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If \( k \) is a field of characteristic \( p \), and \( G \) is a finite \( p \) -group, then the group ring \( k\left\lbrack G\right\rbrack \) is quasilocal. In particular, taking \( k \) finite and \( G \) noncommutative, there exist finite noncommutative quasilocal rings. | To see how this example works out, define \( \epsilon : k\left\lbrack G\right\rbrack \rightarrow k \) via \( \epsilon \left( {\sum {x}_{i}{g}_{i}}\right) = \) \( \sum {x}_{i};\epsilon \) is called the augmentation map, and is a ring homomorphism onto with kernel \( M = \left\{ {\sum {x}_{i}{g}_{i} : \sum {x}_{i} = 0}\r... | Yes |
Lemma 9.35 Suppose \( R \) is quasilocal with maximal ideal \( M \) and quotient division ring \( D = R/M \) . Suppose \( B \) is a finitely generated left \( R \) module, and \( \pi : F \rightarrow B \) is a projective cover as described in Proposition 9.33, with \( K = \ker \pi \) . Then\n\n\[ \n{\operatorname{Tor}}_... | Proof: \( 0 \rightarrow K \rightarrow F \rightarrow B \rightarrow 0 \) is short exact, so we have from the long exact sequence for Tor:\n\n\[ \n0 \rightarrow {\operatorname{Tor}}_{1}\left( {D, B}\right) \rightarrow D \otimes K \rightarrow D \otimes F \rightarrow D \otimes B \rightarrow 0.\n\]\n\nNow\n\n\[ \nD \otimes K... | Yes |
Lemma 9.36 Suppose \( R \) is quasilocal with maximal ideal \( M \) and quotient division ring \( D = R/M \) . Then for any \( B \in {}_{R}\mathbf{M},{\operatorname{Hom}}_{R}\left( {B, D}\right) \approx \) \( {\operatorname{Hom}}_{D}\left( {B/{MB}, D}\right) \) . | Proof: As in the proof of Lemma 9.35, if \( f \in {\operatorname{Hom}}_{R}\left( {B, D}\right) \), then for \( m \in M, x \in B \), we have \( f\left( {mx}\right) = {mf}\left( x\right) = 0 \), since \( f\left( x\right) \in R/M \) . Thus, \( f \) restricts to zero on \( {MB} \), so restriction from \( \operatorname{Hom}... | Yes |
Theorem 9.37 (Universality of D) Suppose \( R \) is quasilocal and left Noetherian with maximal ideal \( M \) and quotient division ring \( D = R/M \) . Let \( B \) be a nonzero finitely generated left \( R \) -module. Then \( {\operatorname{Tor}}_{n}\left( {D, B}\right) \) is nonzero for \( 0 \leq n \leq \mathrm{P} - ... | Proof: The claim is that if \( n \leq \mathrm{P} \) -dim \( B \), then \( {\operatorname{Tor}}_{n}\left( {D, B}\right) \neq 0 \) and the dual of \( {\operatorname{Tor}}_{n}\left( {D, B}\right) \) as a left vector space over \( D \) is \( {\operatorname{Ext}}^{n}\left( {B, D}\right) \) . This is by induction on \( n \) ... | Yes |
Corollary 9.38 Suppose \( R \) is quasilocal and left and right Noetherian with maximal ideal \( M \) and quotient division ring \( D = R/M \) . Then all dimensions (left/right and flat/projective/injective) of \( D \) are equal to the left global dimension of \( R \), and \( {\operatorname{Tor}}_{n}\left( {D, D}\right... | Proof: Theorem 9.37 applies on both sides, giving left/right, flat/injective dimensions \( = \mathrm{{LG}} - \dim R = \mathrm{{RG}} - \dim R = \mathrm{W} - \dim R \) by Corollary 4.21. Since projective dimension \( = \) flat dimension by Proposition 4.20 itself, we have the dimensional result. The rest is now direct fr... | No |
Proposition 9.39 Suppose \( R \) is quasilocal and left Noetherian with maximal ideal \( M \) . Suppose \( \exists x \in M \) with \( x \neq 0 \) but \( {xM} = 0 \) . Then every finitely generated nonprojective left \( R \) -module has infinite projective dimension. | Proof: Suppose not. Suppose some finitely generated \( B \in {}_{R}\mathbf{M} \) has finite projective dimension and is not projective. We can replace \( B \) with \( K \), where \( F \) is free and \( 0 \rightarrow K \rightarrow F \rightarrow B \rightarrow 0 \) is short exact, reducing the dimension if \( n > 1 \), so... | Yes |
