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Corollary 7.3.6 Let \( T \) be a tableau of staircase shape, with \( n \) cells. Then, \[ {p}^{n}\left( T\right) = {T}^{\prime } \] | Proof. Let \( S \) be a tableau that is elementary dual equivalent to \( T \) . Let \( \{ j, j + 1, j + 2\} \) be the entries involved in an elementary dual equivalence of \( S \) and \( T \) . By Theorem 7.3.5 we then conclude that \( {p}^{n}\left( T\right) \) is elementary dual equivalent to \( {p}^{n}\left( S\right)... | Yes |
Proposition 7.4.1 Let \( T \) be a tableau of staircase shape and \( U \) be a miniature final segment of \( T \) . Then, \( U \) is uniquely determined by \( \widehat{p}\left( U\right) \) . | Proof. Since \( U \) is a final segment of a tableau of staircase shape, its shape must be isomorphic (as a poset) to either \( {P}_{2} \) or \( {P}_{3} \) (see Figure 7.7). Suppose it is \( {P}_{2} \) . Then, \( U \) is one of the two tableaux shown in Figure 7.8.\n\n and \( T \) be two elementary dual equivalent tableaux of staircase shape, and let \( \{ j, j + 1, j + 2\} \) be the entries involved in an elementary dual equivalence. Then, \( \widehat{p}\left( S\right) \) and \( \widehat{p}\left( T\right) \) differ at most in positions \( j, j + 1 \), and \... | Proof. The statements before \ | No |
Proposition 7.4.3 Let \( \mu \) be a miniature final segment of \( {\delta }_{n} \), and let \( U \approx \) \( V \) be two tableaux of shape \( \mu \) . Then, \( \{ \widehat{p}\left( U\right) ,\widehat{p}\left( V\right) \} \) is an s-relation, and any two elements of \( \mathcal{P}\left( {\delta }_{n}\right) \) are re... | Proof. The fact that \( \{ \widehat{p}\left( U\right) ,\widehat{p}\left( V\right) \} \) is an \( s \) -relation follows in exactly the same way as in the next to last example. Now, if \( S \) and \( T \) are two elementary dual equivalent tableaux of shape \( {\delta }_{n} \) then by Theorem 7.4.2 (and its proof), \( \... | Yes |
Proposition 7.4.5 Let \( \{ X, Y\} \) be an s-relation. Then, \( \left| {{\widehat{p}}^{-1}\left( {AXB}\right) }\right| = \) \( \left| {{\widehat{p}}^{-1}\left( {AYB}\right) }\right| \) whenever \( {AXB},{AYB} \in {\left\lbrack n\right\rbrack }^{N} \) . In particular, \( {AXB} \in \mathcal{P}\left( {\delta }_{n}\right)... | Proof. We may clearly assume that \( {\widehat{p}}^{-1}\left( {AXB}\right) \neq \varnothing \) (say). We construct a map \( \psi : {\widehat{p}}^{-1}\left( {AXB}\right) \rightarrow {\widehat{p}}^{-1}\left( {AYB}\right) \) as follows. Let \( S \in {\widehat{p}}^{-1}\left( {AXB}\right) \), and let \( U \) be the final se... | Yes |
Theorem 7.4.6 Let \( T \) be a tableau of shape \( {\delta }_{n} \) . Then, \( T \) is uniquely determined by \( \widehat{p}\left( T\right) \) . Equivalently, \( \left| {{\widehat{p}}^{-1}\left( X\right) }\right| = 1 \) for all \( X \in \mathcal{P}\left( {\delta }_{n}\right) \) . | Proof. It follows immediately from Propositions 7.4.4 and 7.4.5 that\n\n\[ \left| {{\widehat{p}}^{-1}\left( X\right) }\right| = \left| {{\widehat{p}}^{-1}\left( Y\right) }\right| \]\n\nfor all \( X, Y \in \mathcal{P}\left( {\delta }_{n}\right) \) . To conclude the proof, we therefore just have to show that there is at ... | Yes |
Theorem 7.4.7 The map \( \widehat{p} \) is a bijection between the set of all tableaux of shape \( {\delta }_{n} \) and \( \mathcal{R}\left( {w}_{0}\right) \), where \( {w}_{0} \) denotes the longest element of \( {S}_{n + 1} \) . | Proof. By the preceding remarks, it is enough to show that\n\n\[ \mathcal{P}\left( {\delta }_{n}\right) = \mathcal{R}\left( {w}_{0}\right) \]\n\n(where we identify, as done in Section 3.4, a sequence \( \left( {{i}_{1},\ldots ,{i}_{k}}\right) \in {\left\lbrack n\right\rbrack }^{k} \) with \( \left( {{s}_{{i}_{1}},\ldot... | Yes |
Corollary 7.4.8\n\n\[ \left| {\mathcal{R}\left( {w}_{0}\right) }\right| = \frac{\left( \begin{matrix} n + 1 \\ 2 \end{matrix}\right) !}{{1}^{n}{3}^{n - 1}{5}^{n - 2}\cdots \left( {{2n} - 1}\right) }.\] | Proof. This follows immediately from Theorem 7.4.7 and the well-known \ | No |
Proposition 7.4.9 Let \( T \) be a tableau of shape \( {\delta }_{n},\widehat{p}\left( T\right) = \left( {{r}_{1},\ldots ,{r}_{N}}\right) \) , and \( k \in \left\lbrack N\right\rbrack \) . Then, the initial segment of \( T \) consisting of the cells of \( T \) containing the entries \( \{ 1,\ldots, k\} \) is uniquely d... | Proof. Let, for brevity, \( A\overset{\text{ def }}{ = }\left( {{r}_{1},\ldots ,{r}_{k}}\right) \), and write \( \widehat{p}\left( T\right) = {AB} \) . Let \( S \) be a tableau of shape \( {\delta }_{n} \) such that \( \widehat{p}\left( S\right) = {AC} \) for some \( C \) . We wish to prove that the initial segments of... | Yes |
Proposition 7.4.10 Let \( w \in {S}_{n + 1} \) and let \( T \) be a tableau with \( \operatorname{sh}\left( T\right) \subseteq {\delta }_{n} \) . Then, \[ \left| {\{ X \in \mathcal{R}\left( w\right) : \theta \left( X\right) = T\} }\right| \] depends only on \( \operatorname{sh}\left( T\right) \) . | Proof. Fix \( E \in \mathcal{R}\left( {{w}^{-1}{w}_{0}}\right) \) . Note that \( X \in \mathcal{R}\left( w\right) \) if and only if \( {XE} \in \) \( \mathcal{R}\left( {w}_{0}\right) \) . Let \( X \in \mathcal{R}\left( w\right) \) be such that \( \theta \left( X\right) = T \) . Let \( U \) be the final segment of \( {\... | Yes |
Theorem 7.4.11 Let \( w \in {S}_{n + 1} \) . Then,\n\n\[ \left| {\mathcal{R}\left( w\right) }\right| = \mathop{\sum }\limits_{\lambda }{a}_{\lambda }\left( w\right) {f}^{\lambda } \]\n\nwhere \( \lambda \) runs over all partitions of \( \ell \left( w\right) \) contained in \( {\delta }_{n} \) . | Proof. By Proposition 7.4.10, we have that\n\n\[ \left| {\mathcal{R}\left( w\right) }\right| = \mathop{\sum }\limits_{T}\left| {\{ X \in \mathcal{R}\left( w\right) : \theta \left( X\right) = T\} }\right| \]\n\n\[ = \mathop{\sum }\limits_{\lambda }\mathop{\sum }\limits_{{\{ T : \operatorname{sh}\left( T\right) = \lambda... | Yes |
Lemma 7.5.1 Let \( w \in {S}_{n + 1} \) and \( X \in \mathcal{R}\left( w\right) \) . Then,\n\n\[ D\left( {\theta \left( X\right) }\right) = D\left( X\right) . \] | Proof. It is a routine case-by-case exercise to verify, using equation (7.25), that if \( T \) is a tableau and \( k, k + 1 \) are entries of \( T, k + 1 < \left| T\right| \), then \( k \in D\left( T\right) \) if and only if \( k \in D\left( {p\left( T\right) }\right) \), where in the promotion tableau \( p\left( T\rig... | No |
