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Theorem 3.2 Suppose \( \Omega \) is a bounded open subset of \( {\mathbb{R}}^{d} \). Given a linear partial differential operator \( L \) with constant coefficients, there exists a bounded linear operator \( K \) on \( {L}^{2}\left( \Omega \right) \) such that whenever \( f \in {L}^{2}\left( \Omega \right) \), then \[ ... | We first prove the theorem assuming the validity of the inequality in the lemma. Consider the pre-Hilbert space \( {\mathcal{H}}_{0} = {C}_{0}^{\infty }\left( \Omega \right) \) equipped with the inner product and norm \[ \langle \varphi ,\psi \rangle = \left( {{L}^{ * }\varphi ,{L}^{ * }\psi }\right) ,\;\parallel \psi ... | Yes |
Lemma 3.4 Suppose \( P\left( z\right) = {z}^{m} + \cdots + {a}_{1}z + {a}_{0} \) is a polynonial of degree \( m \) with leading coefficient 1. If \( F \) is a holomorphic function on \( \mathbb{C} \) , then\n\n\[ \n{\left| F\left( 0\right) \right| }^{2} \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }{\left| P\left( {e}^{i\thet... | Proof. The lemma is a consequence of the special case when \( P = 1 \)\n\n(16)\n\n\[ \n{\left| F\left( 0\right) \right| }^{2} \leq \frac{1}{2\pi }{\int }_{0}^{2\pi }{\int }_{0}^{2\pi }{\left| F\left( {e}^{i\theta }\right) \right| }^{2}{d\theta }\n\]\n\nThis assertion follows directly from the mean-value identity (8) in... | Yes |
Proposition 4.1 Suppose there exists a function \( u \in {C}^{2}\left( \bar{\Omega }\right) \) that minimizes \( \mathcal{D}\left( U\right) \) among all \( U \in {C}^{2}\left( \bar{\Omega }\right) \) with \( {\left. U\right| }_{\partial \Omega } = f \) . Then \( u \) is harmonic in \( \Omega \) . | Proof. For functions \( F \) and \( G \) in \( {C}^{2}\left( \bar{\Omega }\right) \) define the following inner-product\n\n\[ \langle F, G\rangle = {\int }_{\Omega }\left( {\frac{\partial F}{\partial {x}_{1}}\overline{\frac{\partial G}{\partial {x}_{1}}} + \frac{\partial F}{\partial {x}_{2}}\overline{\frac{\partial G}{... | Yes |
Corollary 4.4 Suppose \( \Omega \) is a bounded open set, and let \( \partial \Omega = \bar{\Omega } - \Omega \) denote its boundary. Assume that \( u \) is continuous in \( \bar{\Omega } \) and is harmonic in \( \Omega \) . Then\n\n\[ \mathop{\max }\limits_{{x \in \bar{\Omega }}}\left| {u\left( x\right) }\right| = \ma... | Proof. Since the sets \( \bar{\Omega } \) and \( \partial \Omega \) are compact and \( u \) is continuous, the two maxima above are clearly attained. We suppose that \( \mathop{\max }\limits_{{x \in \bar{\Omega }}}\left| {u\left( x\right) }\right| \) is attained at an interior point \( {x}_{0} \in \Omega \), for otherw... | Yes |
Lemma 4.5 We have the identity\n\n\[ \n{\int }_{B}\left( {v\bigtriangleup u - u\bigtriangleup v}\right) {\eta dx} = {\int }_{B}u\left( {\nabla v \cdot \nabla \eta }\right) - v\left( {\nabla u \cdot \nabla \eta }\right) {dx}. \n\] | Here \( \nabla u \) is the gradient of \( u \), that is, \( \nabla u = \left( {\frac{\partial u}{\partial {x}_{1}},\frac{\partial u}{\partial {x}_{2}},\ldots ,\frac{\partial u}{\partial {x}_{d}}}\right) \) and\n\n\[ \n\nabla v \cdot \nabla \eta = \mathop{\sum }\limits_{{j = 1}}^{d}\frac{\partial v}{\partial {x}_{j}}\fr... | Yes |
Corollary 4.8 Suppose \( \left\{ {u}_{n}\right\} \) is a sequence of harmonic functions in \( \Omega \) that converges to a function \( u \) uniformly on compact subsets of \( \Omega \) as \( n \rightarrow \infty \) . Then \( u \) is also harmonic. | The first of these corollaries was already proved as a consequence of (26). For the second, we use the fact that each \( {u}_{n} \) satisfies the mean-value property\n\n\[ \n{u}_{n}\left( {x}_{0}\right) = \frac{1}{m\left( B\right) }{\int }_{B}{u}_{n}\left( x\right) {dx} \n\]\n\nwhenever \( B \) is a ball with center at... | Yes |
Lemma 4.9 Let \( \Omega \) be an open bounded set in \( {\mathbb{R}}^{d} \). Suppose \( v \) belongs to \( {C}^{1}\left( \bar{\Omega }\right) \) and \( v \) vanishes on \( \partial \Omega \). Then\n\n\[{\int }_{\Omega }{\left| v\left( x\right) \right| }^{2}{dx} \leq {c}_{\Omega }{\int }_{\Omega }{\left| \nabla v\left( ... | Proof. This conclusion could in fact be deduced from the considerations given in Lemma 3.3. We prefer to prove this easy version separately to highlight a simple idea that we shall also use later. It should be noted that the argument yields the estimate \( {c}_{\Omega } \leq d{\left( \Omega \right) }^{2} \), where \( d... | Yes |
Lemma 4.10 Suppose \( \Gamma \) is a compact set in \( {\mathbb{R}}^{d} \), and \( f \) is a continuous function on \( \Gamma \) . Then there exists a sequence \( \left\{ {F}_{n}\right\} \) of smooth functions on \( {\mathbb{R}}^{d} \) so that \( {F}_{n} \rightarrow f \) uniformly on \( \Gamma \) . | To complete the proof of Lemma 4.10, we argue as follows. We regularize the function \( G \) obtained in Lemma 4.11 by defining\n\n\[ {F}_{\epsilon }\left( x\right) = {\epsilon }^{-d}{\int }_{{\mathbb{R}}^{d}}G\left( {x - y}\right) \varphi \left( {y/\epsilon }\right) {dy} = {\int }_{{\mathbb{R}}^{d}}G\left( y\right) {\... | Yes |
Lemma 4.11 Let \( f \) be a continuous function on a compact subset \( \Gamma \) of \( {\mathbb{R}}^{d} \). Then there exists a function \( G \) on \( {\mathbb{R}}^{d} \) that is continuous, and so that \( {\left. G\right| }_{\partial \Gamma } = f \) | Proof. We begin with the observation that if \( {K}_{0} \) and \( {K}_{1} \) are two disjoint compact sets, there exists a continuous function \( 0 \leq g\left( x\right) \leq 1 \) on \( {\mathbb{R}}^{d} \) which takes the value 0 on \( {K}_{0} \) and 1 on \( {K}_{1} \). Indeed, if \( d\left( {x,\Omega }\right) \) denot... | Yes |
Theorem 4.12 Let \( \Omega \) be an open bounded set in \( {\mathbb{R}}^{2} \) that satisfies the outside-triangle condition. If \( f \) is a continuous function on \( \partial \Omega \), then the boundary value problem \( \bigtriangleup u = 0 \) with \( u \) continuous in \( \bar{\Omega } \) and \( {\left. u\right| }_... | We turn to the proof of the theorem. It is based on the following proposition, which may be viewed as a refined version of Lemma 4.9 above.\n\nProposition 4.13 For any bounded open set \( \Omega \) in \( {\mathbb{R}}^{2} \) that satisfies the outside-triangle condition there are two constants \( {c}_{1} < 1 \) and \( {... | No |
Theorem 1.2 If \( {\mu }_{ * } \) is a metric exterior measure on a metric space \( X \) , then the Borel sets in \( X \) are measurable. Hence \( {\mu }_{ * } \) restricted to \( {\mathcal{B}}_{X} \) is a measure. | Proof. By the definition of \( {\mathcal{B}}_{X} \) it suffices to prove that closed sets in \( X \) are Carathéodory measurable. Therefore, let \( F \) denote a closed set and \( A \) a subset of \( X \) with \( {\mu }_{ * }\left( A\right) < \infty \) . For each \( n > 0 \), let\n\n\[ \n{A}_{n} = \left\{ {x \in {F}^{c... | Yes |
Lemma 1.4 If \( {\mu }_{0} \) is a premeasure on an algebra \( \mathcal{A} \), define \( {\mu }_{ * } \) on any subset \( E \) of \( X \) by\n\n\[ \n{\mu }_{ * }\left( E\right) = \inf \left\{ {\mathop{\sum }\limits_{{j = 1}}^{\infty }{\mu }_{0}\left( {E}_{j}\right) : E \subset \mathop{\bigcup }\limits_{{j = 1}}^{\infty... | Proof. Proving that \( {\mu }_{ * } \) is an exterior measure presents no difficulty. To see why the restriction of \( {\mu }_{ * } \) to \( \mathcal{A} \) coincides with \( {\mu }_{0} \), suppose that \( E \in \mathcal{A} \) . Clearly, one always has \( {\mu }_{ * }\left( E\right) \leq {\mu }_{0}\left( E\right) \) sin... | Yes |