Example 37 \( \mathbb{R} = \) real numbers. Set\n\n\[ R = \left\{ {f\left( {t, x, y}\right) \in \mathbb{R}\left( {t, x, y}\right) : f\left( {t,{e}^{-1/{t}^{2}},{e}^{-1/{t}^{4}}}\right) }\right. \text{extends}\n\n\[ \text{to a}{C}^{\infty }\text{function near 0}\} \text{.} \] | Suppose \( f\left( {t, x, y}\right) \in \mathbb{R}\left\lbrack {t, x, y}\right\rbrack \), the polynomial ring, with \( f\left( {t, x, y}\right) \neq 0 \) . There is a lowest power \( n \) of \( y \) that appears so that one may write\n\n\[ f\left( {t, x, y}\right) = {y}^{n}\left( {g\left( {t, x}\right) + {y\phi }\left(... | Yes |
Proposition. The ring \( {\mathbf{Z}}_{p} \) is a principal ideal domain. More precisely, its ideals are the principal ideals \( \{ 0\} \) and \( {p}^{k}{\mathbf{Z}}_{p}\left( {k \in \mathbf{N}}\right) \) . | Proof. Let \( I \neq \{ 0\} \) be a nonzero ideal of \( {\mathbf{Z}}_{p} \) and \( 0 \neq a \in I \) an element of minimal order, say \( k = v\left( a\right) < \infty \) . Write \( a = {p}^{k}u \) with a \( p \) -adic unit \( u \) . Hence \( {p}^{k} = \) \( {u}^{-1}a \in I \) and \( \left( {p}^{k}\right) = {p}^{k}{\mat... | Yes |
Proposition 1. The discrete subgroups of \( \mathbf{R} \) are the subgroups\n\n\[ a\mathbf{Z}\;\left( {0 \leq a \in \mathbf{R}}\right) . \] | Proof. Let \( H \neq \{ 0\} \) be a nontrivial discrete subgroup, hence closed by (3.2). Consider any nonzero \( h \) in \( H \), so that \( 0 < \left| h\right| \left( { = \pm h}\right) \in H \) . The intersection \( H \cap \) \( \left\lbrack {0,\left| h\right| }\right\rbrack \) is compact and discrete, hence finite, a... | Yes |
Proposition 2. Any nondiscrete subgroup of \( \mathbf{R} \) is dense. | Proof. Let \( H \subset \mathbf{R} \) be a nondiscrete subgroup. Then there exists a sequence of distinct elements \( {h}_{n} \in H \) with \( {h}_{n} \rightarrow h \in H \) . Hence \( {\varepsilon }_{n} = \left| {{h}_{n} - h}}\right| \in H \) and \( {\varepsilon }_{n} \rightarrow 0 \) . Since \( H \) is an additive su... | No |
Corollary 1. The quotient of \( {\mathbf{Z}}_{p} \) by a closed subgroup \( H \neq \{ 0\} \) is discrete. | Proof. Indeed, discrete subgroups are closed: We have a complete list of these (being closed in \( {\mathbf{Z}}_{p} \) compact, a discrete subgroup is finite hence trivial). Alternatively, if a subgroup \( H \) contains a nonzero element \( h \), it contains all multiples of \( h \) , and hence \( H \supset \mathbf{N} ... | No |
Corollary 2. The only discrete subgroup of the additive group \( {\mathbf{Z}}_{p} \) is the trivial subgroup \( \{ 0\} \) . | Proof. Indeed, discrete subgroups are closed: We have a complete list of these (being closed in \( {\mathbf{Z}}_{p} \) compact, a discrete subgroup is finite hence trivial). Alternatively, if a subgroup \( H \) contains a nonzero element \( h \), it contains all multiples of \( h \) , and hence \( H \supset \mathbf{N} ... | Yes |
Corollary 1. The topological group \( {\mathbf{Z}}_{p} \) is a completion of the additive group \( \mathbf{Z} \) equipped with the induced topology. | To make the completion process explicit, let us observe that if \( x = \mathop{\sum }\limits_{{i \geq 0}}{a}_{i}{p}^{i} \) is a \( p \) -adic number, then\n\n\[ \n{x}_{n} = \mathop{\sum }\limits_{{0 \leq i < n}}{a}_{i}{p}^{i} \in \mathbf{N} \n\]\n\ndefines a Cauchy sequence converging to \( x \) . | No |