Theorem 7.5.2 Let \( w \in {S}_{n + 1} \) . Then, \( {F}_{w}\left( \mathbf{x}\right) \) is a symmetric function and\n\n\[ \n{F}_{w}\left( \mathbf{x}\right) = \mathop{\sum }\limits_{{\lambda \vdash \ell \left( w\right) }}{a}_{\lambda }\left( w\right) {s}_{\lambda }\left( \mathbf{x}\right) \n\]\n\n(7.38)\n\nwhere \( {a}_... | Proof. From our definitions and Lemma 7.5.1, we have that\n\n\[ \n{F}_{w}\left( \mathbf{x}\right) = \mathop{\sum }\limits_{{E \in \mathcal{R}\left( w\right) }}{Q}_{D\left( E\right), p}\left( \mathbf{x}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{T}\mathop{\sum }\limits_{{\{ E \in \mathcal{R}\left( w\right) : \theta \l... | Yes |
Proposition 8.1.1 Let \( v \in {S}_{n}^{B} \) . Then, \[ {\ell }_{B}\left( v\right) = {\operatorname{inv}}_{B}\left( v\right) \] | Proof. We prove first that \[ {\operatorname{inv}}_{B}\left( v\right) \leq {\ell }_{B}\left( v\right) \] for all \( v \in {S}_{n}^{B} \) . Let \( v \in {S}_{n}^{B} \) . It is easy to see that \( \operatorname{inv}\left( {\left( {v{s}_{0}}\right) \left( 1\right) ,\ldots ,\left( {v{s}_{0}}\right) \left( n\right) }\right)... | Yes |
Proposition 8.1.2 Let \( v \in {S}_{n}^{B} \) . Then,\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {{s}_{i} \in S : v\left( i\right) > v\left( {i + 1}\right) }\right\} \n\]\n\n(8.8)\n\nwhere \( v\left( 0\right) \overset{\text{ def }}{ = }0 \) . | Proof. By Proposition 8.1.1, we have that\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {s \in S : {\operatorname{inv}}_{B}\left( {vs}\right) < {\operatorname{inv}}_{B}\left( v\right) }\right\} ,\n\]\n\nso equation (8.8) follows from equations (8.6) and (8.7). \( ▱ \) | No |
Proposition 8.1.3 \( \\left( {{S}_{n}^{B},{S}_{B}}\\right) \) is a Coxeter system of type \( {B}_{n} \) . | Proof. We proceed exactly as in the proof of Proposition 1.5.4. Let \( i,{i}_{1},\\ldots ,{i}_{p} \\in \\left\\lbrack {0, n - 1}\\right\\rbrack \) be such that\n\n\[ \n{\\ell }_{B}\\left( {{s}_{{i}_{1}}\\ldots {s}_{{i}_{p}}{s}_{i}}\\right) < {\\ell }_{B}\\left( {{s}_{{i}_{1}}\\ldots {s}_{{i}_{p}}}\\right) ,\n\]\n\nand ... | Yes |
Proposition 8.1.5 The set of reflections of \( {S}_{n}^{B} \) is\n\n\[ \n\\{ \\left( {i, j}\\right) \\left( {-i, - j}\\right) : 1 \\leq i < \\left| j\\right| \\leq n\\} \\bigcup \\{ \\left( {i, - i}\\right) : i \\in \\left\\lbrack n\\right\\rbrack \\} .\n\]\n\nIn particular, \( {S}_{n}^{B} \) has \( {n}^{2} \) reflecti... | Proof. Let \( w \\in {S}_{n}^{B} \) and \( i \\in \\left\\lbrack {n - 1}\\right\\rbrack \) . Then, one computes that\n\n\[ \nw{s}_{i}{w}^{-1} = \\left( {w\\left( i\\right), w\\left( {i + 1}\\right) }\\right) \\left( {-w\\left( i\\right) , - w\\left( {i + 1}\\right) }\\right) .\n\]\n\n(8.9)\n\nSimilarly,\n\n\[ \nw{s}_{0... | Yes |
Proposition 8.1.6 Let \( u, v \in {S}_{n}^{B} \). Then, the following are equivalent:\n\n(i) \( u \rightarrow v \).\n\n(ii) There exist \( i, j \in \left\lbrack {\pm n}\right\rbrack, i < j \), such that \( u\left( i\right) < u\left( j\right) \) and either \( v = u\left( {i, j}\right) \left( {-i, - j}\right) \) (if \( \... | Proof. By the definition of the Bruhat graph and Proposition 8.1.5, we only have to check that, given \( i, j, u \), and \( v \) as in (ii), \( {\operatorname{inv}}_{B}\left( v\right) > {\operatorname{inv}}_{B}\left( u\right) \) if and only if \( u\left( i\right) < u\left( j\right) \).\n\nIf \( j = - i \) (note that th... | Yes |
Proposition 8.1.7 Let \( v \in {S}_{n}^{B} \) . Then,\n\n\[ v\left\lbrack {-i - 1, - j + 1}\right\rbrack - v\left\lbrack {i, j}\right\rbrack = j - i - 1, \]\n\nfor all \( i \in \left\lbrack {-n, n - 1}\right\rbrack, j \in \left\lbrack {-n + 1, n}\right\rbrack \) . | Proof. There are several cases to consider. Since they are all similar, we treat only one of them. Suppose that \( i, j > 0 \) . Then we have that\n\n\[ v\left\lbrack {-i - 1, - j + 1}\right\rbrack - v\left\lbrack {i, j}\right\rbrack = \left| {\{ a < - i : - j < v\left( a\right) < j\} }\right| \]\n\n\[ - \left| {\{ a \... | Yes |
Corollary 8.1.9 Let \( u, v \in {S}_{n}^{B} \) . Then, \( u \leq v \) in \( {S}_{n}^{B} \) if and only if \( u \leq v \) in \( {S}_{2n} \) . | Proof. This follows immediately from Theorems 2.1.5 and 8.1.8. | No |
Proposition 8.2.2 Let \( v \in {S}_{n}^{D} \) . Then,\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {{s}_{i} \in S : v\left( i\right) > v\left( {i + 1}\right) }\right\} \n\]\n\n(8.23)\n\nwhere \( v\left( 0\right) \overset{\text{ def }}{ = } - v\left( 2\right) \) and \( v\left( {n + 1}\right) \overset{\text{ def }}{ = }0 \) ... | Proof. By Proposition 8.2.1, we have that\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {s \in S : {\operatorname{inv}}_{D}\left( {vs}\right) < {\operatorname{inv}}_{D}\left( v\right) }\right\} \n\]\n\nso the result follows from equations (8.21) and (8.22). \( ▱ \) | No |
Proposition 8.2.3 \( \left( {{S}_{n}^{D},{S}_{D}}\right) \) is a Coxeter system of type \( {D}_{n} \) . | Proof. We follow the proof of Proposition 1.5.4. Let \( i,{i}_{1},\ldots ,{i}_{p} \in \left\lbrack {0, n - 1}\right\rbrack \) be such that\n\n\[{\ell }_{D}\left( {{s}_{{i}_{1}}\ldots {s}_{{i}_{p}}{s}_{i}}\right) < {\ell }_{D}\left( {{s}_{{i}_{1}}\ldots {s}_{{i}_{p}}}\right)\]\n\nand \( w\overset{\text{ def }}{ = }{s}_{... | Yes |
Proposition 8.2.4 Let \( i \in \left\lbrack {0, n - 1}\right\rbrack \) and \( J\overset{\text{ def }}{ = }S \smallsetminus \left\{ {s}_{i}\right\} \) . Then,\n\n\[{\left( {S}_{n}^{D}\right) }_{J} = \left\{ \begin{array}{ll} \operatorname{Stab}\left( \left\lbrack {i + 1, n}\right\rbrack \right) , & \text{ if }i \neq 1, ... | We illustrate the general case of the preceding proposition with some examples. Let \( n = 9 \), and \( J = \left\{ {{s}_{0},{s}_{1},{s}_{2},{s}_{4},{s}_{5},{s}_{7},{s}_{8}}\right\} \) . Then \( {\left( {S}_{n}^{D}\right) }^{J} = \{ v \in \) \( {S}_{9}^{D} : v\left( {-2}\right) < v\left( 1\right) < v\left( 2\right) < v... | Yes |