Theorem 1.5 Suppose that \( \mathcal{A} \) is an algebra of sets in \( X,{\mu }_{0} \) a premeasure on \( \mathcal{A} \), and \( \mathcal{M} \) the \( \sigma \) -algebra generated by \( \mathcal{A} \) . Then there exists a measure \( \mu \) on \( \mathcal{M} \) that extends \( {\mu }_{0} \) . | Proof. The exterior measure \( {\mu }_{ * } \) induced by \( {\mu }_{0} \) defines a measure \( \mu \) on the \( \sigma \) -algebra of Carathéodory measurable sets. Therefore, by the result in the previous lemma, \( \mu \) is also a measure on \( \mathcal{M} \) that extends \( {\mu }_{0} \) . (We should observe that in... | Yes |
Proposition 3.2 If \( E \) is an arbitrary measurable set in \( X \), then the conclusion of Proposition 3.1 are still valid except that we only assert that \( {E}^{{x}_{2}} \) is \( {\mu }_{1} \) -measurable and \( {\mu }_{1}\left( {E}^{{x}_{2}}\right) \) is defined for almost every \( {x}_{2} \in {X}_{2} \) . | Proof. Consider first the case when \( E \) is a set of measure zero. Then we know by Proposition 1.6 that there is a set \( F \in {\mathcal{A}}_{\sigma \delta } \) such that\n\n\( E \subset F \) and \( \left( {{\mu }_{1} \times {\mu }_{2}}\right) \left( F\right) = 0 \) . Since \( {E}^{{x}_{2}} \subset {F}^{{x}_{2}} \)... | Yes |
Theorem 3.3 In the setting above, suppose \( f\left( {{x}_{1},{x}_{2}}\right) \) is an integrable function on \( \left( {{X}_{1} \times {X}_{2},{\mu }_{1} \times {\mu }_{2}}\right) \) . (i) For almost every \( {x}_{2} \in {X}_{2} \), the slice \( {f}^{{x}_{2}}\left( {x}_{1}\right) = f\left( {{x}_{1},{x}_{2}}\right) \) ... | Proof. Note that if the desired conclusions hold for finitely many functions, they also hold for their linear combinations. In particular it suffices to assume that \( f \) is non-negative. When \( f = {\chi }_{E} \), where \( E \) is a set of finite measure, what we wish to prove is contained in Proposition 3.2. Hence... | Yes |
Proposition 4.1 The total variation \( \left| \nu \right| \) of a signed measure \( \nu \) is itself a (positive) measure that satisfies \( \nu \leq \left| \nu \right| \) . | Proof. Suppose \( {\left\{ {E}_{j}\right\} }_{j = 1}^{\infty } \) is a countable collection of disjoints sets in \( \mathcal{M} \), and let \( E = \bigcup {E}_{j} \) . It suffices to prove:\n\n(11)\n\n\[ \sum \left| \nu \right| \left( {E}_{j}\right) \leq \left| \nu \right| \left( E\right) \;\text{ and }\;\left| \nu \ri... | Yes |
Proposition 4.2 The assertion (14) implies (12). Conversely, if \( \\left| \\nu \\right| \) is a finite measure, then (12) implies (14). | That (12) is a consequence of (14) is obvious because \( \\mu \\left( E\\right) = 0 \) gives \( \\left| {\\nu \\left( E\\right) }\\right| < \\epsilon \) for every \( \\epsilon > 0 \) . To prove the converse, it suffices to consider the case when \( \\nu \) is positive, upon replacing \( \\nu \) by \( \\left| \\nu \\rig... | Yes |
Theorem 4.3 Suppose \( \mu \) is a \( \sigma \) -finite positive measure on the measure space \( \left( {X,\mathcal{M}}\right) \) and \( \nu \) a \( \sigma \) -finite signed measure on \( \mathcal{M} \) . Then there exist unique signed measures \( {\nu }_{a} \) and \( {\nu }_{s} \) on \( \mathcal{M} \) such that \( {\n... | We start with the case when both \( \nu \) and \( \mu \) are positive and finite. Let \( \rho = \nu + \mu \), and consider the transformation on \( {L}^{2}\left( {X,\rho }\right) \) defined by\n\n\[ \ell \left( \psi \right) = {\int }_{X}\psi \left( x\right) {d\nu }\left( x\right) \]\n\nThe mapping \( \ell \) defines a ... | Yes |
Lemma 5.2 The following relations hold among the subspaces \( S,{S}_{ * } \) , and \( \overline{{S}_{1}} \) .\n\n(i) \( S = {S}_{ * } \) .\n\n(ii) The orthogonal complement of \( \overline{{S}_{1}} \) is \( S \) . | Proof. First, since \( T \) is an isometry, we have that \( \left( {{Tf},{Tg}}\right) = \left( {f, g}\right) \) for all \( f, g \in \mathcal{H} \), and thus \( {T}^{ * }T = I \) . (See Exercise 22 in Chapter 4.) So if \( {Tf} = f \) then \( {T}^{ * }{Tf} = {T}^{ * }f \), which means that \( f = {T}^{ * }f \) . To prove... | Yes |
Theorem 5.4 Suppose \( f \) is integrable over \( X \) . Then for almost every \( x \in X \) the averages \( {A}_{m}\left( f\right) = \frac{1}{m}\mathop{\sum }\limits_{{k = 0}}^{{m - 1}}f\left( {{\tau }^{k}\left( x\right) }\right) \) converge to a limit as \( m \rightarrow \infty \) . | The idea of the proof is as follows. We first show that \( {A}_{m}\left( f\right) \) converges to a limit almost everywhere for a set of functions \( f \) that is dense in \( {L}^{1}\left( {X,\mu }\right) \) . We then use the maximal theorem to show that this implies the conclusion for all integrable functions. | No |
Theorem 6.1 Suppose \( T \) is a bounded symmetric operator on a Hilbert space \( \mathcal{H} \). Then there exists a spectral resolution \( \{ E\left( \lambda \right) \} \) such that\n\n\[ T = {\int }_{{a}^{ - }}^{b}{\lambda dE}\left( \lambda \right) \]\n\n in the sense that for every \( f, g \in \mathcal{H} \)\n\n(32... | The integral on the right-hand side is taken in the Lebesgue-Stieltjes sense, as in (iii) and (iv) of Section 3.3.\n\nThe result encompasses the spectral theorem for compact symmetric operators \( T \) in the following sense. Let \( \left\{ {\varphi }_{k}\right\} \) be an orthonormal basis of eigenvectors of \( T \) wi... | Yes |
Proposition 6.2 Suppose \( T \) is symmetric. Then \( \parallel T\parallel \leq M \) if and only if \( - {MI} \leq \) \( T \leq {MI} \) . As a result, \( \parallel T\parallel = \max \left( {\left| a\right| ,\left| b\right| }\right) \) . | This is a consequence of (7) in Chapter 4. | No |
Proposition 6.4 If \( {T}_{1} \) and \( {T}_{2} \) are positive operators that commute, then \( {T}_{1}{T}_{2} \) is also positive. | Indeed, if \( S \) is a square root of \( {T}_{1} \) given in the previous proposition, then \( {T}_{1}{T}_{2} = \) \( {SS}{T}_{2} = S{T}_{2}S \), and hence \( \left( {{T}_{1}{T}_{2}f, f}\right) = \left( {S{T}_{2}{Sf}, f}\right) = \left( {{T}_{2}{Sf},{Sf}}\right) \), since \( S \) is symmetric, and thus the last term i... | Yes |
Proposition 6.5 Suppose \( T \) is symmetric and a and \( b \) are given by (33). If \( p\left( t\right) = \) \( \mathop{\sum }\limits_{{k = 0}}^{n}{c}_{k}{t}^{k} \) is a real polynomial which is positive for \( t \in \left\lbrack {a, b}\right\rbrack \), then the operator \( p\left( T\right) = \mathop{\sum }\limits_{{k... | To see this, write \( p\left( t\right) = c\mathop{\prod }\limits_{j}\left( {t - {\rho }_{j}}\right) \mathop{\prod }\limits_{k}\left( {{\rho }_{k}^{\prime } - t}\right) \mathop{\prod }\limits_{\ell }\left( {{\left( t - {\mu }_{\ell }\right) }^{2} + {\nu }_{\ell }}\right) \), where \( c \) is positive and the third facto... | Yes |
