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Theorem 2. If \( p \) is an odd prime and \( l \in {\mathbb{Z}}^{ + } \), then \( U\left( {\mathbb{Z}/{p}^{l}\mathbb{Z}}\right) \) is cyclic; i.e., there exist primitive roots \( {\;\operatorname{mod}\;{p}^{l}} \) . | Proof. By Theorem 1 there exist primitive roots mod \( p \) . If \( g \in \mathbb{Z} \) is a primitive root \( {\;\operatorname{mod}\;p} \), then so is \( g + p \) . If \( {g}^{p - 1} \equiv 1\left( {p}^{2}\right) \), then \( {\left( g + p\right) }^{p - 1} \equiv {g}^{p - 1} + \left( {p - 1}\right) {g}^{p - 2}p \equiv ... | Yes |
Theorem 3. Let \( n = {2}^{a}{p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{l}^{{a}_{l}} \) be the prime decomposition of \( n \) . Then\n\n\[ U\left( {\mathbb{Z}/n\mathbb{Z}}\right) \approx U\left( {\mathbb{Z}/{2}^{a}\mathbb{Z}}\right) \times U\left( {\mathbb{Z}/{p}_{1}^{{a}_{1}}\mathbb{Z}}\right) \times \cdots \times ... | Proof. Theorems 2, 2', and Theorem 1' of Chapter 3. | No |
Proposition 4.1.3. n possesses primitive roots iff \( n \) is of the form 2,4, \( {p}^{a} \), or \( 2{p}^{a} \) , where \( p \) is an odd prime. | Proof. By Theorem \( {2}^{\prime } \) we can assume that \( n \neq {2}^{l}, l \geq 3 \) . If \( n \) is not of the given form, it is easy to see that \( n \) can be written as a product \( {m}_{1}{m}_{2} \), where \( \left( {{m}_{1},{m}_{2}}\right) \) \( = 1 \) and \( {m}_{1},{m}_{2} > 2 \) . We then have that \( \phi ... | Yes |
Proposition 4.2.1. If \( m \in {\mathbb{Z}}^{ + } \) possesses primitive roots and \( \left( {a, m}\right) = 1 \), then \( a \) is an \( n \) th power residue \( {\;\operatorname{mod}\;m} \) iff \( {a}^{\phi \left( m\right) /d} \equiv 1\left( m\right) \), where \( d = \left( {n,\phi \left( m\right) }\right) \) . | Proof. Let \( g \) be a primitive root \( {\;\operatorname{mod}\;m} \) and \( a = {g}^{b}, x = {g}^{y} \) . Then the congruence \( {x}^{n} \equiv a\left( m\right) \) is equivalent to \( {g}^{ny} \equiv {g}^{b}\left( m\right) \), which in turn is equivalent to \( {ny} \equiv b\left( {\phi \left( m\right) }\right) \) . T... | Yes |
Proposition 4.2.2. Suppose that \( a \) is odd, \( e \geq 3 \), and consider the congruence \( {x}^{n} \equiv a\left( {2}^{e}\right) \) . If \( n \) is odd, a solution always exists and it is unique.\n\nIf \( n \) is even, a solution exists iff \( a \equiv 1\left( 4\right) ,{a}^{2 \bullet - 2/d} \equiv 1\left( {2}^{e}\... | Proof. We leave the proof as an exercise. One begins by writing \( a \equiv {\left( -1\right) }^{s}{5}^{t} \) \( \left( {2}^{e}\right) \) and \( x \equiv {\left( -1\right) }^{y}{5}^{z}\left( {2}^{e}\right) \). | No |
Proposition 4.2.3. If \( p \) is an odd prime, \( p \nmid a \), and \( p \nmid n \), then if \( {x}^{n} \equiv a\left( p\right) \) is solvable, so is \( {x}^{n} \equiv a\left( {p}^{e}\right) \) for all \( e \geq 1 \) . All these congruences have the same number of solutions. | Proof. If \( n = 1 \), the assertion is trivial, so we may assume \( n \geq 2 \) . Suppose that \( {x}^{n} \equiv a\left( {p}^{e}\right) \) is solvable. Let \( {x}_{0} \) be a solution and set \( {x}_{1} = {x}_{0} + b{p}^{e} \) . A short computation shows \( {x}_{1}^{n} \equiv {x}_{0}^{n} + {nb}{p}^{e}{x}_{0}^{n - 1}\l... | Yes |
Proposition 4.2.4. Let \( {2}^{l} \) be the highest power of 2 dividing \( n \) . Suppose that \( a \) is odd and that \( {x}^{n} \equiv a\left( {2}^{{2l} + 1}\right) \) is solvable. Then \( {x}^{n} \equiv a\left( {2}^{e}\right) \) is solvable for all \( e \geq {2l} + 1 \) (and consequently for all \( e \geq 1 \) ). Mo... | Proof. We leave the proof as an exercise. One begins by assuming that \( {x}^{n} \equiv a\left( {2}^{m}\right), m \geq {2l} + 1 \), has a solution \( {x}_{0} \) . Let \( {x}_{1} = {x}_{0} + b{2}^{m - l} \) . One shows, by an appropriate choice of \( b \), that \( {x}_{1}^{n} \equiv a\left( {2}^{m + 1}\right) \) . | No |
Proposition 5.1.1. Let \( m = {2}^{e}{p}_{1}^{{e}_{1}}\cdots {p}_{l}^{{e}_{l}} \) be the prime decomposition of \( m \), and suppose that \( \left( {a, m}\right) = 1 \) . Then \( {x}^{2} \equiv a\left( m\right) \) is solvable iff the following conditions are satisfied:\n\n(a) If \( e = 2 \), then \( a \equiv 1\left( 4\... | Proof. By the Chinese Remainder Theorem the congruence \( {x}^{2} \equiv a\left( m\right) \) is equivalent to the system \( {x}^{2} \equiv a\left( {2}^{e}\right) ,{x}^{2} \equiv a\left( {p}_{1}^{{e}_{1}}\right) ,\ldots ,{x}^{2} \equiv a\left( {p}_{l}^{{a}_{l}}\right) \) .\n\nConsider \( {x}^{2} \equiv a\left( {2}^{e}\r... | Yes |
Corollary 1. There are as many residues as nonresidues mod \( p \) . | Proof. \( {a}^{\left( {p - 1}\right) /2} \equiv 1\left( p\right) \) has \( \left( {p - 1}\right) /2 \) solutions. Thus there are \( \left( {p - 1}\right) /2 \) residues and \( p - 1 - \left( {\left( {p - 1}\right) /2}\right) = \left( {p - 1}\right) /2 \) nonresidues. | Yes |
Corollary 2. The product of two residues is a residue, the product of two nonresidues is a residue, and the product of a residue and a nonresidue is a nonresidue. | Proof. This all follows easily from part (b). | No |
Corollary 3. \( {\left( -1\right) }^{\left( {p - 1}\right) /2} = \left( {-1/p}\right) \). | Proof. Substitute \( a = - 1 \) in part (a). | No |
Proposition 5.1.3. 2 is a quadratic residue of primes of the form \( {8k} + 1 \) and \( {8k} + 7 \) is a quadratic nonresidue of primes of the form \( {8k} + 3 \) and \( {8k} + 5 \) . This information is summarized in the formula\n\n\[ \left( \begin{array}{l} 2 \\ - \\ p \end{array}\right) = {\left( -1\right) }^{\left(... | Proof. We leave to the reader the task of showing that the formula is equivalent to the first two assertions.\n\nLet \( p \) be an odd prime (as usual) and notice that the number \( \mu \) is equal to the number of elements of the set \( 2 \cdot 1,2 \cdot 2,\ldots ,2 \cdot \left( {p - 1}\right) /2 \) that exceed \( \le... | No |