Proposition 8.2.5 The set of reflections of \( {S}_{n}^{D} \) is\n\n\[ \n\\{ \\left( {i, j}\\right) \\left( {-i, - j}\\right) : 1 \\leq \\left| i\\right| < j \\leq n\\} .\n\]\n\nIn particular, \( {S}_{n}^{D} \) has \( {n}^{2} - n \) reflections. | Proof. Let \( w \\in {S}_{n}^{D} \) . Then,\n\n\[ \nw{s}_{0}^{D}{w}^{-1} = \\left( {w\\left( 1\\right) , - w\\left( 2\\right) }\\right) \\left( {-w\\left( 1\\right), w\\left( 2\\right) }\\right) .\n\]\n\nOn the other hand, we have already computed \( w{s}_{i}{w}^{-1} \), for \( i \\in \\left\\lbrack {n - 1}\\right\\rbr... | No |
Proposition 8.2.6 Let \( u, v \in {S}_{n}^{D} \) . Then, the following are equivalent:\n\n(i) \( u \rightarrow v \) .\n\n(ii) There exist \( i, j \in \left\lbrack {\pm n}\right\rbrack ,\left| i\right| < j \), such that \( u\left( i\right) < u\left( j\right) \) and \( v = \) \( u\left( {i, j}\right) \left( {-i, - j}\rig... | Proof. By definition of the Bruhat graph and Proposition 8.2.5, we only have to check that, given \( i, j, u \), and \( v \) as in (ii), \( {\operatorname{inv}}_{D}\left( u\right) < {\operatorname{inv}}_{D}\left( v\right) \) if \( u\left( i\right) < u\left( j\right) \) .\n\nSo, suppose that \( u\left( i\right) < u\left... | Yes |
Theorem 8.2.8 Let \( v, u \in {S}_{n}^{D} \) . Then, \( v \leq u \) if and only if the following hold:\n\n(i) \( v\left\lbrack {i, j}\right\rbrack \leq u\left\lbrack {i, j}\right\rbrack \), for all \( i, j \in \left\lbrack {-n, n}\right\rbrack \) .\n\n(ii) For all \( a, b \in \left\lbrack n\right\rbrack \), if \( \left... | Proof. Suppose that \( v \leq u \) . Then, as we have already observed, \( v \leq u \) in \( {S}_{n}^{B} \) and, hence, by Theorem 8.1.8,(i) holds.\n\nWe now prove (ii). Suppose that \( \left\lbrack {-a, a}\right\rbrack \times \left\lbrack {-b, b}\right\rbrack \) is empty for both \( u \) and \( v \) and that\n\n\[ u\l... | Yes |
Proposition 8.3.2 Let \( v \in {\widetilde{S}}_{n} \) . Then,\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {{s}_{i} \in S : v\left( i\right) > v\left( {i + 1}\right) }\right\} .\n\] | Proof. By Proposition 8.3.1, we have that\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {{s}_{i} \in S : {\operatorname{inv}}_{\widetilde{A}}\left( {v{s}_{i}}\right) < {\operatorname{inv}}_{\widetilde{A}}\left( v\right) }\right\}\n\]\n\nand the result follows from equations (8.34), (8.35), and (8.36). | Yes |
Proposition 8.3.5 The set of reflections of \( {\widetilde{S}}_{n} \) is\n\n\[ \left\{ {{t}_{i, j + {kn}} : 1 \leq i < j \leq n, k \in \mathbb{Z}}\right\} . | Proof. Let \( u \in {\widetilde{S}}_{n} \), and \( i \in \left\lbrack n\right\rbrack \) . Then, we have that\n\n\[ u{s}_{i}{u}^{-1} = \mathop{\prod }\limits_{{r \in \mathbb{Z}}}\left( {u\left( i\right) + {rn}, u\left( {i + 1}\right) + {rn}}\right) .\n\nSince \( u \) is any element of \( {\widetilde{S}}_{n} \) we deduce... | No |
Proposition 8.3.6 Let \( u, v \in {\widetilde{S}}_{n} \). Then, the following are equivalent:\n\n(i) \( u \rightarrow v \).\n\n(ii) There exist \( i, j \in \mathbb{Z}, i < j, i ≢ j\left( {\;\operatorname{mod}\;n}\right) \), such that \( u\left( i\right) < u\left( j\right) \) and \( v = u{t}_{i, j} \) | Proof. By Proposition 8.3.5 and the definition of the Bruhat graph, the equivalence of (i) and (ii) reduces to showing that if the complete notation of \( v \) is obtained from that of \( u \) by interchanging \( u\left( {i + {rn}}\right) \) and \( u\left( {j + {rn}}\right) \) for all \( r \in \mathbb{Z} \), then \( {\... | Yes |
Theorem 8.3.7 Let \( u, v \in {\widetilde{S}}_{n} \) . Then, the following are equivalent:\n\n(i) \( v \leq u \) .\n\n(ii) \( v\left\lbrack {i, j}\right\rbrack \leq u\left\lbrack {i, j}\right\rbrack \), for all \( i, j \in \mathbb{Z} \) . | Proof. Suppose that (i) holds. We may assume that \( v \rightarrow u \) in \( {\widetilde{S}}_{n} \) . This, by Proposition 8.3.6, implies that there exist \( i, j \in \mathbb{Z}, i < j, i ≢ j\left( {\;\operatorname{mod}\;n}\right) \) , such that \( u = v{t}_{i, j} \) and \( u\left( i\right) < u\left( j\right) \) . The... | Yes |
Proposition 8.4.2 Let \( v \in {\widetilde{S}}_{n}^{C} \) . Then,\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {{s}_{i} \in S : v\left( i\right) > v\left( {i + 1}\right) }\right\} .\n\] | Proof. This follows immediately from equations (8.47), (8.48), and (8.50), and the observation that \( v\left( 0\right) = 0 \) and \( v\left( {n + 1}\right) = N - v\left( n\right) \) . \( ▱ \) | Yes |
Proposition 8.4.4 Let \( i \in \left\lbrack {0, n}\right\rbrack \), and \( J\overset{\text{ def }}{ = }S \smallsetminus \left\{ {s}_{i}\right\} \) . Then,\n\n\[{\left( {\widetilde{S}}_{n}^{C}\right) }_{J} = \operatorname{Stab}\left( \left\lbrack {-i, i}\right\rbrack \right) \cap \operatorname{Stab}\left( \left\lbrack {... | The preceding result yields a simple combinatorial rule to compute, given \( u \in {\widetilde{S}}_{n}^{C} \) and \( J \subseteq S \), the minimal coset representative \( {u}^{J} \) . Namely, if \( J = \) \( S \smallsetminus \left\{ {s}_{i}\right\} \), then the complete notation of \( {u}^{J} \) is obtained from that o... | Yes |
Proposition 8.4.5 The set of reflections of \( {\widetilde{S}}_{n}^{C} \) is\n\n\[ \left\{ {{t}_{i, j + {kN}}{t}_{-i, - j - {kN}} : 1 \leq i < \left| j\right| \leq n, k \in \mathbb{Z}}\right\} \cup \left\{ {{t}_{i, - i + {kN}} : i \in \left\lbrack n\right\rbrack, k \in \mathbb{Z}}\right\} . \] | Proof. Let \( u \in {\widetilde{S}}_{n}^{C} \) . Then, we have that\n\n\[ u{s}_{i}{u}^{-1} = \mathop{\prod }\limits_{{r \in \mathbb{Z}}}\left( {u\left( i\right) + {rN}, u\left( {i + 1}\right) + {rN}}\right) \left( {-u\left( i\right) + {rN}, - u\left( {i + 1}\right) + {rN}}\right) ,(8. \]\n\nif \( i \in \left\lbrack {n ... | Yes |
Proposition 8.4.6 Let \( u, v \in {\widetilde{S}}_{n}^{C} \) . Then, the following are equivalent:\n\n(i) \( u \rightarrow v \) .\n\n(ii) There exist \( i, j \in \mathbb{Z}, j ≢ i\left( {\;\operatorname{mod}\;N}\right), i, j ≢ 0\left( {\;\operatorname{mod}\;N}\right), i < j \), such that \( u\left( i\right) < u\left( j... | Proof. By Proposition 8.4.5 and the definition of the Bruhat graph, it is enough to show that if \( v, u, i \), and \( j \) are as in (ii), then \( {\operatorname{inv}}_{\widetilde{C}}\left( v\right) > {\operatorname{inv}}_{\widetilde{C}}\left( u\right) \) if \( u\left( i\right) < u\left( j\right) \) . However, this ca... | No |
