product of \...	Let $ \mathfrak{P} $ be a prime ideal in $ K $ which divides $ \mathfrak{p} $ . Then, by Theorem 107, the relative norm of $ \mathfrak{P} $ is\n\n\[ \mathfrak{P} \cdot s\mathfrak{P}\cdots {s}^{l - 1}\mathfrak{P} = {\mathfrak{p}}^{{f}_{1}}, \]\n\nwhere $ {f}_{1} $ is the relative degree of $ \mathfrak{P} $ ;...	Yes
Theorem 119. Let 1 be a prime factor of $ 1 - \zeta $, which divides $ 1 - \zeta $ exactly to the ath power: $ 1 - \zeta = {\mathrm{I}}^{a}{\mathrm{I}}_{1} $. Suppose that I does not divide $ \mu $. Then I splits into $ l $ factors which are distinct from one another in $ K\left( {\sqrt[1]{\mu };k}\right) $...	I: The solvability of (82) is identical with the decomposition of $ 1 $ into distinct factors in $ K $. Namely from $ \mathfrak{l} = \mathfrak{L} \cdot s\mathfrak{L}\cdots {s}^{l - 1}\mathfrak{L} $, where the conjugates are distinct from one another, it follows as in the proof of Theorem 110 that every integer in...	Yes
Theorem 122. Let $ Z\left( t\right) $ denote the number of integral ideals of the field whose norm is $ \leq t $ . Then\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{Z\left( t\right) }{t} = h\varkappa \]\n\nwhere $ h $ is the class number of the field.	If $ F\left( m\right) $ denotes the number of integral ideals of the field whose norm is equal to the positive number $ m $, then obviously\n\n\[ Z\left( t\right) = \mathop{\sum }\limits_{{m = 1}}^{t}F\left( m\right) \]\n\nHere $ \mathop{\sum }\limits_{{m = 1}}^{t} $ means that the summation index $ m $ runs th...	Yes
Theorem 123. $ {\zeta }_{k}\left( s\right) $ is defined by the convergent series (92) for $ s > 1 $ as a continuous function of $ s $ and as $ s $ approaches 1	\[ \mathop{\lim }\limits_{{s \rightarrow 1}}\left( {s - 1}\right) {\zeta }_{k}\left( s\right) = h\varkappa \]	No
Theorem 124. For $ s > 1 $, the equation\n\n\[ \n{\zeta }_{k}\left( s\right) = \mathop{\prod }\limits_{\mathfrak{p}}\frac{1}{1 - \frac{1}{N{\left( \mathfrak{p}\right) }^{s}}}.\n\]\n\nholds where $ \mathfrak{p} $ runs through all distinct prime ideals $ \mathfrak{p} $ of $ k $ .	To begin with, this product converges, since $ \mathop{\sum }\limits_{\mathfrak{p}}1/N{\left( \mathfrak{p}\right) }^{s} $ converges as the constituent of the series for $ {\zeta }_{k}\left( s\right) $ . For a single factor we obtain a convergent series of positive terms\n\n\[ \n\frac{1}{1 - N{\left( \mathfrak{p}\ri...	Yes
Theorem 125. Following Dedekind the determination of the class number $ h $ is reduced to the determination of the prime ideals of the field by the equation\n\n\[ h \cdot \varkappa = \mathop{\lim }\limits_{{s \rightarrow 1}}\left( {s - 1}\right) \mathop{\prod }\limits_{\mathfrak{p}}\frac{1}{1 - \frac{1}{N{\left( \mat...	This fundamental fact is only another way of writing (88), as has been mentioned already; however, it is more convenient as a starting point for the further calculation than the former equation.	No
Theorem 126. If $ {\mathfrak{p}}_{1} $ runs through the distinct prime ideals of degree one in $ k $ , then, for $ s > 1 $ ,\n\n\[ \mathop{\sum }\limits_{{\mathfrak{p}}_{1}}\frac{1}{N{\left( {\mathfrak{p}}_{1}\right) }^{s}} = \log \frac{1}{s - 1} + {g}_{1}\left( s\right) \]\n\nwhere $ {g}_{1}\left( s\right) $ a...	Proof: Let $ {\mathfrak{p}}_{f} $ run through the distinct prime ideals of degree $ f $ for $ f = $ $ 1,2,\ldots, n $ . (Of course $ {\mathfrak{p}}_{f} $ need not exist for each $ f $ .) Since at most $ n $ distinct prime ideals of $ k $ divide a given rational prime $ p $ then, in any case, for \( s ...	Yes
Theorem 127. There are infinitely many positive rational primes with the property $ p \equiv 1\\left( {\\;\\operatorname{mod}\\;m}\\right) $ .	If $ {n}_{0} $ is the degree of the field of $ m $ th roots of unity (which by $ \\$ {30} $ is no larger than $ \\varphi \\left( m\\right) \$ ), then exactly $ {n}_{0} $ distinct prime ideals of $ k $ divide such a number $ p $, and Equation (97) thus reads\n\n\\[ \n{n}_{0}\\mathop{\\sum }\\limits_{{p \\e...	No
Theorem 128. If $ \chi \left( n\right) $ denotes a residue character of $ n{\;\operatorname{mod}\;m} $, then the Dirichlet series\n\n\[ L\left( {s,\chi }\right) = \mathop{\sum }\limits_{{n = 1}}^{\pi }\frac{\chi \left( n\right) }{{n}^{s}} \]\n\nis absolutely convergent for $ s > 1 $ and as long as $ s > 1 $ we ...	First of all the absolute convergence of the series and the product representation for $ s > 1 $ are obtained immediately from the fact that the coefficients $ \chi \left( n\right) $ are not larger than 1 in absolute value, as $ \chi \left( n\right) $ is either a root of unity, or, in case \( \left( {n, m}\right)...	Yes
Theorem 129. For each character $ \chi {\;\operatorname{mod}\;m} $, if $ s > 1 $ , \[ \log L\left( {s,\chi }\right) = \mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}} + g\left( {s,\chi }\right) \] where $ g\left( {s,\chi }\right) $ remains bounded as $ s $ approaches 1 .	If we define the log function, for $ s > 0 $, by the convergent series \[ \log \frac{1}{1 - \frac{\chi \left( p\right) }{{p}^{s}}} = \frac{\chi \left( p\right) }{{p}^{s}} + \frac{1}{2}\frac{\chi \left( {p}^{2}\right) }{{p}^{2s}} + \frac{1}{3}\frac{\chi \left( {p}^{3}\right) }{{p}^{3s}} + \cdots = \frac{\chi \left( p\...	Yes
Theorem 130. If $ \chi $ is not the principal character, then\n\n\[ L\left( {1,\chi }\right) = \mathop{\lim }\limits_{{s \rightarrow 1}}L\left( {s,\chi }\right) \neq 0. \]	The nonvanishing of the $ L $ -series is now an immediate consequence of the fact that $ {\zeta }_{k}\left( s\right) $ becomes infinite to the first order at $ s = 1 $ . For by (102) it follows for $ a = b = 1 $ that\n\n\[ \mathop{\sum }\limits_{\chi }\log L\left( {s,\chi }\right) = \varphi \left( m\right) \mat...	Yes
Theorem 132. The basis number of the strict class group belonging to 2 is $ {e}_{0}\left( 2\right) = t - 1 $, where $ t $ denotes the number of distinct primes which divide the discriminant $ d $ of $ k $ .	By Theorem 28, we must show that there exist exactly $ {2}^{t - 1} $ classes in $ k $ whose square is the strict principal class. For this purpose we keep in mind that the $ t $ distinct prime ideals $ {\mathfrak{q}}_{1},\ldots ,{\mathfrak{q}}_{t} $, which divide $ d $, have the property that their square is ...	Yes
Theorem 134. If $ d $ is the product of two positive primes $ {q}_{1},{q}_{2} $, which are $ \equiv 3\left( {\;\operatorname{mod}\;4}\right) $, then either $ {q}_{1} $ or $ {q}_{2} $ is the norm of a principal ideal in the strict sense in $ k\left( \sqrt{{q}_{1}{q}_{2}}\right) $ .	To begin with the norm of each unit $ = + 1 $ in such a field. For from $ N\left( \alpha \right) = - 1 $ for $ \alpha = \left( {x + y\sqrt{{q}_{1}{q}_{2}}}\right) /2 $ it would follow that\n\n\[ - 4 \equiv {x}^{2}\left( {{\;\operatorname{mod}\;{q}_{1}}{q}_{2}}\right) \]\n\nthus -1 would be a quadratic residue mod...	Yes
Theorem 135. If $ p $ and $ q $ are odd positive primes, then we have the relations\n\n(I) $ \left( \frac{-1}{p}\right) = {\left( -1\right) }^{\left( {p - 1}\right) /2} $,\n\n(II) \( \left( \frac{p}{q}\right) = \left( \frac{q}{p}\right) {\left( -1\right) }^{\left( {\left( {p - 1}\right) /2}\right) \left( {\left( ...	We obtain the first formula directly from the definition of the residue symbol, Equation (31) in §16. We can also deduce it from field theory, in a somewhat more involved way, but analogous to the subsequent proof of (II) and (III) as follows: If $ \left( \frac{-1}{p}\right) = + 1 $, then $ p $ splits in \( k\left(...	Yes
Theorem 137. If $ d $ is the discriminant of a quadratic field and $ n, m $ are positive integers, then\n\n\[ \left( \frac{d}{n}\right) = \left( \frac{d}{m}\right) \;\text{ if }n \equiv m\left( {\;\operatorname{mod}\;d}\right) \]\n\n(124)\n\n\[ \left( \frac{d}{n}\right) = \left( \frac{d}{m}\right) \cdot \operatorna...	For the proof we must split off the highest power of 2 dividing $ d, n, m $ . Let\n\n\[ d = {2}^{a}{d}^{\prime },\;n = {2}^{b}{n}^{\prime },\;m = {2}^{c}{m}^{\prime } \]\n\nwith odd $ {d}^{\prime },{n}^{\prime },{m}^{\prime } $ .\n\nCase 1: $ a > 0 $ . The case $ b > 0 $ is trivial here, since then, by hypothes...	Yes
Theorem 138. If the odd prime $ p $ does not divide the discriminant $ d $, then each rational integer relatively prime to $ p $ is a norm residue $ {\;\operatorname{mod}\;p} $ for $ k\left( \sqrt{d}\right) $ .	We distinguish two cases in the proof.\n\n1. $ p $ splits into two distinct factors $ \mathfrak{p} $ and $ {\mathfrak{p}}^{\prime } $, of degree 1, in $ k\left( \sqrt{d}\right) $ . Then there is a number $ \pi $ in $ k\left( \sqrt{d}\right) $ which is divisible by $ \mathfrak{p} $ but not by \( {\mathfrak...	Yes
Theorem 139. If the odd prime $ q $ divides the discriminant $ d $ of $ k\left( \sqrt{d}\right) $, then exactly one half of the classes of $ \Re \left( q\right) $ are norm residues mod $ q $, and indeed these are the classes of $ \Re \left( q\right) $ which can be represented as the square of a class.	If $ \mathfrak{q} $ is the prime ideal of $ k\left( \sqrt{d}\right) $ which divides $ q $, then each number $ \alpha $ in $ k $ is congruent to a rational number $ {\;\operatorname{mod}\;q} $, say $ r $ . However, since $ q = {q}^{\prime } $ it follows from \( \alpha \equiv r\left( {\;\operatorname{mod}...	Yes
Theorem 140. If the discriminant $ d $ of $ k\left( \sqrt{d}\right) $ is odd, then each odd number is a norm residue $ {\;\operatorname{mod}\;8} $ . However, if $ d $ is even, then exactly half of all incongruent odd numbers $ {\;\operatorname{mod}\;8} $ are norm residues $ {\;\operatorname{mod}\;8} $ .	For the proof we test the residue classes in $ k\left( \sqrt{d}\right) {\;\operatorname{mod}\;8} $ . We find, with $ \alpha = x + y\sqrt{d} $ and $ d $ odd, that\n\n\[ x = 0,1,2,1 \]\n\n\[ y = 1,0,1,2 \]\n\n\[ N\left( \alpha \right) \equiv 3,1,7,5\left( {\;\operatorname{mod}\;8}\right) ,\;\text{ if }d \equiv 5\le...	Yes
Theorem 142. The $ t $ functions $ {\gamma }_{i}\left( \mathfrak{a}\right) $ are group characters of the genus represented by $ \mathfrak{a} $ .	Now by $ §{10} $, the group of characters of an Abelian group is isomorphic to the group. There are $ u $ independent elements in the group of genera and no more, because this group is of order $ {2}^{u} $ and each element has at most order 2. Consequently, there are also exactly $ u $ independent characters. H...	No
Theorem 143. At least one relation must hold for all ideals $ \mathfrak{a} $ of the field which are coprime to $ d $, namely,\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{t}{\gamma }_{i}^{{c}_{i}}\left( \mathfrak{a}\right) = 1 \]\n\nwhere the rational integers $ {c}_{i} $ are independent of $ \mathfrak{a} $ and are ...	Thus for $ t = 1 $ the equation\n\n\[ {\gamma }_{1}\left( \mathfrak{a}\right) = {\chi }_{1}\left( \left\| {N\left( \mathfrak{a}\right) }\right\| \right) = 1 \]\n\nmust hold. In fact this is exactly one part of the quadratic reciprocity law, which has not been used until now (in §47 and 48). We see that the proof of thi...	No