Theorem 2. Let \( q \) be an odd prime.\n\n(a) If \( q \equiv 1\\left( 4\\right) \), then \( q \) is a quadratic residue \( {\\operatorname{mod}p} \) iff \( p \equiv r\\left( q\\right) \), where \( r \) is a quadratic residue \( {\\operatorname{mod}q} \).\n\n(b) If \( q \equiv 3\\left( 4\\right) \), then \( q \) is a q... | Proof. If \( q \equiv 1\\left( 4\\right) \), then by Theorem 1 we have \( \\left( {q/p}\\right) = \\left( {p/q}\\right) \). Part (a) is thus clear.\n\nIf \( q \equiv 3\\left( 4\\right) \), Theorem 1 yields \( \\left( {q/p}\\right) = {\\left( -1\\right) }^{\\left( {p - 1}\\right) /2}\\left( {p/q}\\right) \). Assume firs... | Yes |
(a) \( \left( {{a}_{1}/b}\right) = \left( {{a}_{2}/b}\right) \) if \( {a}_{1} \equiv {a}_{2}\left( b\right) \) . | Proof. Parts (a) and (b) are immediate from the corresponding properties of the Legendre symbol. | No |
Theorem 3. Let a be a nonsquare integer. Then there are infinitely many primes p for which a is a quadratic nonresidue. | Proof. It is easily seen that we may assume that \( a \) is square-free. Let \( a = {2}^{e}{q}_{1}{q}_{2} \cdots {q}_{n} \), where the \( {q}_{i} \) are distinct odd primes and \( e = 0 \) or 1 . The case \( a = 2 \) has to be dealt with separately. We shall assume to begin with that \( n \geq 1 \), i.e., that \( a \) ... | Yes |
Proposition 5.3.1. If \( n \) is a positive odd integer and \( f\left( z\right) = {e}^{2\pi iz} - {e}^{-{2\pi iz}} \), then\n\n\[ \frac{f\left( {nz}\right) }{f\left( z\right) } = \mathop{\prod }\limits_{{k = 1}}^{{\left( {n - 1}\right) /2}}f\left( {z + \frac{k}{n}}\right) f\left( {z - \frac{k}{n}}\right) . \] | Proof. In the lemma, substitute \( x = {e}^{2\pi iz} \) and \( y = {e}^{-{2\pi iz}} \) . We see that\n\n\[ f\left( {nz}\right) = \mathop{\prod }\limits_{{k = 0}}^{{n - 1}}f\left( {z + \frac{k}{n}}\right) \]\n\nNotice that \( f\left( {z + k/n}\right) = f\left( {z + k/n - 1}\right) = f\left( {z - \left( {n - k}\right) /n... | Yes |
Proposition 5.3.2. If \( p \) is an odd prime, \( a \in \mathbb{Z} \), and \( p \times a \), then\n\n\[ \mathop{\prod }\limits_{{l = 1}}^{{\left( {p - 1}\right) /2}}f\left( \frac{la}{p}\right) = \left( \begin{array}{l} a \\ p \end{array}\right) \mathop{\prod }\limits_{{l = 1}}^{{\left( {p - 1}\right) /2}}f\left( \frac{... | Proof. As in the lemma of Section \( 1,{la} \equiv \pm {m}_{l}\left( p\right) \), where \( 1 \leq {m}_{l} \leq \left( {p - 1}\right) /2 \) . Thus \( {la}/p \) and \( \pm {m}_{l}/p \) differ by an integer. This implies that \( f\left( {{la}/p}\right) = f\left( {\pm {m}_{l}/p}\right) \) \( = \pm f\left( {{m}_{l}/p}\right... | Yes |
Proposition 5.3.3. Let \( p \) and \( q \) be distinct odd primes and \( a \geq 1 \) an integer. Then the following assertions are equivalent:\n\n(a) \( \left( {p/q}\right) \left( {q/p}\right) = {\left( -1\right) }^{\left( {\left( {p - 1}\right) /2}\right) \left( {\left( {q - 1}\right) /2}\right) } \).\n\n(b) If \( p \... | Proof. In order to show (a) implies (b) it is enough, by multiplicativity, to show that (b) holds when \( a \) is prime. For \( a = 2 \) the result follows from Proposition 5.1.3. If \( a \) is an odd prime then by (a) \( \left( {a/p}\right) = {\left( -1\right) }^{\left( {\left( {p - 1}\right) /2}\right) \left( {\left(... | Yes |
Proposition 6.1.1. A rational number \( r \in \mathbb{Q} \) is an algebraic integer iff \( r \in \mathbb{Z} \) . | Proof. If \( r \in \mathbb{Z} \), then \( r \) is a root of \( x - r = 0 \) . Thus \( r \) is an algebraic integer.\n\nSuppose that \( r \in \mathbb{Q} \) and that \( r \) is an algebraic integer; i.e., \( r \) satisfies an equation \( {x}^{n} + {b}_{1}{x}^{n - 1} + \cdots + {b}_{n} = 0 \) with \( {b}_{1},\ldots ,{b}_{... | Yes |
Proposition 6.1.2. Let \( V = \left\lbrack {{\gamma }_{1},{\gamma }_{2},\ldots ,{\gamma }_{l}}\right\rbrack \), and suppose that \( \alpha \in \mathbb{C} \) has the property that \( {\alpha \gamma } \in V \) for all \( \gamma \in V \) . Then \( \alpha \) is an algebraic number. | Proof. \( \alpha {\gamma }_{i} \in V \) for \( i = 1,2,\ldots, l \) . Thus \( \alpha {\gamma }_{i} = \mathop{\sum }\limits_{{j = 1}}^{l}{a}_{ij}{\gamma }_{j} \), where \( {a}_{ij} \in \mathbb{Q} \) . It follows that \( 0 = \mathop{\sum }\limits_{{j = 1}}^{l}\left( {{a}_{ij} - {\delta }_{ij}\alpha }\right) {\gamma }_{j}... | Yes |
Proposition 6.1.3. The set of algebraic numbers forms a field. | Proof. Suppose that \( {\alpha }_{1} \) and \( {\alpha }_{2} \) are algebraic numbers. We shall show that \( {\alpha }_{1}{\alpha }_{2} \) and \( {\alpha }_{1} + {\alpha }_{2} \) are algebraic numbers.\n\nSuppose that \( {\alpha }_{1}^{n} + {r}_{1}{\alpha }_{1}^{n - 1} + {r}_{2}{\alpha }_{1}^{n - 2} + \cdots + {r}_{n} ... | No |
Proposition 6.1.4. Let \( W \) be a \( \mathbb{Z} \) module and suppose that \( \omega \in \mathbb{C} \) is such that \( {\omega \gamma } \in W \) for all \( \gamma \in W \) . Then \( \omega \) is an algebraic integer. | Proof. The proof proceeds exactly as in Proposition 6.1.2, except that now the \( {a}_{ij} \in \mathbb{Z} \) . The equation \( \det \left( {{a}_{ij} - {\delta }_{ij}\omega }\right) = 0 \) when written out shows that \( \omega \) satisfies a monic equation of degree \( l \) with integer coefficients. Thus \( \omega \) i... | Yes |
Proposition 6.1.5. The set of algebraic integers forms a ring. | Proof. The proof follows from Proposition 6.1.4 in exactly the same way in which Proposition 6.1.3 follows from Proposition 6.1.2. We leave the details to the reader. | No |
Proposition 6.1.6. If \( {\omega }_{1},{\omega }_{2} \in \Omega \) and \( p \in \mathbb{Z} \) is a prime, then\n\n\[{\left( {\omega }_{1} + {\omega }_{2}\right) }^{p} \equiv {\omega }_{1}^{p} + {\omega }_{2}^{p}\left( p\right) . | Proof. \( {\left( {\omega }_{1} + {\omega }_{2}\right) }^{p} = \mathop{\sum }\limits_{{k = 0}}^{p}\left( \begin{array}{l} p \\ k \end{array}\right) {\omega }_{1}^{k}{\omega }_{2}^{p - k} \) . By Lemma 2, Chapter 4, we have \( p \mid \left( \begin{array}{l} p \\ k \end{array}\right) \) for \( 1 \leq k \leq p - 1 \) . Th... | Yes |