Proposition 8.5.2 Let \( v \in {\widetilde{S}}_{n}^{B} \) . Then,\n\n\[ \n{D}_{R}\left( v\right) = \left\{ {{s}_{i} \in S : i \in D\left( {v\left( 0\right), v\left( 1\right) ,\ldots, v\left( n\right), v\left( {n + 2}\right) }\right) }\right\} .\n\] | Proof. This follows immediately from equations (8.68), (8.69), and (8.71) and the fact that \( v\left( 0\right) = 0 \) and \( v\left( {n + 2}\right) = N - v\left( {n - 1}\right) \) . \( ▱ \) | Yes |
Proposition 8.5.3 \( \left( {{\widetilde{S}}_{n}^{B},{\widetilde{S}}_{B}}\right) \) is a Coxeter system of type \( {\widetilde{B}}_{n} \) . | Proof. We prove that the exchange condition holds, as in the proofs of the corresponding results in the previous sections. Let \( i,{i}_{1},\ldots ,{i}_{p} \in \left\lbrack {0, n}\right\rbrack \) be such that\n\n\[ \n{\ell }_{\widetilde{B}}\left( {{s}_{{i}_{1}}\ldots {s}_{{i}_{p}}{s}_{i}}\right) < {\ell }_{\widetilde{B... | Yes |
Proposition 8.5.4 Let \( i \in \left\lbrack {0, n}\right\rbrack \), and \( J\overset{\text{ def }}{ = }S \smallsetminus \left\{ {s}_{i}\right\} \) . Then,\n\n\[{\left( {\widetilde{S}}_{n}^{B}\right) }_{J} = \left\{ \begin{array}{ll} \operatorname{Stab}\left( \left\lbrack {-i, i}\right\rbrack \right) \cap \operatorname{... | The preceding proposition yields a simple combinatorial rule for computing the minimal coset representatives of \( {\widetilde{S}}_{n}^{B} \) . Namely, if \( u \in {\widetilde{S}}_{n}^{B} \) and \( J = S \smallsetminus \left\{ {s}_{i}\right\} \), with \( i \neq n - 1 \), then the complete notation of \( {u}^{J} \) is o... | Yes |
Proposition 8.5.5 The set of reflections of \( {\widetilde{S}}_{n}^{B} \) is\n\n\[ \left\{ {{t}_{i, j + {kN}}{t}_{-i, - j - {kN}} : 1 \leq i < \left| j\right| \leq n, k \in \mathbb{Z}}\right\} \cup \left\{ {{t}_{i,{2kN} - i} : i \in \left\lbrack n\right\rbrack, k \in \mathbb{Z}}\right\} . \] | Proof. Let \( w \in {\widetilde{S}}_{n}^{B} \) . We have already computed \( w{s}_{i}{w}^{-1} \) for \( i = 0,1,\ldots \) , \( n - 1 \) in equations (8.53) and (8.54). Furthermore,\n\n\[ u{s}_{n}{u}^{-1} = \mathop{\prod }\limits_{{r \in \mathbb{Z}}}\left( {{rN} + u\left( n\right) ,{rN} + N - u\left( {n - 1}\right) }\ri... | Yes |
Proposition 8.5.6 Let \( u, v \in {\widetilde{S}}_{n}^{B} \) . Then, \( u \rightarrow v \) in \( {\widetilde{S}}_{n}^{B} \) if and only if \( u \rightarrow v \) in \( {\widetilde{S}}_{n}^{C} \) . | Thus, the Bruhat graph of \( {\widetilde{S}}_{n}^{B} \) is the directed subgraph of that of \( {\widetilde{S}}_{n}^{C} \) induced by \( {\widetilde{S}}_{n}^{B} \), and hence a combinatorial description of it is given by Proposition 8.4.6. | Yes |
Proposition 8.6.1 Let \( v \in {\widetilde{S}}_{n}^{D} \) . Then,\n\n\[{\ell }_{\widetilde{D}}\left( v\right) = {\operatorname{inv}}_{\widetilde{D}}\left( v\right)\] | Proof. We prove first that\n\n\[{\operatorname{inv}}_{\widetilde{D}}\left( v\right) \leq {\ell }_{\widetilde{D}}\left( v\right)\]\n\n(8.79)\n\nfor all \( v \in {\widetilde{S}}_{n}^{D} \) . It is clear from definition (8.76) and equation (8.68) that\n\n\[{\operatorname{inv}}_{\widetilde{D}}\left( {v{s}_{i}}\right) = {\o... | Yes |
Proposition 8.6.5 The set of reflections of \( {\widetilde{S}}_{n}^{D} \) is\n\n\[ \left\{ {{t}_{i, j + {kN}}{t}_{-i, - j - {kN}} : 1 \leq i < \left| j\right| \leq n, k \in \mathbb{Z}}\right\} . \] | Proof. The computations for \( {s}_{1},\ldots ,{s}_{n} \) are the same as in equations (8.53) and (8.72). For \( {s}_{0} \), we now obtain that\n\n\[ u{s}_{0}{u}^{-1} = \mathop{\prod }\limits_{{r \in \mathbb{Z}}}\left( {{rN} + u\left( 1\right) ,{rN} - u\left( 2\right) }\right) \left( {{rN} - u\left( 1\right) ,{rN} + u\... | Yes |
Proposition 8.6.6 Let \( u, v \in {\widetilde{S}}_{n}^{D} \) . Then, \( u \rightarrow v \) in \( {\widetilde{S}}_{n}^{D} \) if and only if \( u \rightarrow v \) in \( {\widetilde{S}}_{n}^{B} \) . | Thus, the Bruhat graph of \( {\widetilde{S}}_{n}^{D} \) is the directed subgraph induced on \( {\widetilde{S}}_{n}^{D} \) by the Bruhat graph of \( {\widetilde{S}}_{n}^{B} \) . Hence, a combinatorial criterion for deciding if \( u \rightarrow v \) in \( {\widetilde{S}}_{n}^{D} \) is given by Proposition 8.5.6 (and henc... | Yes |
Theorem 1.2. [Euclid] There are infinitely many primes. | EUCLID'S PROOF IN MODERN FORM. If there are only finitely many primes, we can list them as \( {p}_{1},\ldots ,{p}_{r} \) . Let\n\n\[ N = {p}_{1}\cdots {p}_{r} + 1 > 1 \]\n\nBy the Fundamental Theorem of Arithmetic, \( N \) can be factorized, so it must be divisible by some prime \( {p}_{k} \) of our list. Since \( {p}_... | Yes |
Theorem 1.3. The series \( \mathop{\sum }\limits_{{p \in \mathbb{P}}}\frac{1}{p} \) diverges. | FIRST PROOF OF THEOREM 1.3. We argue by contradiction: Assume that the series converges. Then there is some \( N \) such that\n\n\[ \mathop{\sum }\limits_{{p > N}}\frac{1}{p} < \frac{1}{2} \]\n\nLet\n\n\[ Q = \mathop{\prod }\limits_{{p \leq N}}p \]\n\nbe the product of all the primes less than or equal to \( N \) . The... | Yes |
Theorem 1.5. [Euler Product Representation] For any \( \sigma > 1 \) ,\n\n\[ \zeta \left( \sigma \right) = \mathop{\prod }\limits_{p}{\left( 1 - {p}^{-\sigma }\right) }^{-1}. \] | Proof. For any \( \sigma > 1 \) ,\n\n\[ \left( {1 - {2}^{-\sigma }}\right) \zeta \left( \sigma \right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{\sigma }} - \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{\left( 2n\right) }^{\sigma }} \]\n\n\[ = \mathop{\sum }\limits_{{n\text{ odd }}}\frac{1}{{n}^{\si... | Yes |
Lemma 1.8. For any \( n \geq 1 \) , \[ \mathop{\sum }\limits_{{p \leq n}}\log p < {2n}\log 2 \] | Proof. Let \[ M = \left( \begin{matrix} {2m} + 1 \\ m \end{matrix}\right) = \frac{\left( {{2m} + 1}\right) \left( {2m}\right) \cdots \left( {m + 2}\right) }{m!}. \] This is a binomial coefficient, so it is an integer (see Exercise 1.10 for a stronger form of this). The coefficient \( M \) appears twice in the binomial ... | Yes |
Corollary 1.10. There exists a real number \( \theta \) with the property that\n\n\[ \left\\lfloor {2}^{{2}^{{2}^{{.}^{{.}^{{.}^{\theta }}}}}}\right\\rfloor \]\n\nis a prime number for any number of iterations of the exponential. | Proof. Let \( {q}_{1} \) be any prime, and choose a sequence of primes \( \left( {q}_{n}\right) \) with the property that\n\n\[ {2}^{{q}_{n}} < {q}_{n + 1} < {2}^{{q}_{n} + 1}. \]\n\n(1.20)\n\nThis is possible by Bertrand’s Postulate. Now define functions \( {f}^{\left( 1\right) },{f}^{\left( 2\right) },\ldots \) by \(... | Yes |
