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Lemma 10.3.3. Let \( \gamma \left( s\right) \) be a gamma product. Denote by \( W\left( t\right) \) the inverse Mellin transform of \( \gamma \left( s\right) \), in other words,\n\n\[ W\left( t\right) = \frac{1}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }{t}^{-z}\gamma \left( z\right) {dz} \]\n\nand let \( {... | Proof. (1). Since \( \gamma \left( z\right) \) decreases exponentially when \( \left| {\operatorname{Im}\left( z\right) }\right| \) tends to infinity with \( \operatorname{Re}\left( z\right) \) fixed, it is clear that the integral defining \( W\left( t\right) \) converges absolutely. For the same reason, by integrating... | Yes |
Theorem 10.3.4. For \( i = 1 \) and \( i = 2 \), let \( {L}_{i}\left( s\right) = \mathop{\sum }\limits_{{n \geq 1}}{a}_{i}\left( n\right) {n}^{-s} \) be Dirichlet series such that the \( {a}_{i}\left( n\right) \) have at most polynomial growth (or, equivalently, the series \( {L}_{i}\left( s\right) \) converge in some ... | Proof. Before beginning the proof itself, we will make a few remarks about convergence. First, it is clear that \( {L}_{i}\left( s\right) \) converges in some right half-plane if and only if \( {a}_{i}\left( n\right) \) has at most polynomial growth. Since the functions \( {\Lambda }_{i}\left( s\right) \) have only a f... | Yes |
(1) For all fixed \( r \in \mathbb{R} \), there exists \( e\left( r\right) \) such that as \( \left| T\right| \rightarrow \infty \), we have\n\n\[ \n{L}_{i}\left( {r + {iT}}\right) = O\left( {\left| T\right| }^{e\left( r\right) }\right) \n\] | Proof. We prove (1) for \( {L}_{1} \), the result following by symmetry. For \( r > \sigma \) , we have \( {L}_{1}\left( {r + {iT}}\right) = O\left( 1\right) \) since the series converges absolutely. For \( r < k - \sigma \) , we apply the functional equation, which gives us\n\n\[ \n{L}_{1}\left( {r + {iT}}\right) = w{... | Yes |
For any rational function \( \phi \left( z\right) \) that tends to 0 when \( \left| z\right| \) tends to \( \infty \), we have the identity\n\n\[ \mathop{\sum }\limits_{{a \neq s}}{\operatorname{Res}}_{z = a}\left( \frac{\phi \left( z\right) }{s - z}\right) = \phi \left( s\right) \]\n\nwhere the sum is over all the pol... | Proof. Let \( R \) be a real positive number larger than the modulus of the poles of \( \phi \left( z\right) \) and of \( \left| s\right| \), and let \( {C}_{R} \) be the circle of radius \( R \) centered at the origin. By the residue theorem we have\n\n\[ \mathop{\sum }\limits_{{a \neq s}}{\operatorname{Res}}_{z = a}\... | Yes |
Lemma 2. Let \( \mathfrak{D} \) be a bounded \( {}^{ * } \) -derivation on \( \mathfrak{A} = \mathcal{L}\left( H\right) \) . Then there exists a skew-adjoint operator \( A \in \mathcal{L}\left( H\right) \) such that \[ \mathfrak{D}\left( T\right) = {AT} - {TA}\;\text{ for all }T \in \mathcal{L}\left( H\right) . \] | Proof. For each pair of elements \( x, y \in H \) we define the rank-one operator \( x \otimes y \) by \[ z \mapsto x \otimes y\left( z\right) \mathrel{\text{:=}} \left( {x \mid z}\right) y. \] Take now some fixed \( z \in H,\parallel z\parallel = 1 \), and define \( A \in \mathcal{L}\left( H\right) \) by \[ {Ay} \math... | Yes |
Proposition 2. The resolvent \( R\left( {\lambda, A}\right) \) for \( \operatorname{Re}\lambda > 0 \) of the differentiation operator \( A \) with maximal domain \( D\left( A\right) \) (i.e., of the generator of the left translation semigroup) on any of the above spaces \( X \) is given by | \[ \left( {R\left( {\lambda, A}\right) f}\right) \left( s\right) = {\int }_{s}^{\infty }{\mathrm{e}}^{-\lambda \left( {\tau - s}\right) }f\left( \tau \right) {d\tau }\;\text{ for }f \in X, s \in \mathbb{R}. \] | Yes |
For every \( f \in \mathrm{C}\left\lbrack {0,1}\right\rbrack \), we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{B}_{n}f = f \] | This lemma, which also re-proves the Weierstrass Approximation Theorem 4.11, is one of the fundamental results of classical approximation theory and can be proved in many different ways. A very elegant proof uses Korovkin's theorem, which assures that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{B}_{n}f = f\;\... | Yes |
Lemma 1. The semigroup \( {\left( T\left( t\right) \right) }_{t \geq 0} \) satisfies\n\n\[ T\left( t\right) f\left( s\right) = \left\{ \begin{array}{ll} {\mathrm{e}}^{{\int }_{s}^{0}m\left( \sigma \right) {d\sigma }}\left\lbrack {{\mathrm{e}}^{\left( {s + t}\right) m\left( 0\right) }f\left( 0\right) }\right. & \\ \left... | Proof. For \( f \in D\left( A\right) \) we have\n\n\[ \frac{d}{dr}\left( {{\mathrm{e}}^{{rm}\left( 0\right) }\left( {T\left( {t - r}\right) f}\right) \left( 0\right) + {\int }_{0}^{r}{\mathrm{e}}^{{\tau m}\left( 0\right) }{LT}\left( {t - \tau }\right) {fd\tau }}\right) = 0. \]\n\nThis implies\n\n\[ \left( {T\left( t\ri... | Yes |
Lemma 2. If \( m\left( {-\infty }\right) < 0 \), then the semigroup \( {\left( T\left( t\right) \right) }_{t \geq 0} \) is quasi-compact. | Proof. We define operators \( K\left( t\right) \in \mathcal{L}\left( X\right) \) by\n\n\[ K\left( t\right) f\left( s\right) \mathrel{\text{:=}} \left\{ \begin{array}{ll} {\mathrm{e}}^{{\int }_{s}^{0}m\left( \sigma \right) {d\sigma }}\left\lbrack {{\mathrm{e}}^{\left( {s + t}\right) m\left( 0\right) }f\left( 0\right) }\... | Yes |
Lemma 3. Suppose that \( m\left( {-\infty }\right) < 0 \) . If \( \xi \left( 0\right) \leq 0 \), i.e., \( L{g}_{0} \geq - m\left( 0\right) \), then the characteristic function \( \xi \) has a unique zero \( {\lambda }_{0} \geq 0 \) that is a dominant eigenvalue of the operator \( A \) . | Proof. The function \( \xi : {\mathbb{R}}_{ + } \ni \lambda \mapsto \lambda - {L}_{0}{g}_{\lambda } - a - m\left( 0\right) \) is strictly increasing from \( \xi \left( 0\right) \) to \( \infty \) . Consequently, if \( \xi \left( 0\right) \leq 0 \), it has a unique zero \( {\lambda }_{0} \) that is an eigenvalue of \( A... | Yes |
Theorem 1.1 (Division algorithm) Let \( a \) and \( d \) be integers with \( d \geq 1 \) . There exist unique integers \( q \) and \( r \) such that\n\n\[ a = {dq} + r \]\n\n(1.1)\n\nand\n\n\[ 0 \leq r \leq d - 1 \]\n\n(1.2)\n\n\n\nThe integer \( q \) is called the quotient and the integer \( r \) is called the remaind... | Proof. Consider the set \( S \) of nonnegative integers of the form\n\n\[ a - {dx} \]\n\nwith \( x \in \mathbf{Z} \) . If \( a \geq 0 \), then \( a = a - d \cdot 0 \in S \) . If \( a < 0 \), let \( x = - y \), where \( y \) is a positive integer. Since \( d \) is positive, we have \( a - {dx} = a + {dy} \in S \) if \( ... | Yes |
Theorem 1.3 Let \( H \) be a subgroup of the integers under addition. There exists a unique nonnegative integer \( d \) such that \( H \) is the set of all multiples of \( d \), that is,\n\n\[ H = \{ 0, \pm d, \pm {2d},\ldots \} = d\mathbf{Z}. \] | Proof. We have \( 0 \in H \) for every subgroup \( H \) . If \( H = \{ 0\} \) is the zero subgroup, then we choose \( d = 0 \) and \( H = 0\mathbf{Z} \) . Moreover, \( d = 0 \) is the unique generator of this subgroup.\n\nIf \( H \neq \{ 0\} \), then there exists \( a \in H, a \neq 0 \) . Since \( - a \) also belongs t... | No |