Proposition 6.1.7. If \( \alpha \) is an algebraic number then \( \alpha \) is the root of a unique monic irreducible \( f\left( x\right) \) in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . Furthermore if \( g\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack, g\left( \alpha \right) = 0 \) then \( f\left( x\righ... | Proof. Let \( f\left( x\right) \) be any monic irreducible with \( f\left( \alpha \right) = 0 \) . We prove the second assertion first. If \( f\left( x\right) \smallsetminus g\left( x\right) \) then \( \left( {f\left( x\right), g\left( x\right) }\right) = 1 \) . By Lemma 4, Section 2, Chapter 1 we may write \( f\left( ... | No |
Proposition 6.1.8. If \( \alpha \in \Omega \) then \( \mathbb{Q}\left( \alpha \right) = \mathbb{Q}\left\lbrack \alpha \right\rbrack \) . | Proof. Clearly \( \mathbb{Q}\left\lbrack \alpha \right\rbrack \subset \mathbb{Q}\left( \alpha \right) \) . If \( h\left( \alpha \right) \in \mathbb{Q}\left\lbrack \alpha \right\rbrack, h\left( \alpha \right) \neq 0 \), then by Proposition 6.1.7, \( f\left( x\right) + h\left( x\right) \), where \( f\left( x\right) \) is... | Yes |
Lemma 1. \( \mathop{\sum }\limits_{{t = 0}}^{{p - 1}}{\zeta }^{at} \) is equal to \( p \) if \( a \equiv 0\left( p\right) \) . Otherwise it is zero. | Proof. If \( a \equiv 0\left( p\right) \), then \( {\zeta }^{a} = 1 \), and so \( \mathop{\sum }\limits_{{t = 0}}^{{p - 1}}{\zeta }^{at} = p \) . If \( a ≢ 0\left( p\right) \), then \( {\zeta }^{a} \neq 1 \) and \( \mathop{\sum }\limits_{{t = 0}}^{{p - 1}}{\zeta }^{at} = \left( {{\zeta }^{ap} - 1}\right) /\left( {{\zet... | Yes |
Lemma 2. \( \mathop{\sum }\limits_{t}\left( {t/p}\right) = 0 \), where \( \left( {t/p}\right) \) is the Legendre symbol. | Proof. By definition \( \left( {0/p}\right) = 0 \) . Of the remaining \( p - 1 \) terms in the summation, half are +1 and half are -1, since by Corollary 1 to Proposition 5.1.2, there are as many quadratic residues as quadratic nonresidues mod \( p \) . | Yes |
Proposition 6.3.1. \( {g}_{a} = \left( {a/p}\right) {g}_{1} \) . | Proof. If \( a \equiv 0\left( p\right) \), then \( {\zeta }^{at} = 1 \) for all \( t \), and \( {g}_{a} = \sum \left( {t/p}\right) = 0 \) by Lemma 2 . This gives the result in the case that \( a \equiv 0\left( p\right) \) .\n\nNow suppose that \( a ≢ 0\left( p\right) \) . Then\n\n\[ \left( \frac{a}{p}\right) {g}_{a} = ... | Yes |
Proposition 6.3.2. \( {g}^{2} = {\left( -1\right) }^{\left( {p - 1}\right) /2}p \) . | Proof. The idea of the proof is to evaluate the sum \( \mathop{\sum }\limits_{a}{g}_{a}{g}_{-a} \) in two ways.\n\nIf \( a ≢ 0\left( p\right) \), then \( {g}_{a}{g}_{-a} = \left( {a/p}\right) \left( {-a/p}\right) {g}^{2} = \left( {-1/p}\right) {g}^{2} \) . If follows that\n\n\[ \mathop{\sum }\limits_{a}{g}_{a}{g}_{-a} ... | Yes |
Proposition 6.4.1. The polynomial \( 1 + x + \cdots + {x}^{p - 1} \) is irreducible in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . | Proof. By Exercise 4 at the end of this chapter (\ | No |
Proposition 6.4.2. \( \mathop{\prod }\limits_{{k = 1}}^{{\left( {p - 1}\right) /2}}{\left( {\zeta }^{{2k} - 1} - {\zeta }^{-\left( {{2k} - 1}\right) }\right) }^{2} = {\left( -1\right) }^{\left( {p - 1}\right) /2}p \) . | Proof. One has \( {x}^{p} - 1 = \left( {x - 1}\right) \mathop{\prod }\limits_{{j = 1}}^{{p - 1}}\left( {x - {\zeta }^{j}}\right) \) . Divide by \( x - 1 \) and put \( x = 1 \) to obtain \( p = \mathop{\prod }\limits_{r}\left( {1 - {\zeta }^{r}}\right) \), where the product is over any complete set of representative of ... | Yes |
Proposition 6.4.3.\n\n\\[ \n\\mathop{\\prod }\\limits_{{k = 1}}^{{\\left( {p - 1}\\right) /2}}\\left( {{\\zeta }^{{2k} - 1} - {\\zeta }^{-\\left( {{2k} - 1}\\right) }}\\right) = \\left\\{ \\begin{array}{ll} \\sqrt{p}, & \\text{ if }p \\equiv 1\\left( 4\\right) , \\\\ i\\sqrt{p}, & \\text{ if }p \\equiv 3\\left( 4\\righ... | Proof. By Proposition 6.4.2 we have only to compute the sign of the product.\n\n\\[ \n{i}^{\\left( {p - 1}\\right) /2}\\mathop{\\prod }\\limits_{{k = 1}}^{{\\left( {p - 1}\\right) /2}}2\\sin \\frac{\\left( {{4k} - 2}\\right) \\pi }{p}.\n\\]\n\nBut \\( \\sin \\left( {\\left( {{4k} - 2}\\right) /p}\\right) \\pi < 0 \\) i... | Yes |
Proposition 6.4.4. \( \varepsilon = + 1 \) . | Proof. Consider the polynomial\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{j = 1}}^{{p - 1}}\chi \left( j\right) {x}^{j} - \varepsilon \mathop{\prod }\limits_{{k = 1}}^{{\left( {p - 1}\right) /2}}\left( {{x}^{{2k} - 1} - {x}^{p - \left( {{2k} - 1}\right) }}\right) .\](2)\n\nThen \( f\left( \zeta \right) = 0 \) by ... | Yes |
Proposition 7.1.1.\n\n\[ {x}^{q} - x = \mathop{\prod }\limits_{{a \in F}}\left( {x - \alpha }\right) \] | Proof. Both polynomials are to be considered as elements of \( F\left\lbrack x\right\rbrack \) .\n\nEvery element \( \alpha \in F \) is a root of \( {x}^{q} - x \) . Since \( F \) has \( q \) elements and since the degree of \( {x}^{q} - x \) is \( q \), the result follows. | Yes |
Corollary 1. Let \( F \subset K \), where \( K \) is a field. An element \( \alpha \in K \) is in \( F \) iff \( {\alpha }^{q} = \alpha \) . | Proof. \( {\alpha }^{q} = \alpha \) iff \( \alpha \) is a root of \( {x}^{q} - x \) . By Proposition 7.1.1, the roots of \( {x}^{q} - x \) are precisely the elements of \( F \) . | Yes |
Corollary 2. Iff \( \left( x\right) \) divides \( {x}^{q} - x \), then \( f\left( x\right) \) has \( d \) distinct roots, where \( d \) is the degree of \( f\left( x\right) \) . | Proof. Let \( f\left( x\right) g\left( x\right) = {x}^{q} - x.g\left( x\right) \) has degree \( q - d \) . If \( f\left( x\right) \) has fewer than \( d \) distinct roots, then by Lemma 1 of Chapter \( 4, f\left( x\right) g\left( x\right) \) would have fewer than \( d + \left( {q - d}\right) = q \) distinct roots, whic... | Yes |
Theorem 1. The multiplicative group of a finite field is cyclic. | Proof. This theorem is a generalization of Theorem 1 in Chapter 4. The proof is almost identical.\n\nIf \( d \mid q - 1 \), then \( {x}^{d} - 1 \) divides \( {x}^{q - 1} - 1 \) and it follows from Corollary 2 that \( {x}^{d} - 1 \) had \( d \) distinct roots. Thus the subgroup of \( {F}^{ * } \) consisting of elements ... | Yes |