Corollary 6.6 If \( p\left( t\right) \) is a real polynomial, then\n\n\[ \parallel p\left( T\right) \parallel \leq \mathop{\sup }\limits_{{t \in \left\lbrack {a, b}\right\rbrack }}\left| {p\left( t\right) }\right| \] | This is an immediate consequence using Proposition 6.2, since \( - M \leq p\left( t\right) \leq M \) , where \( M = \mathop{\sup }\limits_{{t \in \left\lbrack {a, b}\right\rbrack }}\left| {p\left( t\right) }\right| \), and thus \( - {MI} \leq p\left( T\right) \leq {MI} \) . | Yes |
Proposition 6.7 Suppose \( \left\{ {T}_{n}\right\} \) is a sequence of positive operators that satisfy \( {T}_{n} \geq {T}_{n + 1} \) for all \( n \) . Then there is a positive operator \( T \), such that \( {T}_{n}f \rightarrow {Tf} \) as \( n \rightarrow \infty \) for every \( f \in \mathcal{H} \) . | Proof. We note that for each fixed \( f \in \mathcal{H} \) the sequence of positive numbers \( \left( {{T}_{n}f, f}\right) \) is decreasing and hence convergent. Now observe that for any positive operator \( S \) with \( \parallel S\parallel \leq M \) we have\n\n(35)\n\n\[ \parallel S\left( f\right) {\parallel }^{2} \l... | Yes |
Proposition 6.8 If \( T \) is symmetric, then \( \sigma \left( T\right) \) is a closed subset of the interval \( \left\lbrack {a, b}\right\rbrack \) given by (33). | Note that if \( z \notin \left\lbrack {a, b}\right\rbrack \), the function \( \Phi \left( t\right) = {\left( t - z\right) }^{-1} \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and \( \Phi \left( T\right) \left( {T - {zI}}\right) = \left( {T - {zI}}\right) \Phi \left( T\right) = I \), so \( \Phi \left( T\rig... | Yes |
Proposition 6.9 For each \( f \in \mathcal{H} \), the Lebesgue-Stieltjes measure corresponding to \( F\left( \lambda \right) = \left( {E\left( \lambda \right) f, f}\right) \) is supported on \( \sigma \left( T\right) \) . | To prove this, let \( J \) be one of the open intervals in the complement of \( \sigma \left( T\right) \) , \( {x}_{0} \in J \), and \( {J}_{0} \) the sub-interval centered at \( {x}_{0} \) of length \( {2\epsilon } \), with \( \epsilon < \begin{Vmatrix}{\left( T - {x}_{0}I\right) }^{-1}\end{Vmatrix} \) . First note th... | Yes |
Property 1 (Monotonicity) If \( {E}_{1} \subset {E}_{2} \), then \( {m}_{\alpha }^{ * }\left( {E}_{1}\right) \leq {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) . | This is straightforward, since any cover of \( {E}_{2} \) is also a cover of \( {E}_{1} \) . | No |
Property 2 (Sub-additivity) \( {m}_{\alpha }^{ * }\left( {\mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j}}\right) \leq \mathop{\sum }\limits_{{j = 1}}^{\infty }{m}_{\alpha }^{ * }\left( {E}_{j}\right) \) for any countable family \( \left\{ {E}_{j}\right\} \) of sets in \( {\mathbb{R}}^{d} \) . | For the proof, fix \( \delta \), and choose for each \( j \) a cover \( {\left\{ {F}_{j, k}\right\} }_{k = 1}^{\infty } \) of \( {E}_{j} \) by sets of diameter less than \( \delta \) such that \( \mathop{\sum }\limits_{k}{\left( \operatorname{diam}{F}_{j, k}\right) }^{\alpha } \leq {\mathcal{H}}_{\alpha }^{\delta }\lef... | Yes |
Property 3 If \( d\left( {{E}_{1},{E}_{2}}\right) > 0 \), then \( {m}_{\alpha }^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) = {m}_{\alpha }^{ * }\left( {E}_{1}\right) + {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) . | It suffices to prove that \( {m}_{\alpha }^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) \geq {m}_{\alpha }^{ * }\left( {E}_{1}\right) + {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) since the reverse inequality is guaranteed by sub-additivity. Fix \( \epsilon > 0 \) with \( \epsilon < \) \( d\left( {{E}_{1},{E}_{2}}\right) ... | Yes |
Property 4 If \( \\left\\{ {E}_{j}\\right\\} \) is a countable family of disjoint Borel sets, and \( E = \\mathop{\\bigcup }\\limits_{{j = 1}}^{\\infty }{E}_{j} \), then | \[ {m}_{\\alpha }\\left( E\\right) = \\mathop{\\sum }\\limits_{{j = 1}}^{\\infty }{m}_{\\alpha }\\left( {E}_{j}\\right) \] | Yes |
Property 5 Hausdorff measure is invariant under translations\n\n\[ \n{m}_{\alpha }\left( {E + h}\right) = {m}_{\alpha }\left( E\right) \;\text{ for all }h \in {\mathbb{R}}^{d}, \n\]\n\nand rotations\n\n\[ \n{m}_{\alpha }\left( {rE}\right) = {m}_{\alpha }\left( E\right) \n\]\n\nwhere \( r \) is a rotation in \( {\mathbb... | Moreover, it scales as follows:\n\n\[ \n{m}_{\alpha }\left( {\lambda E}\right) = {\lambda }^{\alpha }{m}_{\alpha }\left( E\right) \;\text{ for all }\lambda > 0. \n\]\n\nThese conclusions follow once we observe that the diameter of a set \( S \) is invariant under translations and rotations, and satisfies \( \operatorna... | No |
Property 7 If \( E \) is a Borel subset of \( {\mathbb{R}}^{d} \), then \( {c}_{d}{m}_{d}\left( E\right) = m\left( E\right) \) for some constant \( {c}_{d} \) that depends only on the dimension \( d \) . | The constant \( {c}_{d} \) equals \( m\left( B\right) /{\left( \operatorname{diam}B\right) }^{d} \), for the unit ball \( B \) ; note that this ratio is the same for all balls \( B \) in \( {\mathbb{R}}^{d} \), and so \( {c}_{d} = {v}_{d}/{2}^{d} \) (where \( {v}_{d} \) denotes the volume of the unit ball). The proof o... | Yes |
Theorem 2.1 The Cantor set \( \mathcal{C} \) has strict Hausdorff dimension \( \alpha = \) \( \log 2/\log 3 \) . | The inequality\n\n\[ \n{m}_{\alpha }\left( \mathcal{C}\right) \leq 1 \n\]\n\nfollows from the construction of \( \mathcal{C} \) and the definitions. Indeed, recall from Chapter 1 that \( \mathcal{C} = \bigcap {C}_{k} \), where each \( {C}_{k} \) is a finite union of \( {2}^{k} \) intervals of length \( {3}^{-k} \) . Gi... | Yes |
Lemma 2.2 Suppose a function \( f \) defined on a compact set \( E \) satisfies a Lipschitz condition with exponent \( \gamma \) . Then\n\n(i) \( {m}_{\beta }\left( {f\left( E\right) }\right) \leq {M}^{\beta }{m}_{\alpha }\left( E\right) \) if \( \beta = \alpha /\gamma \) .\n\n(ii) \( \dim f\left( E\right) \leq \frac{1... | Proof. Suppose \( \left\{ {F}_{k}\right\} \) is a countable family of sets that covers \( E \) . Then \( \left\{ {f\left( {E \cap {F}_{k}}\right) }\right\} \) covers \( f\left( E\right) \) and, moreover, \( f\left( {E \cap {F}_{k}}\right) \) has diameter less than \( M{\left( \operatorname{diam}{F}_{k}\right) }^{\gamma... | Yes |
Lemma 2.3 The Cantor-Lebesgue function \( F \) on \( \mathcal{C} \) satisfies a Lipschitz condition with exponent \( \gamma = \log 2/\log 3 \) . | Proof. The function \( F \) was constructed in Section 3.1 of Chapter 3 as the limit of a sequence \( \left\{ {F}_{n}\right\} \) of piecewise linear functions. The function \( {F}_{n} \) increases by at most \( {2}^{-n} \) on each interval of length \( {3}^{-n} \) . So the slope of \( {F}_{n} \) is always bounded by \(... | Yes |
Theorem 2.5 The Sierpinski triangle \( \mathcal{S} \) has strict Hausdorff dimension \( \alpha = \log 3/\log 2 \) . | The inequality \( {m}_{\alpha }\left( \mathcal{S}\right) \leq 1 \) follows immediately from the construction. Given \( \delta > 0 \), choose \( K \) so that \( {2}^{-K} < \delta \) . Since the set \( {S}_{K} \) covers \( \mathcal{S} \) and consists of \( {3}^{K} \) triangles each of diameter \( {2}^{-K} < \delta \), we... | Yes |