Theorem 1.4 Let \( A \) be a nonempty set of integers, not all zero. Then \( A \) has a unique greatest common divisor, and there exist integers \( {a}_{1},\ldots ,{a}_{k} \in \) A and \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ \gcd \left( A\right) = {a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} \] | Proof. Let \( H \) be the subset of \( \mathbf{Z} \) consisting of all integers of the form\n\n\[ {a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k}\;\text{ with }{a}_{1},\ldots ,{a}_{k} \in A\text{ and }{x}_{1},\ldots ,{x}_{k} \in \mathbf{Z}. \]\n\nThen \( H \) is a subgroup of \( \mathbf{Z} \) and \( A \subseteq H \) . By Theo... | Yes |
Theorem 1.5 Let \( {a}_{1},\ldots ,{a}_{k} \) be integers, not all zero. Then \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) = \) 1 if and only if there exist integers \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ \n{a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} = 1.\n\] | Proof. This follows immediately from Theorem 1.4. \( ▱ \) | No |
Theorem 1.6 Let \( a \) and \( b \) be integers with \( b \geq 1 \) . If the Euclidean algorithm for \( a \) and \( b \) has length \( n \) with sequence of partial quotients \( {q}_{0},{q}_{1},\ldots ,{q}_{n - 1} \) , then \[ \frac{a}{b} = \left\langle {{q}_{0},{q}_{1},\ldots ,{q}_{n - 1}}\right\rangle \] | Proof. Let \( {r}_{0} = a \) and \( {r}_{1} = b \) . The proof is by induction on \( n \) . If \( n = 1 \) , then \[ {r}_{0} = {r}_{1}{q}_{0} \] and \[ \frac{a}{b} = \frac{{r}_{0}}{{r}_{1}} = {q}_{0} = \left\langle {q}_{0}\right\rangle \] If \( n = 2 \), then \[ {r}_{0} = {r}_{1}{q}_{0} + {r}_{2} \] \[ {r}_{1} = {r}_{2... | Yes |
Theorem 1.7 (Euclid’s lemma) Let \( a, b, c \) be integers. If \( a \) divides \( {bc} \) and \( \left( {a, b}\right) = 1 \), then a divides \( c \) . | Proof. Since \( a \) divides \( {bc} \), we have \( {bc} = {aq} \) for some integer \( q \) . Since \( a \) and \( b \) are relatively prime, Theorem 1.5 implies that there exist integers \( x \) and \( y \) such that\n\n\[ 1 = {ax} + {by}. \]\n\nMultiplying by \( c \), we obtain\n\n\[ c = {acx} + {bcy} = {acx} + {aqy}... | Yes |
Theorem 1.8 Let \( k \geq 2 \), and let \( a,{b}_{1},{b}_{2},\ldots ,{b}_{k} \) be integers. If \( \left( {a,{b}_{i}}\right) = 1 \) for all \( i = 1,\ldots, k \), then \( \left( {a,{b}_{1}{b}_{2}\cdots {b}_{k}}\right) = 1 \) . | Proof. The proof is by induction on \( k \) . Let \( k = 2 \) and \( d = \left( {a,{b}_{1}{b}_{2}}\right) \) . We must show that \( d = 1 \) . Since \( d \) divides \( a \) and \( \left( {a,{b}_{1}}\right) = 1 \), it follows that \( \left( {d,{b}_{1}}\right) = 1 \) . Since \( d \) divides \( {b}_{1}{b}_{2} \), Euclid’s... | Yes |
Theorem 1.9 If a prime number \( p \) divides a product of integers, then \( p \) divides one of the factors. | Proof. Let \( {b}_{1},{b}_{2},\ldots ,{b}_{k} \) be integers such that \( p \) divides \( {b}_{1}\cdots {b}_{k} \) . By Theorem 1.8, we have \( \left( {p,{b}_{i}}\right) > 1 \) for some \( i \) . Since \( p \) is prime, it follows that \( p \) divides \( {b}_{i} \) . | Yes |
Theorem 1.10 (Fundamental theorem of arithmetic) Every positive integer can be written uniquely (up to order) as the product of prime numbers. | Proof. First we prove that every positive integer can be written as a product of primes. Since an empty product is equal to 1 , we can write 1 as the empty product of primes. Let \( n \geq 2 \) . Suppose that every positive integer less than \( n \) is a product of primes. If \( n \) is prime, we are done. If \( n \) i... | Yes |
Theorem 1.11 Let \( {a}_{1},\ldots ,{a}_{k} \) be positive integers. Then\n\n\[\left( {{a}_{1},\ldots ,{a}_{k}}\right) = \mathop{\prod }\limits_{p}{p}^{\min \left\{ {{v}_{p}\left( {a}_{1}\right) ,\ldots ,{v}_{p}\left( {a}_{k}\right) }\right\} }\]\n\nand\n\n\[\left\lbrack {{a}_{1},\ldots ,{a}_{k}}\right\rbrack = \mathop... | Proof. This follows immediately from the fundamental theorem of arithmetic. \( ▱ \) | No |
Theorem 1.12 For every positive integer \( n \) and prime \( p \) , \[ {v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{r = 1}}^{\left\lbrack \frac{\log n}{\log p}\right\rbrack }\left\lbrack \frac{n}{{p}^{r}}\right\rbrack . \] | Proof. Let \( 1 \leq m \leq n \) . If \( {p}^{r} \) divides \( m \), then \( {p}^{r} \leq m \leq n \) and \( r \leq \) \( \log n/\log p \) . Since \( r \) is an integer, we have \( r \leq \left\lbrack {\log n/\log p}\right\rbrack \) and \[ {v}_{p}\left( m\right) = \mathop{\sum }\limits_{\substack{{r = 1} \\ {{p}^{r} \m... | Yes |
Theorem 1.13 Let \( m \) and a be nonzero integers. There exists a positive integer \( k \) such that \( m \) divides \( {a}^{k} \) if and only if \( \operatorname{rad}\left( m\right) \) divides \( \operatorname{rad}\left( a\right) \) . | Proof. We know that \( m \) divides \( {a}^{k} \) if and only if \( {v}_{p}\left( m\right) \leq {v}_{p}\left( {a}^{k}\right) = \) \( k{v}_{p}\left( a\right) \) for every prime \( p \) (Exercise 14). If there exists an integer \( k \) such that \( m \) divides \( {a}^{k} \), then \( {v}_{p}\left( a\right) > 0 \) wheneve... | No |
Theorem 1.14 (Euclid’s theorem) There are infinitely many primes. | Proof. Let \( {p}_{1},\ldots ,{p}_{n} \) be any finite set of prime numbers. Consider the integer\n\n\[ N = {p}_{1}\cdots {p}_{n} + 1 \]\n\nSince \( N > 1 \), it follows from the fundamental theorem of arithmetic that \( N \) is divisible by some prime \( p \) . If \( p = {p}_{i} \) for some \( i = 1,\ldots, n \), then... | Yes |
Theorem 1.15 Let \( {a}_{1},\ldots ,{a}_{k} \) be integers, not all zero. For any integer \( b \) , there exist integers \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ \n{a}_{1}{x}_{1} + \cdots + {a}_{k}{x}_{k} = b \n\]\n\nif and only if \( b \) is a multiple of \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) . In particu... | Proof. Let \( d = \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) . If equation (1.4) is solvable in integers \( {x}_{i} \) , then \( d \) divides \( b \) since \( d \) divides each integer \( {a}_{i} \) . Conversely, if \( d \) divides \( b \), then \( b = {dq} \) for some integer \( q \) . By Theorem 1.4, there exist inte... | Yes |
Theorem 1.16 Let \( {a}_{1},\ldots ,{a}_{k} \) be positive integers such that\n\n\[ \left( {{a}_{1},\ldots ,{a}_{k}}\right) = 1 \]\n\nIf\n\n\[ b \geq \left( {{a}_{k} - 1}\right) \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}{a}_{i} \]\n\nthen there exist nonnegative integers \( {x}_{1},\ldots ,{x}_{k} \) such that\n\n\[ {a}... | Proof. By Theorem 1.15, there exist integers \( {z}_{1},\ldots ,{z}_{k} \) such that\n\n\[ {a}_{1}{z}_{1} + \cdots + {a}_{k}{z}_{k} = b. \]\n\nUsing the division algorithm, we can divide each of the integers \( {z}_{1},\ldots ,{z}_{k - 1} \) by \( {a}_{k} \) so that\n\n\[ {z}_{i} = {a}_{k}{q}_{i} + {x}_{i} \]\n\nand\n\... | Yes |