Proposition 7.1.2. Let \( \alpha \in {F}^{ * } \) . Then \( {x}^{n} = \alpha \) has solutions iff \( {\alpha }^{\left( {q - 1}\right) /d} = 1 \), where \( d = \left( {n, q - 1}\right) \) . If there are solutions, then there are exactly \( d \) solutions. | Proof. Let \( \gamma \) be a generator of \( {F}^{ * } \) and set \( \alpha = {\gamma }^{a} \) and \( x = {\gamma }^{y} \) . Then \( {x}^{n} = \alpha \) is equivalent to the congruence \( {ny} \equiv a\left( {q - 1}\right) \) . The result now follows by applying Proposition 3.3.1. | No |
Lemma 1. Let \( F \) be a finite field. The integer multiples of the identity form a subfield of \( F \) isomorphic to \( \mathbb{Z}/p\mathbb{Z} \) for some prime number \( p \) . | Proof. To avoid confusion, let us temporarily call \( e \) the identity of \( {F}^{ * } \) instead of 1. Map \( \mathbb{Z} \) to \( F \) by taking \( n \) to \( {ne} \) . This is easily seen to be a ring homomorphism. The image is a finite subring of \( F \), and so in particular it is an integral domain. The kernel is... | Yes |
Proposition 7.1.3. The number of elements in a finite field is a power of a prime. | If \( e \) is the identity of the finite field \( F \), let \( p \) be the smallest integer such that \( {pe} = 0 \) . We have seen that \( p \) must be a prime number. It is called the characteristic of \( F \) . For \( \alpha \in F \) we have \( {p\alpha } = p\left( {e\alpha }\right) = \left( {pe}\right) \alpha = 0 \... | No |
Proposition 7.1.4. If \( F \) has characteristic \( p \), then \( {\left( \alpha + \beta \right) }^{{p}^{d}} = {\alpha }^{{p}^{d}} + {\beta }^{{p}^{d}} \) for all \( \alpha ,\beta \in F \) and all positive integers \( d \) . | Proof. The proof is by induction on \( d \) . For \( d = 1 \), we have\n\n\[{\left( \alpha + \beta \right) }^{p} = {\alpha }^{p} + \mathop{\sum }\limits_{{k = 1}}^{{p - 1}}\left( \begin{array}{l} p \\ k \end{array}\right) {\alpha }^{p - k}{\beta }^{k} + {\beta }^{p} = {\alpha }^{p} + {\beta }^{p}.\n\]\n\nAll the interm... | Yes |
Lemma 2. Let \( F \) be a field. Then \( {x}^{l} - 1 \) divides \( {x}^{m} - 1 \) in \( F\left\lbrack x\right\rbrack \) iff \( l \) divides \( m \) . | Proof. Let \( m = {ql} + r \), where \( 0 \leq r < l \) . Then we have\n\n\[ \frac{{x}^{m} - 1}{{x}^{l} - 1} = {x}^{r}\frac{{x}^{ql} - 1}{{x}^{l} - 1} + \frac{{x}^{r} - 1}{{x}^{l} - 1}. \]\n\nSince \( \left( {{x}^{ql} - 1}\right) /\left( {{x}^{l} - 1}\right) = {\left( {x}^{l}\right) }^{q - 1} + {\left( {x}^{l}\right) }... | Yes |
Lemma 3. If \( a \) is a positive integer, then \( {a}^{l} - 1 \) divides \( {a}^{m} - 1 \) iff \( l \) divides \( m \) . | Proof. The proof is analogous to that of Lemma 2 with the number \( a \) playing the role of \( x \) . We leave the details to the reader. | No |
Proposition 7.1.5. Let \( F \) be a finite field of dimension \( n \) over \( \mathbb{Z}/p\mathbb{Z} \). The subfields of \( F \) are in one-to-one correspondence with the divisors of \( n \). | Proof. Suppose that \( E \) is a subfield of \( F \) and let \( d \) be its dimension over \( \mathbb{Z}/p\mathbb{Z} \). We shall show that \( d \mid n \).\n\nSince \( {E}^{ * } \) has \( {p}^{d} - 1 \) elements all satisfying \( {x}^{{p}^{d} - 1} - 1 \), we have that \( {x}^{{p}^{d} - 1} - 1 \) divides \( {x}^{{p}^{n}... | Yes |
Proposition 7.2.1. There exists a field \( K \) containing \( k \) and an element \( \alpha \in K \) such that \( f\left( \alpha \right) = 0 \) . | Proof. We proved in Chapter 1 that \( k\left\lbrack x\right\rbrack \) is a principal ideal domain. It follows that \( \left( {f\left( x\right) }\right) \) is a maximal ideal and thus \( k\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \) is a field. Let \( {K}^{\prime } = k\left\lbrack x\right\rbrack /\l... | Yes |
Proposition 7.2.2. The elements \( 1,\alpha ,{\alpha }^{2},\ldots ,{\alpha }^{n - 1} \) are a vector space basis for \( k\left( \alpha \right) \) over \( k \), where \( n \) is the degree of \( f\left( x\right) \) . | The proof of this proposition is the same as that of Proposition 6.1.8 and its corollary. One replaces \( \mathbb{Q} \) by \( k \) and the complex number \( \alpha \) of that proposition by the above \( \alpha \) . | No |
\[ {x}^{{p}^{n}} - x = \mathop{\prod }\limits_{{d \mid n}}{F}_{d}\left( x\right) \] | Proof. First notice that if \( f\left( x\right) \) divides \( {x}^{{p}^{n}} - x \), then \( f{\left( x\right) }^{2} \) does not divide \( {x}^{{p}^{n}} - x \) . This follows since if \( {x}^{{p}^{n}} - x = f{\left( x\right) }^{2}g\left( x\right) \) we obtain\n\n\[ - 1 = {2f}\left( x\right) {f}^{\prime }\left( x\right) ... | Yes |
Corollary 2. \( {N}_{n} = {n}^{-1}\mathop{\sum }\limits_{{d \mid n}}\mu \left( {n/d}\right) {p}^{d} \) . | Proof. Apply the Möbius inversion formula (Theorem 2 of Chapter 2) to the equation in Corollary 1. | No |
Corollary 3. For each integer \( n \geq 1 \), there exists an irreducible polynomial of degree \( n \) in \( \mathbb{Z}/p\mathbb{Z}\left\lbrack x\right\rbrack \) . | Proof. \( {N}_{n} = {n}^{-1}\left( {{p}^{n} - \cdots + {p\mu }\left( n\right) }\right) \) by Corollary 2 . The term in parentheses cannot be zero since it is the sum of distinct powers of \( p \) with coefficients 1 and -1 . | No |
Proposition 8.1.1. Let \( \chi \) be a multiplicative character and \( a \in {F}_{p}^{ * } \) . Then\n\n(a) \( \chi \left( 1\right) = 1 \) .\n\n(b) \( \chi \left( a\right) \) is \( a\left( {p - 1}\right) \) st root of unity.\n\n(c) \( \chi \left( {a}^{-1}\right) = \chi {\left( a\right) }^{-1} = \overline{\chi \left( a\... | Proof. \( \chi \left( 1\right) = \chi \left( {1 \cdot 1}\right) = \chi \left( 1\right) \chi \left( 1\right) \) . Thus \( \chi \left( 1\right) = 1 \), since \( \chi \left( 1\right) \neq 0 \) .\n\nTo prove part (b), notice that \( {a}^{p - 1} = 1 \) implies that \( 1 = \chi \left( 1\right) = \chi \left( {a}^{p - 1}\right... | Yes |