Proposition 1. Let \( K \) be a valued field. For the topology defined by the metric \( d\left( {x, y}\right) = \left| {x - y}\right|, K \) is a topological field. | Proof. The map \( \left( {x, y}\right) \mapsto x - y \) is continuous. Let us check that the map \( \left( {x, y}\right) \mapsto \) \( x{y}^{-1} \) is continuous on \( {K}^{ \times } \times {K}^{ \times } \) . We have\n\n\[ \frac{x + h}{y + k} - \frac{x}{y} = \frac{{hy} - {kx}}{y\left( {y + k}\right) } \]\n\nHence if \... | Yes |
Proposition 2. Let \( K \) be a valued field. Then the completion \( \widehat{K} \) of \( K \) is again a valued field. | Proof. The completion \( \widehat{K} \) is obviously a topological ring, and inversion is continuous over the subset of invertible elements. We have to show that the completion is a field. Let \( \left( {x}_{n}\right) \) be a Cauchy sequence in \( K \) that defines a nonzero element of the completion \( \widehat{K} \) ... | Yes |
Proposition 1. A projective limit of nonempty compact spaces is nonempty and compact. | Proof. Let \( \left( {{K}_{n},{\varphi }_{n}}\right) \) be a projective system consisting of compact spaces. The product of the \( {K}_{n} \) is a compact space (Tychonoff’s theorem), and the projective limit is a closed subspace of this compact space. Hence \( \lim {K}_{n} \) is compact. Define\n\n\[ \n{K}_{n}^{\prime... | Yes |
Proposition 2. In a projective limit \( E = \mathop{\lim }\limits_{ \leftarrow }{E}_{n} \) of topological spaces, a basis of the topology is furnished by the sets \( {\psi }_{n}^{-1}\left( {U}_{n}\right) \), where \( n \geq 0 \) and \( {U}_{n} \) is an arbitrary open set in \( {E}_{n} \) . | Proof. We take a family \( x = \left( {x}_{i}\right) \) in the projective limit and show that the mentioned open sets containing \( x \) form a basis of neighborhoods of this point. If we take two open sets \( {V}_{n} \subset {E}_{n} \) and \( {V}_{n - 1} \subset {E}_{n - 1} \), the conjunction of the conditions \( {x}... | Yes |
Proposition 3. Let \( A \) be a subset of a projective limit \( E = \lim {E}_{n} \) of topological spaces. Then the closure \( \bar{A} \) of \( A \) is given by\n\n\[ \bar{A} = \mathop{\bigcap }\limits_{{n \geq 0}}{\psi }_{n}^{-1}\left( \overline{{\psi }_{n}\left( A\right) }\right) \] | Proof. It is clear that \( A \) is contained in the above mentioned intersection, and that this intersection is closed. Hence \( \bar{A} \) is also contained in the intersection. Conversely, if \( b \) lies in the intersection, let us show that \( b \) is in the closure of \( A \) . Let \( V \) be a neighborhood of \( ... | Yes |
Proposition 1. When \( p \) is an odd prime, the group of roots of unity in the field \( {\mathbf{Q}}_{p} \) is \( {\mu }_{p - 1} \) . | Proof. We have to prove that the reduction homomorphism \( \varepsilon : \mu \left( {\mathbf{Q}}_{p}\right) \rightarrow {\mathbf{F}}_{p}^{ \times } \) is bijective. It is surjective by Hensel’s lemma. So assume that \( \zeta = 1 + {pt} \in \ker \varepsilon \) \( \left( {t \in {\mathbf{Z}}_{p}}\right) \) is a root of un... | Yes |
Proposition 2. The group of roots of unity in the field \( {\mathbf{Q}}_{2} \) is \( {\mu }_{2} = \{ \pm 1\} \) . | Proof. We have\n\n\[ \n- 1 = 1 + 2 + {2}^{2} + \cdots \in 1 + 2{\mathbf{Z}}_{2} \n\] \n\nand \n\n\[ \n\{ \pm 1\} = {\mu }_{2} \subset {\mathbf{Z}}_{2}^{ \times } = 1 + 2{\mathbf{Z}}_{2} \n\] \n\nOn the other hand, \( {\mathbf{F}}_{2}^{ \times } = \{ 1\} \), and the only roots of unity in \( {\mathbf{Z}}_{2}^{ \times } ... | Yes |