Theorem 1.17 Let \( {a}_{1} \) and \( {a}_{2} \) be relatively prime positive integers. Then\n\n\[ G\left( {{a}_{1},{a}_{2}}\right) = \left( {{a}_{1} - 1}\right) \left( {{a}_{2} - 1}\right) . \] | Proof. We saw in the proof of Theorem 1.15 that for every integer \( b \) there exist integers \( {x}_{1} \) and \( {x}_{2} \) such that\n\n\[ b = {a}_{1}{x}_{1} + {a}_{2}{x}_{2}\;\text{ and }\;0 \leq {x}_{1} \leq {a}_{2} - 1. \]\n\n(1.5)\n\nIf we have another representation\n\n\[ b = {a}_{1}{x}_{1}^{\prime } + {a}_{2}... | Yes |
Theorem 2.2 Let \( m, a, b \) be integers with \( m \geq 1 \) . Let \( d = \left( {a, m}\right) \) be the greatest common divisor of \( a \) and \( m \) . The congruence\n\n\[ \n{ax} \equiv b\;\left( {\;\operatorname{mod}\;m}\right) \n\]\n\n(2.1)\n\nhas a solution if and only if\n\n\[ \nb \equiv 0\;\left( {\;\operatorn... | Proof. Let \( d = \left( {a, m}\right) \) . Congruence (2.1) has a solution if and only if there exist integers \( x \) and \( y \) such that\n\n\[ \n{ax} - b = {my} \n\]\n\nor, equivalently,\n\n\[ \nb = {ax} - {my}. \n\]\n\nBy Theorem 1.15, this is possible if and only if \( b \equiv 0\left( {\;\operatorname{mod}\;d}\... | Yes |
Theorem 2.3 If \( p \) is a prime, then \( \mathbf{Z}/p\mathbf{Z} \) is a field. | Proof. If \( a + p\mathbf{Z} \in \mathbf{Z}/p\mathbf{Z} \) and \( a + p\mathbf{Z} \neq p\mathbf{Z} \), then \( a \) is an integer not divisible by \( p \) . By Theorem 2.2, there exists an integer \( x \) such that \( {ax} \equiv 1 \) \( \left( {\;\operatorname{mod}\;p}\right) \) . This implies that\n\n\[ \left( {a + p... | Yes |
Lemma 2.1 Let \( p \) be a prime number. Then \( {x}^{2} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) if and only if \( x \equiv \pm 1\;\left( {\;\operatorname{mod}\;p}\right) \) . | Proof. If \( x \equiv \pm 1\;\left( {\;\operatorname{mod}\;p}\right) \), then \( {x}^{2} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) . Conversely, if \( {x}^{2} \equiv 1 \) \( \left( {\;\operatorname{mod}\;p}\right) \), then \( p \) divides \( {x}^{2} - 1 = \left( {x - 1}\right) \left( {x + 1}\right) \), and s... | Yes |
Theorem 2.4 (Wilson) If \( p \) is prime, then\n\n\[ \left( {p - 1}\right) ! \equiv - 1\;\left( {\;\operatorname{mod}\;p}\right) \] | Proof. This is true for \( p = 2 \) and \( p = 3 \), since \( 1! \equiv - 1\left( {\;\operatorname{mod}\;2}\right) \) and \( 2! \equiv - 1\;\left( {\;\operatorname{mod}\;3}\right) \) . Let \( p \geq 5 \) . By Theorem 2.2, to each integer \( a \in \) \( \{ 1,2,\ldots, p - 1\} \) there is a unique integer \( {a}^{-1} \in... | Yes |
Theorem 2.5 Let \( m \) and \( d \) be positive integers such that \( d \) divides \( m \) . If a is an integer relatively prime to \( d \), then there exists an integer \( {a}^{\prime } \) such that \( {a}^{\prime } \equiv a\;\left( {\;\operatorname{mod}\;d}\right) \) and \( {a}^{\prime } \) is relatively prime to \( ... | Proof. Let \( m = \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i}^{{r}_{i}} \) and \( d = \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i}^{{s}_{i}} \), where \( {r}_{i} \geq 1 \) and \( 0 \leq {s}_{i} \leq {r}_{i} \) for \( i = 1,\ldots, k \) . Let \( {m}^{\prime } \) be the product of the prime powers that divide \( m \) but ... | Yes |
Theorem 2.6 Let \( m \) and \( n \) be relatively prime positive integers. For every integer \( c \) there exist unique integers \( a \) and \( b \) such that\n\n\[ 0 \leq a \leq n - 1 \]\n\n\[ 0 \leq b \leq m - 1 \]\n\nand\n\n\[ c \equiv {ma} + {nb}\;\left( {\;\operatorname{mod}\;{mn}}\right) .\n\]\n\nMoreover, \( \le... | Proof. If \( {a}_{1},{a}_{2},{b}_{1},{b}_{2} \) are integers such that\n\n\[ m{a}_{1} + n{b}_{1} \equiv m{a}_{2} + n{b}_{2}\;\left( {\;\operatorname{mod}\;{mn}}\right) ,\]\n\nthen\n\n\[ m{a}_{1} \equiv m{a}_{1} + n{b}_{1} \equiv m{a}_{2} + n{b}_{2} \equiv m{a}_{2}\;\left( {\;\operatorname{mod}\;n}\right) .\n\]\n\nSince... | Yes |
Theorem 2.7 The Euler phi function is multiplicative. Moreover, | Proof. Let \( \\left( {m, n}\\right) = 1 \) . There are \( \\varphi \\left( {mn}\\right) \) congruence classes in the ring \( \\mathbf{Z}/{mn}\\mathbf{Z} \) that are relatively prime to \( {mn} \) . By Theorem 2.6, every congruence class modulo \( {mn} \) can be written uniquely in the form \( {ma} + {nb} + {mn}\\mathb... | Yes |
Theorem 2.8 For every positive integer \( m \) , \n\n\[ \n\mathop{\sum }\limits_{{d \mid m}}\varphi \left( d\right) = m \n\] | Proof. We first consider the case where \( m = {p}^{t} \) is a power of a prime \( p \) . The divisors of \( {p}^{t} \) are \( 1, p,{p}^{2},\ldots ,{p}^{t} \), and \n\n\[ \n\mathop{\sum }\limits_{{d \mid {p}^{t}}}\varphi \left( d\right) = \mathop{\sum }\limits_{{r = 0}}^{t}\varphi \left( {p}^{r}\right) = 1 + \mathop{\s... | Yes |
Theorem 2.9 Let \( m \) and \( n \) be positive integers. For any integers \( a \) and \( b \) there exists an integer \( x \) such that\n\n\[ x \equiv a\;\left( {\;\operatorname{mod}\;m}\right) \]\n\nand\n\n\[ x \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nif and only if\n\n\[ a \equiv b\;\left( {\;\operato... | Proof. If \( x \) is a solution of congruence (2.5), then \( x = a + {mu} \) for some integer \( u \) . If \( x \) is also a solution of congruence (2.6), then\n\n\[ x = a + {mu} \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nthat is,\n\n\[ a + {mu} = b + {nv} \]\n\nfor some integer \( v \) . It follows that\n... | Yes |
Theorem 2.10 (Chinese remainder theorem) Let \( k \geq 2 \) . If \( {a}_{1},\ldots ,{a}_{k} \) are integers and \( {m}_{1},\ldots ,{m}_{k} \) are pairwise relatively prime positive integers, then there exists an integer \( x \) such that\n\n\[ \nx \equiv {a}_{i}\;\left( {\;\operatorname{mod}\;{m}_{i}}\right) \;\text{ f... | Proof. We prove the theorem by induction on \( k \) . If \( k = 2 \), then \( \left\lbrack {{m}_{1},{m}_{2}}\right\rbrack = \) \( {m}_{1}{m}_{2} \), and this is a special case of Theorem 2.9.\n\nLet \( k \geq 3 \), and assume that the statement is true for \( k - 1 \) congruences. Then there exists an integer \( z \) s... | Yes |
Theorem 2.11 Let\n\n\\[ \nm = {p}_{1}^{{r}_{1}}\\cdots {p}_{k}^{{r}_{k}}\n\\]\n\nbe the standard factorization of the positive integer \\( m \\) . Let \\( f\\left( x\\right) \\) be a polynomial with integral coefficients. The congruence\n\n\\[ \nf\\left( x\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;m}\\right... | Proof. If \\( f\\left( x\\right) \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;m}\\right) \\) has a solution in integers, then there exists an integer \\( a \\) such that \\( m \\) divides \\( f\\left( a\\right) \\) . Since \\( {p}_{i}^{{r}_{i}} \\) divides \\( m \\), it follows that \\( {p}_{i}^{{r}_{i}} \\) divides \... | Yes |
Theorem 2.12 (Euler) Let \( m \) be a positive integer, and let a be an integer relatively prime to \( m \) . Then\n\n\[ \n{a}^{\varphi \left( m\right) } \equiv 1\;\left( {\;\operatorname{mod}\;m}\right) \n\] | Proof. Let \( \left\{ {{r}_{1},\ldots ,{r}_{\varphi \left( m\right) }}\right\} \) be a reduced set of residues modulo \( m \) . Since \( \left( {a, m}\right) = 1 \), we have \( \left( {a{r}_{i}, m}\right) = 1 \) for \( i = 1,\ldots ,\varphi \left( m\right) \) . Consequently, for every \( i \in \{ 1,\ldots ,\varphi \lef... | Yes |