Proposition 8.1.2. Let \( \chi \) be a multiplicative character. If \( \chi \neq \varepsilon \), then \( \mathop{\sum }\limits_{t}\chi \left( t\right) = 0 \) , where the sum is over all \( t \in {F}_{p} \) . If \( \chi = \varepsilon \), the value of the sum is \( p \) . | Proof. The last assertion is obvious, so we may assume that \( \chi \neq \varepsilon \) . In this case there is an \( a \in {F}_{p}^{ * } \) such that \( \chi \left( a\right) \neq 1 \) . Let \( T = \mathop{\sum }\limits_{t}\chi \left( t\right) \) . Then\n\n\[ \chi \left( a\right) T = \mathop{\sum }\limits_{t}\chi \left... | Yes |
Proposition 8.1.3. The group of characters is a cyclic group of order \( p - 1 \) . If \( a \in {F}_{p}^{ * } \) and \( a \neq 1 \), then there is a character \( \chi \) such that \( \chi \left( a\right) \neq 1 \) . | Proof. We know that \( {F}_{p}^{ * } \) is cyclic (see Theorem 1 of Chapter 4). Let \( g \in {F}_{p}^{ * } \) be a generator. Then every \( a \in {F}_{p}^{ * } \) is equal to a power of \( g \) . If \( a = {g}^{l} \) and \( \chi \) is a character, then \( \chi \left( a\right) = \chi {\left( g\right) }^{l} \) . This sho... | Yes |
Proposition 8.1.4. If \( a \in {F}_{p}^{ * }, n \mid p - 1 \), and \( {x}^{n} = a \) is not solvable, then there is a character \( \chi \) such that\n\n(a) \( {\chi }^{n} = \varepsilon \) .\n\n(b) \( \chi \left( a\right) \neq 1 \) . | Proof. Let \( g \) and \( \lambda \) be as in Proposition 8.1.3 and set \( \chi = {\lambda }^{\left( {p - 1}\right) /n} \) . Then \( \chi \left( g\right) = {\lambda }^{\left( {p - 1}\right) /n}\left( g\right) = \lambda {\left( g\right) }^{\left( {p - 1}\right) /n} = {e}^{{2\pi i}/n} \) . Now \( a = {g}^{l} \) for some ... | Yes |
Proposition 8.1.5. \( N\left( {{x}^{n} = a}\right) = \mathop{\sum }\limits_{{{x}^{n} = \varepsilon }}\chi \left( a\right) \) where the sum is over all characters of order dividing \( n \) . | Proof. We claim first that there are exactly \( n \) characters of order dividing \( n \) . Since the value of \( \chi \left( g\right) \) for such a character must be an \( n \) th root of unity, there are at most \( n \) such characters. In Proposition 8.1.4, we found a character \( \chi \) such that \( \chi \left( g\... | Yes |
Proposition 8.2.1. If \( a \neq 0 \) and \( \chi \neq \varepsilon \), we have \( {g}_{a}\left( \chi \right) = \chi \left( {a}^{-1}\right) {g}_{1}\left( \chi \right) \) . If \( a \neq 0 \) and \( \chi = \varepsilon \) we have \( {g}_{a}\left( \varepsilon \right) = 0 \) . If \( a = 0 \) and \( \chi \neq \varepsilon \), w... | Proof. Suppose that \( a \neq 0 \) and that \( \chi \neq \varepsilon \) . Then\n\n\[ \chi \left( a\right) {g}_{a}\left( \chi \right) = \chi \left( a\right) \mathop{\sum }\limits_{t}\chi \left( t\right) {\zeta }^{at} = \mathop{\sum }\limits_{t}\chi \left( {at}\right) {\zeta }^{at} = {g}_{1}\left( \chi \right) . \]\n\nTh... | Yes |
Proposition 8.2.2. If \( \chi \neq \varepsilon \), then \( \left| {g\left( \chi \right) }\right| = \sqrt{p} \) . | Proof. The idea is to evaluate the sum \( \mathop{\sum }\limits_{a}{g}_{a}\left( \chi \right) \overline{{g}_{a}\left( \chi \right) } \) in two ways.\n\nIf \( a \neq 0 \), then by Proposition 8.2.1, \( \overline{{g}_{a}\left( \chi \right) } = \overline{\chi \left( {a}^{-1}\right) g\left( \chi \right) } = \chi \left( a\r... | Yes |
Theorem 1. Let \( \chi \) and \( \lambda \) be nontrivial characters. Then\n\n(a) \( J\left( {\varepsilon ,\varepsilon }\right) = p \) .\n\n(b) \( J\left( {\varepsilon ,\chi }\right) = 0 \) .\n\n(c) \( J\left( {\chi ,{\chi }^{-1}}\right) = - \chi \left( {-1}\right) \) .\n\n(d) If \( {\chi \lambda } \neq \varepsilon \),... | Proof. Part (a) is immediate, and part (b) is an immediate consequence of Proposition 8.1.2.\n\nTo prove part (c), notice that\n\n\[ J\left( {\chi ,{\chi }^{-1}}\right) = \mathop{\sum }\limits_{{a + b = 1}}\chi \left( a\right) {\chi }^{-1}\left( b\right) = \mathop{\sum }\limits_{\substack{{a + b = 1} \\ {b \neq 0} }}\c... | Yes |
Proposition 8.3.1. If \( p \equiv 1\\left( 4\\right) \), then there exist integers \( a \) and \( b \) such that \( {a}^{2} + {b}^{2} = p \) . | Proof. If \( p \equiv 1\\left( 4\\right) \), there is a character \( \\chi \) of order 4 (if \( \\lambda \) has order \( p - 1 \), let \( \\chi = {\\lambda }^{\\left( {p - 1}\\right) /4} \) ). The values of \( \\chi \) are in the set \( \\{ 1, - 1, i, - i\\} \), where \( i = \\sqrt{-1} \) . Thus \( J\\left( {\\chi ,\\c... | Yes |
Proposition 8.3.2. If \( p \equiv 1\\left( 3\\right) \), then there are integers \( A \) and \( B \) such that \( {4p} = {A}^{2} + {27}{B}^{2} \). In this representation of \( {4p}, A \) and \( B \) are uniquely determined up to sign. | Proof. The proof of the uniqueness is left to the Exercises. | No |
Proposition 8.3.3. Suppose that \( p \equiv 1\left( n\right) \) and that \( \chi \) is a character of order \( n > 2 \) . Then\n\n\[ g{\left( \chi \right) }^{n} = \chi \left( {-1}\right) {pJ}\left( {\chi ,\chi }\right) J\left( {\chi ,{\chi }^{2}}\right) \cdots J\left( {\chi ,{\chi }^{n - 2}}\right) . \] | Proof. Using part (d) of Theorem 1 we have \( g{\left( \chi \right) }^{2} = J\left( {\chi ,\chi }\right) g\left( {\chi }^{2}\right) \) . Multiply both sides by \( g\left( \chi \right) \) and we get \( g{\left( \chi \right) }^{3} = J\left( {\chi ,\chi }\right) J\left( {\chi ,{\chi }^{2}}\right) g\left( {\chi }^{3}\right... | Yes |
Proposition 8.3.4. Suppose that \( p \equiv 1\left( 3\right) \) and that \( \chi \) is a cubic character. Set \( J\left( {\chi ,\chi }\right) = a + {b\omega } \) as above. Then\n\n(a) \( b \equiv 0\left( 3\right) \).\n\n(b) \( a \equiv - 1\left( 3\right) \). | Proof. We shall work with congruences in the ring of algebraic integers as in Chapter 6:\n\n\[ g{\left( \chi \right) }^{3} = {\left( \mathop{\sum }\limits_{t}\chi \left( t\right) {\zeta }^{t}\right) }^{3} \equiv \mathop{\sum }\limits_{t}\chi {\left( t\right) }^{3}{\zeta }^{3t}\left( 3\right) .\n\]\n\nSince \( \chi \lef... | Yes |
Theorem 2. Suppose that \( p \equiv 1\\left( 3\\right) \) . Then there are integers \( A \) and \( B \) such that \( {4p} = {A}^{2} + {27}{B}^{2} \) . If we require that \( A \equiv 1\\left( 3\\right), A \) is uniquely determined, and\n\n\\[ \nN\\left( {{x}^{3} + {y}^{3} = 1}\\right) = p - 2 + A.\n\\] | Proof. We have already shown that \( N\\left( {{x}^{3} + {y}^{3} = 1}\\right) = p - 2 + 2\\operatorname{Re}J\\left( {\\chi ,\\chi }\\right) \) . Since \( J\\left( {\\chi ,\\chi }\\right) = a + {b\\omega } \) as above, we have \( \\operatorname{Re}J\\left( {\\chi ,\\chi }\\right) = \\left( {{2a} - b}\\right) /2 \) . Thu... | No |