Proposition 1. For each positive integer \( m \geq 1 \) prime to \( p \) the solenoid \( {\mathbf{S}}_{p} \) has a unique cyclic subgroup \( {C}_{m} \) of order \( m \) . | Proof. Let us denote temporarily by \( {C}_{m}^{n} \) the cyclic subgroup of order \( m \) of the circle \( \mathbf{R}/{p}^{n}\mathbf{Z} \) (it is the subgroup \( {m}^{-1}\mathbf{Z}/{p}^{n}\mathbf{Z} \) ). Since the transition maps\n\n\[ \n{\varphi }_{n} : \mathbf{R}/{p}^{n + 1}\mathbf{Z} \rightarrow \mathbf{R}/{p}^{n}... | Yes |
Proposition 2. The p-adic solenoid \( {S}_{p} \) has no p-torsion. | Proof. Let \( \sigma : \mathbf{Z}/p\mathbf{Z} \rightarrow {\mathbf{S}}_{p} \) be any homomorphism of a cyclic group of order \( p \) into the solenoid. I claim that all composites\n\n\[ \n{\varphi }_{n} \circ {\psi }_{n + 1} \circ \sigma : \mathbf{Z}/p\mathbf{Z} \rightarrow {\mathbf{S}}_{p} \rightarrow \mathbf{R}/{p}^{... | Yes |
Corollary 1. The solenoid can also be viewed as a quotient of \( \\mathbf{R} \\times {\\mathbf{Z}}_{p} \) by the discrete subgroup \( {\\Delta }_{\\mathbf{Z}} = \\{ \\left( {m, - m}\\right) : m \\in \\mathbf{Z}\\} \) | Proof. Since the restriction of the sum homomorphism \( f : \\mathbf{R} \\times {\\mathbf{Q}}_{p} \\rightarrow {\\mathbf{S}}_{p} \) to the subgroup \( \\mathbf{R} \\times {\\mathbf{Z}}_{p} \) is already surjective, this restriction gives a (topological and algebraic) isomorphism\n\n\[ {f}^{\\prime } : \\left( {\\mathbf... | Yes |
Corollary 2. The solenoid can also be viewed as a quotient of the topological space \( \left\lbrack {0,1}\right\rbrack \times {\mathbf{Z}}_{p} \) by the equivalence relation identifying \( \left( {1, x}\right) \) to \( \left( {0, x + 1}\right) \;\left( {x \in {\mathbf{Z}}_{p}}\right) . | Proof. This follows immediately from the previous corollary, since the restriction of the sum homomorphism to \( \left\lbrack {0,1}\right\rbrack \times {\mathbf{Z}}_{p} \) is already surjective, whereas its restriction to \( \lbrack 0,1) \times {\mathbf{Z}}_{p} \) is bijective. | Yes |
Proposition 1. Let \( U \) be any proper subset of the circle \( \mathbf{R}/\mathbf{Z} \) . Then the subspace \( {\psi }^{-1}\left( U\right) \subset {\mathbf{S}}_{p} \) of the solenoid is homeomorphic to \( U \times {\mathbf{Z}}_{p} \) . The map | \[ \left( {t, x}\right) = \left( {t - \left\lbrack t\right\rbrack, x + \left\lbrack t\right\rbrack }\right) \mapsto \left( {0, x + \left\lbrack t\right\rbrack }\right) \] furnishes by restriction a continuous retraction of \( {\psi }^{-1}\left( \left\lbrack {0,\eta }\right\rbrack \right) \subset {\mathbf{S}}_{p} \) ont... | No |
Proposition 2. The solenoid \( {\mathbf{S}}_{p} \) is an indecomposable compact connected topological space. | Proof. Let us take two compact connected subsets \( A \) and \( B \) covering \( {\mathbf{S}}_{p} \) . We have to show that if \( A \neq {\mathbf{S}}_{p} \), then \( B = {\mathbf{S}}_{p} \) . Thus we assume \( A \neq {\mathbf{S}}_{p} \) from now on: \( B \neq \varnothing \) . Since we have\n\n\[ K = \mathop{\bigcap }\l... | Yes |
Lemma 1. (a) Any point of a ball is a center of the ball. | Proof. (a) If \( b \in {B}_{ < r}\left( a\right) \), then \( d\left( {a, b}\right) < r \) and\n\n\[ x \in {B}_{ < r}\left( a\right) \Leftrightarrow d\left( {x, a}\right) < r\overset{d\left( {a, b}\right) < r}{ \Leftrightarrow }d\left( {x, b}\right) < r \Leftrightarrow x \in {B}_{ < r}\left( b\right) \]\n\nproving \( {B... | Yes |