Theorem 2.13 (Fermat) Let \( p \) be a prime number. If the integer \( a \) is not divisible by \( p \), then\n\n\[ \n{a}^{p - 1} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nMoreover,\n\n\[ \n{a}^{p} \equiv a\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nfor every integer \( a \) . | Proof. If \( p \) is prime and does not divide \( a \), then \( \left( {a, p}\right) = 1,\varphi \left( p\right) = p - 1 \) , and\n\n\[ \n{a}^{p - 1} = {a}^{\varphi \left( p\right) } \equiv 1\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nby Euler’s theorem. Multiplying this congruence by \( a \), we obtain\n\n\[ \n{a... | Yes |
Theorem 2.14 Let \( m \) be a positive integer and a an integer relatively prime to \( m \) . If \( d \) is the order of a modulo \( m \), then \( {a}^{k} \equiv {a}^{\ell }\;\left( {\;\operatorname{mod}\;m}\right) \) if and only if \( k \equiv \ell \;\left( {\;\operatorname{mod}\;d}\right) \) . In particular, \( {a}^{... | Proof. Since \( a \) has order \( d \) modulo \( m \), we have \( {a}^{d} \equiv 1\left( {\;\operatorname{mod}\;m}\right) \) . If \( k \equiv \ell \;\left( {\;\operatorname{mod}\;d}\right) \), then \( k = \ell + {dq} \), and so\n\n\[ \n{a}^{k} = {a}^{\ell + {dq}} = {a}^{\ell }{\left( {a}^{d}\right) }^{q} \equiv {a}^{\e... | Yes |
Theorem 2.15 (Lagrange’s theorem) If \( G \) is a finite group and \( H \) is a subgroup of \( G \), then the order of \( H \) divides the order of \( G \) . | Proof. Let \( G \) be a group, written multiplicatively, and let \( X \) be a nonempty subset of \( G \) . For every \( a \in G \) we define the set\n\n\[ \n{aX} = \{ {ax} : x \in X\} .\n\]\n\nThe map \( f : X \rightarrow {aX} \) defined by \( f\left( x\right) = {ax} \) is a bijection, and so \( \left| X\right| = \) \(... | Yes |
Theorem 2.16 Let \( G \) be a finite group, and \( a \in G \) . Then the order of the element a divides the order of the group \( G \) . | Proof. This follows immediately from Theorem 2.15, since the order of \( a \) is the order of the cyclic subgroup that \( a \) generates. \( ▱ \) | Yes |
Theorem 2.17 Let \( G \) be a cyclic group of order \( m \), and let \( H \) be a subgroup of \( G \). If \( a \) is a generator of \( G \), then there exists a unique divisor \( d \) of \( m \) such that \( H \) is the cyclic subgroup generated by \( {a}^{d} \), and \( H \) has order \( m/d \). | Proof. Let \( S \) be the set of all integers \( u \) such that \( {a}^{u} \in H \). If \( u, v \in S \), then \( {a}^{u},{a}^{v} \in H \). Since \( H \) is a subgroup, it follows that \( {a}^{u}{a}^{v} = {a}^{u + v} \in H \) and \( {a}^{u}{\left( {a}^{v}\right) }^{-1} = {a}^{u - v} \in H \). Therefore, \( u \pm v \in ... | Yes |
Theorem 2.18 Let \( G \) be a cyclic group of order \( m \), and let \( a \) be a generator of \( G \) . For every integer \( k \), the cyclic subgroup generated by \( {a}^{k} \) has order \( m/d \), where \( d = \left( {m, k}\right) \), and \( \left\langle {a}^{k}\right\rangle = \left\langle {a}^{d}\right\rangle \) . ... | Proof. Since \( d = \left( {k, m}\right) \), there exist integers \( x \) and \( y \) such that \( d = \) \( {kx} + {my} \) . Then\n\n\[ \n{a}^{d} = {a}^{{kx} + {my}} = {\left( {a}^{k}\right) }^{x}{\left( {a}^{m}\right) }^{y} = {\left( {a}^{k}\right) }^{x}, \n\]\n\nand so \( {a}^{d} \in \left\langle {a}^{k}\right\rangl... | Yes |
Theorem 2.19 Let \( m \) be an integer that is the product of two prime numbers. The prime divisors of \( m \) are the roots of the quadratic equation\n\n\[ {x}^{2} - \left( {m + 1 - \varphi \left( m\right) }\right) x + m = 0, \]\n\nand so \( \varphi \left( m\right) \) determines the prime factors of \( m \) . | Proof. If \( m = {pq} \), then\n\n\[ \varphi \left( m\right) = \left( {p - 1}\right) \left( {q - 1}\right) = {pq} - p - q + 1 = m - p - \frac{m}{p} + 1, \]\n\nand so\n\n\[ p - \left( {m + 1 - \varphi \left( m\right) }\right) + \frac{m}{p} = 0. \]\n\nEquivalently, \( p \) and \( q \) are the solutions of the quadratic e... | Yes |
Theorem 3.1 (Division algorithm for polynomials) Let \( F \) be a field. If \( f\left( x\right) \) and \( d\left( x\right) \) are polynomials in \( F\left\lbrack x\right\rbrack \) and if \( d\left( x\right) \neq 0 \), then there exist unique polynomials \( q\left( x\right) \) and \( r\left( x\right) \) such that \( f\l... | Proof. Let \( d\left( x\right) = {b}_{m}{x}^{m} + \cdots + {b}_{1}x + {b}_{0} \), where \( {b}_{m} \neq 0 \) and \( \deg \left( d\right) = \) \( m \) . If \( d\left( x\right) \) does not divide \( f\left( x\right) \), then \( f - {dq} \neq 0 \) and \( \deg \left( {f - {dq}}\right) \) is a nonnegative integer for every ... | Yes |
Theorem 3.2 Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack, f\left( x\right) \neq 0 \), and let \( {N}_{0}\left( f\right) \) denote the number of distinct zeros of \( f\left( x\right) \) in \( F \) . Then \( {N}_{0}\left( f\right) \) does not exceed the degree of \( f\left( x\right) \), that is,\n\n\[ \n{N}_{... | Proof. We use the division algorithm for polynomials. Let \( \alpha \in F \) . Dividing \( f\left( x\right) \) by \( x - \alpha \), we obtain\n\n\[ \nf\left( x\right) = \left( {x - \alpha }\right) q\left( x\right) + r\left( x\right) \n\]\n\nwhere \( r\left( x\right) = 0 \) or \( \deg \left( r\right) < \deg \left( {x - ... | Yes |
Theorem 3.3 Let \( G \) be a finite subgroup of the multiplicative group of a field. Then \( G \) is cyclic. | Proof. Let \( \left| G\right| = m \) . By Theorem 2.15, if \( a \in G \), then the order of \( a \) is a divisor of \( m \) . For every divisor \( d \) of \( m \), let \( \psi \left( d\right) \) denote the number of elements of \( G \) of order \( d \) . If \( \psi \left( d\right) \neq 0 \), then there exists an elemen... | Yes |
For every prime \( p \), the multiplicative group of the finite field \( \mathbf{Z}/p\mathbf{Z} \) is cyclic. This group has \( \varphi \left( {p - 1}\right) \) generators. Equivalently, for every prime \( p \), there exist \( \varphi \left( {p - 1}\right) \) pairwise incongruent primitive roots modulo \( p \) . | This follows immediately from Theorem 3.3, since \( \left| {\left( \mathbf{Z}/p\mathbf{Z}\right) }^{ \times }\right| = \) \( p - 1 \) . \( ▱ \) | No |
Theorem 3.5 Let \( m \) be a positive integer that is not a power of 2 . If \( m \) has a primitive root, then \( m = {p}^{k} \) or \( 2{p}^{k} \), where \( p \) is an odd prime and \( k \) is a positive integer. | Proof. Let \( a \) and \( m \) be integers such that \( \left( {a, m}\right) = 1 \) and \( m \geq 3 \) . Suppose that\n\n\[ m = {m}_{1}{m}_{2},\;\text{ where }\left( {{m}_{1},{m}_{2}}\right) = 1\text{ and }{m}_{1} \geq 3,{m}_{2} \geq 3. \]\n\n(3.2)\n\nThen \( \left( {a,{m}_{1}}\right) = \left( {a,{m}_{2}}\right) = 1 \)... | No |