Theorem 144. Let $ {e}_{1},{e}_{2},\ldots ,{e}_{t} $ be t numbers $ \pm 1 $ such that $ {e}_{1} \cdot {e}_{2}\cdots {e}_{t} = 1 $ . Then there are infinitely many prime ideals $ \mathfrak{p} $ of degree 1 in $ k\left( \sqrt{d}\right) $ for which \[ {\gamma }_{i}\left( \mathfrak{p}\right) = {e}_{i}\;\left( {i ...	If we set $ N\left( \mathfrak{p}\right) = p $, then the assertion obviously states that there are infinitely many rational primes $ p $ which satisfy the conditions and \[ {\chi }_{i}\left( p\right) = {e}_{i}\;\left( {i = 1,\ldots, t}\right) \] \[ \left( \frac{d}{p}\right) = + 1 \] By (136), the last condition is n...	Yes
Theorem 147. Let $ {a}_{1},{a}_{2},\ldots ,{a}_{r} $ be any rational integers such that a product of powers\n\n\[ \n{a}_{1}^{{u}_{1}}{a}_{2}^{{u}_{2}}\cdots {a}_{r}^{{u}_{r}} \n\]\n\nis a rational square only if all $ {u}_{i} $ are even. Moreover, let $ {c}_{1},{c}_{2},\ldots ,{c}_{r} $ take arbitrary values \( \...	For the proof we set, for the sake of symmetry,\n\n\[ \nL\left( {s;1}\right) = \mathop{\sum }\limits_{{p > 2}}\frac{1}{{p}^{s}} \n\]\n\n(a function which grows beyond all bounds as $ s \rightarrow 1 $ by $ §{43} $ ), and we form the sum\n\n\[ \n\mathop{\sum }\limits_{{{u}_{1},\ldots ,{u}_{r}}}{c}_{1}^{{u}_{1}}{c}_{...	Yes
Theorem 148. For each natural number $ n $ we have\n\n\[ F\left( n\right) = \mathop{\sum }\limits_{{m \mid n}}\left( \frac{d}{m}\right) \]\n\nwhere $ m $ runs through all distinct positive divisors of $ n $ .	If we decompose $ n $ into its distinct prime factors\n\n\[ n = {p}_{1}^{{k}_{1}} \cdot {p}_{2}^{{k}_{2}}\cdots {p}_{r}^{{k}_{r}} \]\n\nthen\n\n\[ F\left( n\right) = F\left( {p}_{1}^{{k}_{1}}\right) \cdot F\left( {p}_{2}^{{k}_{2}}\right) \cdots F\left( {p}_{r}^{{k}_{r}}\right) = \mathop{\prod }\limits_{{i = 1}}^{r}\m...	Yes
Theorem 149. The expression\n\n\[ \n{\varepsilon }^{2h} = {e}^{\sqrt{d}\mathop{\sum }\limits_{{n = 1}}^{\infty }\left( \frac{d}{n}\right) /n} \n\]\n\nrepresents a unit of infinite degree in $ k\left( \sqrt{d}\right) $ with $ d > 0 $ . And\n\n\[ \n{\varepsilon }^{2h} + {\varepsilon }^{\prime {2h}} = {\varepsilon }^{...	The rational integer $ A $ can thus also be calculated numerically by estimating the remainder of the convergent series $ L\left( 1\right) $ and with this we have a transcendental method for finding a unit in real quadratic fields.	Yes
Theorem 150. For each rational integer $ n $\n\n\[ G\left( {n, d}\right) = \chi \left( n\right) G\left( {1, d}\right) ,\;{c}_{n} = \chi \left( n\right) \frac{G\left( {1, d}\right) }{d}. \]	For if $ n $ and $ d $ are not coprime, then by (154) both sides of the first equation are equal to 0 . However, if $ \left( {n, d}\right) = 1 $, then we choose $ c $ in (155) in such way that $ {cn} \equiv 1\left( {\;\operatorname{mod}\;d}\right) $ ; thus $ \chi \left( c\right) = \chi \left( n\right) $ .	No
Theorem 152. The class number $ h $ of the quadratic field with discriminant $ d $ has the value\n\n\[ h = - \frac{\rho }{\left\| d\right\| }\mathop{\sum }\limits_{{n = 1}}^{{\left\| d\right\| - 1}}n\left( \frac{d}{n}\right) ,\;\rho = \frac{-{iG}\left( {1, d}\right) }{\left\| \sqrt{d}\right\| } = \pm 1\;\text{ for }d < -...	In the second expression $ a $ and $ b $ run through those numbers $ 1,2,\ldots, d - 1 $ for which\n\n\[ \left( \frac{d}{a}\right) = - 1,\;\left( \frac{d}{b}\right) = + 1 \]\n\nrespectively. The final result will be that we always have $ \rho = + 1 $ (§58). The formula for the class number of an imaginary field...	Yes
Theorem 153. If $ D $ is the discriminant of $ F $ and $ D > 0 $, then $ F\left( {x, y}\right) $ may take positive as well as negative values for appropriate real values of $ x, y $ . If $ D < 0 $ , then either the value of $ F $ is $ \geq 0 $ for all real $ x, y $ or \( F\left( {x, y}\right) \leq 0 \...	For the proof we consider the decomposition\n\n\[ A \cdot F\left( {x, y}\right) = {\left( Ax + \frac{B}{2}y\right) }^{2} - \frac{D}{4}{y}^{2}. \]\n\nNow if $ D = {B}^{2} - {4AC} < 0 $, then we must have $ A \neq 0 $ and it follows from the equation that\n\n\[ {AF}\left( {x, y}\right) \geq 0, \]\n\nwhere the equalit...	Yes
Theorem 155. Suppose that the denominator $ \mathfrak{a} $ of $ \mathfrak{d}\omega $ is an odd ideal. Then for every integer $ \varkappa $ which is relatively prime to $ \mathfrak{a} $\n\n\[ C\left( {\varkappa \omega }\right) = \left( \frac{\varkappa }{a}\right) C\left( \omega \right) \]	To begin with, the theorem is true if $ \mathfrak{a} $ is a prime ideal $ \mathfrak{p} $, since if we apply Lemma (a), we have\n\n\[ \mathop{\sum }\limits_{{\mu {\;\operatorname{mod}\;\mathfrak{p}}}}\left( \frac{\mu }{\mathfrak{p}}\right) {e}^{{2\pi iS}\left( {\mu \omega }\right) } = \mathop{\sum }\limits_{{\mu {\;...	Yes
Theorem 158. The n-tuple theta-function\n\n\\[ T\\left( {{u}_{1},\\ldots ,{u}_{n}}\\right) = \\mathop{\\sum }\\limits_{{{m}_{1},\\ldots ,{m}_{n} = - \\infty }}^{{+\\infty }}{e}^{-{\\pi Q}\\left( {{m}_{1} + {u}_{1},\\ldots ,{m}_{n} + {u}_{n}}\\right) }\n\\]\n\n---	admits the representation\n\n\\[ T\\left( {{u}_{1},\\ldots ,{u}_{n}}\\right) = \\mathop{\\sum }\\limits_{{{m}_{1},\\ldots ,{m}_{n} = - \\infty }}^{{+\\infty }}a\\left( \\left( m\\right) \\right) {e}^{-{2\\pi i}\\left( {{m}_{1}{u}_{1} + \\cdots + {m}_{n}{u}_{n}}\\right) }\n\\]\n\nwhere\n\n\\[ a\\left( \\left( m\\right) ...	Yes
Theorem 159. The theta-series defined by (175) also admits the representation\n\n\$ \\theta \\left( {t, z;\\mathfrak{a}}\\right) \$\n\n\\[ = \\frac{1}{N\\left( \\mathfrak{a}\\right) \\left\| \\sqrt{d}\\right\| \\sqrt{{t}_{1} \\cdot {t}_{2}\\cdots {t}_{n}}}\\mathop{\\sum }\\limits_{{\\lambda \\text{ in }1/\\mathfrak{a}\...	We now see at once that this equation also holds for nonreal \$ t \$, provided only that the real part of each \$ {t}_{p} \$ is positive. For then the real part of \$ 1/{t}_{p} \$ is also positive and the series on both sides of the formula represent analytic functions of \$ {t}_{1},\\ldots ,{t}_{n} \$, which a...	Yes
Theorem 163. The reciprocity\n\n\\[ \frac{C\\left( \\omega \\right) }{\\left\| \\sqrt{N\\left( \\mathfrak{a}\\right) }\\right\| } = \\left\| \\frac{\\sqrt{N\\left( {2\\mathfrak{b}}\\right) }}{N\\left( {\\mathfrak{b}}_{1}\\right) }\\right\| {e}^{\\left( {{\\pi i}/4}\\right) S\\left( {\\operatorname{sgn}\\omega }\\right) }C\...	Thus if \$ a, b \$ are relatively prime rational integers, then\n\n\\[ C\\left( \\frac{b}{a}\\right) = \\mathop{\\sum }\\limits_{{n{\\;\\operatorname{mod}\\;a}}}{e}^{{2\\pi i}\\left( {{n}^{2}b/a}\\right) }.\\]\n\nFor odd \$ a \$ the reciprocity formula, Theorem 163, asserts\n\n\\[ \\frac{C\\left( \\frac{1}{a}\\righ...	Yes
Theorem 165 (Law of Quadratic Reciprocity). For two odd relatively prime integers $ \alpha ,\beta $ of which at least one is primary	\[ \left( \frac{\alpha }{\beta }\right) \cdot \left( \frac{\beta }{\alpha }\right) = {\left( -1\right) }^{\mathop{\sum }\limits_{{p = 1}}^{{r}_{1}}\left( {\left( {\operatorname{sgn}{\alpha }^{\left( p\right) } - 1}\right) /2}\right) \left( {\left( {\operatorname{sgn}{\beta }^{\left( p\right) } - 1}\right) /2}\right) } ...	Yes
Theorem 169. Let $ {\mu }_{1},{\mu }_{2},\ldots ,{\mu }_{m} $ be integers in $ k $ such that a product of powers $ {\mu }_{1}^{{x}_{1}}\cdots {\mu }_{m}^{{x}_{m}} $ is the square of a number in $ k $ only if all the exponents $ {x}_{1},\ldots ,{x}_{m} $ are even. Let $ {c}_{1},{c}_{2},\ldots ,{c}_{m} $ be a...	For by the hypothesis, the square root of each of the $ {2}^{m} - 1 $ products of powers $ \mu = {\mu }_{1}^{{x}_{1}}\cdots {\mu }_{m}^{{x}_{m}}\left( {{x}_{i} = 0}\right. $ or 1, not all $ \left. {{x}_{i} = 0}\right) $ defines a relative quadratic field $ K\left( {\sqrt{\mu }, k}\right) $ . However, it now obv...	Yes
Theorem 175. Let $ \lambda = \operatorname{Ir} $ be a totally positive number, $ \mathrm{r} $ an odd ideal, and let I be a prime factor of 2 . In order that the primary integer $ \alpha $, which is relatively prime to $ \lambda $, be hyperprimary, it is necessary and sufficient that\n\n\[ \chi \left( \alpha \ri...	Theorem 167 asserts that the condition is necessary. The proof of the preceding theorem shows in the following way that it is sufficient. First suppose that $ 1 $ is equivalent in the strict sense to the square of an ideal. Then we can find integers $ \beta ,\rho ,\lambda $ such that\n\n\[ \lambda {\beta }^{2} = {\...	Yes
Theorem 179. Let $ \omega $ be a singular primary number. Then there is a subgroup $ \mathfrak{G}\left( \omega \right) $ of order $ {h}_{0}/2 $ in the group of the $ {h}_{0} $ ideal classes in the strict sense such that a prime ideal $ \mathfrak{p} $ splits in the field \( K\left( {\sqrt{\omega }, k}\right) \...	The set of odd ideals $ \mathfrak{r} $, for which $ Q\left( {\omega ,\mathfrak{r}}\right) = + 1 $, determines a subgroup of order $ {2}^{{e}_{0} - 1} $ in the group of class complexes in the strict sense, by Theorem 172. Since each class complex consists of $ {h}_{0}/{2}^{{e}_{0}} $ classes in the strict sense,...	Yes
Theorem 0.2. If $ \mathcal{S} $ is a semi-algebra of subsets of $ X $ and $ \tau : \mathcal{S} \rightarrow {R}^{ + } $ is finitely additive then there is a unique finitely additive function $ {\tau }_{1} : \mathcal{L}\left( \mathcal{S}\right) \rightarrow {R}^{ + } $ which is an extension of $ \tau $ (i.e. \( ...	It could be that $ X \notin \mathcal{S} $ but we always have that $ X $ is a disjoint union of a finite number of members $ {E}_{1},\ldots ,{E}_{n} $ of $ \mathcal{S} $ so $ {\tau }_{1}\left( X\right) = 1 $ if $ \mathop{\sum }\limits_{{i = 1}}^{n}\tau \left( {E}_{i}\right) = 1 $ .	No
Theorem 0.5. Fix $ k \geq 1 $ and let $ Y = \{ 0,1,\ldots, k - 1\} $ and $ \left( {X,\mathcal{D}}\right) = \mathop{\prod }\limits_{{j = x}}^{\infty }(Y $ , $ \left. {2}^{Y}\right) $ . For each natural number $ n $ and $ {a}_{0},\ldots ,{a}_{n} \in Y $ suppose a non-negative real number \( {p}_{n}\left( {{a}...	The proof boils down to showing that the function naturally defined on the algebra of all finite unions of elementary rectangles is countable additive by using Theorem 0.4.	Yes
Theorem 0.7. Let $ \left( {X,\mathcal{B}, m}\right) $ be a probability space and let $ \mathcal{A} $ be an algebra of subsets of $ X $ with $ \mathcal{B}\left( \mathcal{A}\right) = \mathcal{B} $ . Then for each $ \varepsilon > 0 $ and each $ B \in \mathcal{B} $ there is some $ A \in \mathcal{A} $ with \( ...	Note that when $ m\left( {A\bigtriangleup B}\right) < \varepsilon $ then $ \left\| {m\left( A\right) - m\left( B\right) }\right\| < \varepsilon $ because $ m\left( A\right) = $ $ m\left( {A \smallsetminus B}\right) + m\left( {A \cap B}\right) $ and \( m\left( B\right) = m\left( {B \smallsetminus A}\right) + m\lef...	No
Theorem 0.12. Let $ \left( {X,\mathcal{D}, m}\right) $ be a probability space and let $ \mathcal{C} $ be a sub- $ \sigma $ -algebra of $ \mathcal{B} $ . The restriction of the conditional expectation operator $ E\left( {\cdot /\mathcal{C}}\right) $ to $ {L}^{2}\left( {X,\mathcal{B}, m}\right) $ is the ortho...	Proof. For $ f \in {L}^{1}\left( {X,\mathcal{B}, m}\right) $ we know $ E\left( {f/\mathcal{C}}\right) $ is the only $ \mathcal{C} $ -measurable function $ h $ such that $ {\int }_{C}{hdm} = {\int }_{C}{fdm}\forall C \in \mathcal{C} $ . Let $ P $ denote the orthogonal projection of \( {L}^{2}\left( {X,\mathc...	Yes