Lemma 3. (a) The spheres \( {S}_{r}\left( a\right) \left( {r > 0}\right) \) are both open and closed. | Proof. (a) The spheres are closed in all metric spaces, since the distance function \( x \mapsto d\left( {x, a}\right) \) is continuous. A sphere of positive radius is open in an ultrametric space by part \( \left( c\right) \) of the previous lemma. | No |
Lemma 4. (a) A sequence \( {\left( {x}_{n}\right) }_{n \geq 0} \) with \( d\left( {{x}_{n},{x}_{n + 1}}\right) \rightarrow 0\left( {n \rightarrow \infty }\right) \) is a Cauchy sequence. | Proof. (a) Observe that if \( d\left( {{x}_{n},{x}_{n + 1}}\right) < \varepsilon \) for all \( n \geq N \), then also\n\n\[ d\left( {{x}_{n},{x}_{n + m}}\right) \leq \mathop{\max }\limits_{{0 \leq i < m}}d\left( {{x}_{n + i},{x}_{n + i + 1}}\right) < \varepsilon \]\n\nfor all \( n \geq N \) and \( m \geq 0 \) . | No |
Theorem 1 (Eisenstein). Let \( f\left( X\right) \in {\mathbf{Z}}_{p}\left\lbrack X\right\rbrack \) be a monic polynomial of degree \( n \geq 1 \) with \( f\left( X\right) \equiv {X}^{n}{\;\operatorname{mod}\;p}, f\left( 0\right) ≢ 0{\;\operatorname{mod}\;{p}^{2}} \) . In other words,\n\n\[ f\left( X\right) = {X}^{n} + ... | Proof. Take a factorization \( f = g \cdot h \) in \( {\mathbf{Z}}_{p}\left\lbrack X\right\rbrack \) - or in \( {\mathbf{Q}}_{p}\left\lbrack X\right\rbrack \) ; this is the same by an elementary lemma attributed to Gauss - say\n\n\[ g = {b}_{\ell }{X}^{\ell } + \cdots + {b}_{0},\;h = {c}_{m}{X}^{m} + \cdots + {c}_{0}. ... | Yes |
Theorem 2. Let \( K \) be a finite, totally ramified extension of \( {\mathbf{Q}}_{p} \). Then \( K \) is generated by a root of an Eisenstein polynomial. | Proof. The maximal ideal \( P \) of the subring \( R = {B}_{ \leq 1} \) of \( K \) is principal and generated by an element \( \pi \) with \( {\left| \pi \right| }^{e} = \left| p\right| \). Since \( n = \left\lbrack {K : {\mathbf{Q}}_{p}}\right\rbrack = e \) by assumption, the linearly independent powers \( {\left( {\p... | Yes |
Proposition 1. Let \( K \) be any ultrametric extension of \( {\mathbf{Q}}_{p} \) . Then\n\n\[ \n{\mu }_{{p}^{\infty }}\left( K\right) = \mu \left( K\right) \cap \left( {1 + M}\right) .\n\] | Proof. First, if \( \zeta \in \mu \left( K\right) \) has order a power of \( p \), denote by \( \widetilde{\zeta } = \varepsilon \left( \zeta \right) \in k \) its reduction. Then\n\n\[ \n{\zeta }^{{p}^{f}} = 1 \Rightarrow {\widetilde{\zeta }}^{{p}^{f}} = \widetilde{1} \in k \Rightarrow \widetilde{\zeta } = \widetilde{1... | Yes |
Proposition 2. Assume that the extension \( K \) of \( {\mathbf{Q}}_{p} \) is complete with residue field \( k \) algebraic over \( {\mathbf{F}}_{p} \) . Then we have a split exact sequence\n\n\[\n\\left( 1\\right) \\rightarrow {\\mu }_{{p}^{\\infty }}\\left( K\\right) \\rightarrow \\mu \\left( K\\right) \\rightarrow {... | Proof. Let \( \\varepsilon : \\mu \\rightarrow {k}^{ \\times } \) be the group homomorphism obtained by restriction of the reduction (ring) homomorphism \( A \\rightarrow A/M \) . It will be enough to show that \( \\varepsilon \) induces an isomorphism \( {\\mu }_{\\left( p\\right) }\\left( K\\right) \\cong {k}^{ \\tim... | Yes |
Corollary 1. If the ramification index \( e = e\left( K\right) \) is finite, then the group \( {\mu }_{{p}^{\infty }}\left( K\right) \) of roots of unity in \( K \) having order a power of \( p \) is finite. More precisely,\n\n\[ \n\# \left( {{\mu }_{{p}^{\infty }}\left( K\right) }\right) \leq \frac{e}{1 - 1/p}.\n\] | Proof. In general, if the field \( K \) has a root of order \( {p}^{t} \), the preceding theorem shows that the ramification index \( e \) is a multiple of \( \varphi \left( {p}^{t}\right) = {p}^{t} - {p}^{t - 1} \) . Hence\n\n\[ \n{p}^{t}\left( {1 - 1/p}\right) \leq e\n\]\n\nThis gives a bound for the order \( {p}^{t}... | Yes |