Theorem 3.6 Let \( p \) be an odd prime, and let \( a \neq \pm 1 \) be an integer not divisible by \( p \) . Let \( d \) be the order of a modulo \( p \) . Let \( {k}_{0} \) be the largest integer such that \( {a}^{d} \equiv 1\left( {\;\operatorname{mod}\;{p}^{{k}_{0}}}\right) \) . Then the order of a modulo \( {p}^{k}... | Proof. There exists an integer \( {u}_{0} \) such that\n\n\[ \n{a}^{d} = 1 + {p}^{{k}_{0}}{u}_{0}\;\text{ and }\;\left( {{u}_{0}, p}\right) = 1.\n\]\n\n(3.3)\n\nLet \( 1 \leq k \leq {k}_{0} \), and let \( e \) be the order of \( a \) modulo \( {p}^{k} \) . If \( {a}^{e} \equiv 1\;\left( {\;\operatorname{mod}\;{p}^{k}}\... | Yes |
Theorem 3.7 Let \( p \) be an odd prime. If \( g \) is a primitive root modulo \( p \) , then either \( g \) or \( g + p \) is a primitive root modulo \( {p}^{k} \) for all \( k \geq 2 \) . If \( g \) is a primitive root modulo \( {p}^{k} \) and \( {g}_{1} \in \left\{ {g, g + {p}^{k}}\right\} \) is odd, then \( {g}_{1}... | Proof. Let \( g \) be a primitive root modulo \( p \) . The order of \( g \) modulo \( p \) is \( p - 1 \) . Let \( {k}_{0} \) be the largest integer such that \( {p}^{{k}_{0}} \) divides \( {g}^{p - 1} - 1 \) . By Theorem 3.6, if \( {k}_{0} = 1 \), then the order of \( g \) modulo \( {p}^{k} \) is \( \left( {p - 1}\ri... | Yes |
Theorem 3.8 There exists a primitive root modulo \( m = {2}^{k} \) if and only if \( m = 2 \) or 4 . | Proof. We note that 1 is a primitive root modulo 2, and 3 is a primitive root modulo 4 . We shall prove that if \( k \geq 3 \), then there is no primitive root modulo \( {2}^{k} \) . Since \( \varphi \left( {2}^{k}\right) = {2}^{k - 1} \), it suffices to show that\n\n\[ \n{a}^{{2}^{k - 2}} \equiv 1\;\left( {\;\operator... | Yes |
Theorem 3.9 For every positive integer \( k \) ,\n\n\[ \n{5}^{{2}^{k}} \equiv 1 + 3 \cdot {2}^{k + 2}\;\left( {\;\operatorname{mod}\;{2}^{k + 4}}\right) .\n\] | Proof. The proof is by induction on \( k \) . For \( k = 1 \) we have\n\n\[ \n{5}^{{2}^{1}} = {25} \equiv 1 + 3 \cdot {2}^{3}\;\left( {\;\operatorname{mod}\;{2}^{5}}\right) .\n\]\n\nSimilarly, for \( k = 2 \) we have\n\n\[ \n{5}^{{2}^{2}} = {625} = 1 + {48} + {576} \equiv 1 + 3 \cdot {2}^{4}\;\left( {\;\operatorname{mo... | Yes |
Theorem 3.10 If \( k \geq 3 \), then 5 has order \( {2}^{k - 2} \) modulo \( {2}^{k} \) . If \( a \equiv 1 \) \( \left( {\;\operatorname{mod}\;4}\right) \), then there exists a unique integer \( i \in \left\{ {0,1,\ldots ,{2}^{k - 2} - 1}\right\} \) such that\n\n\[ a \equiv {5}^{i}\;\left( {\;\operatorname{mod}\;{2}^{k... | Proof. In the case \( k = 3 \), we observe that 5 has order 2 modulo 8, and\n\n\[ 1 \equiv {5}^{0}\;\left( {\;\operatorname{mod}\;8}\right) \]\n\n\[ 3 \equiv - {5}^{1}\;\left( {\;\operatorname{mod}\;8}\right) \]\n\n\[ 5 \equiv {5}^{1}\;\left( {\;\operatorname{mod}\;8}\right) \]\n\n\[ 7 \equiv - {5}^{0}\left( {\;\operat... | Yes |
Theorem 3.11 Let \( p \) be prime, \( k \geq 2 \), and \( d = \left( {k, p - 1}\right) \) . Let a be an integer not divisible by \( p \) . Let \( g \) be a primitive root modulo \( p \), Then \( a \) is a kth power residue modulo \( p \) if and only if\n\n\[{\operatorname{ind}}_{g}\left( a\right) \equiv 0\;\left( {\;\o... | Proof. Let \( \ell = {\operatorname{ind}}_{g}\left( a\right) \), where \( g \) is a primitive root modulo \( p \) . Congruence (3.7) is solvable if and only if there exists an integer \( y \) such that\n\n\[{g}^{y} \equiv x\;\left( {\;\operatorname{mod}\;p}\right)\]\n\nand\n\n\[{g}^{ky} \equiv {x}^{k} \equiv a \equiv {... | Yes |
Theorem 3.12 Let \( p \) be an odd prime. For every integer \( a \) ,\n\n\[ \left( \frac{a}{p}\right) \equiv {a}^{\left( {p - 1}\right) /2}\;\left( {\;\operatorname{mod}\;p}\right) \] | Proof. If \( p \) divides \( a \), then both sides of the congruence are 0 . If \( p \) does not divide \( a \), then, by Fermat’s theorem,\n\n\[ {\left( {a}^{\left( {p - 1}\right) /2}\right) }^{2} \equiv {a}^{p - 1} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \]\n\nand so\n\n\[ {a}^{\left( {p - 1}\right) /2} \eq... | Yes |
Theorem 3.13 Let \( p \) be an odd prime, and let \( a \) and \( b \) be integers. Then\n\n\[ \left( \frac{ab}{p}\right) = \left( \frac{a}{p}\right) \left( \frac{b}{p}\right) \] | Proof. If \( p \) divides \( a \) or \( b \), then \( p \) divides \( {ab} \), and\n\n\[ \left( \frac{ab}{p}\right) = 0 = \left( \frac{a}{p}\right) \left( \frac{b}{p}\right) . \]\n\nIf \( p \) does not divide \( {ab} \), then, by Theorem 3.12,\n\n\[ \left( \frac{ab}{p}\right) \equiv {\left( ab\right) }^{\left( {p - 1}\... | Yes |
Theorem 3.14 Let \( p \) be an odd prime number. Then\n\n\[ \left( \frac{-1}{p}\right) = \begin{cases} 1 & \text{ if } & p \equiv 1 & \left( {\;\operatorname{mod}\;4}\right) , \\ - 1 & \text{ if } & p \equiv 3 & \left( {\;\operatorname{mod}\;4}\right) . \end{cases} \]\n\nEquivalently,\n\n\[ \left( \frac{-1}{p}\right) =... | Proof. We observe that\n\n\[ {\left( -1\right) }^{\left( {p - 1}\right) /2} = \begin{cases} 1 & \text{ if } & p \equiv 1 & \left( {\;\operatorname{mod}\;4}\right) , \\ - 1 & \text{ if } & p \equiv 3 & \left( {\;\operatorname{mod}\;4}\right) . \end{cases} \]\n\nApplying Theorem 3.12 with \( a = - 1 \), we obtain\n\n\[ \... | Yes |
Theorem 3.15 (Gauss's lemma) Let \( p \) be an odd prime, and a an integer not divisible by \( p \) . Let \( S \) be a Gaussian set modulo \( p \) . For every \( s \in S \) there exist unique integers \( {u}_{a}\left( s\right) \in S \) and \( {\varepsilon }_{a}\left( s\right) \in \{ 1, - 1\} \) such that\n\n\[ \n{as} \... | Proof. Since \( S \) is a Gaussian set, for every \( s \in S \) there exist unique integers \( {u}_{a}\left( s\right) \in S \) and \( {\varepsilon }_{a}\left( s\right) \in \{ 1, - 1\} \) such that\n\n\[ \n{as} \equiv {\varepsilon }_{a}\left( s\right) {u}_{a}\left( s\right) \;\left( {\;\operatorname{mod}\;p}\right) .\n\... | Yes |
Theorem 3.16 Let \( p \) be an odd prime. Then\n\n\[ \left( \frac{2}{p}\right) = \left\{ \begin{array}{lll} 1 & \text{ if }p \equiv \pm 1 & \left( {\;\operatorname{mod}\;8}\right) , \\ - 1 & \text{ if }p \equiv \pm 3 & \left( {\;\operatorname{mod}\;8}\right) . \end{array}\right. \]\n\nEquivalently,\n\n\[ \left( \frac{2... | Proof. We apply Gauss's lemma (Theorem 3.15) to the Gaussian set \( S = \{ 1,2,3,\ldots ,\left( {p - 1}\right) /2\} \) . Then\n\n\[ \{ {2s} : s \in S\} = \{ 2,4,6,\ldots, p - 1\} \]\n\nand\n\n\[ \left( \frac{2}{p}\right) = {\left( -1\right) }^{m} \]\n\nwhere \( m \) is the number of integers \( s \in S \) such that \( ... | Yes |
Theorem 3.18 Let \( R \) be a ring and \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \) a polynomial with coefficients in \( R \) . Then\n\n\[ f\left( {x + h}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) h + r\left( {x, h}\right) {h}^{2}. \]\n\nwhere \( r\left( {x, h}\right) \) is ... | Proof. This is a standard calculation. Expanding \( f\left( {x + h}\right) \) by the binomial theorem, we obtain\n\n\[ f\left( {x + h}\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{\left( x + h\right) }^{i} \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}\mathop{\sum }\limits_{{j = 0}}^{i}\left( \begin{array... | Yes |