Lemma 1.11. If \( {2}^{n} - 1 \) is prime, then \( n \) is prime. | Proof. We prove the contrapositive statement that \( n \) being composite forces \( {2}^{n} - 1 \) to be composite. If \( n = {ab} \) with \( a, b > 1 \), then\n\n\[ \n{2}^{n} - 1 = \left( {{2}^{a} - 1}\right) \left( {{2}^{n - a} + {2}^{n - {2a}} + \cdots + {2}^{a} + 1}\right) ,\n\]\n\nso \( {2}^{n} - 1 \) is composite... | Yes |
Theorem 1.12. [Fermat's Little Theorem] For any prime \( p \) and any integer \( a \) ,\n\n\[ {a}^{p} \equiv a\;\left( {\;\operatorname{mod}\;p}\right) \] | Combinatorial Proof. It is enough to prove the statement when \( a \) is a positive integer, so we use induction. The result is true for \( a = 1 \) because both sides are 1. Assume it is true for \( a = b \) . Now\n\n\[ {\left( b + 1\right) }^{p} = {b}^{p} + p{b}^{p - 1} + \cdots + {pb} + 1 = \mathop{\sum }\limits_{{j... | Yes |
Lemma 1.13. Suppose \( p \) is a prime and \( q \) is a nontrivial prime divisor of \( {M}_{p} \) . Then \( q \equiv 1 \) modulo \( p \) . | Proof Using the Euclidean Algorithm. The condition that \( q \) divides \( {M}_{p} \) amounts to\n\n\[ \n{2}^{p} \equiv 1\;\left( {\;\operatorname{mod}\;q}\right) \n\]\n\nBy Fermat’s Little Theorem, \( {2}^{q - 1} \equiv 1 \) modulo \( q \) . Let \( d = \gcd \left( {p, q - 1}\right) \) . If \( d = p \), then \( p \mid ... | Yes |
Example 1.14. Lemma 1.13 is a significant help in factorizing \( {M}_{n} \) . To see how this works, we present Fermat’s proof from 1640 that \( {2}^{23} - 1 \) is not prime. If \( q \) is a prime dividing \( {2}^{23} - 1 \), then \( q \equiv 1 \) modulo 23 . Now \( {23n} + 1 \) is a prime smaller than \( \sqrt{{2}^{23... | \[ n = 2,{12},{20},{26},{30},{36},{42},{44},{50},{56},{60},{62},{72},{84},{86},{102},{104},{110}. \] Trial division shows that \( {M}_{23} \) is divisible by the first of the resulting numbers, 47. In general, there is no reason to expect the smallest possible candidate to be a divisor, but even if the largest were the... | Yes |
Theorem 1.15. [Zsigmondy] Let \( {M}_{n} = {2}^{n} - 1 \) . Then for every \( n \neq 6, n > 1 \) , the term \( {M}_{n} \) has a primitive divisor. | The proof of Theorem 1.15 is presented in Section 8.3.1 on p. 167, after we have proved the Möbius inversion formula (Theorem 8.15). | Yes |
Example 1.18. When \( n = 5 \), Lemma 1.17 shows that if \( p \) is a prime dividing \( {F}_{5} \) , then \( p = {2}^{6}k + 1 = {64k} + 1 \) for some \( k \) . Thus the list of possible divisors is greatly reduced. We only have to test \( {F}_{5} \) for divisibility by\n\n\[ \n{65},{129},{193},{257},{321},{385},{449},{... | Proof of Lemma 1.17. Suppose \( p \) is a prime with \( p \mid {F}_{n} \), so \( {2}^{{2}^{n}} \equiv - 1 \) modulo \( p \) and \( p \) is odd. Hence\n\n\[ \n{2}^{{2}^{n + 1}} = {\left( {2}^{{2}^{n}}\right) }^{2} \equiv {\left( -1\right) }^{2} \equiv 1\;\left( {\;\operatorname{mod}\;p}\right) .\n\] \n\nLet \( d = \gcd ... | Yes |
Theorem 1.19. An integer \( n > 1 \) is prime if and only if\n\n\[ \left( {n - 1}\right) ! \equiv - 1\;\left( {\;\operatorname{mod}\;n}\right) \] | Proof of 'only if' direction. We prove that the congruence is satisfied when \( n \) is prime and leave the converse as an exercise. Assume that \( n = p \) is an odd prime. (The congruence is clear for \( n = 2 \) .)\n\nEach of the integers \( 1 < a < p - 1 \) has a unique multiplicative inverse distinct from \( a \) ... | No |
Testing the congruence \( {2}^{n - 1} \equiv 1 \) modulo \( n \) fails to detect the fact that \( n = {341} = {11} \cdot {31} \) is composite. | By Fermat’s Little Theorem, \( {2}^{10} \equiv 1 \) modulo \( {11} \) so \( {2}^{340} \equiv {1}^{34} \equiv 1 \) modulo 11. Also \( {2}^{5} = {32} \equiv 1 \) modulo 31, so\n\n\[ \n{2}^{340} = {\left( {2}^{5}\right) }^{68} \equiv {1}^{68} = 1\;\left( {\;\operatorname{mod}\;{31}}\right) \n\]\n\nThus \( {2}^{340} - 1 \)... | Yes |
Theorem 1.23. If \( d = \gcd \left( {a, b}\right) \) with \( a, b \in \mathbb{Z} \) not both zero, then there are numbers \( x, y \in \mathbb{Z} \) with\n\n\[ d = {ax} + {by}. \] | Proof. The idea is to work your way up the chain of equations in the Euclidean Algorithm, always expressing the remainder in terms of the previous two remainders. Writing \( * \) for an integer, we get\n\n\[ \gcd \left( {a, b}\right) = {r}_{n} = {r}_{n - 2} - {r}_{n - 1}{q}_{n} \]\n\n\[ = {r}_{n - 2}\left( {1 + {q}_{n}... | Yes |
Example 1.24. Using the equations from Example 1.21 we find that | \[ 1 = 6 - 5 \] \[ = 6 - \left( {{11} - 6}\right) \] \[ = 2 \cdot 6 - {11} \] \[ = 2\left( {{17} - {11}}\right) - {11} \] \[ = 2 \cdot {17} - 3 \cdot {11}\text{.} \] | Yes |
Corollary 1.25. Let \( n > 1 \) and a denote elements of \( \mathbb{Z} \) . Then a and \( n \) are coprime if and only if there exists \( x \) with\n\n\[{ax} \equiv 1\;\left( {\;\operatorname{mod}\;n}\right)\]\n\nThat is, \( \gcd \left( {a, n}\right) = 1 \) if and only if \( a \) is invertible modulo \( n \) . | Proof. The congruence is equivalent to the existence of an integer \( y \) with\n\n\[{ax} + {ny} = 1\text{.}\]\n\nIf \( a \) and \( n \) have a factor in common then that factor will also divide 1, so the congruence implies \( a \) and \( n \) are coprime. Conversely, if \( a \) and \( n \) are coprime then 1 is a grea... | Yes |
Theorem 1.26. The prime numbers are more than any assigned multitude of prime numbers. | Proof. Let \( A, B \), and \( C \) be the assigned prime numbers. I say that there are more prime numbers than \( A, B \), and \( C \) . Take the least number \( {DE} \) measured by \( A, B \), and \( C \) . Add the unit \( {DF} \) to \( {DE} \) .\n\nThen \( {EF} \) is either prime or not.\n\nFirst, let it be prime. Th... | Yes |
(1) Let \( R = \mathbb{Z}\left\lbrack \mathrm{i}\right\rbrack \) denote the Gaussian integers, so\n\n\[ R = \{ x + \mathrm{i}y \mid x, y \in \mathbb{Z}\} \]\n\nwhere \( {\mathrm{i}}^{2} = - 1 \) . Setting \( N\left( {x + \mathrm{i}y}\right) = {x}^{2} + {y}^{2} \) shows that \( R \) is a Euclidean ring. | Proof THAT ℤ[i] Is Euclidean. Condition (1) of Definition 2.2 is easily verified by direct computation. For property (2), let \( a, b \neq 0 \in R \) and write \( a{b}^{-1} = p + \mathrm{i}q \) with \( p, q \in \mathbb{Q} \) . Now define \( m, n \in \mathbb{Z} \) by\n\n\[ m \in \lbrack p - 1/2, p + 1/2), n \in \lbrack ... | Yes |