Corollary 2. The group of roots of unity in \( {\mathbf{Q}}_{p} \) is precisely\n\n\[ \mu \left( {\mathbf{Q}}_{p}\right) = {\mu }_{\left( p\right) }\left( {\mathbf{Q}}_{p}\right) = {\mu }_{p - 1}\;p\text{ odd prime,}\]\n\n\[ \mu \left( {\mathbf{Q}}_{2}\right) = {\mu }_{2}\left( {\mathbf{Q}}_{2}\right) = \{ \pm 1\} \] | Example. Let \( K \) be the extension generated over \( {\mathbf{Q}}_{p} \) by a primitive \( p \) th root of unity and \( {K}^{\prime } \) the extension of \( K \) generated by a primitive root of unity of order \( {p}^{2} \) . Both extensions are totally ramified. The degrees of these cyclotomic extensions are determ... | No |
Corollary 1. The balls \( {B}_{r}\left( {r > 0}\right) \) make up a fundamental system of neighborhoods of 0 in \( K \) . In particular,\n\n\[ \n{a}^{n} \rightarrow 0\text{ in }K \Leftrightarrow m\left( a\right) < 1.\n\] | Proof. If \( V \) is any compact neighborhood of 0 in \( K \), put \( r = \mathop{\max }\limits_{V}m\left( x\right) \) in order to have \( V \subset {B}_{r} \) . Since 0 is not in the closure of \( {B}_{r} - V \), the minimum \( {r}^{\prime } \) of \( m\left( x\right) \) on the closure \( \Omega \) of \( {B}_{r} - V \)... | Yes |
Corollary 2. Any discrete subfield of \( K \) is finite. | Proof. Let \( F \) be a discrete subfield of \( K \) . Choose any \( a \in K \) with \( m\left( a\right) > 1 \) . Then we have \( m\left( {a}^{-n}\right) = m{\left( a\right) }^{-n} \rightarrow 0 \), whence \( {a}^{-n} \rightarrow 0 \), and since \( F \) is discrete it shows \( a \notin F \) . This proves \( F \subset {... | Yes |
Theorem 1. A one-dimensional topological vector space \( V \) over \( K \) is isomorphic as a topological vector space to \( K \) . More precisely, for each \( 0 \neq v \in V \) , the map \( a \mapsto {av} : K \rightarrow V \) is a bijective linear homeomorphism. | Proof. Fix \( 0 \neq v \in V \) . The one-to-one linear map \( a \mapsto {av} : K \rightarrow V \) is continuous, since \( V \) is a topological vector space over \( K \) . We have to show the continuity of the inverse, namely\n\n\[ \n\forall \varepsilon > 0\exists U\text{ neighborhood of }0\text{ in }V\text{ such that... | Yes |
Theorem 2. Assume that the field \( K \) is complete. Then a finite-dimensional topological vector space \( V \) over \( K \) is isomorphic as a topological vector space to a Cartesian product \( {K}^{d} \) . More precisely, for any basis \( \left( {e}_{i}\right) \) of \( V \), the linear map\n\n\[ \left( {\lambda }_{i... | Proof. We proceed by induction on the dimension of the vector space \( V \) : The dimension-1 case is covered by the first theorem. Assume that the statement is true up to dimension \( d - 1 \) . If \( {\dim }_{K}V = d \), select a basis \( {e}_{1},\ldots ,{e}_{d} \) of \( V \) and consider the linear span \( W \) of t... | Yes |
Theorem 1 (Krasner’s Lemma). Let \( K \subset {\mathbf{Q}}_{p}^{a} \) be a finite extension of \( {\mathbf{Q}}_{p} \) and let \( a \in {\mathbf{Q}}_{p}^{a} \) (so that \( a \) is algebraic over \( {\mathbf{Q}}_{p} \) ). Denote by \( {a}^{\sigma } \) the conjugates of a over \( K \) and put \( r = \mathop{\min }\limits_... | Proof. Take any algebraic element \( b \) such that \( a \notin K\left( b\right) \) . Since we are in characteristic 0, Galois theory asserts that there is a conjugate \( {a}^{\sigma } \neq a \) of \( a \) over \( K\left( b\right) \) (the automorphism \( \sigma \) fixes \( K\left( b\right) \) elementwise) and we can es... | Yes |