Theorem 3.20 Let \( p \) be an odd prime, and let \( a \) be an integer not divisible by \( p \) . If \( a \) is a quadratic residue modulo \( p \), then \( a \) is a quadratic residue modulo \( {p}^{k} \) for every \( k \geq 1 \) . | Proof. Consider the polynomial \( f\left( x\right) = {x}^{2} - a \) and its derivative \( {f}^{\prime }\left( x\right) = \) \( {2x} \) . If \( a \) is a quadratic residue modulo \( p \), then there exists an integer \( {x}_{1} \) such that \( {x}_{1} ≢ 0\;\left( {\;\operatorname{mod}\;p}\right) \) and \( {x}_{1}^{2} \e... | Yes |
Theorem 4.1 Let \( G \) be a finite abelian group, written additively, and let \( \left| G\right| = m \) . For every prime number \( p \), let \( G\left( p\right) \) be the set of all elements \( g \in G \) whose order is a power of \( p \) . Then\n\n\[ G = {\bigoplus }_{p \mid m}G\left( p\right) \] | Proof. Let \( m = \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i}^{{r}_{i}} \) be the standard factorization of \( m \), and let \( {m}_{i} = m{p}_{i}^{-{r}_{i}} \) for \( i = 1,\ldots, k \) . Then \( \left( {{m}_{1},\ldots ,{m}_{k}}\right) = 1 \) by Exercise 15 in Section 1.4, and so there exist integers \( {u}_{1},\ldots... | No |
Lemma 4.1 Let \( G \) be a finite abelian p-group. Let \( {g}_{1} \in G \) be an element of maximum order \( {p}^{{r}_{1}} \), and let \( {G}_{1} = \left\langle {g}_{1}\right\rangle \) be the cyclic subgroup generated by \( {g}_{1} \) . Consider the quotient group \( G/{G}_{1} \) . Let \( h \in G \) . If \( h + {G}_{1}... | Proof. If \( h + {G}_{1} \) has order \( {p}^{r} \) in \( G/{G}_{1} \), then the order of \( h \) in \( G \) is at most \( {p}^{{r}_{1}} \) (since \( {p}^{{r}_{1}} \) is the maximum order in \( G \) ) and at least \( {p}^{r} \) (by Exercise 7). Since \( {G}_{1} = {p}^{r}\left( {h + {G}_{1}}\right) = {p}^{r}h + {G}_{1} ... | Yes |
Theorem 4.3 Every finite abelian group is a direct sum of cyclic groups. | Proof. This follows immediately from Theorem 4.1 and Theorem 4.2. \( ▱ \) | No |
Lemma 4.2 The dual of a cyclic group of order \( n \) is also a cyclic group of order \( n \) . | Proof. We introduce the exponential functions\n\n\[ e\left( x\right) = {e}^{2\pi ix} \]\n\nand\n\n\[ {e}_{n}\left( x\right) = e\left( {x/n}\right) = {e}^{{2\pi ix}/n}. \]\n\nThe \( n \) th roots of unity are the complex numbers \( {e}_{n}\left( a\right) \) for \( a = 0,1,\ldots, n - 1 \) . Let \( G \) be a finite cycli... | No |
Lemma 4.3 Let \( G \) be a finite abelian group and let \( {G}_{1},\ldots ,{G}_{k} \) be subgroups of \( G \) such that \( G = {G}_{1} \oplus \cdots \oplus {G}_{k} \) . For every character \( \chi \in \widehat{G} \) there exist unique characters \( {\chi }_{i} \in \widehat{{G}_{i}} \) such that if \( g \in G \) and \( ... | Proof. If \( {\chi }_{i} \in \widehat{{G}_{i}} \) for \( i = 1,\ldots, k \), then we can construct a map \( \chi : G \rightarrow \) \( {\mathbf{C}}^{ \times } \) as follows. Let \( g \in G \) . There exist unique elements \( {g}_{i} \in {G}_{i} \) such that \( g = {g}_{1} + \cdots + {g}_{k} \) . Define\n\n\[ \n\chi \le... | Yes |
Theorem 4.5 A finite abelian group \( G \) is isomorphic to its dual, that is,\n\n\[ G \cong \widehat{G} \] | Proof. By Lemma 4.2, the dual of a finite cyclic group of order \( n \) is also a finite cyclic group of order \( n \) . By Theorem 4.3, a finite abelian group \( G \) has cyclic subgroups \( {G}_{1},\ldots ,{G}_{k} \) such that\n\n\[ G = {G}_{1} \oplus \cdots \oplus {G}_{k} \]\n\nBy Lemma 4.3 and Exercise 5 in Section... | No |
Theorem 4.6 (Orthogonality relations) Let \( G \) be a finite abelian group of order \( n \), and let \( \widehat{G} \) be its dual group. If \( \chi \in \widehat{G} \), then\n\n\[ \mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) = \left\{ \begin{array}{ll} n & \text{ if }\chi = {\chi }_{0} \\ 0 & \text{ if }\chi ... | Proof. For \( \chi \in \widehat{G} \), let\n\n\[ S\left( \chi \right) = \mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) \]\n\nIf \( \chi = {\chi }_{0} \), then \( S\left( {\chi }_{0}\right) = \left| G\right| = n \) . If \( \chi \neq {\chi }_{0} \), then \( \chi \left( b\right) \neq 1 \) for some \( b \in G \) , a... | Yes |
Theorem 4.7 (Orthogonality relations) Let \( G \) be a finite abelian group of order \( n \), and let \( \widehat{G} \) be its dual group. If \( {\chi }_{1},{\chi }_{2} \in \widehat{G} \), then \[ \mathop{\sum }\limits_{{a \in G}}{\chi }_{1}\left( a\right) \overline{{\chi }_{2}}\left( a\right) = \left\{ \begin{array}{l... | Proof. These identities follow immediately from Theorem 4.6, since \[ {\chi }_{1}\left( a\right) \overline{{\chi }_{2}}\left( a\right) = {\chi }_{1}{\chi }_{2}^{-1}\left( a\right) \] and \[ \chi \left( a\right) \bar{\chi }\left( b\right) = \chi \left( {a - b}\right) \] This completes the proof. | Yes |
Theorem 4.8 (Fourier inversion) Let \( G \) be a finite abelian group of order \( n \) with dual group \( \widehat{G} \) . If \( f \in {L}^{2}\left( G\right) \), then\n\n\[ f = \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \chi \]\n\nand (4.7) is the unique representation of \(... | Proof. This is a straightforward calculation. Let \( a \in G \) . Defining the Fourier transform by (4.6), we have\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \chi \left( a\right) = \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\left( {\mathop{\sum }\limits_{... | No |
Theorem 4.9 (Plancherel’s formula) If \( G \) is a finite abelian group of order \( n \) and \( f \in {L}^{2}\left( G\right) \), then \[ \parallel \widehat{f}{\parallel }_{2} = \sqrt{n}\parallel f{\parallel }_{2} \] | Proof. We have \[ \parallel \widehat{f}{\parallel }_{2}^{2} = \left( {\widehat{f},\widehat{f}}\right) \] \[ = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \overline{\widehat{f}\left( \chi \right) } \] \[ = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\left( {\mathop{\sum }\limits_{{b \i... | Yes |
Theorem 4.10 (Uncertainty principle) If \( G \) is a finite abelian group and \( f \in {L}^{2}\left( G\right), f \neq 0 \), then\n\n\[ \left| {\operatorname{supp}\left( f\right) }\right| \left| {\operatorname{supp}\left( \widehat{f}\right) }\right| \geq \left| G\right| \text{.} \] | Proof. Let \( a \in G \) . By Theorem 4.8,\n\n\[ f\left( a\right) = \frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\widehat{f}\left( \chi \right) \chi \left( a\right) \]\n\nSince \( \left| {\chi \left( a\right) }\right| = 1 \) for all \( \chi \in \widehat{G} \), it follows that\n\n\[ \left| {f\left( a\right) ... | Yes |
Lemma 4.4 Let \( G \) be a finite abelian group with subgroup \( H \) . Then\n\n\[ \n{\widehat{G}}^{H} = \widehat{G} \cap {L}^{2}{\left( G\right) }^{H} \n\] | Proof. If \( \chi \in {\widehat{G}}^{H} \subseteq \widehat{G} \), then \( \chi \left( {x + h}\right) = \chi \left( x\right) \chi \left( h\right) = \chi \left( x\right) \) for all \( x \in G \) and \( h \in H \), and so \( \chi \in \widehat{G} \cap {L}^{2}\left( {G/H}\right) \) . Conversely, if \( \chi \in \widehat{G} \... | Yes |