Theorem 2.5. Every Euclidean ring has the Fundamental Theorem of Arithmetic. | Proof. Clearly, every irreducible \( \mu \) has \( N\left( \mu \right) \geq 2 \) . Arguing as we did in \( \mathbb{Z} \) shows we cannot keep factorizing into irreducibles forever, so the existence part is easy. To complete the argument, we just need to show that every irreducible is prime. This follows easily from The... | Yes |
Theorem 2.6. The prime \( p \) can be written as the sum of two squares if and only if \( p = 2 \) or \( p \) is congruent to 1 modulo 4. | To prove this, we are going to use the Fundamental Theorem of Arithmetic in the ring of Gaussian integers \( R = \mathbb{Z}\left\lbrack \mathrm{i}\right\rbrack \) with norm function \( N : R \rightarrow \mathbb{N} \) defined by \( N\left( {x + \mathrm{i}y}\right) = {x}^{2} + {y}^{2} \) as in Example 2.3(1). | Yes |
Lemma 2.7. If \( p \) is 2 or a prime congruent to 1 modulo 4, then the congruence\n\n\[ \n{T}^{2} + 1 \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \n\]\n\nis solvable in integers. | Proof. This is clear for \( p = 2 \) so suppose \( p = {4n} + 1 \) for some integer \( n > 0 \) . Using al-Haytham's Theorem (Theorem 1.19),\n\n\[ \n\left( {p - 1}\right) ! = \left( {p - 1}\right) \left( {p - 2}\right) \cdots 3 \cdot 2 \cdot 1 \equiv - 1\;\left( {\;\operatorname{mod}\;p}\right) . \n\]\n\nNow\n\n\[ \n{4... | Yes |
Lemma 2.9. Let \( p \) be an odd prime. Then there are integers \( a \) and \( b \) with\n\n\[ \n{a}^{2} + {b}^{2} + 1 \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \n\] | Proof. Define the sets\n\n\[ \nA = \left\{ {{a}^{2}\left| {\;0 \leq a \leq \frac{p - 1}{2}}\right. }\right\} \n\]\n\nand\n\n\[ \nB = \left\{ {-{b}^{2} - 1 \mid 0 \leq b \leq \frac{p - 1}{2}}\right\} . \n\]\n\nNo two elements of \( A \) are congruent modulo \( p \), and no two elements of \( B \) are congruent modulo \(... | Yes |
Theorem 2.11. The only integral solution of the equation\n\n\[ \n{y}^{2} = {x}^{3} + x \n\]\n\n(2.7)\n\nis \( x = 0, y = 0 \) . | Proof. Let \( x \) and \( y \) be integers with \( {y}^{2} = {x}^{3} + x \) . Write the right-hand side of the equation as \( {x}^{3} + x = x\left( {{x}^{2} + 1}\right) \) . Any factor of \( x \) will divide \( {x}^{2} \), so any factor common to \( x \) and \( {x}^{2} + 1 \) will also divide 1 . Thus \( x \) and \( {x... | Yes |
Theorem 2.12. The only integral solution of the equation\n\n\\[ \n{y}^{2} = {x}^{3} - 1 \n\\]\n\n(2.8)\n\nis \\( x = 1, y = 0 \\) . | Proof of Theorem 2.12. Rewrite the equation as\n\n\\[ \n{y}^{2} + 1 = {x}^{3} \n\\]\n\nand then factorize the left-hand side as \\( \\left( {y + \\mathrm{i}}\\right) \\left( {y - \\mathrm{i}}\\right) \\) in \\( \\mathbb{Z}\\left\\lbrack \\mathrm{i}\\right\\rbrack \\) . We claim that the two factors \\( y \\pm \\mathrm{... | Yes |
Corollary 3.6. If \( n \) is factorized into powers of distinct primes, \( n = \mathop{\prod }\limits_{p}{p}^{{e}_{p}} \) , then | \[ \phi \left( n\right) = \mathop{\prod }\limits_{{p \mid n}}\left( {p - 1}\right) {p}^{{e}_{p} - 1} = n\mathop{\prod }\limits_{{p \mid n}}\frac{p - 1}{p}. \] | Yes |
Theorem 3.7. [Chinese Remainder Theorem] Suppose \( m, n \in \mathbb{N} \) are co-prime. Then the simultaneous congruences\n\n\[ x \equiv a\;\left( {\;\operatorname{mod}\;m}\right) \]\n\n\[ x \equiv b\;\left( {\;\operatorname{mod}\;n}\right) \]\n\nhave a solution \( x \in \mathbb{N} \) for any \( a, b \in \mathbb{Z} \)... | Proof of the Chinese Remainder Theorem. The coprimality condition guarantees that there exist \( {m}^{\prime },{n}^{\prime } \) such that\n\n\[ m{m}^{\prime } = 1\;\left( {\;\operatorname{mod}\;n}\right) \text{ and }n{n}^{\prime } = 1\;\left( {\;\operatorname{mod}\;m}\right) \]\n\n(3.3)\n\nby Corollary 1.25. Then \( x ... | Yes |
Solve the simultaneous congruences \( x \equiv 2 \) modulo 17 and \( x \equiv 8 \) modulo 11. | We find \( {m}^{\prime } = 2 \) and \( {n}^{\prime } = {14} \) in the proof of the Chinese Remainder Theorem. Then\n\n\[ x = 8 \cdot \left( {{17} \cdot 2}\right) + 2 \cdot \left( {{11} \cdot {14}}\right) = {580} \]\n\nsatisfies the two congruences. (The smallest solution is the remainder of 580 divided by \( {11} \cdot... | Yes |
Theorem 3.9. For any \( n \in \mathbb{N} \) , \[ \mathop{\sum }\limits_{{d \mid n}}\phi \left( d\right) = n \] | Proof. First check the equality when \( n = {p}^{r} \) is a prime power. The left-hand side is \[ 1 + \mathop{\sum }\limits_{{i = 1}}^{r}\left( {p - 1}\right) {p}^{i - 1} = 1 + \left( {{p}^{r} - 1}\right) = n \] by summing the geometric progression or noticing that it is a telescoping sum. Next, observe that both sides... | Yes |
Example 3.10. The distinct powers of 3 in \( {\mathbb{F}}_{7}^{ * } \) are\n\n\[ \n{3}^{0} \equiv 1,{3}^{1} \equiv 3,{3}^{2} \equiv 2,{3}^{3} \equiv 6,{3}^{4} \equiv 4,{3}^{5} \equiv 5.\n\] | The only values of \( j,1 \leq j \leq 6 \) with \( \gcd \left( {j,6}\right) = 1 \) are 1 and 5 . Since \( {3}^{5} \equiv 5 \) modulo 7,5 is another generator of \( {\mathbb{F}}_{7}^{ * } \) . | No |
Theorem 3.13. [Euler's Criterion] Let \( p \) be an odd prime. Then\n\n\[ \left( \frac{a}{p}\right) \equiv {a}^{\left( {p - 1}\right) /2}\;\left( {\;\operatorname{mod}\;p}\right) \] | Proof. The statement is obvious if \( a \equiv 0 \) modulo \( p \), so assume that \( a \) is coprime to \( p \) . Notice that the only square roots of 1 modulo \( p \) are congruent to \( \pm 1 \) since \( {x}^{2} - 1 = \left( {x - 1}\right) \left( {x + 1}\right) \) in any field.\n\nNow\n\n\[ {\left( {a}^{\left( {p - ... | Yes |
Corollary 3.14. The Legendre symbol satisfies\n\n\[ \left( \frac{ab}{p}\right) = \left( \frac{a}{p}\right) \left( \frac{b}{p}\right) \]\n\nThat is, the Legendre symbol viewed as an arithmetic function\n\n\[ \left( \begin{array}{l} \cdot \\ - \\ p \end{array}\right) : \mathbb{Z} \rightarrow \{ 0, \pm 1\} \]\n\nis comple... | The proof follows immediately from Theorem 3.13 because the right-hand side of Equation (3.6) is completely multiplicative. | Yes |