Lemma 2.6 Suppose \( B \) is a ball in the covering \( \mathcal{B} \) that satisfies\n\n\[ \n{2}^{-\ell } \leq \operatorname{diam}B < {2}^{-\ell + 1}\;\text{ for some }\ell \leq k.\n\]\n\nThen \( B \) contains at most \( c{3}^{k - \ell } \) vertices of the \( {k}^{\text{th }} \) generation. | Proof of Lemma 2.6. Let \( {B}^{ * } \) denote the ball with same center as \( B \) but three times its diameter, and let \( {\bigtriangleup }_{k} \) be a triangle of the \( {k}^{\text{th }} \) generation whose vertex \( v \) lies in \( B \) . If \( {\bigtriangleup }_{\ell }^{\prime } \) denotes the triangle of the \( ... | Yes |
Lemma 2.8 Suppose \( \\left\\{ {f}_{j}\\right\\} \) is a sequence of continuous functions on the interval \( \\left\\lbrack {0,1}\\right\\rbrack \) that satisfy\n\n\[ \n\\left| {{f}_{j}\\left( t\\right) - {f}_{j}\\left( s\\right) }\\right| \\leq {A}^{j}\\left| {t - s}\\right| \\;\\text{ for some }A > 1,\n\]\n\nand\n\n\... | Proof. The continuous limit \( f \) is given by the uniformly convergent series\n\n\[ \nf\\left( t\\right) = {f}_{1}\\left( t\\right) + \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }\\left( {{f}_{k + 1}\\left( t\\right) - {f}_{k}\\left( t\\right) }\\right)\n\]\n\nand therefore\n\n\[ \n\\left| {f\\left( t\\right) - {f}_{... | Yes |
Theorem 2.9 Suppose \( {S}_{1},{S}_{2},\ldots ,{S}_{m} \) are \( m \) similartities, each with the same ratio \( r \) that satisfies \( 0 < r < 1 \) . Then there exists a unique nonempty compact set \( F \) such that\n\n\[ F = {S}_{1}\left( F\right) \cup \cdots \cup {S}_{m}\left( F\right) \] | The proof of this theorem is in the nature of a fixed point argument. We shall begin with some large ball \( B \) and iteratively apply the mappings \( {S}_{1},\ldots ,{S}_{m} \) . The fact that the similarities have ratio \( r < 1 \) will suffice to imply that this process contracts to a unique set \( F \) with the de... | Yes |
Lemma 2.10 There exists a closed ball \( B \) so that \( {S}_{j}\left( B\right) \subset B \) for all \( j = 1,\ldots, m \) . | Proof. Indeed, we note that if \( S \) is a similarity with ratio \( r \), then\n\n\[ \left| {S\left( x\right) }\right| \leq \left| {S\left( x\right) - S\left( 0\right) }\right| + \left| {S\left( 0\right) }\right| \]\n\n\[ \leq r\left| x\right| + \left| {S\left( 0\right) }\right| \text{.} \]\n\nIf we require that \( \l... | Yes |
Lemma 2.11 The distance function dist defined on compact subsets of \( {\mathbb{R}}^{d} \) satisfies\n\n(i) \( \operatorname{dist}\left( {A, B}\right) = 0 \) if and only if \( A = B \) .\n\n(ii) \( \operatorname{dist}\left( {A, B}\right) = \operatorname{dist}\left( {B, A}\right) \) .\n\n(iii) \( \operatorname{dist}\lef... | The proof of the lemma is simple and may be left to the reader. | No |
Theorem 2.12 Suppose \( {S}_{1},{S}_{2},\ldots ,{S}_{m} \) are \( m \) separated similarities with the common ratio \( r \) that satisfies \( 0 < r < 1 \) . Then the set \( F \) has Hausdorff dimension equal to \( \log m/\log \left( {1/r}\right) \) . | We now turn to the proof of Theorem 2.12, which will follow the same approach used in the case of the Sierpinski triangle. If \( \alpha = \log m/\log \left( {1/r}\right) \) , we claim that \( {m}_{\alpha }\left( F\right) < \infty \), hence \( \dim F \leq \alpha \) . Moreover, this inequality holds even without the sepa... | Yes |
Lemma 2.13 Suppose \( B \) is a ball in the covering \( \mathcal{B} \) that satisfies\n\n\[ \n{r}^{\ell } \leq \operatorname{diam}B < {r}^{\ell - 1}\;\text{ for some }\ell \leq k.\n\]\n\nThen \( B \) contains at most \( {\mathrm{{cm}}}^{k - \ell } \) vertices of the \( {k}^{\text{th }} \) generation. | Proof. If \( v \) is a vertex of the \( {k}^{\text{th }} \) generation with \( v \in B \), and \( \mathcal{O}\left( v\right) \) denotes the corresponding open set of the \( {k}^{\text{th }} \) generation, then, for some fixed dilate \( {B}^{ * } \) of \( B \), properties (a) and (b) above guarantee that \( \mathcal{O}\... | Yes |
Corollary 3.2 There are subsets \( {Z}_{1} \subset \left\lbrack {0,1}\right\rbrack \) and \( {Z}_{2} \subset \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \), each of measure zero, such that \( \mathcal{P} \) is bijective from | \[ \left\lbrack {0,1}\right\rbrack - {Z}_{1}\;\text{to}\;\left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack - {Z}_{2} \] and measure preserving. In other words, \( E \) is measurable if and only if \( \mathcal{P}\left( E\right) \) is measurable, and \[ {m}_{1}\left( E\right) = {m}_{2}\left( {\mathca... | Yes |
Proposition 3.3 Chains of quartic intervals satisfy the following properties:\n\n(i) If \( \\left\\{ {I}^{k}\\right\\} \) is a chain of quartic intervals, then there exists a unique \( t \\in \\left\\lbrack {0,1}\\right\\rbrack \) such that \( t \\in \\mathop{\\bigcap }\\limits_{k}{I}^{k} \) . | Proof. Part (i) follows from the fact that \( \\left\\{ {I}^{k}\\right\\} \) is a decreasing sequence of compact sets whose diameters go to 0 . | Yes |
Lemma 3.6 Let\n\n\\[ \n{E}_{0} = \\left\\{ {x = \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }{a}_{k}/{4}^{k},\\;}\\right. \\text{where}\\left. {{a}_{k} \\neq {f}_{k}\\text{for all sufficiently large}k}\\right\\} \\text{.} \n\\]\n\nThen \\( m\\left( {E}_{0}\\right) = 0 \\) . | Indeed, if we fix \\( r \\), then \\( m\\left( \\left\\{ {x : {a}_{r} \\neq {f}_{r}}\\right\\} \\right) = 3/4 \\), and\n\n\\[ \nm\\left( \\left\\{ {x : {a}_{r} \\neq {f}_{r}\\text{ and }{a}_{r + 1} \\neq {f}_{r + 1}}\\right\\} \\right) = {\\left( 3/4\\right) }^{2},\\;\\text{ etc. } \n\\]\n\nThus \\( m\\left( \\left\\{ ... | Yes |
Lemma 3.8 If \( \Phi \) is the dyadic correspondence in Lemma 3.7, then \( {\Phi }^{ * }\left( t\right) = \) \( \mathcal{P}\left( t\right) \) for every \( 0 \leq t \leq 1 \) . | Proof. First, we observe that \( {\Phi }^{ * }\left( t\right) \) is unambiguously defined for every \( t \) . Indeed, suppose \( t \in \mathop{\bigcap }\limits_{k}{I}^{k} \) and \( t \in \mathop{\bigcap }\limits_{k}{J}^{k} \) are two chains of quartic intervals; then \( {I}^{k} \) and \( {J}^{k} \) must be adjacent for... | Yes |
Theorem 4.3 There exists a set \( \mathcal{B} \) in \( {\mathbb{R}}^{2} \) that:\n\n(i) is compact,\n\n(ii) has Lebesgue measure zero,\n\n(iii) contains a translate of every unit line segment. | Note that with \( F = \mathcal{B} \) and \( \gamma \in {S}^{1} \) one has \( {m}_{1}\left( {F \cap {\mathcal{P}}_{{t}_{0},\gamma }}\right) \geq 1 \) for some \( {t}_{0} \) . If \( {m}_{1}\left( {F \cap {\mathcal{P}}_{t,\gamma }}\right) \) were continuous in \( t \), then this measure would be strictly positive for an i... | No |
Lemma 4.7 If \( f \) is continuous with compact support, then for every \( \gamma \in {S}^{d - 1} \) we have\n\n\[ \widehat{\mathcal{R}}\left( f\right) \left( {\lambda ,\gamma }\right) = \widehat{f}\left( {\lambda \gamma }\right) \] | Proof. For each unit vector \( \gamma \) we use the adapted coordinate system described above: \( x = \left( {{x}_{1},\ldots ,{x}_{d}}\right) \) where \( \gamma \) coincides with the \( {x}_{d} \) direction. We can then write each \( x \in {\mathbb{R}}^{d} \) as \( x = \left( {u, t}\right) \) with \( u \in {\mathbb{R}}... | Yes |