Theorem 0.13. Let $ G $ be a compact topological group. There exists a probability measure $ m $ defined on the $ \sigma $ -algebra $ \mathcal{B}\left( G\right) $ of Borel subsets of $ G $ such that $ m\left( {xE}\right) = m\left( E\right) \forall x \in G\forall E \in \mathcal{B}\left( G\right) $ and \( m \...	This unique measure is called Haar measure. Notice we have required Haar measure to be a probability measure. The Haar measure $ m $ also satisfies $ m\left( {Ex}\right) = m\left( E\right) \forall x \in G,\forall E \in \mathcal{B}\left( G\right) $, because for each fixed $ x \in G $ the measure $ {m}_{x} $ defi...	Yes
Theorem 0.17. Let $ A $ be an irreducible and aperiodic non-negative matrix. Let\n\n\[ u = \left( {{u}_{1},\ldots ,{u}_{k}}\right) ,\;v = \left( \begin{matrix} {v}_{1} \\ \vdots \\ {v}_{k} \end{matrix}\right) \]\n\nbe the strictly positive eigenvectors corresponding to the largest eigenvalue $ \lambda $ as in Theor...	This theorem follows from the renewal theorem which we shall need for another purpose (see Feller [1], p. 291).	No
Theorem 0.19. Let $ X $ be a compact Hausdorff space. The following are equivalent:\n\n(1) $ X $ is metrisable.\n\n(2) $ X $ has a countable base.\n\n(3) $ C\left( X\right) $ (the space of all complex-valued continuous functions on $ X $ ) has a countable dense subset.	See Kelley [1] for the proof.	No
Theorem 0.20 (Lebesgue Covering Lemma). If $ \\left( {X, d}\\right) $ is a compact metric . space and $ \\alpha $ is an open cover of $ X $ then there exists a $ \\delta > 0 $ such that each subset of $ X $ of diameter less than or equal to $ \\delta $ lies in some member of $ \\alpha $ . (Such a \( \\dot...	Proof. We may as well suppose $ \\alpha $ is a finite cover.\n\nLet $ \\alpha = \\left\\{ {{A}_{1},\\ldots ,{A}_{p}}\\right\\} $ . Assume the theorem is false. Then for each $ n \\geq 1 $ there exists $ {B}_{n} \\subseteq X $ such that $ \\operatorname{diam}\\left( {B}_{n}\\right) \\leq 1/n $ and \( {B}_{n} \...	Yes
Theorem 1.1. Suppose $ \left( {{\lambda }_{1},{\mathcal{B}}_{1},{m}_{1}}\right) ,\left( {{X}_{2},{\mathcal{D}}_{2},{m}_{2}}\right) $ are probability spaces and $ T : {X}_{1} \rightarrow {X}_{2} $ is a transformation. Let $ {\mathcal{S}}_{2} $ be a semi-algebra which generates - $ {\mathcal{B}}_{2} $ . If for ea...	Proof. Let $ {\mathcal{C}}_{2} = \left\{ {B \in {\mathcal{B}}_{2} : {T}^{-1}\left( B\right) \in {\mathcal{B}}_{1},{m}_{1}\left( {{T}^{-1}\left( B\right) }\right) = {m}_{2}\left( B\right) }\right\} $ . We want to show that $ {\mathcal{C}}_{2} = {\mathcal{D}}_{2} $ . However \( {\mathcal{S}}_{2} \subseteq {\mathcal{C...	Yes
Lemma 1.2. Let $ \left( {{X}_{i},{\mathcal{A}}_{i},{m}_{i}}\right) i = 1,2 $ be a probability space and let $ T : {X}_{1} \rightarrow {X}_{2} $ be measure-preserving. If $ F \in {L}^{0}\left( {{X}_{2},{\mathcal{B}}_{2},{m}_{2}}\right) $ then $ \int {U}_{T}{Fd}{m}_{1} = \int {Fd}{m}_{2} $ (where if one side does...	Proof. It suffices to prove the result when $ F $ is real-valued and, by considering positive and negative parts of $ F $, it suffices to consider non-negative functions. So suppose $ F \geq 0 $ . If $ F $ is a simple function then the result is true because $ T $ is measure-preserving. Choose simple function...	Yes
Theorem 1.3. Let $ p \geq 1 $ . With the above notation $ {U}_{T}{L}^{P}\left( {{X}_{2},{\mathcal{B}}_{2},{m}_{2}}\right) \subset $ $ {L}^{P}\left( {{X}_{1},{\mathcal{B}}_{1},{m}_{1}}\right) $ and \( {\begin{Vmatrix}{U}_{T}f\end{Vmatrix}}_{p} = \parallel f{\parallel }_{p}\forall f \in {L}^{P}\left( {{X}_{2},{\mat...	Proof. Let $ f \in {L}^{p}\left( {{X}_{2},{\mathcal{B}}_{2},{m}_{2}}\right) $ . Put $ F\left( x\right) = {\left\| f\left( x\right) \right\| }^{p} $ in Lemma 1.2 to get $ {\begin{Vmatrix}{U}_{T}f\end{Vmatrix}}_{p} = \parallel f{\parallel }_{p}. $ Therefore a measure-preserving transformation \( T : {X}_{1} \rightarr...	Yes
Theorem 1.4 (Poincaré’s Recurrence Theorem). Let $ T : X \rightarrow X $ be a measure-preserving transformation of a probability space $ \left( {X,\mathcal{B}, m}\right) $ . Let $ E \in \mathcal{B} $ with $ m\left( E\right) > 0 $ . Then almost all points of $ E $ return infinitely often to $ E $ under posit...	Proof. For $ N \geq 0 $ let $ {E}_{N} = \mathop{\bigcup }\limits_{{n = N}}^{\prime }{T}^{-n}E $ . Then $ \mathop{\bigcap }\limits_{{N = 0}}^{\prime }{E}_{N} $ is the set of all points of $ X $ which enter $ E $ infinitely often under positive iteration by $ T $ . Hence the set \( F = E \cap \mathop{\bigcap ...	Yes
Theorem 1.5. If $ T : X \rightarrow X $ is a measure-preserving transformation of the probability space $ \left( {X,\mathcal{A}, m}\right) $ then the following statements are equivalent:\n\n(i) $ T $ is ergodic.\n\n(u) The only members $ B $ of $ \mathcal{B} $ with \( m\left( {{T}^{-1}B\bigtriangleup B}\right...	## Proof\n\n(i) $ \Rightarrow $ (ii) Let $ B \in \mathcal{B} $ and $ m\left( {{T}^{-1}B\bigtriangleup B}\right) = 0 $ . We shall construct a set $ B $, with $ {T}^{-1}{B}_{r} = B $, and $ m\left( {B\bigtriangleup {B}_{r}}\right) = 0 $ . For each $ n \geq 0 $ we have \( m\left( {{T}^{-n}B\bigtriangleup B}\...	No
Theorem 1.6. If $ \left( {X,\mathcal{A}, m}\right) $ is a probability space and $ T : X \rightarrow X $ is measure-preserving then the following statements are equivalent:\n\n(i) $ T $ is ergodic.\n\n(ii) Whenever $ f $ is measurable and \( \left( {f \cdot T}\right) \left( x\right) = f\left( x\right) \forall x ...	Proof. Trivially we have (iii) $ \Rightarrow $ (ii),(ii) $ \Rightarrow $ (iv),(v) $ \Rightarrow $ (iv), and (iii) $ \Rightarrow $ (v). So it remains to show (i) $ \Rightarrow $ (iii) and (iv) $ \Rightarrow $ (i). We first show (i) $ \Rightarrow $ (iii). Let $ T $ be ergodic and suppose $ f $ is measur...	Yes
Theorem 1.7. Let $ X $ be a compact metric space, $ \mathcal{B}\left( X\right) $ the $ \sigma $ -algebra of Borel subsets of $ X $ and let $ m $ be a probability measure on $ \left( {X,\mathcal{B}\left( X\right) }\right) $ such that $ m\left( U\right) > 0 $ for every non-empty open set $ U $ . Suppose \...	Proof. Let $ {\left\{ {U}_{n}\right\} }_{n = 1}^{\infty } $ be a base for the topology of $ X $ . Then $ \left\{ {{T}^{n}x \mid n \geq 0}\right\} $ is dense in $ X $ iff $ x \in \mathop{\bigcap }\limits_{{n = 1}}^{\infty }\mathop{\bigcup }\limits_{{k = 0}}^{\infty }{T}^{-k}{U}_{n} $ . Since \( {T}^{-1}\left( ...	Yes
Theorem 1.8. The rotation $ T\left( z\right) = {az} $ of the unit circle $ K $ is ergodic (relative to Haar measure $ m $ ) iff $ a $ is not a root of unity.	Proof. Suppose $ a $ is a root of unity, then $ {a}^{p} = 1 $ for some $ p \neq 0 $ . Let $ f\left( z\right) = {z}^{p} $ . Then $ f \circ T = f $ and $ f $ is not constant a.e. Therefore $ T $ is not ergodic by Theorem 1.6(ii). Conversely, suppose $ a $ is not a root of unity and $ f \circ T = f $ , \...	Yes
Theorem 1.9. Let $ G $ be a compact group and $ T\left( x\right) = {ax} $ a rotation of $ G $ . Then $ T $ is ergodic iff $ \left\{ {a}^{n}\right\} = 0 $ is dense in $ G $ . In particular, if $ T $ is ergodic, then $ G $ is abelian.	Proor. Suppose $ T $ is ergodic. Let $ H $ denote the closure of the subgroup $ \left\{ {a}^{n}\right\} = c $ of $ G $ . If $ H \neq G $ then by (7) of $ \$ {0.7} $ there exists $ \gamma \in \widehat{G} $ with $ \gamma ≢ 1 $ but $ \gamma \left( h\right) = 1\forall h \in H $ . Then \( \gamma \left( {Tx...	Yes
Theorem 1.10. If $ G $ is a compact abelian group (equipped with normalized Haar measure) and $ A : G \rightarrow G $ is a surjective continuous endomorphism of $ G $ then $ A $ is ergodic iff the trivial character $ \gamma \equiv 1 $ is the only $ \gamma \in \widehat{G} $ that satisfies \( \gamma \circ {A}...	Proof. Suppose that whenever $ \gamma {A}^{n} = \gamma $ for some $ n \geq 1 $ we have $ \gamma \equiv 1 $ . Let $ f \circ A = f $ with $ f \in {L}^{2}\left( m\right) $ . Let $ f\left( x\right) $ have the Fourier series $ \sum {a}_{n}{\gamma }_{n} $ where $ {\gamma }_{n} \in \widehat{G} $ and \( \sum {\...	Yes

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Theorem 50. If \( \alpha ,\beta \) are algebraic numbers over \( k \), then the same is true for \( \alpha + \beta ,\alpha - \beta ,{\alpha \beta } \), and, if \( \beta \neq 0 \), for \( \alpha /\beta \) .	If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are the conjugates of \( \alpha \) and \( {\beta }_{1},\ldots ,{\beta }_{m} \) are those of \( \beta \) with respect to \( k \), then the elementary symmetric functions of \( \alpha \), as well as those of \( \beta \), are numbers in \( k \) . The product\n\n\[ \nH\left( x\r...	Yes
Theorem 51. If \( \omega \) is a root of a polynomial\n\n\[ \varphi \left( x\right) = {x}^{m} + \alpha {x}^{m - 1} + \beta {x}^{m - 2} + \cdots + \lambda \]\n\nwhose coefficients are algebraic numbers over \( k \), then \( \omega \) is also an algebraic number over \( k \) .	Suppose \( {\alpha }_{i} \) runs through the conjugates of \( \alpha ,{\beta }_{k} \) runs through the conjugates of \( \beta \) etc. Then by the theorem on symmetric functions, the polynomial\n\n\[ F\left( x\right) = \mathop{\prod }\limits_{{i, k,\ldots, s}}\left( {{x}^{m} + {\alpha }_{i}{x}^{m - 1} + {\beta }_{k}{x}^...	Yes
Theorem 53. Every number in \( K\left( \theta \right) \) is obtained exactly once in the form\n\n\[ \alpha = {c}_{0} + {c}_{1}\theta + {c}_{2}{\theta }^{2} + \cdots + {c}_{n - 1}{\theta }^{n - 1} \]\n\n(34)\n\nwhere the \( {c}_{0},\ldots ,{c}_{n - 1} \) run through all numbers of the ground field \( k \) .	To prove this suppose \( \alpha = P\left( \theta \right) /Q\left( \theta \right), Q\left( \theta \right) \neq 0 \) . Then \( Q\left( x\right) \) does not have the root \( \theta \) in common with the function \( f\left( x\right) \) belonging to \( \theta \) which is irreducible over \( k \) ; hence by Theorem 49, \( Q\...	Yes
Theorem 54. Every number \( g\left( \theta \right) \) of the field \( K\left( \theta \right) \) is likewise an algebraic number over \( k \) of degree at most \( n \) . The relative conjugates of a number \( \alpha = g\left( \theta \right) \) are the distinct numbers among the numbers \( g\left( {\theta }_{i}\right) \l...	For if \( {\theta }_{1},\ldots ,{\theta }_{n} \) are the conjugates of \( \theta \) with respect to \( k \), we form the product\n\n\[ F\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {x - g\left( {\theta }_{i}\right) }\right) \]\n\nThe coefficients of this polynomial are integral rational combinations of ...	Yes
Theorem 55. Each rational equation \( R\left( {\alpha ,\beta ,\gamma ,\ldots }\right) = 0 \) between numbers \( \alpha ,\beta ,\gamma ,\ldots \) in \( K\left( \theta \right) \) with coefficients in \( k \) remains true if \( \alpha ,\beta ,\gamma ,\ldots \) are replaced by the conjugates with the same index.	As a rational function of \( \alpha ,\beta ,\gamma ,\ldots, R \) is identical to the quotient of two integeral rational expressions \( P \) and \( Q \)\n\n\[ R\left( {\alpha ,\beta ,\gamma ,\ldots }\right) = \frac{P\left( {\alpha ,\beta ,\gamma ,\ldots }\right) }{Q\left( {\alpha ,\beta ,\gamma ,\ldots }\right) } \]\n\n...	Yes