Theorem 3.15. Let \( p \) and \( q \) denote odd primes. If \( p \equiv q \equiv 3 \) modulo 4, then\n\n\[ \left( \frac{q}{p}\right) = - \left( \frac{p}{q}\right) \]\n\nIf at least one of \( p \) or \( q \) is 1 modulo 4, then the symbols are equal. | Theorem 3.15 can be stated as a neater formula, and this is what we will prove. If \( p \) and \( q \) are odd primes, then\n\n\[ \left( \frac{q}{p}\right) = {\left( -1\right) }^{\left( {p - 1}\right) /2 \cdot \left( {q - 1}\right) /2}\left( \frac{p}{q}\right) . \]\n\n(3.7) | Yes |
Theorem 3.16. If \( p \) is an odd prime, then\n\n\[\n\\left( \\frac{2}{p}\\right) = 1\\text{ if and only if }p \\equiv \\pm 1\\;\\left( {\\;\\operatorname{mod}\\;8}\\right) .\n\] | Proof of Theorem 3.16. The prime \( p \) is odd, so \( {p}^{2} - 1 \\equiv 0 \) modulo 8 . Let \( \\mathbb{F} \) denote the field with \( {p}^{2} \) elements. Then \( {\\mathbb{F}}^{ * } \) is a cyclic group of order \( {p}^{2} - 1 \) by Theorem 3.3. Since \( {p}^{2} - 1 \) is divisible by 8, this implies that \( {\\ma... | No |
Lemma 3.17. For all \( j \in \mathbb{Z}, f\left( p\right) f\left( {jp}\right) = f\left( j\right) \) . | Clearly, this is true if \( j \) is even, so suppose that \( j \) is odd. The statement is true for any odd pair \( j \) and \( p \) . This can be checked by examining all the possibilities for \( j \) and \( p \) modulo 16. Alternatively, notice that\n\n\[{\left( -1\right) }^{\left( {{\left( jp\right) }^{2} - 1}\right... | Yes |
Compute the Legendre symbol \( \left( \frac{91}{167}\right) \) using the Quadratic Reciprocity Law. | First notice that\n\n\[ \left( \frac{91}{167}\right) = \left( \frac{7}{167}\right) \left( \frac{13}{167}\right) = - \left( \frac{167}{7}\right) \left( \frac{167}{13}\right) = - \left( \frac{6}{7}\right) \left( \frac{11}{13}\right) .\n\]\n\nThe problem has become more manageable and is readily finished by noting that\n\... | Yes |
Consider the equation\n\n\[ \n{x}^{2} + 5{y}^{2} = p \n\] | In order to understand this, we expect to use the Quadratic Reciprocity Law to solve\n\n\[ \n{T}^{2} + 5 \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) \n\]\n\nfor \( T \) . | No |
Lemma 3.20. There are infinitely many coprime pairs of integers \( p \) and \( q > 0 \) with\n\n\[ \left| {q\sqrt{d} - p}\right| < \frac{1}{q} \] | Proof. Let \( Q > 1 \) denote an integer. Divide the interval \( \lbrack 0,1) \) into \( Q \) subintervals \( \lbrack 0,1/Q),\lbrack 1/Q,2/Q),\ldots \) and consider the \( \left( {Q + 1}\right) \) numbers\n\n\[ 0,\{ \sqrt{d}\} ,\{ 2\sqrt{d}\} ,\ldots ,\{ Q\sqrt{d}\} .\n\nThere are \( \left( {Q + 1}\right) \) of them si... | Yes |
Theorem 3.21. If \( d > 1 \) is a square-free integer, then\n\n\[ \n{x}^{2} - d{y}^{2} = 1 \n\]\n\nhas infinitely many solutions in integers \( \left( {x, y}\right) \) . Moreover, each solution corresponds to a unit in \( R = \mathbb{Z}\left\lbrack \sqrt{d}\right\rbrack \) with norm 1. Any unit with norm 1 has the form... | Proof. Using Lemma 3.20, choose \( p, q > 0 \) with\n\n\[ \n\left| {q\sqrt{d} - p}\right| < \frac{1}{q} \n\]\n\n(3.16)\n\nThen\n\n\[ \np - \frac{1}{q} < q\sqrt{d} < p + \frac{1}{q} \n\]\n\nso\n\[ \n\left| {q\sqrt{d} + p}\right| < {2q}\sqrt{d} + \frac{1}{q} \n\]\n\n(3.17)\n\nMultiplying the inequalities (3.16) and (3.17... | Yes |
Theorem 3.22. [LAGRANGE] Let \( \Delta \) be a nonsquare integer. Then there is a quadratic form \( a{x}^{2} + {bxy} + c{y}^{2} \) of discriminant \( \Delta \) with a primitive solution to\n\n\[ a{x}^{2} + {bxy} + c{y}^{2} = n \]\n\nif and only if the congruence\n\n\[ {z}^{2} + {\rho z} - \left( \frac{\Delta - \rho }{4... | Proof. Assume that \( \left( {\alpha ,\beta }\right) \) is a primitive integral solution to Equation (3.18). By Theorem 1.23, there are integers \( \gamma ,\delta \) with\n\n\[ {\alpha \gamma } + {\beta \delta } = 1. \]\n\nLet\n\n\[ \left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack = \left\lbrack \begin{mat... | Yes |
Let \( a = 1, b = 0, c = 5 \), and \( n = 7 \), so \( \rho = 0 \) and \( \Delta = - {20} \). Theorem 3.22 applies to say that there is a quadratic form representing 7 with discriminant -20 if and only if\n\n\[ \n{z}^{2} + 5 \equiv 0\;\left( {\;\operatorname{mod}\;7}\right) \n\]\n\nhas a solution. | We know that -5 is a quadratic residue modulo 7, so there is such a form. The proof constructs the form\n\n\[ \n7{x}^{2} + {6xy} + 2{y}^{2} \n\]\n\nand of course this represents 7 when \( x = 1 \) and \( y = 0 \) . | Yes |
Theorem 4.4. Let \( I \) denote an ideal in \( {O}_{\mathbb{K}} \) . Then there are elements \( \alpha ,\beta \) in \( I \) with \( I = \left( {\alpha ,\beta }\right) \) . | Proof. Since \( {O}_{\mathbb{K}} \) as an additive group is a subgroup of \( {\mathbb{Q}}^{2} \), it follows that \( I \) can be generated as an additive group by two elements. We will first show that one of these elements can be chosen to lie in \( \mathbb{Z} \) . Let\n\n\[ B = \{ b \in \mathbb{Z} \mid a + {b\delta } ... | Yes |
Corollary 4.6. Let \( I \) denote any ideal of \( {O}_{\mathbb{K}} \) . Then \( I{I}^{ * } is a principal ideal \( k\mathbb{Z} \) of \( \mathbb{Z} \) . | Proof. We know that \( I = \left( {\alpha ,\beta }\right) \) for some \( \alpha ,\beta \), so\n\n\[ I{I}^{ * } = \left( {\alpha ,\beta }\right) \left( {{\alpha }^{ * },{\beta }^{ * }}\right) = \left( {\alpha {\alpha }^{ * },\alpha {\beta }^{ * },\beta {\alpha }^{ * },\beta {\beta }^{ * }}\right) .\n\]\n\nThis means tha... | Yes |
Corollary 4.8. (1) For \( I = \left( {\alpha ,\beta }\right) \), \[ N\left( I\right) = \gcd \left( {N\left( \alpha \right), N\left( \beta \right), T\left( {\alpha {\beta }^{ * }}\right) }\right) . \] | Proof.(1) This appeared in the proof of Corollary 4.6. | No |
Corollary 4.9. If \( I, J \) and \( K \) are ideals of \( {O}_{\mathbb{K}} \) with \( I \neq \{ 0\} \) and \( {IJ} = {IK} \) , then \( J = K \) . | Proof. This is obvious if \( I = \left( \alpha \right) \) is principal because in that case \( {IJ} = {\alpha J} \) so \( J = {\alpha }^{-1}\left( {IJ}\right) \) . Similarly, \( K = {\alpha }^{-1}\left( {IK}\right) = {\alpha }^{-1}\left( {IJ}\right) = J \) . In general, the identity \( {IJ} = {IK} \) implies that\n\n\[... | Yes |