(a) \( {J}_{0}\left( {\varepsilon ,\varepsilon ,\ldots ,\varepsilon }\right) = J\left( {\varepsilon ,\varepsilon ,\ldots ,\varepsilon }\right) = {p}^{l - 1} \) . | Proof. If \( {t}_{1},{t}_{2},\ldots ,{t}_{l - 1} \) are chosen (arbitrarily) in \( {F}_{p} \), then \( {t}_{l} \) is uniquely determined by the condition \( {t}_{1} + {t}_{2} + \cdots + {t}_{l - 1} + {t}_{l} = 0 \) . Thus \( {J}_{0}\left( {\varepsilon ,\varepsilon ,\ldots ,\varepsilon }\right) = \) \( {p}^{l - 1} \) . ... | Yes |
Theorem 3. Assume that \( {\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{r} \) are nontrivial and also that \( {\chi }_{1}{\chi }_{2}\cdots {\chi }_{r} \) is nontrivial. Then\n\n\[ g\left( {\chi }_{1}\right) g\left( {\chi }_{2}\right) \cdots g\left( {\chi }_{r}\right) = J\left( {{\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{r}}\r... | Proof. Let \( \psi : {F}_{p} \rightarrow \mathbb{C} \) be defined by \( \psi \left( t\right) = {\zeta }^{t} \) . Then \( \psi \left( {{t}_{1} + {t}_{2}}\right) = \psi \left( {t}_{1}\right) \psi \left( {t}_{2}\right) \) , and \( g\left( \chi \right) = \sum \chi \left( t\right) \psi \left( t\right) \) . The introduction ... | Yes |
Corollary 1. Suppose that \( {\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{r} \) are nontrivial and that \( {\chi }_{1}{\chi }_{2}\cdots {\chi }_{r} \) is trivial. Then\n\n\[ g\left( {\chi }_{1}\right) g\left( {\chi }_{2}\right) \cdots g\left( {\chi }_{r}\right) = {\chi }_{r}\left( {-1}\right) {pJ}\left( {{\chi }_{1},{\chi ... | Proof. \( g\left( {\chi }_{1}\right) g\left( {\chi }_{2}\right) \cdots g\left( {\chi }_{r - 1}\right) = J\left( {{\chi }_{1},\ldots ,{\chi }_{r - 1}}\right) g\left( {{\chi }_{1}{\chi }_{2}\cdots {\chi }_{r - 1}}\right) \) by Theorem 3. Multiply both sides by \( g\left( {\chi }_{r}\right) \) . Since \( {\chi }_{1}{\chi ... | Yes |
Corollary 2. Let the hypotheses be as in Corollary 1. Then\n\n\[ J\left( {{\chi }_{1},\ldots ,{\chi }_{r}}\right) = - {\chi }_{r}\left( {-1}\right) J\left( {{\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{r - 1}}\right) . \]\n\n\( \left\lbrack {\text{If}r = 2\text{, we set}J\left( {\chi }_{1}\right) = 1\text{.}}\right\rbrack ... | Proof. If \( r = 2 \), this is the assertion of part (c) of Theorem 1 .\n\nSuppose that \( r > 2 \) . In the proof of Theorem 3 use the hypothesis that \( {\chi }_{1}{\chi }_{2}\cdots {\chi }_{r} = \varepsilon \) . This yields\n\n\[ g\left( {\chi }_{1}\right) g\left( {\chi }_{2}\right) \cdots g\left( {\chi }_{r}\right)... | Yes |
Theorem 4. Assume that \( {\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{r} \) are nontrivial.\n\n(a) If \( {\chi }_{1}{\chi }_{2}\cdots {\chi }_{r} \neq \varepsilon \), then\n\n\[ \left| {J\left( {{\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{r}}\right) }\right| = {p}^{\left( {r - 1}\right) /2}. \]\n\n(b) If \( {\chi }_{1}{\chi ... | Proof. If \( \chi \) is nontrivial, \( \left| {g\left( \chi \right) }\right| = \sqrt{p} \) . Part (a) follows directly from Theorem 3. Part (b) follows similarly from part (c) of Proposition 8.5.1 and from Corollary 2 to Theorem 3. | Yes |
Proposition 9.1.1. \( \alpha \in D \) is a unit iff \( {N\alpha } = 1 \) . The units in \( D \) are \( 1, - 1,\omega \) , \( - \omega ,{\omega }^{2} \), and \( - {\omega }^{2} \) . | Proof. If \( {N\alpha } = 1,\alpha \bar{\alpha } = 1 \), which implies that \( \alpha \) is a unit since \( \bar{\alpha } \in D \) . If \( \alpha \) is a unit, there is a \( \beta \in D \) such that \( {\alpha \beta } = 1 \) . Thus \( {N\alpha N\beta } = 1 \) . Since \( {N\alpha } \) and \( {N\beta } \) are positive in... | Yes |
Proposition 9.1.2. If \( \pi \) is a prime in \( D \), then there is a rational prime \( p \) such that \( {N\pi } = p \) or \( {p}^{2} \). In the former case \( \pi \) is not associate to a rational prime; in the latter case \( \pi \) is associate to \( p \). | Proof. We have \( {N\pi } = n > 1 \), or \( \pi \bar{\pi } = n.n \) is a product of rational primes. Thus \( \pi \mid p \) for some rational prime \( p \). If \( p = {\pi \gamma },\gamma \in D \), then \( {N\pi N\gamma } = {Np} = {p}^{2} \). Thus either \( {N\pi } = {p}^{2} \) and \( {N\gamma } = 1 \) or \( {N\pi } = p... | Yes |
Proposition 9.1.3. If \( \pi \in D \) is such that \( {N\pi } = p \), a rational prime, then \( \pi \) is a prime in \( D \) . | Proof. If \( \pi \) were not prime in \( D \), then we could write \( \pi = {\rho \gamma } \) with \( {N\rho } \) , \( {N}_{\gamma } > 1 \) . Then \( p = {N\pi } = {N\rho N\gamma } \), which cannot be true since \( p \) is prime in \( \mathbb{Z} \) . Thus \( \pi \) is a prime in \( D \) . | Yes |
Proposition 9.1.4. Suppose that \( p \) and \( q \) are rational primes. If \( q \equiv 2\left( 3\right) \), then \( q \) is prime in D. If \( p \equiv 1\left( 3\right) \), then \( p = \pi \bar{\pi } \), where \( \pi \) is prime in D. Finally \( 3 = - {\omega }^{2}{\left( 1 - \omega \right) }^{2} \), and \( 1 - \omega ... | Proof. Suppose that \( p \) were not a prime. Then \( p = {\pi \gamma } \), with \( {N\pi } > 1,{N\gamma } > 1 \) . Thus \( {p}^{2} = {N\pi N\gamma } \) and \( {N\pi } = p \) . Let \( \pi = a + {b\omega } \) . Then \( p = {a}^{2} - {ab} + {b}^{2} \) or \( {4p} = {\left( 2a - b\right) }^{2} + 3{b}^{2} \), yielding \( p ... | Yes |
Proposition 9.2.1. Let \( \pi \in D \) be a prime. Then \( D/{\pi D} \) is a finite field with \( {N\pi } \) elements. | Proof. We first show that \( D/{\pi D} \) is a field. Let \( \alpha \in D \) be such that \( \alpha ≢ 0\left( \pi \right) \) . By Corollary 1 to Proposition 1.3.2 there exist elements \( \beta ,\gamma \in D \) such that \( {\beta \alpha } + {\gamma \pi } = 1 \) . Thus \( {\beta \alpha } \equiv 1\left( \pi \right) \), w... | Yes |