Theorem 56. A number \( \alpha \) in \( K\left( \theta \right) \) belongs to the ground field \( k \) if and only if it is equal to its \( n \) conjugates. A number \( \alpha \) in \( K\left( \theta \right) \) has degree \( n \) with respect to \( k \) if and only if it is distinct from all its conjugates. The latter c...	The first two statements follow immediately from Theorem 54 and the definition which follows it. Moreover, if \( \alpha \) in \( K\left( \theta \right) \) is to generate the field \( K\left( \theta \right) \) , thus if \( K\left( \theta \right) = K\left( \alpha \right) \) is to hold, then the degree of \( \alpha \) mus...	Yes
Theorem 57. In order that the numbers\n\n\[ \n{\omega }^{\left( i\right) } = \mathop{\sum }\limits_{{k = 1}}^{n}{c}_{ik}{\theta }^{k - 1}\;\left( {{c}_{ik}\text{ numbers in }k}\right) \n\]\n\n(35)\n\nform a fundamental system of \( K\left( \theta \right) \), it is necessary and sufficient that the determinant \( \begin...	Obviously we need only investigate when the numbers \( 1,\theta ,\ldots ,{\theta }^{n - 1} \) can be represented in terms of the \( {\omega }^{\left( i\right) } \) as\n\n\[ \n{\theta }^{p - 1} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{pi}{\omega }^{\left( i\right) }\;\left( {p = 1,\ldots, n}\right) \left( {a}_{pi}\righ...	Yes
Theorem 58. The n numbers \( {\omega }^{\left( 1\right) },\ldots ,{\omega }^{\left( n\right) } \) in \( K\left( \theta \right) \) form a fundamental system if and only if there is no linear relation\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i}{\omega }^{\left( i\right) } = 0 \]\n\n(37)\n\nwith coefficients in \( k ...	The \( n \) numbers \( {\omega }^{\left( i\right) } \) of this type are said to be linearly independent. For, in the notation used above it would follow from (37) that\n\n\[ 0 = \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i}\mathop{\sum }\limits_{{k = 1}}^{n}{c}_{ik}{\theta }^{k - 1} \]\n\nand as before, if the \( {u}_{i} ...	Yes
Theorem 60. If \( \alpha \) satisfies any equation at all with integral coefficients whose leading coefficient is equal to 1, then \( \alpha \) is an integer.	Let \( \varphi \left( x\right) = {x}^{N} + {a}_{1}{x}^{N - 1} + \cdots + {a}_{N} \) with rational integral \( a \)’s and \( \varphi \left( \alpha \right) = 0 \). Moreover let\n\n\[ f\left( x\right) = {c}_{0}{x}^{n} + {c}_{1}{x}^{n - 1} + \cdots + {c}_{n} \]\n\nbe the irreducible polynomial in \( k\left( 1\right) \) whi...	Yes
Theorem 61. The sum, difference, and product of two integers is again an integer. Hence every rational integral function (polynomial) of integers with rational integral coefficients is again an integer.	For if \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are the conjugates of a number \( \alpha \) and if \( {\beta }_{1},\ldots ,{\beta }_{m} \) are the conjugates of a number \( \beta \), then\n\n\[ F\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\mathop{\prod }\limits_{{k = 1}}^{m}\left( {x - \left( {{\alpha }_{i} ...	No
Theorem 63. Every algebraic number \( \alpha \) can be transformed into an integer by multiplication by a suitable nonzero rational number.	To prove this assume that\n\n\[ \n{c}_{0}{x}^{n} + {c}_{1}{x}^{n - 1} + \cdots + {c}_{n - 1}x + {c}_{n} = 0 \n\]\n\nis an equation for \( \alpha \) with rational integral coefficients and \( {c}_{0} \neq 0 \) . Then by multiplication by \( {c}_{0}^{n - 1} \) we obtain an integer equation for \( y = {c}_{0}x \) with lea...	Yes
Theorem 65. Every ideal a can be written in the form \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) with the \( \alpha \) suitably chosen integers in \( K \) . Moreover, we may even take \( r \leq n \) .	The numbers of an ideal \( \mathfrak{a} \) which is not \( \left( 0\right) \) (the case \( \mathfrak{a} = \left( 0\right) \) is trivial) obviously form an infinite Abelian group, under composition by addition, which is a subgroup of the group of all integers in \( K \) . Consequently by Theorem 34 the ideal \( \mathfra...	Yes
Theorem 68. If \( \mathfrak{a}\mathfrak{b} = \mathfrak{a}\mathfrak{c} \), then if \( \mathfrak{a} \neq 0,\mathfrak{b} = \mathfrak{c} \) .	To see this we determine an ideal \( m \) such that \( {am} = \left( \delta \right) \) is a principal ideal. Then\n\n\[ \n{amb} = {amc},\;\left( \alpha \right) b = \left( \alpha \right) c.\n\]\n\nThe latter equation asserts that \( \alpha \) times every number from \( \mathfrak{b} \) is of the form \( \alpha \) times a...	No
Theorem 69. An ideal \( \mathfrak{c} = \left( {{\gamma }_{1},\ldots ,{\gamma }_{r}}\right) \) is a divisor of \( \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right) \) if and only if every number of \( \mathfrak{a} \) belongs to \( \mathfrak{c} \) .	If \( \mathfrak{c} \mid \mathfrak{a} \), then there is \( \mathrm{a}\mathfrak{b} = \left( {{\beta }_{1},\ldots ,{\beta }_{p}}\right) \) for which \( \mathfrak{b} \neq \left( 0\right) \) and\n\n\[ \left( {{\alpha }_{1},\ldots ,{\alpha }_{m}}\right) = \left( {\beta ,\ldots ,{\beta }_{p}}\right) \cdot \left( {{\gamma }_{1...	Yes
Theorem 70. For every two ideals \( \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) ,\mathfrak{b} = \left( {{\beta }_{1},\ldots ,{\beta }_{s}}\right) \) which are not both \( = \left( 0\right) \), there is a uniquely determined greatest common divisor \( \mathfrak{d} = \left( {\mathfrak{a},\mathfrak{...	We show that \( \mathfrak{d} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r},{\beta }_{1},\ldots ,{\beta }_{s}}\right) \) has the stated properties of divisibility. Since every sum \	No
Theorem 71. If \( \mathfrak{p} \) is a prime ideal and \( \mathfrak{p} \mid \mathfrak{a}\mathfrak{b} \), then \( \mathfrak{p} \) divides either \( \mathfrak{a} \) or \( \mathfrak{b} \) or both.	For if \( \mathfrak{p} \) does not divide the factor \( \mathfrak{b} \), then\n\n\[ \left( {p, b}\right) = \left( 1\right) \]\n\nsince, as a prime ideal, \( \mathfrak{p} \) has no factors except (1) and \( \mathfrak{p} \) . It follows from (41) that\n\n\[ \mathfrak{a} = \mathfrak{a}\left( 1\right) = \mathfrak{a}\left( ...	Yes
Theorem 73. There are infinitely many prime ideals in each field.	Each rational prime \( p \) defines an ideal \( \left( p\right) \), and moreover if \( p \) and \( q \) are distinct positive primes, then \( \left( {p, q}\right) = 1 \) in the sense of our ideal theory, since the number 1 occurs in the form \( {px} + {qy} \) in \( \left( {p, q}\right) \) . Consequently, the same prime...	Yes
Theorem 74. If \( \mathfrak{a} \) and \( \mathfrak{b} \) are ideals distinct from (0), then there is always a number \( \omega \) for which\n\n\[ \left( {\omega ,\mathfrak{a}\mathfrak{b}}\right) = \mathfrak{a}. \]\n\nThis \( \omega \) then obviously has a decomposition \( \omega = \mathfrak{{ac}} \) where \( \left( {\m...	For a proof, let \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{r} \) be all the distinct prime ideals which divide \( \mathfrak{a}\mathfrak{b} \), and let \( \mathfrak{a} = {\mathfrak{p}}_{1}^{{a}_{1}}\cdots {\mathfrak{p}}_{r}^{{a}_{r}}\left( {{a}_{i} \geq 0}\right) \) . We define the \( r \) ideals \( {\mathfrak{d}}_{...	Yes
Theorem 76. The number of residue classes \( {\;\operatorname{mod}\;\mathfrak{a}} \) is finite. If the number of residue classes is denoted by \( N\left( \mathfrak{a}\right) \) and if \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is a basis for \( \mathfrak{a} \), then \( N\left( \mathfrak{a}\right) = \left\| {\Delta \left(...	The numbers of a form a subgroup of the group \( \mathfrak{G} \) of all integers of the field. The different cosets in \( \mathfrak{G} \) determined by a obviously form the different residue classes mod a. Hence the number of distinct residue classes mod a is the index of \( \mathfrak{a} \) in \( \mathfrak{G} \) . This...	Yes
Theorem 78. For given \( \alpha \) and \( \beta \) the congruence\n\n\[ \n{\alpha \xi } \equiv \beta \left( {\;\operatorname{mod}\;a}\right)\n\]\n\ncan be solved by an integer \( \xi \) in \( K \) if and only if \( \left( {\alpha ,\mathfrak{a}}\right) \mid \beta \) . If \( \left( {\alpha ,\mathfrak{a}}\right) = 1 \), t...	If we assume \( \left( {\alpha ,\mathfrak{a}}\right) = 1 \) to begin with, and let \( \xi \) run through a system of \( N\left( \mathfrak{a}\right) \) numbers which are incongruent \( {\;\operatorname{mod}\;\mathfrak{a}} \), then \( {\alpha \xi } \) runs through all the residue classes \( {\;\operatorname{mod}\;\mathfr...	Yes
Theorem 79. For two ideals \( \mathfrak{a} \) and \( \mathfrak{b} \), we always have\n\n\[ N\left( \mathfrak{{ab}}\right) = N\left( \mathfrak{a}\right) \cdot N\left( \mathfrak{b}\right) \]	Let \( \alpha \) be a number divisible by \( \mathfrak{a} \) such that \( \left( {\alpha ,\mathfrak{a}\mathfrak{b}}\right) = \mathfrak{a} \) . If we let \( {\xi }_{i} \) \( \left( {i = 1,2,\ldots, N\left( \mathrm{\;b}\right) }\right) \) run through a complete system of residues mod \( \mathrm{b} \) and let \( {\eta }_{...	Yes
Theorem 80. If \( \left( {\mathfrak{a},\mathfrak{b}}\right) = 1 \), then \( \varphi \left( {\mathfrak{a}\mathfrak{b}}\right) = \varphi \left( \mathfrak{a}\right) \cdot \varphi \left( \mathfrak{b}\right) \) and in general\n\n\[ \varphi \left( \mathfrak{a}\right) = N\left( \mathfrak{a}\right) \mathop{\prod }\limits_{{\ma...	To see this let \( \alpha \) be chosen so that \( \left( {\alpha ,{ab}}\right) = \alpha \) and \( \beta \) so that \( \left( {\beta ,{ab}}\right) = b \) . Then if \( \xi \) runs through a complete system of residues \( {\;\operatorname{mod}\;b} \) and \( \eta \) runs through such a system mod a in \( {\alpha \xi } + {\...	Yes
Theorem 81. The norm of a prime ideal \( \mathfrak{p} \) is a power of a certain rational prime \( p, N\left( \mathfrak{p}\right) = {p}^{f}.f \) is called the degree of \( \mathfrak{p} \) . Every ideal \( \left( p\right) \), where \( p \) is a rational prime, can be decomposed into at most \( n \) factors.	For each prime ideal \( \mathfrak{p} \) divides certain rational numbers and consequently also certain rational primes \( p \) . Suppose that \( \mathfrak{p} \mid p, p = \mathfrak{{pa}} \) . Then \( N\left( p\right) = \) \( N\left( \mathfrak{p}\right) \cdot N\left( \mathfrak{a}\right) \) and consequently the rational i...	Yes
Theorem 83. The group of residue classes \( {\;\operatorname{mod}\;\mathfrak{p}} \) is an Abelian group \( \mathfrak{G}\left( \mathfrak{p}\right) \) of order \( N\left( \mathfrak{p}\right) = {p}^{f} \) under composition by addition and the number of basis elements is equal to the degree \( f \) of the prime ideal \( \m...	For since \( \mathfrak{p} \mid p \), the number of residue classes whose elements \( \alpha \) satisfy the congruence\n\n\[ \n{p\alpha } \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\right)\n\]\n\nis equal to the number of all residue classes, thus \( {p}^{f} \). Consequently, by Theorem \( {27}, f \) is equal to...	Yes