Theorem 2 (Continuity of Roots of Equations). Let \( K \) be a finite extension of the p-adic field \( {\mathbf{Q}}_{p} \) and fix an algebraic element \( a \in {\mathbf{Q}}_{p}^{a} \) of degree \( n \) over \( K \) corresponding to a monic irreducible polynomial \( f \in K\left\lbrack X\right\rbrack \) (of degree \( n... | Proof. Let us factorize the polynomial \( g \) in the algebraic closure \( {\mathbf{Q}}_{p}^{a} \) of \( K \), say \( g\left( X\right) = \Pi \left( {X - {b}_{i}}\right) \), and evaluate it at the root \( a \) of \( f \) :\n\n\[ \n\prod \left( {a - {b}_{i}}\right) = g\left( a\right) = g\left( a\right) - f\left( a\right)... | Yes |
Corollary 1. Let \( f \in K\left\lbrack X\right\rbrack \) be a monic irreducible polynomial, \( a \in {\mathbf{Q}}_{p}^{a} \) a root of \( f \), and \( {\left( {g}_{i}\right) }_{i \in \mathbf{N}} \) a sequence of monic polynomials with coefficients in \( K \) of the same degree as \( f \) . If \( {g}_{i} \rightarrow f ... | Proof. As soon as \( \begin{Vmatrix}{{g}_{i} - f}\end{Vmatrix} < \varepsilon \) is small enough, the above result is applicable and shows that \( \left| {a - {x}_{i}}\right| \) is small for at least one root \( {x}_{i} \) of \( {g}_{i} \) . More precisely, the inequality\n\n\[ \left| {a - {x}_{i}}\right| \leq {\begin{V... | Yes |
Corollary 2. The algebraic closure \( {\mathbf{Q}}_{p}^{a} \) of \( {\mathbf{Q}}_{p} \) is a separable metric space. | Proof. Take \( a \in {\mathbf{Q}}_{p}^{a} \) and let \( f \) be its minimal polynomial over \( {\mathbf{Q}}_{p} \). Since \( \mathbf{Q} \) is dense in \( {\mathbf{Q}}_{p} \), we can find monic polynomials \( g \in \mathbf{Q}\left\lbrack X\right\rbrack \) as close to \( f \) as we want. If we choose a sequence \( {g}_{n... | Yes |
Proposition 2. Let \( K \) be a nondiscrete ultrametric field and put\n\n\[ A = \{ x \in K : \\left| x\\right| \\leq 1\} : \\text{ maximal subring of }K \]\n\n\[ M = \{ x \in K : \\left| x\\right| < 1\} : \\text{ maximal ideal of }A\\text{. } \]\n\nThen, either \( M \) is principal, or \( M = {M}^{2} \) and the ring \(... | Proof. By hypothesis \( \\Gamma = \\left| {K}^{ \\times }\\right| \\neq \\{ 1\\} \), and either \( \\Gamma \\cap \\left( {0,1}\\right) \) has a maximal element \( \\theta \) or it has a sequence tending to 1 . In the first case we can choose \( \\pi \\in M \) with \( \\left| \\pi \\right| = \\theta \), and \( M = {\\pi... | Yes |
Proposition 3. With the same notation as before:\n\n(a) If \( K \) is algebraically closed, so is the residue field \( k \) .\n\n(b) If \( L \) is an algebraic extension of \( K \), the residue field \( {k}_{L} \) of \( L \) is also an algebraic extension of the residue field \( k \) of \( K \) . | Proof. In any ultrametric field, \( \left| \xi \right| > 1,\left| {a}_{i}\right| \leq 1\left( {i < n}\right) \) implies\n\n\[ \left| {\xi }^{n}\right| > \left| {\xi }^{i}\right| \geq \left| {{a}_{i}{\xi }^{i}}\right| \;\left( {i < n}\right) ,\]\n\n\[ {\left| \xi \right| }^{n} > \left| {\mathop{\sum }\limits_{{i < n}}{a... | Yes |