Lemma 4.11. Given two ideals \( I \) and \( J \) in \( {O}_{\mathbb{K}}, I \mid J \) if and only if \( J \subseteq I \) . | Proof. One direction is already proved, so assume that \( J \subseteq I \) . Then\n\n\[ J{I}^{ * } \subseteq I{I}^{ * } = \left( {N\left( I\right) }\right) \]\n\nso\n\n\[ K = \frac{1}{N\left( I\right) }J{I}^{ * } \]\n\nis an ideal contained in \( {O}_{\mathbb{K}} \) . It follows that\n\n\[ {IK} = \frac{1}{N\left( I\rig... | Yes |
Theorem 4.13. [Fundamenta Theorem of Arithmetic for Ideals] Any nonzero proper ideal in \( {O}_{\mathbb{K}} \) can be written as a product of prime ideals, and that factorization is unique up to order. | Proof. If \( I \) is not maximal, it can be written as a product of two nontrivial ideals. Comparing norms shows these ideals must have norms smaller than \( I \) . Keep going: The sequence of norms is descending, so it must terminate, resulting in a finite factorization of \( I \) . By Exercise 4.12, every maximal ide... | No |
Theorem 4.14. There are only finitely many equivalence classes of ideals in \( {O}_{\mathbb{K}} \) under \( \sim \) . | One class is easy to spot - namely the one consisting of all principal ideals. Of course, \( {O}_{\mathbb{K}} \) is a principal ideal domain if and only if there is only one class under the relation. One can define a multiplication on classes: If \( \left\lbrack I\right\rbrack \) denotes the class containing \( I \), t... | No |
Corollary 4.15. The set of classes under \( \sim \) forms a finite Abelian group. | Proof of Corollary 4.15. In the class group, associativity of multiplication is inherited from \( {O}_{\mathbb{K}} \) . The element \( \left\lbrack {O}_{\mathbb{K}}\right\rbrack \) acts as the identity. Finally, given any nonzero ideal \( I \), the relation \( I{I}^{ * } = \left( {N\left( I\right) }\right) \) shows tha... | Yes |
Theorem 5.2. There is a constant \( h > 0 \) for which\n\n\[ \frac{1}{{n}^{2}}\log {B}_{n} \rightarrow h\\text{ as }n \rightarrow \infty \]\n\nwhere \( \\left( {B}_{n}\\right) \) is the sequence in Table 5.1. | We are not going to prove this result. | No |
There is a right-angled triangle with rational sides and area 5. | The triple \( \left( {1\frac{1}{2},6\frac{2}{3},6\frac{5}{6}}\right) \) is Pythagorean: Expressing these as fractions over 6 and checking that\n\n\[ \n{9}^{2} + {40}^{2} = {41}^{2} \n\]\n\nconfirms the triangle with the sides given is right-angled. The area of the triangle is easily computed to be 5. | Yes |
Theorem 5.8. Suppose \( n \) is a positive integer and \( \left( {x, y}\right) \) denotes a rational point on the elliptic curve \( {y}^{2} = {x}^{3} - {n}^{2}x \) with \( x \) equal to the square of a rational with an even denominator. Then \( n \) is a congruent number. | Proof. The proof uses the characterization of Pythagorean triples from Theorem 2.1. Initially we retrace some of the steps used earlier, but there is an ingenious twist at the end of the proof. Let \( u = \sqrt{x} > 0 \) ; by assumption \( u \in \mathbb{Q} \) . Write \( v = y/u \) so\n\n\[ \n{v}^{2} = {y}^{2}/{u}^{2} =... | Yes |
Consider the curve \( {y}^{2} = {x}^{3} - {36x} \) and the point \( P = \left( {\frac{25}{4},\frac{35}{8}}\right) \) on the curve. Then \( P \) is a rational point of infinite order, and we compute that | \[ {2P} = \left( {\frac{1442401}{19600}, - \frac{1726556399}{2744000}}\right) \] and \[ {4P} = \left( {\frac{4386303618090112563849601}{233710164715943220558400}, - \frac{870369109085580828275935650626254401}{11298385812463619737216684496448000}}\right) . \] | Yes |
Theorem 5.11. [Mordell's Theorem] Let \( E \) denote an elliptic curve defined over \( \mathbb{Q} \) . Then \( E\left( \mathbb{Q}\right) \) is a finitely generated Abelian group. | A complete proof of this theorem may be found in the references at the end of the chapter. In Section 7.2 we will show how it follows from the so-called weak Mordell Theorem, and then in Section 7.3 will prove the weak Mordell Theorem in a special case. | No |
Example 5.12. Consider the curve \( E : {y}^{2} = {x}^{3} + 1 \), and let \( P = \left( {2,3}\right) \). Using the formulas, we find | \[ {2P} = \left( {0,1}\right) ,{3P} = {2P} + P = \left( {-1,0}\right) ,{4P} = {3P} + P = \left( {0, - 1}\right) = - {2P}\text{.} \] It follows that \( {6P} = 0 \) (so \( P \) is a torsion point with respect to the group structure on the curve), and since \( P,{2P},{3P} \neq 0 \), the point \( P \) has order 6 (see Figu... | Yes |
The curve\n\n\[ \n{y}^{2} = {x}^{3} + {x}^{2} - {4x} \n\]\n\nis 2-isogenous to the curve in Weierstrass form,\n\n\[ \nE : {y}^{2} = {x}^{3} + {x}^{2} + {16x} + {16}. \n\] | The generator \( \left( {-2,2}\right) \) maps to the generator \( P = \left( {0,4}\right) \) on \( E \) . Thus the sequence of denominators for \( P \) on \( E \) contains only a finite number of prime powers. | No |
Lemma 6.1. The series\n\n\[ \n{\wp }_{L}\left( z\right) = \frac{1}{{z}^{2}} + \mathop{\sum }\limits_{{0 \neq \ell \in L}}\left\{ {\frac{1}{{\left( z - \ell \right) }^{2}} - \frac{1}{{\ell }^{2}}}\right\} \n\]\n\nis absolutely convergent for any \( z \notin L \) . The series defines a meromorphic function whose only sin... | Proof. Let \( z \) be any point not in \( L \) . Write\n\n\[ \n\frac{1}{{\left( z - \ell \right) }^{2}} - \frac{1}{{\ell }^{2}} = \frac{1}{{\ell }^{3}} \cdot \frac{{2z} - {z}^{2}/\ell }{{\left( z/\ell - 1\right) }^{2}}. \n\]\n\nSince \( \left| {z/\ell - 1}\right| \) is bounded below by a positive constant, there is a c... | Yes |
Lemma 6.2. The Weierstrass \( \wp \) -function \( {\wp }_{L} \) is periodic with respect to \( L \) . | Proof. We want to prove that\n\n\[ \n{\wp }_{L}\left( {z + {\omega }_{1}}\right) = {\wp }_{L}\left( {z + {\omega }_{2}}\right) = {\wp }_{L}\left( z\right) \n\] \n\nfor all \( z \notin L \) . First, notice that by Equation (6.1) and Equation (6.2), \n\n\[ \n{\wp }_{L}\left( {-z}\right) = {\wp }_{L}\left( z\right) \text{... | Yes |
Lemma 6.4. An elliptic function with no poles in its fundamental domain is constant. | Proof. The first statement is clear: Any such function would be bounded on \( \Pi \), and therefore on all of \( \mathbb{C} \), by periodicity, so it is a bounded entire function and therefore must be constant by Liouville's Theorem. | Yes |
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