Theorem 86. If \( \mathfrak{p} \) is a prime ideal, and if for two polynomials \( P \) and \( Q \) the product\n\n\[ P\left( {{x}_{1},\ldots ,{x}_{m}}\right) \cdot Q\left( {{x}_{1},\ldots ,{x}_{m}}\right) \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) ,\]\n\nthen at least one of the polynomials is \( \equiv...	The theorem is true for polynomials of 0 variables, that is, for constants. We show that it is correct in general by passing from \( m \) to \( m + 1 \) . Assume it is already proven for all polynomials with \( m \) or fewer variables. Each polynomial of \( m + 1 \) variables can be put into the form\n\n\[ P\left( {{x}...	Yes
Theorem 88. For each ideal \( \mathfrak{a} \) of a Galois field the principal ideal \( \left( {N\left( \mathfrak{a}\right) }\right) = \) \( {\mathfrak{a}}^{\left( 1\right) }{\mathfrak{a}}^{\left( 2\right) }\cdots {\mathfrak{a}}^{\left( n\right) } \) (compare with Theorem 107).	For the proof, we form the polynomial \( P\left( x\right) = {\alpha }_{1}x + {\alpha }_{2}{x}^{2} + \cdots + {\alpha }_{r}{x}^{r} \) from a new variable \( x \) and \( \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) where the GCD of the coefficients \( = \mathfrak{a} \) . The product of the conjug...	Yes
Theorem 89. Let \( p \) be a rational prime which does not divide \( d \) . Then \( p \) splits in the field \( K\left( \sqrt{d}\right) \) into two distinct prime ideals \( \mathfrak{p},{\mathfrak{p}}^{\prime } \) provided the congruence\n\n\[ \n{x}^{2} \equiv d\left( {\;\operatorname{mod}\;{4p}}\right) \n\]\n\ncan be ...	If the prime \( p \) which does not divide \( d \) splits in \( K\left( \sqrt{d}\right) \), then \( p \) can only split into prime factors \( \mathfrak{p},{\mathfrak{p}}^{\prime } \) which are of degree 1 . By Theorem 85, each integer in \( K \) is congruent \( {\;\operatorname{mod}\;\mathfrak{p}} \) to a rational numb...	Yes
Theorem 91. If the prime \( p \) does not divide \( D \cdot m \), then \( p \) is not divisible by the square of a prime ideal in \( K\left( \zeta \right) \) .	For if \( {\mathfrak{p}}^{2} \mid p \), then let us choose a number \( \omega \) which is divisible by \( \mathfrak{p} \) but not by \( {\mathfrak{p}}^{2} \) . It follows from the lemma that\n\n\[ \n{\omega }^{pf} \equiv \omega \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{2}}\right)\n\]\n\nSince \( {p}^{f} \geq 2 \), ...	Yes
Theorem 92. If the prime \( p \) does not divide \( D \cdot m \), and if \( f \) is the smallest positive exponent such that \( {p}^{f} \equiv 1\\left( {\\operatorname{mod}m}\\right) \), then \( p \) splits into exactly \( e = h/f \) distinct prime factors in \( K\\left( \\zeta \\right) \). Each factor has degree \( f ...	Let \( \\mathfrak{p} \) be a prime factor of \( p \) of degree \( {f}_{1} \). Then, by (43), for each integer \( \\omega \) in \( K\\left( \\zeta \\right) \), \[ {\\omega }^{{p}^{{f}_{1}}} \equiv \\omega \\left( {\\operatorname{mod}\\mathfrak{p}}\\right) \] (46) and this congruence holds for each integer \( \\omega \) ...	Yes
Theorem 93. Each ideal \( \mathfrak{g} \) is the range of values of a linear form\n\n\[{\xi }_{1}{\rho }_{1} + \cdots + {\xi }_{r}{\rho }_{r}\]\n\nwhere \( {\rho }_{1},\ldots ,{\rho }_{r} \) are certain integers or fractions in \( \mathfrak{g} \), while the \( {\xi }_{i} \) run through all integers in \( K \) . We writ...	Let \( v \) be chosen for \( g \) according to (2). Then all products of \( v \) with the numbers in \( \mathfrak{g} \) obviously form an integral ideal \( \mathfrak{a} = \left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) and then \( \mathrm{g} = \left( {{\alpha }_{1}/v,\ldots ,{\alpha }_{r}/v}\right) \).\n\nIf \( ...	No
Theorem 94. Assume we are given \( n \) linear homogeneous expressions\n\n\[ \n{L}_{p}\left( x\right) = \mathop{\sum }\limits_{{q = 1}}^{n}{a}_{pq}{x}_{q}\;\left( {p = 1,2,\ldots, n}\right) ,\n\]\n\nwith real coefficients \( {a}_{pq} \), whose determinant \( D = \left\| {a}_{pq}\right\| \) is different from zero, as well...	The proof is along the lines of Minkowski's contribution to the geometry of numbers. To begin with we ask: \	No
Theorem 95. Let \( n \) linear forms \( {L}_{p}\left( x\right) = \mathop{\sum }\limits_{{q = 1}}^{n}{a}_{pq}{x}_{q}\left( {p = 1,\ldots, n}\right) \) be given with real or complex coefficients whose determinant \( D \neq 0 \) . Moreover if one of the forms is not real, we assume the complex conjugate of a form also occ...	To prove this we replace the system \( {L}_{p}\left( x\right) \) by that system of real forms \( {L}^{\prime }\left( x\right) \) which arises if the real and imaginary components of the \( {L}_{p}\left( x\right) \) are considered by themselves. We take \( {L}_{p}^{\prime }\left( x\right) = {L}_{p}\left( x\right) \) if ...	Yes
Theorem 96. In each ideal class of \( K \) there is an integral ideal whose norm is \( \leq \left\| \sqrt{d}\right\| \) . Thus the number of ideal classes in \( K \) is finite.	To prove this let \( \mathfrak{a} \) be an integral ideal in the class \( {B}^{-1} \), where \( B \) is an arbitrarily given class. If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) denotes a basis of \( a \), then, by Theorem 95, there are rational integers \( {x}_{1},\ldots ,{x}_{n} \), not all vanishing, such that\n\n\[ ...	Yes
Theorem 98. For each ideal \( \mathfrak{a} \) in \( K \) there is a number \( A \) which generally does not belong to the field \( K \), such that the numbers of \( \mathfrak{a} \) are identical with those numbers of the field \( K \) which are divisible by \( A \) .	By Theorem \( {97}{\mathfrak{a}}^{h} \) is equal to a principal ideal \( \left( \omega \right) \) . The number \( A = \sqrt[h]{\omega } \) has the asserted property for if \( \alpha \) is a number in \( \mathfrak{a} \), then \( {\alpha }^{h} \) belongs to \( {\mathfrak{a}}^{h} \) and therefore \( {\alpha }^{h}/\omega \...	Yes
Theorem 99. The group \( \mathfrak{W} \) of all roots of unity in \( K \) is finite, and indeed it is a cyclic group of order \( w \geq 2 \) .	Since all roots of unity, including all conjugates, have absolute value 1, the first assertion follows from the lemma. Moreover if \( p \) is a prime dividing the order of \( \mathfrak{W} \), then the number of solutions of \( {x}^{p} = 1 \) is equal to \( {p}^{1} \), and thus, by Theorem 28, the basis number of the gr...	No
Theorem 100. The group \( \mathfrak{E} \) of all units in \( K \) has a finite basis. Furthermore this basis consists of precisely \( r = {r}_{1} + {r}_{2} - 1 \) elements of infinite order, while the remaining basis elements are roots of unity.	Thus this means:\n\nThere are \( r + 1 \) units \( \zeta ,{\eta }_{1},{\eta }_{2},\ldots ,{\eta }_{r} \), where \( \zeta \) is a wth root of unity, such that each unit of the field is obtained exactly once in the form\n\n\[ \varepsilon = {\zeta }^{a}{\eta }_{1}^{{a}_{1}}\cdots {\eta }_{r}^{{a}_{r}} \]\n\nwhere \( {a}_{...	Yes
Theorem 102. If \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are basis elements of the ideal \( \mathfrak{a} \), then the \( n \) sequences of numbers \( {\beta }_{1}^{\left( p\right) },\ldots ,{\beta }_{n}^{\left( p\right) }\left( {p = 1,\ldots, n}\right) \), which are defined by (65), are conjugate sequences of numbers ...	Since, moreover,\n\n\[ \n{\Delta }^{2}\left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) = \frac{1}{{\Delta }^{2}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) } = \frac{1}{d{N}^{2}\left( \mathfrak{a}\right) }\n\]\n\nand by (47)\n\n\[ \n{\Delta }^{2}\left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) = {N}^{2}\left( ...	No
Theorem 104. All numbers of the ideal \( \mathfrak{f} = {F}^{\prime }\left( \theta \right) /\mathfrak{d} \) belong to the ring \( R\left( \theta \right) \), and if all numbers of an ideal a belong to the ring \( R\left( \theta \right) \), then \( \mathfrak{a} \) is divisible by \( \mathfrak{f} \) .	If \( \omega \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{f}}\right) \), then \( \alpha = \omega /{F}^{\prime }\left( \theta \right) \) is a number with denominator \( \mathfrak{d} \), and by Lemma (a), \( \alpha {F}^{\prime }\left( \theta \right) \) must be a number of this ring. Hence the first part of our theorem...	Yes
Theorem 107. The ideal \( {N}_{k}\left( \mathfrak{A}\right) \) is an ideal in \( k \) . If \( k \) is the field of rational numbers, then \( {N}_{k}\left( \mathfrak{A}\right) = \left( {N\left( \mathfrak{A}\right) }\right) \) .	To begin let \( \mathfrak{A} = \left( {{A}_{1},\ldots ,{A}_{s}}\right) \) be an integral ideal, where the \( {A}_{i} \) are numbers in \( K \) . Then, by \( \$ {28} \), for any variables \( {u}_{1},\ldots ,{u}_{s} \), the content of the conjugate polynomials\n\n\[ \n{F}^{\left( i\right) }\left( u\right) = {A}_{1}^{\lef...	Yes
Theorem 108. For each prime ideal \( \mathfrak{P} \) of \( K \) there is exactly one prime ideal \( \mathfrak{p} \) in \( k \) which is divisible by \( \mathfrak{P} \) . Then\n\n\[ \n{N}_{k}\left( \mathfrak{P}\right) = {\mathfrak{p}}^{{f}_{1}} \n\]\n\nwhere \( {f}_{1} \) is a natural number \( \leq m.{f}_{1} \) is call...	By Theorem \( {107}{N}_{k}\left( \mathfrak{P}\right) \) is an ideal in \( k \) which, by definition, is divisible by \( \mathfrak{P} \) . If \( {N}_{k}\left( \mathfrak{P}\right) \) is decomposed into its prime factors, then by the fundamental theorem \( \mathfrak{P} \) must divide at least one of these prime ideals in ...	Yes
Theorem 109. If \( N \) denotes the norm in \( K \) and \( n \) denotes the norm in \( k \), then for each ideal \( \mathfrak{A} \) in \( K \)\n\n\[ N\left( \mathfrak{A}\right) = n\left( {{N}_{k}\left( \mathfrak{A}\right) }\right) \]	To begin with, this assertion follows immediately for each number \( A \) in \( K \), by (74). By Theorem 107, or also by consideration of the principal ideal \( {\mathfrak{A}}^{h} \), the result is also obtained for each ideal in \( K \) .	Yes
Theorem 110. If the relative degree of the prime ideal \( \mathfrak{P} \) is equal to 1, then each number in \( K \) is congruent modulo \( \mathfrak{P} \) to a number in \( k \) .	By Theorem 108, \( N\left( \mathfrak{P}\right) = n{\left( \mathfrak{p}\right) }^{{f}_{1}} \) ; hence for \( {f}_{1} = 1 \) the number of residue classes \( {\;\operatorname{mod}\;\mathfrak{P}} \) in \( K \) is equal to the number of residue classes \( {\;\operatorname{mod}\;\mathfrak{p}} \) in \( k \) . However if a nu...	Yes
Theorem 111. If \( \mathfrak{D} \) and \( \mathfrak{d} \) are the differents of \( K \) and \( k \) respectively, then for the relative differents \( {\mathfrak{D}}_{k} \) the relation\n\n\[ \mathfrak{D} = {\mathfrak{D}}_{k}\mathfrak{d} \]\n\nholds.	Proof. If \( \Delta \) is a number in \( K \) such that \( \Delta {\mathfrak{D}}_{k}\mathfrak{d} \) is integral, then by the definition of \( {\mathfrak{D}}_{k} \),\n\n\[ \mathfrak{d}{S}_{k}\left( {\Delta A}\right) \text{is integral} \]\n\nfor each integer \( A \), since for each number \( \xi \) in \( k \) which is di...	Yes
Theorem 112. The relative different of \( K \) is the GCD of all relative differents of integers of \( K \) with respect to \( k \) .	For the proof of this theorem we must proceed almost exactly as in the proof of Theorem 105. If \( \theta \) is an integer generating the field \( K \), then the relative ring \( {R}_{k}\left( \theta \right) \) is the set of all numbers \[ {\alpha }_{0} + {\alpha }_{1}\theta + \cdots + {\alpha }_{m - 1}{\theta }^{m - 1...	Yes