Proposition 9.3.1. If \( \pi \nmid \alpha \), then\n\n\[{\alpha }^{{N\pi } - 1} \equiv 1\left( \pi \right)\] | If the norm of \( \pi \) is different from 3, then the residue classes of \( 1,\omega \), and \( {\omega }^{2} \) are distinct in \( D/{\pi D} \) . To see this, suppose, for example, that \( \omega \equiv 1\left( \pi \right) \) . Then \( \pi \mid \left( {1 - \omega }\right) \), and since \( 1 - \omega \) is prime, \( \... | No |
Proposition 9.3.2. Suppose that \( \pi \) is a prime such that \( {N\pi } \neq 3 \) and that \( \pi {\chi \alpha } \) . Then there is a unique integer \( m = 0,1 \), or 2 such that \( {\alpha }^{\left( {{N\pi } - 1}\right) /3} \equiv {\omega }^{m}\left( \pi \right) \) . | Proof. We know that \( \pi \) divides \( {\alpha }^{{N\pi } - 1} - 1 \) . Now,\n\n\[ \n{\alpha }^{{N\pi } - 1} - 1 = \left( {{\alpha }^{\left( {{N\pi } - 1}\right) /3} - 1}\right) \left( {{\alpha }^{\left( {{N\pi } - 1}\right) /3} - \omega }\right) \left( {{\alpha }^{\left( {{N\pi } - 1}\right) /3} - {\omega }^{2}}\rig... | Yes |
(a) \( {\left( \alpha /\pi \right) }_{3} = 1 \) iff \( {x}^{3} \equiv \alpha \left( \pi \right) \) is solvable, i.e., iff \( \alpha \) is a cubic residue. | Proof. Part (a) is a special case of Proposition 7.1.2. Take \( F = D/{\pi D}, q = {N\pi } \) , and \( n = 3 \) in that proposition. | No |
(a) \( \overline{{\chi }_{\pi }\left( \alpha \right) } = {\chi }_{\pi }{\left( \alpha \right) }^{2} = {\chi }_{\pi }\left( {\alpha }^{2}\right) \) . | (a) \( {\chi }_{\pi }\left( \alpha \right) \) is by definition \( 1,\omega \), or \( {\omega }^{2} \), and each of these numbers squared is equal to its conjugate. | Yes |
Proposition 9.3.5. Suppose that \( {N\pi } = p \equiv 1\left( 3\right) \) . Among the associates of \( \pi \) exactly one is primary. | Proof. Write \( \pi = a + {b\omega } \) . The associates of \( \pi \) are \( \pi ,{\omega \pi },{\omega }^{2}\pi , - \pi , - {\omega \pi } \), and \( - {\omega }^{2}\pi \) . In terms of \( a \) and \( b \) these elements can be expressed as\n\n(a) \( a + {b\omega } \) .\n\n(b) \( - b + \left( {a - b}\right) \omega \) .... | Yes |
Lemma 1. \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = \pi \) . | Proof. Let \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = {\pi }^{\prime } \) . Since \( \pi \bar{\pi } = p = {\pi }^{\prime }{\bar{\pi }}^{\prime } \) we have \( \pi \left| {{\pi }^{\prime }\text{or}\pi }\right| {\bar{\pi }}^{\prime } \) .\n\nSince all the primes involved are primary we must have \( \pi = {\pi }^... | Yes |
Proposition 9.6.1. \( {x}^{3} \equiv 2\left( \pi \right) \) is solvable iff \( \pi \equiv 1\left( 2\right) \), i.e., iff \( a \equiv 1\left( 2\right) \) and \( b \equiv 0\left( 2\right) \) . | It is possible to formulate this proposition in another way. Let \( \pi = a + {b\omega } \) be a primary complex prime and \( p = {N\pi } = {a}^{2} - {ab} + {b}^{2} \) . Then \( {4p} = \) \( {\left( 2a - b\right) }^{2} + 3{b}^{2} \) . If we set \( A = {2a} - b \) and \( B = b/3 \), then \( {4p} = {A}^{2} + {27}{B}^{2} ... | No |
Proposition 9.6.2. If \( p \equiv 1\\left( 3\\right) \), then \( {x}^{3} \equiv 2\\left( p\\right) \) is solvable iff there are integers \( C \) and \( D \) such that \( p = {C}^{2} + {27}{D}^{2} \) . | Proof. If \( {x}^{3} \equiv 2\\left( p\\right) \) is solvable, so is \( {x}^{3} \equiv 2\\left( \\pi \\right) \) and thus \( \\pi \equiv 1\\left( 2\\right) \) by Proposition 9.6.1. We have\n\n\\[ \n{4p} = {A}^{2} + {27}{B}^{2},\\;\\text{ where }A = {2a} - b, B = \\frac{b}{3}.\n\\]\n\nSince \( b \) is even, so are \( B ... | Yes |
Lemma 1. If \( \pi \) is irreducible then there is a prime \( p \in \mathbb{Z} \) such that \( \pi \mid p \) . | Proof. \( N\left( \pi \right) = \pi \bar{\pi } = n = {p}_{1}\cdots {p}_{s},{p}_{i} \) prime, \( {p}_{i} \in \mathbb{Z} \) . Thus \( \pi \mid {p}_{i} \) for some \( i \) . | Yes |
Lemma 2. If \( \alpha \in D \), and \( N\left( \alpha \right) \) is prime then \( \alpha \) is irreducible. | Proof. If \( \alpha = {\mu \lambda } \) then \( N\left( \alpha \right) = N\left( \mu \right) N\left( \lambda \right) \) . Since \( N\left( \alpha \right) \) is prime it follows that \( N\left( \mu \right) = 1 \) or \( N\left( \lambda \right) = 1 \) . Thus either \( \mu \) or \( \lambda \) is a unit. | Yes |
Lemma 3. \( 1 + i \) is irreducible and \( 2 = - i{\left( 1 + i\right) }^{2} \) is the prime factorization of 2 in \( D \) . | Proof. \( \;N\left( {1 + i}\right) = 2 \) and so the first assertion follows from Lemma 2. The second assertion results from a direct calculation. | No |
Lemma 4. If \( q \equiv 3\left( 4\right) \) is a prime in \( \mathbb{Z} \), then \( q \) is irreducible considered as an element of \( D \) . | Proof. If \( q \) were not irreducible in \( D \), then \( q = {\alpha \beta } \) with \( N\left( \alpha \right) > 1 \) and \( N\left( \beta \right) > 1 \) . Taking norms we find \( {q}^{2} = N\left( \alpha \right) N\left( \beta \right) \) . It follows that \( q = N\left( \alpha \right) \) . If \( \alpha = a + {bi} \) ... | Yes |
Lemma 5. If \( p \) is prime, \( p \equiv 1\\left( 4\\right) \) then there is an irreducible \( \\pi \) such that \( p = \\pi \\bar{\\pi } \) . Furthermore \( \\left( \\pi \\right) \\neq \\left( \\bar{\\pi }\\right) \) . | Proof. The first statement is part (a) of Proposition 8.3.1. Another proof not using Jacobi sums is the following. Since \( p \\equiv 1\\left( 4\\right) \) there is, by Proposition 5.1.2, an integer \( a \) with \( {a}^{2} \\equiv - 1\\left( p\\right) \) . Thus \( p \\mid {a}^{2} + 1 = \\left( {a + i}\\right) \\left( {... | No |
Lemma 6. A nonunit \( \alpha \) is primary iff either \( a \equiv 1\\left( 4\\right), b \equiv 0\\left( 4\\right) \) or \( a \equiv 3\\left( 4\\right) \) , \( b \equiv 2\\left( 4\\right) \) . | Proof. Since \( {\\left( 1 + i\\right) }^{3} = {2i}\\left( {1 + i}\\right) \) it follows that \( a + {bi} \) is primary iff\n\n\[ \n\\frac{\\left( {a - 1}\\right) + {bi}}{2 + {2i}} = \\frac{a + b - 1}{4} + \\frac{b - a + 1}{4}i \\in D.\n\]\n\nThis is equivalent to the congruences \( a + b \\equiv 1\\left( 4\\right), a ... | Yes |