Lemma 4.8 If \( f \) is continuous with compact support, then\n\n\[ \n{\int }_{{S}^{d - 1}}\left( {{\int }_{-\infty }^{\infty }{\left| \widehat{\mathcal{R}}\left( f\right) \left( \lambda ,\gamma \right) \right| }^{2}{\left| \lambda \right| }^{d - 1}{d\lambda }}\right) {d\sigma }\left( \gamma \right) = 2{\int }_{{\mathb... | Proof. The Plancherel formula in Chapter 5 guarantees that\n\n\[ \n2{\int }_{{\mathbb{R}}^{d}}{\left| f\left( x\right) \right| }^{2}{dx} = 2{\int }_{{\mathbb{R}}^{d}}{\left| \widehat{f}\left( \xi \right) \right| }^{2}{d\xi }.\n\]\n\nChanging to polar coordinates \( \xi = {\lambda \gamma } \) where \( \lambda > 0 \) and... | Yes |
Lemma 4.9 Suppose\n\n\\[ F\\left( t\\right) = {\\int }_{-\\infty }^{\\infty }\\widehat{F}\\left( \\lambda \\right) {e}^{2\\pi i\\lambda t}{d\\lambda } \\]\n\nwhere\n\n\\[ \\mathop{\\sup }\\limits_{{\\lambda \\in \\mathbb{R}}}\\left| {\\widehat{F}\\left( \\lambda \\right) }\\right| \\leq A\\;\\text{ and }\\;{\\int }_{-\... | Proof. The first inequality is obtained by considering separately the two cases \\( \\left| \\lambda \\right| \\leq 1 \\) and \\( \\left| \\lambda \\right| > 1 \\) . We write\n\n\\[ F\\left( t\\right) = {\\int }_{\\left| \\lambda \\right| \\leq 1}\\widehat{F}\\left( \\lambda \\right) {e}^{2\\pi i\\lambda t}{d\\lambda }... | Yes |
Theorem 4.10 If \( f \) is continuous with compact support, then\n\n\[ \n{\int }_{{S}^{1}}{\mathcal{R}}_{\delta }^{ * }\left( f\right) \left( \gamma \right) {d\sigma }\left( \gamma \right) \leq c{\left( \log 1/\delta \right) }^{1/2}\left( {\parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{2}\right) } + \parallel f{\... | The same argument as in the proof of Theorem 4.5 applies here, except that we need a modified version of Lemma 4.9. More precisely, let us set\n\n\[ \n{F}_{\delta }\left( t\right) = {\int }_{-\infty }^{\infty }\widehat{F}\left( \lambda \right) \left( \frac{{e}^{{2\pi i}\left( {t + \delta }\right) \lambda } - {e}^{{2\pi... | Yes |
Theorem 4.12 The set \( F \) is compact and of two-dimensional measure zero. It contains a translate of any unit line segment whose slope is a number \( s \) that lies outside the intervals \( \left( {-1,2}\right) \) . | The proof of the required properties of the set \( F \) amounts to showing the following paradoxical facts about the set \( \mathcal{C} + \lambda \mathcal{C} \), for \( \lambda > 0 \) . Here \( \mathcal{C} + \lambda \mathcal{C} = \left\{ {{x}_{1} + \lambda {x}_{2} : {x}_{1} \in \mathcal{C},{x}_{2} \in \mathcal{C}}\righ... | Yes |
Lemma 4.14 For every \( {\lambda }_{0} \) there is a pair \( 1 \leq {i}_{1},{i}_{2} \leq 4 \), with \( {i}_{1} \neq {i}_{2} \) such that \( {\mathcal{K}}_{{i}_{1}}\left( {\lambda }_{0}\right) \) and \( {\mathcal{K}}_{{i}_{2}}\left( {\lambda }_{0}\right) \) intersect. | Proof. Indeed, if the \( {\mathcal{K}}_{i} \) are disjoint for \( 1 \leq i \leq 4 \) then for sufficiently small \( \delta \) the \( {\mathcal{K}}_{i}^{\delta } \) are also disjoint. Here we have used the notation that \( {F}^{\delta } \) denotes the set of points of distance less than \( \delta \) from \( F \) . (See ... | Yes |
Corollary 1.2.9. Let \( {S}_{0} \) be a finite set of prime ideals of \( K \), let \( {\left( {e}_{\mathfrak{p}}\right) }_{\mathfrak{p} \in {S}_{0}} \) be a set of integers indexed by \( {S}_{0} \), and let \( {\left( {s}_{\sigma }\right) }_{\sigma \in {S}_{\infty }} \) be a set of signs \( \pm 1 \) indexed by the set ... | Proof. Set \( S = {S}_{0} \cup {S}_{\infty } \) considered as a set of places of \( K \) thanks to Ostrowski’s theorem. For \( \mathfrak{p} \in {S}_{0} \), we choose\n\n\[ \n{y}_{\mathfrak{p}} \in {\mathfrak{p}}^{{e}_{\mathfrak{p}}} \smallsetminus {\mathfrak{p}}^{{e}_{\mathfrak{p}} + 1}\;\text{ and }\;{\varepsilon }_{\... | Yes |
Corollary 1.2.10. Let \( \mathfrak{m} \) be any nonzero ideal. There exists \( \alpha \in \mathfrak{m} \) such that for every prime ideal \( \mathfrak{p} \) such that \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) \neq 0 \) we have \( {v}_{\mathfrak{p}}\left( \alpha \right) = {v}_{\mathfrak{p}}\left( \mathfrak{m}\righ... | Proof. This is an immediate consequence of Corollary 1.2.9. | No |
Corollary 1.2.11. Let \( \mathfrak{m} \) be any (nonzero) integral ideal, and let \( \mathfrak{a} \) be an ideal of \( R \) . There exists \( \alpha \in {K}^{ * } \) such that \( \alpha \mathfrak{a} \) is an integral ideal coprime to \( \mathfrak{m} \) ; in other words, in any ideal class there exists an integral ideal... | Proof. Indeed, apply the weak approximation theorem to the set of prime ideals \( \mathfrak{p} \) that divide \( \mathfrak{m} \) or such that \( {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) < 0 \), taking \( {e}_{\mathfrak{p}} = - {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) \) . Then, if \( \alpha \) is such that \( {... | Yes |
Lemma 1.2.17. If \( f \) is a surjective map from any module \( F \) onto a projective module \( P \) and if \( h \) is a section of \( f \) (so that \( f \circ h = i{d}_{P} \) ), then \( F = h\left( P\right) \oplus \operatorname{Ker}\left( f\right) \) . | Proof. Indeed, if \( x \in F \), then \( y = x - h\left( {f\left( x\right) }\right) \) is clearly in \( \operatorname{Ker}\left( f\right) \) since \( f \circ h = i{d}_{P} \) ; hence \( x \in h\left( P\right) + \operatorname{Ker}\left( f\right) \), so \( F = h\left( P\right) + \operatorname{Ker}\left( f\right) \) . Furt... | Yes |
Corollary 1.2.18. A projective module is torsion-free. | Proof. Indeed, the third characterization of projective modules shows that a projective module is isomorphic to a submodule of a free module and hence is torsion-free since a free module is evidently torsion-free. | Yes |
Theorem 1.2.19. Let \( M \) be a finitely generated, torsion-free module of rank \( n \) over a Dedekind domain \( R \) . Then \( M \) is a projective module. In addition, there exists an ideal \( I \) of \( R \) such that\n\n\[ M \simeq {R}^{n - 1} \oplus I \] | Proof of Theorem 1.2.19. We prove the theorem by induction on the rank of \( M \) . If the rank of \( M \) is zero, then \( M \) is torsion, and since \( M \) is torsion-free, \( M = \{ 0\} \) . Assume the theorem proved up to rank \( n - 1 \), and let \( M \) be a torsion-free module of rank \( n \) . Let \( e \) be a... | Yes |
Lemma 1.2.20. If \( I \) and \( J \) are any fractional ideals of \( R \), we have an isomorphism of \( R \) -modules:\n\n\[ I \oplus J \simeq R \oplus {IJ}. \] | Proof. Since \( I \simeq {kI} \) for any \( k \in R \), we can always reduce to the case where \( I \) and \( J \) are integral ideals. By Corollary 1.2.11, in the ideal class of \( J \) there exists an integral ideal \( {J}_{1} \) coprime to \( I \) . Thus, there exists \( \alpha \in {K}^{ * } \) such that \( {J}_{1} ... | Yes |
Corollary 1.2.21. Every fractional ideal is a projective R-module. | Proof. Simply apply the preceding lemma to \( J = {I}^{-1} \) and use Proposition 1.2.16 (3). | No |
Corollary 1.2.24. If \( I \) and \( J \) are two (fractional) ideals of \( R \) and \( {R}^{m - 1} \oplus \) \( I \simeq {R}^{n - 1} \oplus J \), then \( m = n \) and \( J \) and \( I \) are in the same ideal class (in other words, there exists \( \alpha \in {K}^{ * } \) such that \( J = {\alpha I} \) ). | Proof. Since \( I \) and \( J \) are of rank 1, it is clear that \( m = n \) . From the given isomorphism, we deduce that\n\n\[ \n{R}^{n - 1} \oplus I \oplus {I}^{-1} \simeq {R}^{n - 1} \oplus J \oplus {I}^{-1}.\n\]\n\nUsing Lemma 1.2.20, we obtain\n\n\[ \n{R}^{n + 1} \simeq {R}^{n} \oplus J{I}^{-1}.\n\]\n\nThus Theore... | Yes |