Theorem 114. If a prime ideal \( \mathfrak{P} \) in \( K \) divides a prime ideal \( \mathfrak{p} \) in \( k \) to a higher power than the first, then \( \mathfrak{P} \) is a factor of the relative different of \( K \) with respect to \( k \) . Thus there can only exist finitely many prime ideals \( \mathfrak{P} \) of ...	For a proof, let the decomposition of \( \mathfrak{p} \) in \( K \) be\n\n\[ \mathfrak{p} = {\mathfrak{P}}^{e}\mathfrak{A},\;\text{ where }\left( {\mathfrak{A},\mathfrak{P}}\right) = 1, e \geq 2. \]\n\nFor each integer \( A \) in \( K \) we now have, by the often used properties of the binomial coefficients \( \left( \...	Yes
Theorem 115. Suppose that \( K \) is identical with all relative conjugate fields with respect to \( k \) (that is, suppose that \( K \) is a relative Galois field). Then the only prime ideals of \( K \) dividing the relative different \( {\mathfrak{D}}_{k} \) of \( K \) are those which divide a prime ideal of \( k \) ...	For the proof of Theorem 115 it suffices, by Theorem 112, to display a number \( A \) in \( K \) whose different is not divisible by this \( \mathfrak{P} \) . We choose \( A \) to be a primitive root \( {\;\operatorname{mod}\;\mathfrak{P}} \) which is divisible by \( \mathfrak{p}{\mathfrak{P}}^{-1} \) . Now, by the abo...	Yes
Theorem 116. If \( \mu \) is a number in \( k \) which is not the lth power of a number in \( k \), then the field \( K\left( {\sqrt[1]{\mu };k}\right) \) has the relative degree \( l \) with respect to \( k \) . The field \( K\left( {\sqrt[1]{\mu };k}\right) \) is identical with its relative-conjugate fields.	The number \( M = \sqrt[1]{\mu } \) (suppose that the value of the root is somehow fixed) satisfies the equation \( {x}^{i} - \mu = 0 \), whose roots are the \( l \) numbers\n\n\[{\zeta }^{a}M\;\left( {a = 0,1,\ldots, l - 1}\right) .\n\]\n\nIn any case all relative conjugates of \( M \) must occur among them. Let these...	Yes
Theorem 117. Only the following possibilities exist for the behavior of a prime ideal \( \mathfrak{p} \) of \( k \) under passage to \( K \) :\n\n\( \mathfrak{p} \) remains a prime ideal in \( K \) ,\n\n\( \mathfrak{p} \) becomes the lth power of a prime ideal in \( k \) ,\n\n\( \mathfrak{p} \) becomes the product of \...	Let \( \mathfrak{P} \) be a prime ideal in \( K \) which divides \( \mathfrak{p} \) . Then, by Theorem 107, the relative norm of \( \mathfrak{P} \) is\n\n\[ \mathfrak{P} \cdot s\mathfrak{P}\cdots {s}^{l - 1}\mathfrak{P} = {\mathfrak{p}}^{{f}_{1}}, \]\n\nwhere \( {f}_{1} \) is the relative degree of \( \mathfrak{P} \) ;...	Yes
Theorem 119. Let 1 be a prime factor of \( 1 - \zeta \), which divides \( 1 - \zeta \) exactly to the ath power: \( 1 - \zeta = {\mathrm{I}}^{a}{\mathrm{I}}_{1} \). Suppose that I does not divide \( \mu \). Then I splits into \( l \) factors which are distinct from one another in \( K\left( {\sqrt[1]{\mu };k}\right) \)...	I: The solvability of (82) is identical with the decomposition of \( 1 \) into distinct factors in \( K \). Namely from \( \mathfrak{l} = \mathfrak{L} \cdot s\mathfrak{L}\cdots {s}^{l - 1}\mathfrak{L} \), where the conjugates are distinct from one another, it follows as in the proof of Theorem 110 that every integer in...	Yes
Theorem 122. Let \( Z\left( t\right) \) denote the number of integral ideals of the field whose norm is \( \leq t \) . Then\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{Z\left( t\right) }{t} = h\varkappa \]\n\nwhere \( h \) is the class number of the field.	If \( F\left( m\right) \) denotes the number of integral ideals of the field whose norm is equal to the positive number \( m \), then obviously\n\n\[ Z\left( t\right) = \mathop{\sum }\limits_{{m = 1}}^{t}F\left( m\right) \]\n\nHere \( \mathop{\sum }\limits_{{m = 1}}^{t} \) means that the summation index \( m \) runs th...	Yes
Theorem 123. \( {\zeta }_{k}\left( s\right) \) is defined by the convergent series (92) for \( s > 1 \) as a continuous function of \( s \) and as \( s \) approaches 1	\[ \mathop{\lim }\limits_{{s \rightarrow 1}}\left( {s - 1}\right) {\zeta }_{k}\left( s\right) = h\varkappa \]	No
Theorem 124. For \( s > 1 \), the equation\n\n\[ \n{\zeta }_{k}\left( s\right) = \mathop{\prod }\limits_{\mathfrak{p}}\frac{1}{1 - \frac{1}{N{\left( \mathfrak{p}\right) }^{s}}}.\n\]\n\nholds where \( \mathfrak{p} \) runs through all distinct prime ideals \( \mathfrak{p} \) of \( k \) .	To begin with, this product converges, since \( \mathop{\sum }\limits_{\mathfrak{p}}1/N{\left( \mathfrak{p}\right) }^{s} \) converges as the constituent of the series for \( {\zeta }_{k}\left( s\right) \) . For a single factor we obtain a convergent series of positive terms\n\n\[ \n\frac{1}{1 - N{\left( \mathfrak{p}\ri...	Yes
Theorem 125. Following Dedekind the determination of the class number \( h \) is reduced to the determination of the prime ideals of the field by the equation\n\n\[ h \cdot \varkappa = \mathop{\lim }\limits_{{s \rightarrow 1}}\left( {s - 1}\right) \mathop{\prod }\limits_{\mathfrak{p}}\frac{1}{1 - \frac{1}{N{\left( \mat...	This fundamental fact is only another way of writing (88), as has been mentioned already; however, it is more convenient as a starting point for the further calculation than the former equation.	No
Theorem 126. If \( {\mathfrak{p}}_{1} \) runs through the distinct prime ideals of degree one in \( k \) , then, for \( s > 1 \) ,\n\n\[ \mathop{\sum }\limits_{{\mathfrak{p}}_{1}}\frac{1}{N{\left( {\mathfrak{p}}_{1}\right) }^{s}} = \log \frac{1}{s - 1} + {g}_{1}\left( s\right) \]\n\nwhere \( {g}_{1}\left( s\right) \) a...	Proof: Let \( {\mathfrak{p}}_{f} \) run through the distinct prime ideals of degree \( f \) for \( f = \) \( 1,2,\ldots, n \) . (Of course \( {\mathfrak{p}}_{f} \) need not exist for each \( f \) .) Since at most \( n \) distinct prime ideals of \( k \) divide a given rational prime \( p \) then, in any case, for \( s ...	Yes
Theorem 127. There are infinitely many positive rational primes with the property \( p \equiv 1\\left( {\\;\\operatorname{mod}\\;m}\\right) \) .	If \( {n}_{0} \) is the degree of the field of \( m \) th roots of unity (which by \( \\$ {30} \) is no larger than \( \\varphi \\left( m\\right) \\) ), then exactly \( {n}_{0} \) distinct prime ideals of \( k \) divide such a number \( p \), and Equation (97) thus reads\n\n\\[ \n{n}_{0}\\mathop{\\sum }\\limits_{{p \\e...	No
Theorem 128. If \( \chi \left( n\right) \) denotes a residue character of \( n{\;\operatorname{mod}\;m} \), then the Dirichlet series\n\n\[ L\left( {s,\chi }\right) = \mathop{\sum }\limits_{{n = 1}}^{\pi }\frac{\chi \left( n\right) }{{n}^{s}} \]\n\nis absolutely convergent for \( s > 1 \) and as long as \( s > 1 \) we ...	First of all the absolute convergence of the series and the product representation for \( s > 1 \) are obtained immediately from the fact that the coefficients \( \chi \left( n\right) \) are not larger than 1 in absolute value, as \( \chi \left( n\right) \) is either a root of unity, or, in case \( \left( {n, m}\right)...	Yes
Theorem 129. For each character \( \chi {\;\operatorname{mod}\;m} \), if \( s > 1 \) , \[ \log L\left( {s,\chi }\right) = \mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}} + g\left( {s,\chi }\right) \] where \( g\left( {s,\chi }\right) \) remains bounded as \( s \) approaches 1 .	If we define the log function, for \( s > 0 \), by the convergent series \[ \log \frac{1}{1 - \frac{\chi \left( p\right) }{{p}^{s}}} = \frac{\chi \left( p\right) }{{p}^{s}} + \frac{1}{2}\frac{\chi \left( {p}^{2}\right) }{{p}^{2s}} + \frac{1}{3}\frac{\chi \left( {p}^{3}\right) }{{p}^{3s}} + \cdots = \frac{\chi \left( p\...	Yes
Theorem 130. If \( \chi \) is not the principal character, then\n\n\[ L\left( {1,\chi }\right) = \mathop{\lim }\limits_{{s \rightarrow 1}}L\left( {s,\chi }\right) \neq 0. \]	The nonvanishing of the \( L \) -series is now an immediate consequence of the fact that \( {\zeta }_{k}\left( s\right) \) becomes infinite to the first order at \( s = 1 \) . For by (102) it follows for \( a = b = 1 \) that\n\n\[ \mathop{\sum }\limits_{\chi }\log L\left( {s,\chi }\right) = \varphi \left( m\right) \mat...	Yes
Theorem 132. The basis number of the strict class group belonging to 2 is \( {e}_{0}\left( 2\right) = t - 1 \), where \( t \) denotes the number of distinct primes which divide the discriminant \( d \) of \( k \) .	By Theorem 28, we must show that there exist exactly \( {2}^{t - 1} \) classes in \( k \) whose square is the strict principal class. For this purpose we keep in mind that the \( t \) distinct prime ideals \( {\mathfrak{q}}_{1},\ldots ,{\mathfrak{q}}_{t} \), which divide \( d \), have the property that their square is ...	Yes
Theorem 134. If \( d \) is the product of two positive primes \( {q}_{1},{q}_{2} \), which are \( \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), then either \( {q}_{1} \) or \( {q}_{2} \) is the norm of a principal ideal in the strict sense in \( k\left( \sqrt{{q}_{1}{q}_{2}}\right) \) .	To begin with the norm of each unit \( = + 1 \) in such a field. For from \( N\left( \alpha \right) = - 1 \) for \( \alpha = \left( {x + y\sqrt{{q}_{1}{q}_{2}}}\right) /2 \) it would follow that\n\n\[ - 4 \equiv {x}^{2}\left( {{\;\operatorname{mod}\;{q}_{1}}{q}_{2}}\right) \]\n\nthus -1 would be a quadratic residue mod...	Yes
Theorem 135. If \( p \) and \( q \) are odd positive primes, then we have the relations\n\n(I) \( \left( \frac{-1}{p}\right) = {\left( -1\right) }^{\left( {p - 1}\right) /2} \),\n\n(II) \( \left( \frac{p}{q}\right) = \left( \frac{q}{p}\right) {\left( -1\right) }^{\left( {\left( {p - 1}\right) /2}\right) \left( {\left( ...	We obtain the first formula directly from the definition of the residue symbol, Equation (31) in §16. We can also deduce it from field theory, in a somewhat more involved way, but analogous to the subsequent proof of (II) and (III) as follows: If \( \left( \frac{-1}{p}\right) = + 1 \), then \( p \) splits in \( k\left(...	Yes
Theorem 137. If \( d \) is the discriminant of a quadratic field and \( n, m \) are positive integers, then\n\n\[ \left( \frac{d}{n}\right) = \left( \frac{d}{m}\right) \;\text{ if }n \equiv m\left( {\;\operatorname{mod}\;d}\right) \]\n\n(124)\n\n\[ \left( \frac{d}{n}\right) = \left( \frac{d}{m}\right) \cdot \operatorna...	For the proof we must split off the highest power of 2 dividing \( d, n, m \) . Let\n\n\[ d = {2}^{a}{d}^{\prime },\;n = {2}^{b}{n}^{\prime },\;m = {2}^{c}{m}^{\prime } \]\n\nwith odd \( {d}^{\prime },{n}^{\prime },{m}^{\prime } \) .\n\nCase 1: \( a > 0 \) . The case \( b > 0 \) is trivial here, since then, by hypothes...	Yes
Theorem 138. If the odd prime \( p \) does not divide the discriminant \( d \), then each rational integer relatively prime to \( p \) is a norm residue \( {\;\operatorname{mod}\;p} \) for \( k\left( \sqrt{d}\right) \) .	We distinguish two cases in the proof.\n\n1. \( p \) splits into two distinct factors \( \mathfrak{p} \) and \( {\mathfrak{p}}^{\prime } \), of degree 1, in \( k\left( \sqrt{d}\right) \) . Then there is a number \( \pi \) in \( k\left( \sqrt{d}\right) \) which is divisible by \( \mathfrak{p} \) but not by \( {\mathfrak...	Yes