Lemma 7. Let \( \alpha \in D \) be a nonunit, \( \left( {1 + i}\right) \times \alpha \) . Then there is a unique unit \( u \) such that \( {u\alpha } \) is primary. | Proof. There is a unit \( \varepsilon \) such that \( {\varepsilon \alpha } = a + {bi} \) where \( a \) is odd and \( b \) is even. Multiplying if necessary by -1, Lemma 6 shows that \( \alpha \) has a primary associate. If \( {u}_{1} \) and \( {u}_{2} \) are units such that \( {u}_{1}\alpha \) and \( {u}_{2}\alpha \) ... | Yes |
Lemma 8. A primary element can be written as the product of primary irreducibles. | Proof. Let \( \alpha \in D \) be primary. Then there are rational primes \( {q}_{i} \equiv 3\left( 4\right) \) , primary irreducibles \( {\pi }_{i}, N\left( {\pi }_{i}\right) \equiv 1\left( 4\right) \) and a unit \( u \) such that \( \alpha = u{\pi }_{1}\cdots \) \( {\pi }_{t}\left( {-{q}_{1}}\right) \cdots \left( {-{q... | No |
Proposition 9.8.1. The residue class ring \( D/{\pi D} \) is a finite field with \( N\left( \pi \right) \) elements. | Proof. The proof proceeds in exactly the same way as Proposition 9.2.1, replacing the classification of irreducibles in \( \mathbb{Z}\left\lbrack \omega \right\rbrack \) by the corresponding classification in \( D = \mathbb{Z}\left\lbrack i\right\rbrack \) . | No |
Proposition 9.8.2. If \( \pi \times \alpha ,\left( \pi \right) \neq \left( {1 + i}\right) \) there exists a unique integer \( j \) , \( 0 \leq j \leq 3 \) such that\n\n\[{\alpha }^{\left( {N\left( \pi \right) - 1}\right) /4} \equiv {i}^{j}\left( \pi \right)\] | Proof. It is easy to see that the residue classes of \( 1, - 1, i, - i \) are distinct. They are the roots of \( {x}^{4} \equiv 1\left( \pi \right) \) . However the residue class of \( {\alpha }^{\left( {N\left( \pi \right) - 1}\right) /4} \) is also a solution to \( {x}^{4} \equiv 1\left( \pi \right) \) by the above c... | No |
(a) If \( \pi \downarrow \alpha \) then \( {\chi }_{\pi }\left( \alpha \right) = 1 \Leftrightarrow {x}^{4} \equiv \alpha \left( \pi \right) \) has a solution in \( D \) . | Proof. Part (a) follows from Proposition 7.1.2. | No |
Proposition 9.8.4. Let \( q \) be prime, \( q \equiv 3\left( 4\right) \) . Then \( {\chi }_{q}\left( a\right) = 1 \) for \( a \in \mathbb{Z}, q \times a \) . | Proof. \( N\left( q\right) = {q}^{2} \) . Thus\n\n\[ \n{\chi }_{q}\left( a\right) \equiv {a}^{\left( {{q}^{2} - 1}\right) /4} = {\left( {a}^{q - 1}\right) }^{\left( {q + 1}\right) /4} \equiv 1\left( q\right) , \n\] \n\nby Fermat's Little Theorem. | Yes |
Proposition 9.8.5. Let \( \\alpha \\in \\mathbb{Z},\\alpha \\neq 0 \\), and \( a \\in \\mathbb{Z} \) be an odd nonunit. If \( \\left( {a,\\alpha }\\right) = 1 \\) , then\n\n\[ \n{\\chi }_{a}\\left( \\alpha \\right) = 1\\text{.} \n\] | Proof. We may assume \( a > 0 \) . Write \( a = \\prod {p}_{i}\\prod {q}_{i} \) where \( {p}_{i},{q}_{i} \) are prime, \( {p}_{i} \\equiv 1\\left( 4\\right) \) and \( {q}_{i} \\equiv 3\\left( 4\\right) \) . By Proposition 9.8.4 we need only verify that \( {\\chi }_{{p}_{i}}\\left( \\alpha \\right) = 1 \) . If \( {p}_{i... | Yes |
Proposition 9.8.6. If \( n \neq 1 \) is an integer \( n \equiv 1\\left( 4\\right) \), then \( {\\chi }_{n}\\left( i\\right) = {\\left( -1\\right) }^{\\left( {n - 1}\\right) /4} \) . | Proof. Note that \( n \) may be negative. If \( n \) is a positive prime \( p \equiv 1\\left( 4\\right) \) then writing \( p = \\pi \\bar{\\pi } \) one has\n\n\[ \n{\\chi }_{p}\\left( i\\right) = {\\chi }_{\\pi }\\left( i\\right) {\\chi }_{\\dot{\\pi }}\\left( i\\right) = {\\left( {i}^{\\left( {p - 1}\\right) /4}\\righ... | No |
Theorem 2. \( {\chi }_{\pi }\left( \lambda \right) = {\chi }_{\lambda }\left( \pi \right) {\left( -1\right) }^{\left( {\left( {N\left( \lambda \right) - 1}\right) /4}\right) \left( {\left( {N\left( \pi \right) - 1}\right) /4}\right) } \) . | If \( \lambda \) and \( \pi \) are primary, where \( \lambda = c + {di} \) and \( \pi = a + b\mathrm{i} \), it is simple to see that \( \left( {\left( {N\left( \lambda \right) - 1}\right) /4}\right) \left( {\left( {N\left( \pi \right) - 1}\right) /4}\right) \) and \( \left( {\left( {a - 1}\right) /2}\right) (\left( {c ... | Yes |
Proposition 9.9.1. \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = {\chi }_{\pi }\left( {-1}\right) J\left( {{\chi }_{\pi },\psi }\right) \) | Proof. By Theorem 1, Chapter 8, one has \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = g{\left( {\chi }_{\pi }\right) }^{2}/g\left( \psi \right) \). Thus\n\n\[ J{\left( {\chi }_{\pi },{\chi }_{\pi }\right) }^{2} = \frac{g{\left( {\chi }_{\pi }\right) }^{4}}{g{\left( \psi \right) }^{2}} = {\chi }_{\pi }\left( {-1}\... | Yes |
Proposition 9.9.3. \( - {\chi }_{\pi }\left( {-1}\right) J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) \) is primary. | Proof. Clearly\n\n\[ J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = 2\mathop{\sum }\limits_{{t = 2}}^{{\left( {p - 1}\right) /2}}{\chi }_{\pi }\left( t\right) {\chi }_{\pi }\left( {1 - t}\right) + {\chi }_{\pi }{\left( \frac{p + 1}{2}\right) }^{2}. \]\n\nBut any unit in \( D \) is congruent to 1 modulo \( 1 + i \) . ... | Yes |
Proposition 9.9.4. \( - {\chi }_{\pi }\left( {-1}\right) J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = \pi \) . | Proof. By Lemma 7 of Section 7 it is enough to show that the left- and righthand sides differ by a unit. Now \( J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) \equiv \mathop{\sum }\limits_{{t = 1}}^{{p - 1}}{t}^{\left( {p - 1}\right) /4}{\left( 1 - t\right) }^{\left( {p - 1}\right) /4}\left( \pi \right) \) . By Exercis... | Yes |
Proposition 9.9.6. Let \( q > 0 \) be a real irreducible in D. Then\n\n\[{\chi }_{\pi }\left( {-q}\right) = {\chi }_{q}\left( \pi \right)\] | Proof. Since \( q \equiv 3\left( 4\right) \) one has\n\n\[g{\left( {\chi }_{\pi }\right) }^{q} \equiv \mathop{\sum }\limits_{{j = 1}}^{{p - 1}}{\chi }_{\pi }{\left( j\right) }^{q}{\zeta }^{qj} \equiv \sum {\chi }_{\pi }^{3}\left( j\right) {\zeta }^{qj}\left( q\right)\]\n\n\[ \equiv {\chi }_{\pi }\left( q\right) g\left(... | Yes |
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