Corollary 1.2.25. Let \( M \) be a finitely generated, torsion-free module. There exist elements \( {\omega }_{1},\ldots ,{\omega }_{n} \) in \( M \) and fractional ideals \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) of \( R \) such that\n\n\[ M = {\mathfrak{a}}_{1}{\omega }_{1} \oplus \cdots \oplus {\mathfrak{a... | Proof. From Theorem 1.2.19, we know that \( M \) is isomorphic to \( {R}^{n - 1} \oplus I \) for some ideal \( I \) whose ideal class is the Steinitz class of \( M \) . Replacing if necessary \( I \) by \( I/\alpha \) for some nonzero element \( \alpha \) of \( I \), we may assume that \( 1 \in \) \( I \) . Let \( f \)... | Yes |
Corollary 1.2.26. Let \( M, N \), and \( P \) be three finitely generated, torsion-free modules. Assume that \( P \oplus M \simeq P \oplus N \) . Then \( M \simeq N \) . | Proof. Using Theorem 1.2.19, we have \( M \simeq {R}^{m - 1} \oplus \operatorname{St}\left( M\right), N \simeq {R}^{n - 1} \oplus \operatorname{St}\left( N\right), P \simeq {R}^{p - 1} \oplus \operatorname{St}\left( P\right) \), so that\n\n\[{\mathbb{R}}^{p + m - 2} \oplus \mathrm{{St}}\left( P\right) \oplus \mathrm{{S... | Yes |
Proposition 1.2.27. Let\n\n\[ 0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \]\n\nbe an exact sequence of finitely generated, torsion-free modules. Then\n\n\[ M \simeq {M}^{\prime } \oplus {M}^{\prime \prime }\;\text{ and }\;\operatorname{St}\left( M\right) = \operatorname{St}... | Proof. The isomorphism follows immediately from Lemma 1.2.17: if \( f \) is the map from \( M \) to \( {M}^{\prime \prime } \), there exists a map \( h \) from \( {M}^{\prime \prime } \) to \( M \) such that \( f \circ h = i{d}_{{M}^{\prime \prime }} \) and \( M = h\left( {M}^{\prime \prime }\right) \oplus \operatornam... | Yes |
Proposition 1.2.28. If \( R \) is a Dedekind domain with only a finite number of prime ideals, then \( R \) is a principal ideal domain. | Proof. Let \( \mathfrak{b} \) be the product of the (nonzero) prime ideals of \( R \), which are finite in number. If \( \mathfrak{c} \) is an ideal of \( R \), by Corollary 1.2.11 we can find an \( x \in {K}^{ * } \) such that \( x\mathfrak{c} \) is an integral ideal coprime to \( \mathfrak{b} \) . But this means that... | Yes |
Proposition 1.2.29. Let \( M \) be a finitely generated \( R \) -module, and let \( {M}_{\text{tors }} \) be the torsion submodule of \( M \) . Then there exists a torsion-free submodule \( N \) of \( M \) such that\n\n\[ M = {M}_{\text{tors }} \oplus N \] | Proof. If \( P = M/{M}_{\text{tors }} \), then \( P \) is torsion-free. Indeed, if \( \bar{y} \in {P}_{\text{tors }} \), there exists \( a \in R \smallsetminus \{ 0\} \) such that \( {ay} \in {M}_{\text{tors }} \), and hence there exists \( b \in R \smallsetminus \{ 0\} \) such that \( {bay} = 0 \), so \( y \in {M}_{\t... | Yes |
Lemma 1.2.31. Let \( S \) be a finite set of prime ideals of \( R \) and let \( x \in {K}^{ * } \) such that \( {v}_{\mathfrak{p}}\left( x\right) \geq 0 \) for all \( \mathfrak{p} \in S \) . There exist \( n \) and \( d \) in \( R \) such that \( x = n/d \) and \( d \) not divisible by any \( \mathfrak{p} \) in \( S \)... | Proof. Let \( x = n/d \) with \( n \) and \( d \) in \( R \), for the moment arbitrary. By the approximation theorem, there exists \( b \in K \) such that\n\n\[ \forall \mathfrak{p} \in S,{v}_{\mathfrak{p}}\left( b\right) = - {v}_{\mathfrak{p}}\left( d\right) \;\text{and}\;\forall \mathfrak{p} \notin S,{v}_{\mathfrak{p... | Yes |
Lemma 1.2.32. Let \( \mathfrak{a} \) be a nonzero integral ideal of \( R \) and set\n\n\[ B = \\left\\{ {x \\in K/\\forall \\mathfrak{p} \\mid \\mathfrak{a},{v}_{\\mathfrak{p}}\\left( x\\right) \\geq 0}\\right\\} .\n\]\n\nThen\n\n(1)\n\n\[ B = \\left\\{ {x = \\frac{n}{d}/n, d \\in R,\\left( {{dR},\\mathfrak{a}}\\right)... | Proof. (1). It is clear that if \( \\left( {{dR},\\mathfrak{a}}\\right) = 1 \), then \( {v}_{\\mathfrak{p}}\\left( {n/d}\\right) = {v}_{\\mathfrak{p}}\\left( n\\right) \\geq 0 \) for all \( \\mathfrak{p} \\mid \\mathfrak{a} \), and hence \( n/d \\in B \) . Conversely, let \( x \\in B \) . Taking for \( S \) the set of ... | Yes |
Proposition 1.2.34. Assume that there exist nonzero ideals \( {\mathfrak{a}}_{i} \) such that an \( R \) -module \( M \) satisfies \( M \simeq {\bigoplus }_{1 \leq i \leq k}R/{\mathfrak{a}}_{i} \) . Then the order-ideal of \( M \) is equal to \( \mathop{\prod }\limits_{{1 \leq i \leq k}}{\mathfrak{a}}_{i} \) . | Proof. This immediately follows from the fact that the order-ideal is unchanged by module isomorphism, and that the order-ideal of a product of two modules is equal to the product of the order-ideals. | No |
Theorem 1.2.35. Let \( M \) and \( N \) be two torsion-free (or projective) modules of rank \( m \) and \( n \), respectively, such that \( N \subset M\left( {\text{so}n \leq m}\right) \). There exist fractional ideals \( {\mathfrak{b}}_{1},\ldots ,{\mathfrak{b}}_{m} \) of \( R \), a basis \( \left( {{e}_{1},\ldots ,{e... | Proof. Let us first prove uniqueness, so let \( {\mathfrak{d}}_{i} \) and \( {\mathfrak{b}}_{i} \) be ideals as in the theorem. Since \( {\mathfrak{b}}_{i}/{\mathfrak{d}}_{i}{\mathfrak{b}}_{i} \simeq R/{\mathfrak{d}}_{i} \), we have \[ M/N \simeq R/{\mathfrak{d}}_{1} \oplus \cdots R/{\mathfrak{d}}_{n} \oplus {R}^{m - n... | Yes |
Proposition 1.3.1. Given two coprime integral ideals \( \mathfrak{a} \) and \( \mathfrak{b} \) in \( R \), we can find in polynomial time elements \( a \in \mathfrak{a} \) and \( b \in \mathfrak{b} \) such that \( a + b = 1 \) . | Proof. Since this is a very simple but basic proposition, we give the proof as an algorithm.\n\nAlgorithm 1.3.2 (Extended Euclid in Dedekind Domains). Let \( R \) be a Dedekind domain in which one can compute, and let \( {\left( {\omega }_{i}\right) }_{1 \leq i \leq n} \) be an integral basis chosen so that \( {\omega ... | Yes |
Theorem 1.3.3. Let \( \\mathfrak{a} \) and \( \\mathfrak{b} \) be two (fractional) ideals in \( R \), let \( a \) and \( b \) be two elements of \( K \) not both equal to zero, and set \( \\mathfrak{d} = a\\mathfrak{a} + b\\mathfrak{b} \). There exist \( u \\in {\\mathfrak{{ad}}}^{-1} \) and \( v \\in {\\mathfrak{{bd}}... | Proof. If \( a \) (resp., \( b \) ) is equal to zero, we can take \( \\left( {u, v}\\right) = \\left( {0,1/b}\\right) \) (resp., \( \\left( {u, v}\\right) = \\left( {1/a,0}\\right) ) \), since in that case we have \( 1/b \\in {\\mathfrak{{bd}}}^{-1} = R/b \) (resp., \( 1/a \\in \\mathfrak{a}{\\mathfrak{d}}^{-1} = R/a) ... | Yes |
Proposition 1.3.4. Let \( \mathfrak{a},\mathfrak{b},\mathfrak{c},\mathfrak{d} \) be fractional ideals of \( R \), and let \( a, b, c \) , \( d \) be elements of \( K \) . Set \( e = {ad} - {bc} \), and assume that\n\n\[ \mathfrak{a}\mathfrak{b} = e\mathfrak{c}\mathfrak{d},\;a \in \mathfrak{a}{\mathfrak{c}}^{-1},\;b \in... | Proof. We have \( {x}^{\prime } = {ax} + {by} \) and \( {y}^{\prime } = {cx} + {dy} \) ; hence\n\n\[ \mathfrak{c}{x}^{\prime } + \mathfrak{d}{y}^{\prime } \subset \left( {a\mathfrak{c} + c\mathfrak{d}}\right) x + \left( {b\mathfrak{c} + d\mathfrak{d}}\right) y \subset \mathfrak{a}x + \mathfrak{b}y. \]\n\nConversely, we... | Yes |