Theorem 139. If the odd prime \( q \) divides the discriminant \( d \) of \( k\left( \sqrt{d}\right) \), then exactly one half of the classes of \( \Re \left( q\right) \) are norm residues mod \( q \), and indeed these are the classes of \( \Re \left( q\right) \) which can be represented as the square of a class.	If \( \mathfrak{q} \) is the prime ideal of \( k\left( \sqrt{d}\right) \) which divides \( q \), then each number \( \alpha \) in \( k \) is congruent to a rational number \( {\;\operatorname{mod}\;q} \), say \( r \) . However, since \( q = {q}^{\prime } \) it follows from \( \alpha \equiv r\left( {\;\operatorname{mod}...	Yes
Theorem 140. If the discriminant \( d \) of \( k\left( \sqrt{d}\right) \) is odd, then each odd number is a norm residue \( {\;\operatorname{mod}\;8} \) . However, if \( d \) is even, then exactly half of all incongruent odd numbers \( {\;\operatorname{mod}\;8} \) are norm residues \( {\;\operatorname{mod}\;8} \) .	For the proof we test the residue classes in \( k\left( \sqrt{d}\right) {\;\operatorname{mod}\;8} \) . We find, with \( \alpha = x + y\sqrt{d} \) and \( d \) odd, that\n\n\[ x = 0,1,2,1 \]\n\n\[ y = 1,0,1,2 \]\n\n\[ N\left( \alpha \right) \equiv 3,1,7,5\left( {\;\operatorname{mod}\;8}\right) ,\;\text{ if }d \equiv 5\le...	Yes
Theorem 142. The \( t \) functions \( {\gamma }_{i}\left( \mathfrak{a}\right) \) are group characters of the genus represented by \( \mathfrak{a} \) .	Now by \( §{10} \), the group of characters of an Abelian group is isomorphic to the group. There are \( u \) independent elements in the group of genera and no more, because this group is of order \( {2}^{u} \) and each element has at most order 2. Consequently, there are also exactly \( u \) independent characters. H...	No
Theorem 143. At least one relation must hold for all ideals \( \mathfrak{a} \) of the field which are coprime to \( d \), namely,\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{t}{\gamma }_{i}^{{c}_{i}}\left( \mathfrak{a}\right) = 1 \]\n\nwhere the rational integers \( {c}_{i} \) are independent of \( \mathfrak{a} \) and are ...	Thus for \( t = 1 \) the equation\n\n\[ {\gamma }_{1}\left( \mathfrak{a}\right) = {\chi }_{1}\left( \left\| {N\left( \mathfrak{a}\right) }\right\| \right) = 1 \]\n\nmust hold. In fact this is exactly one part of the quadratic reciprocity law, which has not been used until now (in §47 and 48). We see that the proof of thi...	No
Theorem 144. Let \( {e}_{1},{e}_{2},\ldots ,{e}_{t} \) be t numbers \( \pm 1 \) such that \( {e}_{1} \cdot {e}_{2}\cdots {e}_{t} = 1 \) . Then there are infinitely many prime ideals \( \mathfrak{p} \) of degree 1 in \( k\left( \sqrt{d}\right) \) for which \[ {\gamma }_{i}\left( \mathfrak{p}\right) = {e}_{i}\;\left( {i ...	If we set \( N\left( \mathfrak{p}\right) = p \), then the assertion obviously states that there are infinitely many rational primes \( p \) which satisfy the conditions and \[ {\chi }_{i}\left( p\right) = {e}_{i}\;\left( {i = 1,\ldots, t}\right) \] \[ \left( \frac{d}{p}\right) = + 1 \] By (136), the last condition is n...	Yes
Theorem 147. Let \( {a}_{1},{a}_{2},\ldots ,{a}_{r} \) be any rational integers such that a product of powers\n\n\[ \n{a}_{1}^{{u}_{1}}{a}_{2}^{{u}_{2}}\cdots {a}_{r}^{{u}_{r}} \n\]\n\nis a rational square only if all \( {u}_{i} \) are even. Moreover, let \( {c}_{1},{c}_{2},\ldots ,{c}_{r} \) take arbitrary values \( \...	For the proof we set, for the sake of symmetry,\n\n\[ \nL\left( {s;1}\right) = \mathop{\sum }\limits_{{p > 2}}\frac{1}{{p}^{s}} \n\]\n\n(a function which grows beyond all bounds as \( s \rightarrow 1 \) by \( §{43} \) ), and we form the sum\n\n\[ \n\mathop{\sum }\limits_{{{u}_{1},\ldots ,{u}_{r}}}{c}_{1}^{{u}_{1}}{c}_{...	Yes
Theorem 148. For each natural number \( n \) we have\n\n\[ F\left( n\right) = \mathop{\sum }\limits_{{m \mid n}}\left( \frac{d}{m}\right) \]\n\nwhere \( m \) runs through all distinct positive divisors of \( n \) .	If we decompose \( n \) into its distinct prime factors\n\n\[ n = {p}_{1}^{{k}_{1}} \cdot {p}_{2}^{{k}_{2}}\cdots {p}_{r}^{{k}_{r}} \]\n\nthen\n\n\[ F\left( n\right) = F\left( {p}_{1}^{{k}_{1}}\right) \cdot F\left( {p}_{2}^{{k}_{2}}\right) \cdots F\left( {p}_{r}^{{k}_{r}}\right) = \mathop{\prod }\limits_{{i = 1}}^{r}\m...	Yes
Theorem 149. The expression\n\n\[ \n{\varepsilon }^{2h} = {e}^{\sqrt{d}\mathop{\sum }\limits_{{n = 1}}^{\infty }\left( \frac{d}{n}\right) /n} \n\]\n\nrepresents a unit of infinite degree in \( k\left( \sqrt{d}\right) \) with \( d > 0 \) . And\n\n\[ \n{\varepsilon }^{2h} + {\varepsilon }^{\prime {2h}} = {\varepsilon }^{...	The rational integer \( A \) can thus also be calculated numerically by estimating the remainder of the convergent series \( L\left( 1\right) \) and with this we have a transcendental method for finding a unit in real quadratic fields.	Yes
Theorem 150. For each rational integer \( n \)\n\n\[ G\left( {n, d}\right) = \chi \left( n\right) G\left( {1, d}\right) ,\;{c}_{n} = \chi \left( n\right) \frac{G\left( {1, d}\right) }{d}. \]	For if \( n \) and \( d \) are not coprime, then by (154) both sides of the first equation are equal to 0 . However, if \( \left( {n, d}\right) = 1 \), then we choose \( c \) in (155) in such way that \( {cn} \equiv 1\left( {\;\operatorname{mod}\;d}\right) \) ; thus \( \chi \left( c\right) = \chi \left( n\right) \) .	No
Theorem 152. The class number \( h \) of the quadratic field with discriminant \( d \) has the value\n\n\[ h = - \frac{\rho }{\left\| d\right\| }\mathop{\sum }\limits_{{n = 1}}^{{\left\| d\right\| - 1}}n\left( \frac{d}{n}\right) ,\;\rho = \frac{-{iG}\left( {1, d}\right) }{\left\| \sqrt{d}\right\| } = \pm 1\;\text{ for }d < -...	In the second expression \( a \) and \( b \) run through those numbers \( 1,2,\ldots, d - 1 \) for which\n\n\[ \left( \frac{d}{a}\right) = - 1,\;\left( \frac{d}{b}\right) = + 1 \]\n\nrespectively. The final result will be that we always have \( \rho = + 1 \) (§58). The formula for the class number of an imaginary field...	Yes
Theorem 153. If \( D \) is the discriminant of \( F \) and \( D > 0 \), then \( F\left( {x, y}\right) \) may take positive as well as negative values for appropriate real values of \( x, y \) . If \( D < 0 \) , then either the value of \( F \) is \( \geq 0 \) for all real \( x, y \) or \( F\left( {x, y}\right) \leq 0 \...	For the proof we consider the decomposition\n\n\[ A \cdot F\left( {x, y}\right) = {\left( Ax + \frac{B}{2}y\right) }^{2} - \frac{D}{4}{y}^{2}. \]\n\nNow if \( D = {B}^{2} - {4AC} < 0 \), then we must have \( A \neq 0 \) and it follows from the equation that\n\n\[ {AF}\left( {x, y}\right) \geq 0, \]\n\nwhere the equalit...	Yes
Theorem 155. Suppose that the denominator \( \mathfrak{a} \) of \( \mathfrak{d}\omega \) is an odd ideal. Then for every integer \( \varkappa \) which is relatively prime to \( \mathfrak{a} \)\n\n\[ C\left( {\varkappa \omega }\right) = \left( \frac{\varkappa }{a}\right) C\left( \omega \right) \]	To begin with, the theorem is true if \( \mathfrak{a} \) is a prime ideal \( \mathfrak{p} \), since if we apply Lemma (a), we have\n\n\[ \mathop{\sum }\limits_{{\mu {\;\operatorname{mod}\;\mathfrak{p}}}}\left( \frac{\mu }{\mathfrak{p}}\right) {e}^{{2\pi iS}\left( {\mu \omega }\right) } = \mathop{\sum }\limits_{{\mu {\;...	Yes
Theorem 158. The n-tuple theta-function\n\n\\[ T\\left( {{u}_{1},\\ldots ,{u}_{n}}\\right) = \\mathop{\\sum }\\limits_{{{m}_{1},\\ldots ,{m}_{n} = - \\infty }}^{{+\\infty }}{e}^{-{\\pi Q}\\left( {{m}_{1} + {u}_{1},\\ldots ,{m}_{n} + {u}_{n}}\\right) }\n\\]\n\n---	admits the representation\n\n\\[ T\\left( {{u}_{1},\\ldots ,{u}_{n}}\\right) = \\mathop{\\sum }\\limits_{{{m}_{1},\\ldots ,{m}_{n} = - \\infty }}^{{+\\infty }}a\\left( \\left( m\\right) \\right) {e}^{-{2\\pi i}\\left( {{m}_{1}{u}_{1} + \\cdots + {m}_{n}{u}_{n}}\\right) }\n\\]\n\nwhere\n\n\\[ a\\left( \\left( m\\right) ...	Yes
Theorem 159. The theta-series defined by (175) also admits the representation\n\n\\( \\theta \\left( {t, z;\\mathfrak{a}}\\right) \\)\n\n\\[ = \\frac{1}{N\\left( \\mathfrak{a}\\right) \\left\| \\sqrt{d}\\right\| \\sqrt{{t}_{1} \\cdot {t}_{2}\\cdots {t}_{n}}}\\mathop{\\sum }\\limits_{{\\lambda \\text{ in }1/\\mathfrak{a}\...	We now see at once that this equation also holds for nonreal \\( t \\), provided only that the real part of each \\( {t}_{p} \\) is positive. For then the real part of \\( 1/{t}_{p} \\) is also positive and the series on both sides of the formula represent analytic functions of \\( {t}_{1},\\ldots ,{t}_{n} \\), which a...	Yes
Theorem 163. The reciprocity\n\n\\[ \frac{C\\left( \\omega \\right) }{\\left\| \\sqrt{N\\left( \\mathfrak{a}\\right) }\\right\| } = \\left\| \\frac{\\sqrt{N\\left( {2\\mathfrak{b}}\\right) }}{N\\left( {\\mathfrak{b}}_{1}\\right) }\\right\| {e}^{\\left( {{\\pi i}/4}\\right) S\\left( {\\operatorname{sgn}\\omega }\\right) }C\...	Thus if \\( a, b \\) are relatively prime rational integers, then\n\n\\[ C\\left( \\frac{b}{a}\\right) = \\mathop{\\sum }\\limits_{{n{\\;\\operatorname{mod}\\;a}}}{e}^{{2\\pi i}\\left( {{n}^{2}b/a}\\right) }.\\]\n\nFor odd \\( a \\) the reciprocity formula, Theorem 163, asserts\n\n\\[ \\frac{C\\left( \\frac{1}{a}\\righ...	Yes
Theorem 165 (Law of Quadratic Reciprocity). For two odd relatively prime integers \( \alpha ,\beta \) of which at least one is primary	\[ \left( \frac{\alpha }{\beta }\right) \cdot \left( \frac{\beta }{\alpha }\right) = {\left( -1\right) }^{\mathop{\sum }\limits_{{p = 1}}^{{r}_{1}}\left( {\left( {\operatorname{sgn}{\alpha }^{\left( p\right) } - 1}\right) /2}\right) \left( {\left( {\operatorname{sgn}{\beta }^{\left( p\right) } - 1}\right) /2}\right) } ...	Yes
Theorem 169. Let \( {\mu }_{1},{\mu }_{2},\ldots ,{\mu }_{m} \) be integers in \( k \) such that a product of powers \( {\mu }_{1}^{{x}_{1}}\cdots {\mu }_{m}^{{x}_{m}} \) is the square of a number in \( k \) only if all the exponents \( {x}_{1},\ldots ,{x}_{m} \) are even. Let \( {c}_{1},{c}_{2},\ldots ,{c}_{m} \) be a...	For by the hypothesis, the square root of each of the \( {2}^{m} - 1 \) products of powers \( \mu = {\mu }_{1}^{{x}_{1}}\cdots {\mu }_{m}^{{x}_{m}}\left( {{x}_{i} = 0}\right. \) or 1, not all \( \left. {{x}_{i} = 0}\right) \) defines a relative quadratic field \( K\left( {\sqrt{\mu }, k}\right) \) . However, it now obv...	Yes
Theorem 175. Let \( \lambda = \operatorname{Ir} \) be a totally positive number, \( \mathrm{r} \) an odd ideal, and let I be a prime factor of 2 . In order that the primary integer \( \alpha \), which is relatively prime to \( \lambda \), be hyperprimary, it is necessary and sufficient that\n\n\[ \chi \left( \alpha \ri...	Theorem 167 asserts that the condition is necessary. The proof of the preceding theorem shows in the following way that it is sufficient. First suppose that \( 1 \) is equivalent in the strict sense to the square of an ideal. Then we can find integers \( \beta ,\rho ,\lambda \) such that\n\n\[ \lambda {\beta }^{2} = {\...	Yes