Lemma 4.5 Let \( G \) be a finite abelian group with subgroup \( H \), and let \( \pi \) : \( G \rightarrow G/H \) be the natural map onto the quotient group. For \( {f}^{\sharp } \in {L}^{2}\left( {G/H}\right) \) , define the map \( {\pi }^{\sharp }\left( {f}^{\sharp }\right) \in {L}^{2}\left( G\right) \) by\n\n\[ \n{... | Proof. Let \( {f}^{\sharp } \in {L}^{2}\left( {G/H}\right) \) . If \( x \in G \) and \( h \in H \), then\n\n\[ \n{\pi }^{\sharp }\left( {f}^{\sharp }\right) \left( {x + h}\right) = {f}^{\sharp }\pi \left( {x + h}\right) = {f}^{\sharp }\pi \left( x\right) = {\pi }^{\sharp }\left( {f}^{\sharp }\right) \left( x\right) , \... | No |
Theorem 4.11 (Poisson summation formula) Let \( G \) be a finite abelian group and \( H \) a subgroup of \( G \) . If \( f \in {L}^{2}\left( G\right) \), then\n\n\[ \frac{1}{\left| H\right| }\mathop{\sum }\limits_{{y \in H}}f\left( y\right) = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{\chi \in {\widehat{G}}^{H}}... | Proof. Let \( f \in {L}^{2}\left( G\right) \) and \( \chi \in {\widehat{G}}^{H} \) . We define the function \( {f}^{\sharp } \in \) \( {L}^{2}\left( {G/H}\right) \) by\n\n\[ {f}^{\sharp }\left( {x + H}\right) = \mathop{\sum }\limits_{{y \in H}}f\left( {x + y}\right) \]\n\nWe define the character \( {\chi }^{\sharp } \i... | Yes |
Theorem 4.13 Let \( G = \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) be an abelian group of order \( n \) . Let \( K \in {L}^{2}\left( {G \times G}\right) \) and let \( {\Phi }_{K} \) be the associated integral operator on \( {L}^{2}\left( G\right) \) . The matrix of \( {\Phi }_{K} \) with respect to the orthonormal ba... | Proof. The matrix of the operator \( {\Phi }_{K} \) is \( \left( {c}_{ij}\right) \), where \( {c}_{ij} \) is defined by\n\n\[ \n{\Phi }_{K}\left( {\delta }_{{x}_{j}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{c}_{ij}{\delta }_{{x}_{i}} \n\]\n\nThen\n\n\[ \n{c}_{ij} = {\Phi }_{K}\left( {\delta }_{{x}_{j}}\right) \left... | Yes |
Theorem 4.15 (Trace formula) For \( h \in {L}^{2}\left( G\right) \), let \( {C}_{h} \) be the convolution operator on \( {L}^{2}\left( G\right) \), that is, \( {C}_{h}\left( f\right) = h * f \) for \( f \in {L}^{2}\left( G\right) \). The dual group \( \widehat{G} \) is a basis of eigenvectors for \( {C}_{h} \). If \( \... | Proof. This is a straightforward calculation. For \( x \in G \), we have\n\n\[ \n{C}_{h}\left( \chi \right) \left( x\right) = h * \chi \left( x\right) = \chi * h\left( x\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{y \in G}}\chi \left( {x - y}\right) h\left( y\right) \n\]\n\n\[ \n= \left( {\mathop{\sum }\limits_{{y \i... | Yes |
Theorem 4.17 If \( p \) is an odd prime and \( \left( {a, p}\right) = 1 \), then\n\n\[ \n\tau \left( {{\ell }_{p}, a}\right) = \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}{e}_{p}\left( {a{x}^{2}}\right) .\n\]\n\nIn particular,\n\n\[ \n\tau \left( p\right) = \mathop{\sum }\limits_{{x = 0}}^{{p - 1}}{e}^{{2\pi i}{x}^{2}/p}.... | Proof. The set \( R = \left\{ {k \in \{ 1,\ldots, p - 1\} : {\ell }_{p}\left( {k + p\mathbf{Z}}\right) = 1}\right\} \) is a set of representatives of the congruence classes of quadratic residues modulo \( p \) , and \( N = \left\{ {k \in \{ 1,\ldots, p - 1\} : {\ell }_{p}\left( {k + p\mathbf{Z}}\right) = - 1}\right\} \... | Yes |
Theorem 4.18 If \( p \) is prime and \( \left( {a, p}\right) = 1 \), then\n\n\[ \tau {\left( {\ell }_{p}, a\right) }^{2} = \left( \frac{-1}{p}\right) p = {\left( -1\right) }^{\frac{p - 1}{2}}p. \] | Proof. If \( p \) does not divide \( a \), then\n\n\[ \tau {\left( {\ell }_{p}, a\right) }^{2} = \mathop{\sum }\limits_{{x = 1}}^{{p - 1}}\left( \frac{x}{p}\right) {e}_{p}\left( {ax}\right) \mathop{\sum }\limits_{{y = 1}}^{{p - 1}}\left( \frac{y}{p}\right) {e}_{p}\left( {ay}\right) \]\n\n\[ = \mathop{\sum }\limits_{{x ... | Yes |
Theorem 4.20 If \( p \) and \( q \) are distinct odd primes, then\n\n\[ \left( \widehat{{\widehat{{\ell }_{p}}}^{q}}\right) \left( {\Delta \left( {-q + p\mathbf{Z}}\right) }\right) = {p\tau }{\left( p\right) }^{q - 1}\left( \frac{q}{p}\right) . | Proof. The function on the left side of the equation is a bit complicated. Let \( G = \mathbf{Z}/p\mathbf{Z} \) . Since \( {\ell }_{p} \in {L}^{2}\left( G\right) \), it follows that the Fourier transform \( {\ell }_{p} \in {L}^{2}\left( \widehat{G}\right) \), and also its \( q \) th power \( {\widehat{\ell }}_{p}^{q} \... | Yes |
Theorem 4.21 If \( p \) and \( q \) are distinct odd primes, then\n\n\[ \left( \widehat{{\widehat{{\ell }_{p}}}^{q}}\right) \left( {\Delta \left( {-q + p\mathbf{Z}}\right) }\right) = p\mathop{\sum }\limits_{\substack{{{x}_{1} + \cdots + {x}_{q} \equiv q\;\left( {\;\operatorname{mod}\;p}\right) } \\ {1 \leq {x}_{i} \leq... | Proof. Let \( k \) be a positive integer. By Exercise 10 in Section 4.3, a product of Fourier transforms is the Fourier transform of the convolution,\n\nand so\n\[ {\widehat{{\ell }_{p}}}^{k} = \underset{k\text{ times }}{\underbrace{\widehat{{\ell }_{p}} * \cdots * \widehat{{\ell }_{p}}}} = \underset{k\text{ times }}{\... | Yes |
Theorem 4.22 For all functions \( f \in {L}^{2}\left( {\mathbf{Z}/n\mathbf{Z}}\right) \) , \[ {\mathcal{F}}^{2}\left( f\right) \left( {a + n\mathbf{Z}}\right) = {nf}\left( {-a + n\mathbf{Z}}\right) . \] | Proof. This is similar to the proof of (4.8) in Theorem 4.8. Writing \( \mathcal{F}\left( f\right) = g \), we have \[ g\left( {x + n\mathbf{Z}}\right) = \mathop{\sum }\limits_{{y = 0}}^{{n - 1}}f\left( {y + n\mathbf{Z}}\right) {\omega }^{-{xy}} \] and \[ {\mathcal{F}}^{2}\left( f\right) \left( {a + n\mathbf{Z}}\right) ... | Yes |
Theorem 5.1 The spectrum of the ring of integers is\n\n\[ \operatorname{Spec}\left( \mathbf{Z}\right) = \{ p\mathbf{Z} : p\text{ is prime or }p = 0\} . \] | Proof. Since \( \mathbf{Z} \) is principal, every ideal is of the form \( d\mathbf{Z} \) for some nonnegative integer \( d \) . If \( d = 0 \), then \( d\mathbf{Z} = \{ 0\} \), and the zero ideal is prime, since \( {ab} = 0 \) if and only if \( a = 0 \) or \( b = 0 \) . Let \( d \geq 1 \) . If \( d = p \) is prime and ... | Yes |
Theorem 5.2 For \( m \geq 2 \), let \( \mathbf{Z}/m\mathbf{Z} \) be the ring of congruence classes modulo \( m \) . Then\n\n(i) \( \mathbf{Z}/m\mathbf{Z} \) is principal, and the ideals of \( \mathbf{Z}/m\mathbf{Z} \) are the ideals generated by the congruence classes \( d + m\mathbf{Z} \), where \( d \) is a divisor o... | Proof. Let \( J \) be an ideal in the ring \( R = \mathbf{Z}/m\mathbf{Z} \) . Consider the union of congruence classes\n\n\[ I = \mathop{\bigcup }\limits_{{a + m\mathbf{Z} \in J}}\left( {a + m\mathbf{Z}}\right) \]\n\nThe set \( I \) is an ideal in \( \mathbf{Z} \) . Since \( \mathbf{Z} \) is principal, \( I = d\mathbf{... | Yes |
Theorem 5.3 The ring \( \mathbf{C}\left\lbrack t\right\rbrack \) of polynomials with coefficients in the field \( \mathbf{C} \) of complex numbers is a principal ring. | Proof. This is a special case of Exercise 18 in Section 3.1. \( ▱ \) | No |
Theorem 5.4 Let \( f\left( t\right) \in \mathbf{C}\left\lbrack t\right\rbrack \) and \( R = \mathbf{C}\left\lbrack t\right\rbrack /I \), where \( I = \langle f\left( t\right) \rangle \) is the principal ideal generated by \( f\left( t\right) \) . The radical of \( R \) is the principal ideal generated by \( \operatorna... | Proof. This follows immediately from the observation that if \( f\left( t\right) \) and \( g\left( t\right) \) are polynomials with complex coefficients, then there exists a positive integer \( k \) such that \( f\left( t\right) \) divides \( g{\left( t\right) }^{k} \) if and only if \( \operatorname{rad}\left( f\right... | Yes |