Corollary 1.3.5. Let \( \mathfrak{a} \) and \( \mathfrak{b} \) be two ideals, \( a \) and \( b \) be two elements of \( K \) not both zero, \( \mathfrak{d} = a\mathfrak{a} + b\mathfrak{b} \), and \( u \in \mathfrak{a}{\mathfrak{d}}^{-1}, v \in \mathfrak{b}{\mathfrak{d}}^{-1} \) such that \( {au} + {bv} = 1 \) as given ... | Proof. Since \( b \in {\mathfrak{b}}^{-1}\mathfrak{d} \) and \( a \in {\mathfrak{a}}^{-1}\mathfrak{d} \), this is clearly a special case of Proposition 1.3.4 with \( \mathfrak{c} = \mathfrak{a}{\mathfrak{{bd}}}^{-1} \). | Yes |
Corollary 1.3.6. Let \( \mathfrak{a},\mathfrak{b} \) be two ideals. Assume that \( a, b, c \), and \( d \) are four elements of \( K \) such that\n\n\[ \n{ad} - {bc} = 1,\;a \in \mathfrak{a},\;b \in \mathfrak{b},\;c \in {\mathfrak{b}}^{-1},\;d \in {\mathfrak{a}}^{-1}.\n\]\n\nLet \( x \) and \( y \) be two elements of a... | Proof. This is also a special case of Proposition 1.3.4 with \( \mathfrak{c} = R \) and \( \mathfrak{d} = \mathfrak{{ab}} \) . We will see in Proposition 1.3.12 how to find \( a, b, c \), and \( d \), given \( \mathfrak{a} \) and \( \mathfrak{b} \) . | No |
Proposition 1.3.7. Given ideals \( {\mathfrak{a}}_{i} \) for \( 1 \leq i \leq k \) whose sum is equal to \( R \) , we can in polynomial time find elements \( {a}_{i} \in {\mathfrak{a}}_{i} \) such that \( \mathop{\sum }\limits_{i}{a}_{i} = 1 \) . | Proof. Same proof as for Proposition 1.3.1, except that we concatenate the \( k \) HNF matrices of the ideals and we split \( Z \) into \( k \) pieces at the end. Note that the matrix \( U \) will be an \( {nk} \times {nk} \) unimodular matrix, which can become quite large. | Yes |
Proposition 1.3.8. Let \( S \) be a finite set of prime ideals of \( R \) and let \( {\left( {e}_{\mathfrak{p}}\right) }_{\mathfrak{p} \in S} \in {\mathbb{Z}}^{S} \) . There exists a polynomial-time algorithm that finds \( a \in K \) such that \( {v}_{\mathfrak{p}}\left( a\right) = {e}_{\mathfrak{p}} \) for \( \mathfra... | Proof. We can write \( {e}_{\mathfrak{p}} = {f}_{\mathfrak{p}} - {g}_{\mathfrak{p}} \) with \( {f}_{\mathfrak{p}} \geq 0 \) and \( {g}_{\mathfrak{p}} \geq 0 \) . If we can find \( n \) (resp., \( d \) ) such that the conditions are satisfied with \( {e}_{\mathfrak{p}} \) replaced by \( {f}_{\mathfrak{p}} \) (resp., \( ... | Yes |
Corollary 1.3.9. Given two integral ideals \( \mathfrak{a} \) and \( \mathfrak{b} \) of \( R \) such that the factorization of the norm of \( \mathfrak{b} \) is known, there exists a polynomial-time algorithm that finds \( x \in K \) such that \( x\mathfrak{a} \) is an integral ideal coprime to \( \mathfrak{b} \), and ... | Proof. For \( x \), apply Proposition 1.3.8 to \( S \) equal to the prime ideal factors of \( \mathfrak{b} \) and to \( {e}_{\mathfrak{p}} = - {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) \) for all \( \mathfrak{p} \in S \) . For \( y \), apply Proposition 1.3.8 to \( S \) equal to the prime ideal factors of \( \mathfr... | Yes |
Proposition 1.3.10. Let \( \mathfrak{a} \) be an integral ideal of \( R \) and \( a \in \mathfrak{a}, a \neq 0 \) . Assume that the prime ideal factorization of a is known. Then there exists a polynomial-time algorithm that finds \( b \in \mathfrak{a} \) such that \( \mathfrak{a} = {aR} + {bR} \) . | Proof. Write \( {aR} = \mathop{\prod }\limits_{\mathfrak{p}}{\mathfrak{p}}^{{e}_{\mathfrak{p}}} \) with \( {e}_{\mathfrak{p}} \geq 0 \) . Thus, \( \mathfrak{a} = \mathop{\prod }\limits_{\mathfrak{p}}{\mathfrak{p}}^{{v}_{\mathfrak{p}}\left( \mathfrak{a}\right) } \) with \( 0 \leq \) \( {v}_{\mathfrak{p}}\left( \mathfrak... | Yes |
Proposition 1.3.11. Let \( S \) be a finite set of prime ideals of \( R \), let \( {\left( {e}_{\mathfrak{p}}\right) }_{\mathfrak{p} \in S} \in \) \( {\mathbb{Z}}^{S} \), and let \( {\left( {x}_{\mathfrak{p}}\right) }_{\mathfrak{p} \in S} \in {K}^{S} \) . Then there exists a polynomial-time algorithm that finds \( x \i... | Proof. Assume first that the \( {e}_{\mathfrak{p}} \) are nonnegative and \( {x}_{\mathfrak{p}} \in R \) . Then we introduce the same ideals \( I \) and \( {\mathfrak{a}}_{\mathfrak{p}} \) and elements \( {a}_{\mathfrak{p}} \) as in the proof of Proposition 1.3.8. If we set\n\n\[ x = \mathop{\sum }\limits_{{\mathfrak{p... | Yes |
Proposition 1.3.12. Let \( \mathfrak{a} \) and \( \mathfrak{b} \) be two (fractional) ideals in \( R \) . Assume that the prime ideal factorization of \( \mathfrak{a} \) or of \( \mathfrak{b} \) is known. Then it is possible to find in polynomial time elements \( a \in \mathfrak{a}, b \in \mathfrak{b}, c \in {\mathfrak... | Proof. Multiplying if necessary \( \mathfrak{a} \) and \( \mathfrak{b} \) by an element of \( {\mathbb{Q}}^{ * } \), we can reduce to the case where \( \mathfrak{a} \) and \( \mathfrak{b} \) are integral ideals. Assume, for example, that the factorization of \( \mathfrak{b} \) is known. According to Corollary 1.2.11, w... | Yes |
Corollary 1.4.3. Let \( M \) be a finitely generated, torsion-free \( R \) -module together with a nondegenerate, bilinear pairing \( T\left( {x, y}\right) \) from \( M \times M \) to \( R \) (for example, \( M = {\mathbb{Z}}_{L} \), where \( L \) is a number field containing \( K \), and \( T\left( {x, y}\right) = {\o... | Proof. Note that, since in general \( {\omega }_{j} \notin M \), in the above definition we extend the bilinear form \( T \) to \( V \times V \) (where \( V = {KM} \) ) by bilinearity.\n\nLet \( U \) be the matrix expressing the \( {\eta }_{j} \) in terms of the \( {\omega }_{i} \) . We know that \( \mathfrak{a} = \det... | Yes |
Proposition 1.4.4. Let \( {\left( {\omega }_{i},{\mathfrak{a}}_{i}\right) }_{i} \) be a pseudo-basis for a finitely generated, torsion-free module \( M \), and similarly \( {\left( {\omega }_{j}^{\prime },{\mathfrak{a}}_{j}^{\prime }\right) }_{j} \) for a module \( {M}^{\prime } \) . Let \( f \) be a \( K \) -linear ma... | Proof. The (very easy) proof is left to the reader (Exercise 11). The matrix \( A \) will of course be called the matrix of the map \( f \) on the chosen pseudo-bases of \( {M}^{\prime } \) and \( M \) . Note that we need only a matrix and not a pseudo-matrix (see Definition 1.4.5) to represent a map. Thus, we will rep... | No |
Theorem 1.4.6 (Hermite Normal Form in Dedekind Domains). Let \( \left( {A, I}\right) \) be a pseudo-matrix, where \( I = \left( {\mathfrak{a}}_{i}\right) \) is a list of \( k \) fractional ideals, and \( A = \left( {a}_{i, j}\right) \) is an \( n \times k \) matrix. Assume that \( A \) is of rank \( n\left( {\text{so}k... | Proof. We give the proof of the existence of the HNF as an algorithm, very similar to [Coh0, Algorithm 2.4.5], which is the naive HNF algorithm.\n\nAlgorithm 1.4.7 (HNF Algorithm in Dedekind Domains). Given an \( n \times k \) matrix \( A = \left( {a}_{i, j}\right) \) of rank \( n \), and \( k \) (fractional) ideals \(... | Yes |