Theorem 179. Let \( \omega \) be a singular primary number. Then there is a subgroup \( \mathfrak{G}\left( \omega \right) \) of order \( {h}_{0}/2 \) in the group of the \( {h}_{0} \) ideal classes in the strict sense such that a prime ideal \( \mathfrak{p} \) splits in the field \( K\left( {\sqrt{\omega }, k}\right) \...	The set of odd ideals \( \mathfrak{r} \), for which \( Q\left( {\omega ,\mathfrak{r}}\right) = + 1 \), determines a subgroup of order \( {2}^{{e}_{0} - 1} \) in the group of class complexes in the strict sense, by Theorem 172. Since each class complex consists of \( {h}_{0}/{2}^{{e}_{0}} \) classes in the strict sense,...	Yes
Theorem 0.2. If \( \mathcal{S} \) is a semi-algebra of subsets of \( X \) and \( \tau : \mathcal{S} \rightarrow {R}^{ + } \) is finitely additive then there is a unique finitely additive function \( {\tau }_{1} : \mathcal{L}\left( \mathcal{S}\right) \rightarrow {R}^{ + } \) which is an extension of \( \tau \) (i.e. \( ...	It could be that \( X \notin \mathcal{S} \) but we always have that \( X \) is a disjoint union of a finite number of members \( {E}_{1},\ldots ,{E}_{n} \) of \( \mathcal{S} \) so \( {\tau }_{1}\left( X\right) = 1 \) if \( \mathop{\sum }\limits_{{i = 1}}^{n}\tau \left( {E}_{i}\right) = 1 \) .	No
Theorem 0.5. Fix \( k \geq 1 \) and let \( Y = \{ 0,1,\ldots, k - 1\} \) and \( \left( {X,\mathcal{D}}\right) = \mathop{\prod }\limits_{{j = x}}^{\infty }(Y \) , \( \left. {2}^{Y}\right) \) . For each natural number \( n \) and \( {a}_{0},\ldots ,{a}_{n} \in Y \) suppose a non-negative real number \( {p}_{n}\left( {{a}...	The proof boils down to showing that the function naturally defined on the algebra of all finite unions of elementary rectangles is countable additive by using Theorem 0.4.	Yes
Theorem 0.7. Let \( \left( {X,\mathcal{B}, m}\right) \) be a probability space and let \( \mathcal{A} \) be an algebra of subsets of \( X \) with \( \mathcal{B}\left( \mathcal{A}\right) = \mathcal{B} \) . Then for each \( \varepsilon > 0 \) and each \( B \in \mathcal{B} \) there is some \( A \in \mathcal{A} \) with \( ...	Note that when \( m\left( {A\bigtriangleup B}\right) < \varepsilon \) then \( \left\| {m\left( A\right) - m\left( B\right) }\right\| < \varepsilon \) because \( m\left( A\right) = \) \( m\left( {A \smallsetminus B}\right) + m\left( {A \cap B}\right) \) and \( m\left( B\right) = m\left( {B \smallsetminus A}\right) + m\lef...	No
Theorem 0.12. Let \( \left( {X,\mathcal{D}, m}\right) \) be a probability space and let \( \mathcal{C} \) be a sub- \( \sigma \) -algebra of \( \mathcal{B} \) . The restriction of the conditional expectation operator \( E\left( {\cdot /\mathcal{C}}\right) \) to \( {L}^{2}\left( {X,\mathcal{B}, m}\right) \) is the ortho...	Proof. For \( f \in {L}^{1}\left( {X,\mathcal{B}, m}\right) \) we know \( E\left( {f/\mathcal{C}}\right) \) is the only \( \mathcal{C} \) -measurable function \( h \) such that \( {\int }_{C}{hdm} = {\int }_{C}{fdm}\forall C \in \mathcal{C} \) . Let \( P \) denote the orthogonal projection of \( {L}^{2}\left( {X,\mathc...	Yes
Theorem 0.13. Let \( G \) be a compact topological group. There exists a probability measure \( m \) defined on the \( \sigma \) -algebra \( \mathcal{B}\left( G\right) \) of Borel subsets of \( G \) such that \( m\left( {xE}\right) = m\left( E\right) \forall x \in G\forall E \in \mathcal{B}\left( G\right) \) and \( m \...	This unique measure is called Haar measure. Notice we have required Haar measure to be a probability measure. The Haar measure \( m \) also satisfies \( m\left( {Ex}\right) = m\left( E\right) \forall x \in G,\forall E \in \mathcal{B}\left( G\right) \), because for each fixed \( x \in G \) the measure \( {m}_{x} \) defi...	Yes
Theorem 0.17. Let \( A \) be an irreducible and aperiodic non-negative matrix. Let\n\n\[ u = \left( {{u}_{1},\ldots ,{u}_{k}}\right) ,\;v = \left( \begin{matrix} {v}_{1} \\ \vdots \\ {v}_{k} \end{matrix}\right) \]\n\nbe the strictly positive eigenvectors corresponding to the largest eigenvalue \( \lambda \) as in Theor...	This theorem follows from the renewal theorem which we shall need for another purpose (see Feller [1], p. 291).	No
Theorem 0.19. Let \( X \) be a compact Hausdorff space. The following are equivalent:\n\n(1) \( X \) is metrisable.\n\n(2) \( X \) has a countable base.\n\n(3) \( C\left( X\right) \) (the space of all complex-valued continuous functions on \( X \) ) has a countable dense subset.	See Kelley [1] for the proof.	No
Theorem 0.20 (Lebesgue Covering Lemma). If \( \\left( {X, d}\\right) \) is a compact metric . space and \( \\alpha \) is an open cover of \( X \) then there exists a \( \\delta > 0 \) such that each subset of \( X \) of diameter less than or equal to \( \\delta \) lies in some member of \( \\alpha \) . (Such a \( \\dot...	Proof. We may as well suppose \( \\alpha \) is a finite cover.\n\nLet \( \\alpha = \\left\\{ {{A}_{1},\\ldots ,{A}_{p}}\\right\\} \) . Assume the theorem is false. Then for each \( n \\geq 1 \) there exists \( {B}_{n} \\subseteq X \) such that \( \\operatorname{diam}\\left( {B}_{n}\\right) \\leq 1/n \) and \( {B}_{n} \...	Yes
Theorem 1.1. Suppose \( \left( {{\lambda }_{1},{\mathcal{B}}_{1},{m}_{1}}\right) ,\left( {{X}_{2},{\mathcal{D}}_{2},{m}_{2}}\right) \) are probability spaces and \( T : {X}_{1} \rightarrow {X}_{2} \) is a transformation. Let \( {\mathcal{S}}_{2} \) be a semi-algebra which generates - \( {\mathcal{B}}_{2} \) . If for ea...	Proof. Let \( {\mathcal{C}}_{2} = \left\{ {B \in {\mathcal{B}}_{2} : {T}^{-1}\left( B\right) \in {\mathcal{B}}_{1},{m}_{1}\left( {{T}^{-1}\left( B\right) }\right) = {m}_{2}\left( B\right) }\right\} \) . We want to show that \( {\mathcal{C}}_{2} = {\mathcal{D}}_{2} \) . However \( {\mathcal{S}}_{2} \subseteq {\mathcal{C...	Yes
Lemma 1.2. Let \( \left( {{X}_{i},{\mathcal{A}}_{i},{m}_{i}}\right) i = 1,2 \) be a probability space and let \( T : {X}_{1} \rightarrow {X}_{2} \) be measure-preserving. If \( F \in {L}^{0}\left( {{X}_{2},{\mathcal{B}}_{2},{m}_{2}}\right) \) then \( \int {U}_{T}{Fd}{m}_{1} = \int {Fd}{m}_{2} \) (where if one side does...	Proof. It suffices to prove the result when \( F \) is real-valued and, by considering positive and negative parts of \( F \), it suffices to consider non-negative functions. So suppose \( F \geq 0 \) . If \( F \) is a simple function then the result is true because \( T \) is measure-preserving. Choose simple function...	Yes
Theorem 1.3. Let \( p \geq 1 \) . With the above notation \( {U}_{T}{L}^{P}\left( {{X}_{2},{\mathcal{B}}_{2},{m}_{2}}\right) \subset \) \( {L}^{P}\left( {{X}_{1},{\mathcal{B}}_{1},{m}_{1}}\right) \) and \( {\begin{Vmatrix}{U}_{T}f\end{Vmatrix}}_{p} = \parallel f{\parallel }_{p}\forall f \in {L}^{P}\left( {{X}_{2},{\mat...	Proof. Let \( f \in {L}^{p}\left( {{X}_{2},{\mathcal{B}}_{2},{m}_{2}}\right) \) . Put \( F\left( x\right) = {\left\| f\left( x\right) \right\| }^{p} \) in Lemma 1.2 to get \( {\begin{Vmatrix}{U}_{T}f\end{Vmatrix}}_{p} = \parallel f{\parallel }_{p}. \) Therefore a measure-preserving transformation \( T : {X}_{1} \rightarr...	Yes
Theorem 1.4 (Poincaré’s Recurrence Theorem). Let \( T : X \rightarrow X \) be a measure-preserving transformation of a probability space \( \left( {X,\mathcal{B}, m}\right) \) . Let \( E \in \mathcal{B} \) with \( m\left( E\right) > 0 \) . Then almost all points of \( E \) return infinitely often to \( E \) under posit...	Proof. For \( N \geq 0 \) let \( {E}_{N} = \mathop{\bigcup }\limits_{{n = N}}^{\prime }{T}^{-n}E \) . Then \( \mathop{\bigcap }\limits_{{N = 0}}^{\prime }{E}_{N} \) is the set of all points of \( X \) which enter \( E \) infinitely often under positive iteration by \( T \) . Hence the set \( F = E \cap \mathop{\bigcap ...	Yes
Theorem 1.5. If \( T : X \rightarrow X \) is a measure-preserving transformation of the probability space \( \left( {X,\mathcal{A}, m}\right) \) then the following statements are equivalent:\n\n(i) \( T \) is ergodic.\n\n(u) The only members \( B \) of \( \mathcal{B} \) with \( m\left( {{T}^{-1}B\bigtriangleup B}\right...	## Proof\n\n(i) \( \Rightarrow \) (ii) Let \( B \in \mathcal{B} \) and \( m\left( {{T}^{-1}B\bigtriangleup B}\right) = 0 \) . We shall construct a set \( B \), with \( {T}^{-1}{B}_{r} = B \), and \( m\left( {B\bigtriangleup {B}_{r}}\right) = 0 \) . For each \( n \geq 0 \) we have \( m\left( {{T}^{-n}B\bigtriangleup B}\...	No
Theorem 1.6. If \( \left( {X,\mathcal{A}, m}\right) \) is a probability space and \( T : X \rightarrow X \) is measure-preserving then the following statements are equivalent:\n\n(i) \( T \) is ergodic.\n\n(ii) Whenever \( f \) is measurable and \( \left( {f \cdot T}\right) \left( x\right) = f\left( x\right) \forall x ...	Proof. Trivially we have (iii) \( \Rightarrow \) (ii),(ii) \( \Rightarrow \) (iv),(v) \( \Rightarrow \) (iv), and (iii) \( \Rightarrow \) (v). So it remains to show (i) \( \Rightarrow \) (iii) and (iv) \( \Rightarrow \) (i). We first show (i) \( \Rightarrow \) (iii). Let \( T \) be ergodic and suppose \( f \) is measur...	Yes
Theorem 1.7. Let \( X \) be a compact metric space, \( \mathcal{B}\left( X\right) \) the \( \sigma \) -algebra of Borel subsets of \( X \) and let \( m \) be a probability measure on \( \left( {X,\mathcal{B}\left( X\right) }\right) \) such that \( m\left( U\right) > 0 \) for every non-empty open set \( U \) . Suppose \...	Proof. Let \( {\left\{ {U}_{n}\right\} }_{n = 1}^{\infty } \) be a base for the topology of \( X \) . Then \( \left\{ {{T}^{n}x \mid n \geq 0}\right\} \) is dense in \( X \) iff \( x \in \mathop{\bigcap }\limits_{{n = 1}}^{\infty }\mathop{\bigcup }\limits_{{k = 0}}^{\infty }{T}^{-k}{U}_{n} \) . Since \( {T}^{-1}\left( ...	Yes
Theorem 1.8. The rotation \( T\left( z\right) = {az} \) of the unit circle \( K \) is ergodic (relative to Haar measure \( m \) ) iff \( a \) is not a root of unity.	Proof. Suppose \( a \) is a root of unity, then \( {a}^{p} = 1 \) for some \( p \neq 0 \) . Let \( f\left( z\right) = {z}^{p} \) . Then \( f \circ T = f \) and \( f \) is not constant a.e. Therefore \( T \) is not ergodic by Theorem 1.6(ii). Conversely, suppose \( a \) is not a root of unity and \( f \circ T = f \) , \...	Yes
Theorem 1.9. Let \( G \) be a compact group and \( T\left( x\right) = {ax} \) a rotation of \( G \) . Then \( T \) is ergodic iff \( \left\{ {a}^{n}\right\} = 0 \) is dense in \( G \) . In particular, if \( T \) is ergodic, then \( G \) is abelian.	Proor. Suppose \( T \) is ergodic. Let \( H \) denote the closure of the subgroup \( \left\{ {a}^{n}\right\} = c \) of \( G \) . If \( H \neq G \) then by (7) of \( \$ {0.7} \) there exists \( \gamma \in \widehat{G} \) with \( \gamma ≢ 1 \) but \( \gamma \left( h\right) = 1\forall h \in H \) . Then \( \gamma \left( {Tx...	Yes
Theorem 1.10. If \( G \) is a compact abelian group (equipped with normalized Haar measure) and \( A : G \rightarrow G \) is a surjective continuous endomorphism of \( G \) then \( A \) is ergodic iff the trivial character \( \gamma \equiv 1 \) is the only \( \gamma \in \widehat{G} \) that satisfies \( \gamma \circ {A}...	Proof. Suppose that whenever \( \gamma {A}^{n} = \gamma \) for some \( n \geq 1 \) we have \( \gamma \equiv 1 \) . Let \( f \circ A = f \) with \( f \in {L}^{2}\left( m\right) \) . Let \( f\left( x\right) \) have the Fourier series \( \sum {a}_{n}{\gamma }_{n} \) where \( {\gamma }_{n} \in \widehat{G} \) and \( \sum {\...	Yes