Theorem 5.5 Let \( R \) be a ring and \( R\\left\\lbrack t\\right\\rbrack \) the ring of polynomials with coefficients in \( R \) . Define \( D : R\\left\\lbrack t\\right\\rbrack \\rightarrow R\\left\\lbrack t\\right\\rbrack \) by\n\n\[ D\\left( {\\mathop{\\sum }\\limits_{{i = 0}}^{m}{a}_{i}{t}^{i}}\\right) = \\mathop{... | Proof. Let \( f = f\\left( t\\right) = \\mathop{\\sum }\\limits_{{i = 0}}^{m}{a}_{i}{t}^{i} \) and \( g = g\\left( t\\right) = \\mathop{\\sum }\\limits_{{j = 0}}^{n}{b}_{j}{t}^{j} \) . It is immediate that \( D\\left( {f + g}\\right) = D\\left( f\\right) + D\\left( g\\right) \), and so \( D \) is a homomorphism of the ... | Yes |
Theorem 5.6 Let \( R \) be an integral domain with quotient field \( F \), and let \( D \) be a derivation on \( R \) . There exists a unique derivation \( {D}_{F} \) on \( F \) such that \( {D}_{F}\left( x\right) = D\left( x\right) \) for all \( x \in R \) . | Proof. Suppose that there exists a derivation \( {D}_{F} \) on \( F \) such that \( {D}_{F}\left( a\right) = \) \( D\left( a\right) \) for all \( a \in R \) . Let \( x \in F, x \neq 0 \) . There exist \( a, b \in R \) with \( b \neq 0 \) and \( x = a/b \) . Since \( a = {bx} \in R \), it follows that\n\n\[ D\left( a\ri... | No |
Theorem 5.8 If \( n \geq 3 \), then the Fermat equation \( {x}^{n} + {y}^{n} = {z}^{n} \) has no solution in nonzero, relatively prime polynomials, not all constant. | Proof. Let \( n \geq 3 \), and suppose that \( x, y \), and \( z \) are nonzero, relatively prime polynomials, not all constant, such that \( {x}^{n} + {y}^{n} = {z}^{n} \) . We apply Mason’s theorem with \( a = {x}^{n}, b = {y}^{n} \), and \( c = {z}^{n} \) . Then\n\n\[ \operatorname{rad}\left( {abc}\right) = \operato... | Yes |
Theorem 5.9 (Asymptotic Fermat theorem) The abc conjecture implies that there exists an integer \( {n}_{0} \) such that the Fermat equation has no solution in relatively prime integers for any exponent \( n \geq {n}_{0} \) . | Proof. Let \( x, y \), and \( z \) be relatively prime positive integers such that\n\n\[ \n{x}^{n} + {y}^{n} = {z}^{n} \n\] \n\nWe note that \n\n\[ \n\operatorname{rad}\left( {{x}^{n}{y}^{n}{z}^{n}}\right) = \operatorname{rad}\left( {xyz}\right) \leq {xyz} \leq {z}^{3}. \n\] \n\nIf \( n \geq 2 \), then \( z \geq 3 \) .... | Yes |
Theorem 5.10 (Asymptotic Catalan theorem) The abc conjecture implies that the Catalan equation has only finitely many solutions. | Proof. Let \( \left( {x, y, m, n}\right) \) be a solution of the Catalan equation with \( \min \left( {m, n}\right) \geq \) 3. Then \( x \) and \( y \) are relatively prime. It follows from the \( {abc} \) conjecture with \( \varepsilon = 1/4 \) that there exists a constant \( {K}_{2} = K\left( {1/4}\right) \) such tha... | Yes |
Lemma 5.1 Let \( p \) be an odd prime. If there exists a positive integer \( n \) such that \( {2}^{n} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) \) but \( {2}^{n} ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), then \( p \) is a Wieferich prime. | Proof. Let \( d \) be the order of 2 modulo \( p \) . Then \( d \) divides \( n \) . Since \( {2}^{n} ≢ 1 \) \( \left( {\;\operatorname{mod}\;{p}^{2}}\right) \), it follows that \( {2}^{d} ≢ 1\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) . Then \( {2}^{d} = 1 + {kp} \), where \( \left( {k, p}\right) = 1 \) . Moreo... | Yes |
Theorem 5.11 The abc conjecture implies that there exist infinitely many Wieferich primes. | Proof. Let \( W \) be the set of Wieferich primes. For every positive integer \( n \), we write\n\n\[ {2}^{n} - 1 = {u}_{n}{v}_{n} \]\n\nwhere \( {v}_{n} \) is the maximal powerful divisor of \( {2}^{n} - 1 \) . Then \( {u}_{n} \) is a square-free integer,\n\n\[ {u}_{n} = \mathop{\prod }\limits_{\substack{{p \mid n} \\... | Yes |
Lemma 5.2 Let \( a, b, c \) be relatively prime positive integers such that\n\n\[ a < b < c \]\n\nand\n\n\[ a + b = c.\]\n\nLet \( n \geq 2 \) . If \( c \) is odd, define\n\n\[ {A}_{n} = {\left( b - a\right) }^{n} \]\n\n\[ {B}_{n} = {c}^{n} - {\left( b - a\right) }^{n} \]\n\n\[ {C}_{n} = {c}^{n}. \]\n\nIf \( c \) is ev... | Proof. It is left to the reader to show that \( {A}_{n},{B}_{n},{C}_{n} \) are distinct, relatively prime positive integers such that \( {A}_{n} + {B}_{n} = {C}_{n} \) (Exercises 1,2, and 3).\n\nLet \( m \geq 3 \) and \( n = \varphi \left( m\right) \) . Then \( n \geq 2 \) . We must prove that\n\n\[ {A}_{n}{B}_{n}{C}_{... | No |
Theorem 6.1 The set of all complex-valued arithmetic functions, with addition defined by pointwise sum and multiplication defined by Dirichlet convolution, is a commutative ring with additive identity \( 0\left( n\right) \) and multiplicative identity \( \delta \left( n\right) \) . | Proof. It is easy to check that the set of arithmetic functions is an additive abelian group with the zero function as the additive identity.\n\nWe shall prove that Dirichlet convolution is commutative, associative, and distributes over addition, that is,\n\n\[ f * g = g * f \]\n\n\[ \left( {f * g}\right) * h = f * \le... | Yes |
Theorem 6.2 Consider the arithmetic function \( L\left( n\right) \) defined by\n\n\[ L\left( n\right) = \log n\;\text{ for all }n \geq 1. \]\n\nPointwise multiplication by \( L\left( n\right) \) is a derivation on the ring of arithmetic functions. | Proof. Observe that if \( d \) is a positive divisor of \( n \), then\n\n\[ L\left( n\right) = L\left( d\right) + L\left( {n/d}\right) \]\n\nWe must prove that\n\n\[ L \cdot \left( {f * g}\right) = \left( {L \cdot f}\right) * g + f * \left( {L \cdot g}\right) \]\n\nfor all arithmetic functions \( f \) and \( g \) . We ... | Yes |
Theorem 6.3 Let \( a \) and \( b \) be integers with \( a < b \), and let \( f\left( t\right) \) be a function that is monotonic on the interval \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \min \left( {f\left( a\right), f\left( b\right) }\right) \leq \mathop{\sum }\limits_{{n = a}}^{b}f\left( n\right) - {\int }... | Proof. If \( f\left( t\right) \) is increasing on \( \left\lbrack {n, n + 1}\right\rbrack \), then\n\n\[ f\left( n\right) \leq {\int }_{n}^{n + 1}f\left( t\right) {dt} \leq f\left( {n + 1}\right) \]\n\nIf \( f\left( t\right) \) is increasing on the interval \( \left\lbrack {a, b}\right\rbrack \), then\n\n\[ f\left( a\r... | Yes |
Theorem 6.4 For \( x \geq 2 \) ,\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\log n = x\log x - x + O\left( {\log x}\right) . \] | Proof. The function \( f\left( t\right) = \log t \) is increasing on \( \left\lbrack {1, x}\right\rbrack \) . By Theorem 6.3,\n\n\[ {\int }_{1}^{x}\log {tdt} \leq \mathop{\sum }\limits_{{n \leq x}}\log n \leq {\int }_{1}^{x}\log {tdt} + \log x \]\n\nand so\n\n\[ \mathop{\sum }\limits_{{n \leq x}}\log n = x\log x - x + ... | Yes |
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