Theorem 1.4.9. With the notation of Theorem 1.4.6, for \( 1 \leq j \leq n \), set \( {\mathfrak{c}}_{j} = {\mathfrak{b}}_{k - n + j} \) . Then the ideals \( {\mathfrak{c}}_{j} \) are unique. More precisely, if we call \( {\mathfrak{g}}_{j} = \) \( {\mathfrak{g}}_{j}\left( A\right) \) the ideal generated by all the \( \... | Proof. One easily checks that the ideals \( {\mathfrak{g}}_{m}\left( A\right) \) are invariant under the elementary transformations of the type used in Algorithm 1.4.7. In particular, \( {\mathfrak{g}}_{j}\left( A\right) = {\mathfrak{g}}_{j}\left( {AU}\right) \) . But in the last \( n + 1 - j \) rows of \( {AU} \) ther... | Yes |
Proposition 1.4.10. If \( {AU} \) is of the form given by Theorem 1.4.6, a necessary and sufficient condition for \( {AV} \) to be of the same form with the same ideals \( {\mathfrak{b}}_{j} \) for \( j > k - n \) is that \( {U}^{-1}V \) be a block matrix \( \left( \begin{matrix} B & C \\ 0 & D \end{matrix}\right) \) w... | Proof. Trivial and left to the reader. | No |
Corollary 1.4.11. For each \( i \) and \( j \) with \( 1 \leq i < j \leq n \), let \( {S}_{i, j} \) be a system of representatives of \( K/{\mathfrak{c}}_{i}{\mathfrak{c}}_{j}^{-1} \) . Write \( {AU} = \left( {0 \mid H}\right) \) as in Theorem 1.4.6. Then in that theorem, we may assume that for every \( i \) and \( j \... | Proof. For \( i < j \), let \( {h}_{i, j} \) be the entry in row \( i \) and column \( j \) of the matrix \( H \) . There exists a unique \( {h}_{i, j}^{\prime } \in {S}_{i, j} \) such that\n\n\[ q = {h}_{i, j}^{\prime } - {h}_{i, j} \in {\mathfrak{c}}_{i}{\mathfrak{c}}_{j}^{-1}. \]\n\nIf the \( {H}_{j} \) are the colu... | Yes |
Theorem 1.7.2 (Smith Normal Form in Dedekind Domains). Let \( \\left( {A, I, J}\\right) \) be an integral pseudo-matrix as above, with \( A = \\left( {a}_{i, j}\\right) \) an \( n \\times n \) matrix and \( I = \\left( {{\\mathfrak{b}}_{1},\\ldots ,{\\mathfrak{b}}_{n}}\\right) \), and \( J = \\left( {{\\mathfrak{a}}_{1... | Proof. Again we prove this theorem by giving an explicit algorithm for constructing the Smith normal form. We follow closely [Coh0, Algorithm 2.4.14], except that we do not work modulo the determinant (although such a modular version of the Smith normal form algorithm is easily written).\n\nAlgorithm 1.7.3 (SNF Algorit... | Yes |
Proposition 2.1.1. Let \( K \) be a number field, and let \( A \) be a finite-dimensional commutative \( K \) -algebra (in other words, a finite-dimensional \( K \) -vector space with an additional commutative ring structure with unit, compatible with the vector space structure). The following three properties are equi... | Proof. That (1) implies (2) is trivial. Let us prove that (2) implies (3), so assume (2), and let \( a \in A \) . The set \( {I}_{a} \) of polynomials \( P \in K\left\lbrack X\right\rbrack \) such that \( P\left( a\right) = 0 \) is clearly an ideal of \( K\left\lbrack X\right\rbrack \) . Furthermore, since \( A \) is o... | Yes |
Proposition 2.1.3. Let \( B \) be a commutative ring with unit, and let \( {T}_{1} \) and \( {T}_{2} \) be polynomials in \( B\left\lbrack X\right\rbrack \) . There exist polynomials \( {U}_{1}\left( X\right) \) and \( {U}_{2}\left( X\right) \) in \( B\left\lbrack X\right\rbrack \) such that \[ {U}_{1}\left( X\right) {... | Proof. Let \( M \) be the Sylvester matrix associated to the polynomials \( {T}_{1} \) and \( {T}_{2} \) (see [Coh0, Lemma 3.3.4]). If \( \deg \left( {T}_{i}\right) = {n}_{i} \), let \( {U}_{1}\left( X\right) = \mathop{\sum }\limits_{{0 \leq i < {n}_{2}}}{x}_{i}{X}^{i} \) and \( {U}_{2}\left( X\right) = \mathop{\sum }\... | Yes |
Theorem 2.1.5 (Primitive Element Theorem). Let \( K \) be a number field and \( A \) be an étale algebra of dimension \( n \) over \( K \) . There exists \( \theta \in A \) (called a primitive element) such that \( A = K\left\lbrack \theta \right\rbrack \), in other words such that \( 1,\theta ,\ldots ,{\theta }^{n - 1... | Proof. Since \( A \) is finite-dimensional over \( K \), there exist elements \( {\theta }_{1},\ldots ,{\theta }_{m} \) such that \( A = K\left\lbrack {{\theta }_{1},\ldots ,{\theta }_{m}}\right\rbrack \) . For example, we can take for the \( {\theta }_{i} \) a \( K \) -basis of \( A \) . We prove the theorem by induct... | Yes |
Corollary 2.1.6. Let \( A \) be an étale algebra over \( K \) .\n\n(1) There exists a squarefree monic polynomial \( T\left( X\right) \in K\left\lbrack X\right\rbrack \) (called as above a defining polynomial for \( A/K \) ) such that \( A \) is isomorphic to \( K\left\lbrack X\right\rbrack /T\left( X\right) K\left\lbr... | Proof. (1). By Theorem 2.1.5, we know that \( A = K\left\lbrack \theta \right\rbrack \) for some \( \theta \in A \) . If \( T \) is the minimal monic polynomial of \( \theta \) in \( K\left\lbrack X\right\rbrack \), then by definition of an étale algebra the polynomial \( T\left( X\right) \) is squarefree, and the map ... | Yes |
(1) There exists an integer \( k \in \mathbb{Z} \) such that the polynomial \( R\left( {X, k}\right) \) is square-free. | Proof. By definition, \( L = K\left( {{\theta }_{1},{\theta }_{2}}\right) \) . By the proof of Lemma 2.1.4, there exists \( k \in \mathbb{Z} \) such that \( R\left( {X, k}\right) \) is squarefree, and if \( \theta = {\theta }_{2} + k{\theta }_{1} \), then \( L = K\left( \theta \right) \) with \( \theta \) a root of \( ... | Yes |
Theorem 2.1.10. Let \( {L}_{1} = K\left( {\theta }_{1}\right) \) and \( {L}_{2} = {L}_{1}\left( {\theta }_{2}\right) \) be two number fields, where \( {\theta }_{1} \) is a root of the irreducible polynomial \( {T}_{1}\left( X\right) \in K\left\lbrack X\right\rbrack \) of degree \( {n}_{1} \) , and \( {\theta }_{2} \) ... | Proof. The proof is very close to that of Lemma 2.1.4 and Proposition 2.1.7 (see also [Coh0, Lemma 3.6.2]).\n\n(1). Let \( \Omega = \overline{{L}_{2}} \) be some algebraic closure of \( {L}_{2} \) . Then \( \Omega \) is also an algebraic closure of \( K \) and of \( {L}_{1} \) . We denote by \( {\theta }_{1}^{\left( i\... | Yes |
Lemma 2.1.13. With the above notation, we have\n\n\[ D\left( k\right) = {k}^{{n}_{1}{n}_{2}\left( {{n}_{1} - 1}\right) }\operatorname{disc}{\left( {T}_{1}\right) }^{{n}_{2}}\operatorname{disc}{\left( {T}_{2}\right) }^{{n}_{1}}\mathop{\prod }\limits_{{s \in G, s \neq {1}_{G}}}{D}_{s}\left( k\right) . \] | Furthermore, for all \( s \), we have \( {D}_{{s}^{-1}} = {D}_{s} \), and if \( {s}^{2} = {1}_{G} \) and \( s \neq {1}_{G} \), then \( {D}_{s} \) is the square of a rational integer.\n\nHence, we have split our large discriminant \( D\left( k\right) \) as a product of smaller pieces \( {D}_{s}\left( k\right) \) . This ... | Yes |
Theorem 2.1.14. Let \( {K}_{1} = \mathbb{Q}\left( {\theta }_{1}\right) \) and \( {K}_{2} = \mathbb{Q}\left( {\theta }_{2}\right) \) be number fields of respective degrees \( {n}_{1} \) and \( {n}_{2} \), and let \( {T}_{1}\left( X\right) \) and \( {T}_{2}\left( X\right) \) be the minimal monic polynomials of \( {\theta... | For \( s \in G, s \neq {1}_{G} \), define \( {A}_{s}\left( X\right) \) to be the polynomial expressing \( s\left( {\theta }_{2}\right) \) in terms of \( {\theta }_{2} \), and set \( {V}_{s}\left( X\right) = {\mathcal{R}}_{Y}\left( {{T}_{2}\left( Y\right), X + Y - {A}_{s}\left( Y\right) }\right) \) (this depends only on... | Yes |
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