Proposition 1. The subset \( \mathcal{J} = {\varphi }^{-1}\left( 0\right) \) is a maximal ideal of the ring \( R \), and the field \( {\Omega }_{p} = R/\mathcal{J} \) is an extension of the field \( {\mathbf{Q}}_{p}^{a} \) . | Proof. Let us show that each element \( \alpha \notin \mathcal{J} \) is invertible \( {\;\operatorname{mod}\;\mathcal{J}} \) . But if \( \alpha = \left( {\alpha }_{n}\right) \) is not in the ideal \( \mathcal{J} \), the limit \( r = \varphi \left( \alpha \right) > 0 \) does not vanish, so we can find a subset \( A \in ... | Yes |
Proposition 2. The absolute value \( {\left. \mid .\right| }_{\Omega } \) coincides with the quotient norm of \( R/\mathcal{J} \), namely for \( a = \left( {\alpha {\;\operatorname{mod}\;\mathcal{J}}}\right) \), \[ {\left| a\right| }_{\Omega } = \parallel \alpha {\;\operatorname{mod}\;\mathcal{J}}{\parallel }_{R/\mathc... | Proof. We have \( \mathop{\lim }\limits_{\mathcal{U}}\left| {\gamma }_{i}\right| \leq \sup \left| {\gamma }_{i}\right| \left( {\gamma \in R}\right) \), and hence \[ \mathop{\lim }\limits_{\mathcal{U}}\left| {\alpha }_{i}\right| = \mathop{\lim }\limits_{\mathcal{U}}\left| {{\alpha }_{i} - {\beta }_{i}}\right| \leq \sup ... | Yes |
Proposition 3. We have\n\n\[ \left| {\Omega }_{p}^{ \times }\right| = {\mathbf{R}}_{ > 0} \] | Proof. This is a simple consequence of the fact that \( \left| {\mathbf{Q}}_{p}^{a}\right| \) is dense in \( {\mathbf{R}}_{ \geq 0} \) . Indeed, each positive real number \( r \) is limit of a sequence \( \left( {r}_{n}\right) \) of elements \( {r}_{n} \in \left| {\mathbf{Q}}_{p}^{a}\right| \), say \( {r}_{n} = \left| ... | Yes |
Theorem 2. Let \( \Omega \) be any algebraically closed extension of \( {\mathbf{Q}}_{p} \) and \( K \subset \Omega \) any complete subfield. Select an algebraic element \( a\left( { \in \Omega }\right) \) over \( K \) and denote by \( {a}^{\sigma } \) its conjugates over \( K \) . Let \( r = \mathop{\min }\limits_{{{a... | Proof. We can proceed as in (1.5), since we now have uniqueness of the extension of absolute values for finite extensions of \( K \) . For any algebraic element \( b \) such that \( a \notin K\left( b\right), a \) has a conjugate \( {a}^{\sigma } \neq a \) over \( K\left( b\right) \) (the automorphism \( \sigma \) leav... | Yes |
Proposition 1. The group \( 1 + {\mathbf{M}}_{p} \) is divisible. For each \( m \geq 2 \) prime to \( p \), it is uniquely \( m \) -divisible. | Proof. It is enough to prove that the group \( 1 + {\mathbf{M}}_{p} \) is \( p \) -divisible and uniquely \( m \) -divisible for each \( m \) prime to \( p \) .\n\n(1) Let \( 1 + t \in 1 + {\mathbf{M}}_{p} \) and select a root \( x \in {\mathbf{C}}_{p} \) of \( {X}^{p} - \left( {1 + t}\right) \) : this is possible, sin... | Yes |
Proposition 2. For \( x \in {\mathbf{C}}_{p} \) we have\n\n\[ x \in 1 + {\mathbf{M}}_{p} \Leftrightarrow {x}^{{p}^{n}} \rightarrow 1\;\left( {n \rightarrow \infty }\right) . \] | Proof. If \( x = 1 + t \in 1 + {\mathbf{M}}_{p} \), the sequence\n\n\[ {x}^{{p}^{n}} - 1 = {\left( 1 + t\right) }^{{p}^{n}} - 1 \]\n\ntends to 0 by the fundamental inequality (4.3) (second form). Conversely, assume that \( {x}^{{p}^{n}} \rightarrow 1 \) (for some \( x \in {\mathbf{C}}_{p} \) ) and take an integer \( n ... | Yes |
Corollary 1. Let \( \mathcal{U} \) be an ultrafilter on a set \( X \) . If \( {A}_{1},\ldots ,{A}_{n} \) is a finite family of subsets of \( X \) such that \( \mathop{\bigcup }\limits_{{1 \leq i \leq n}}{A}_{i} \in \mathcal{U} \), then there exists at least one index \( i \) for which \( {A}_{i} \in \mathcal{U} \) . | Proof. It is enough to prove the assertion for two subsets (by induction). If \( A \notin \mathcal{U} \) and \( B \notin \mathcal{U} \), we infer from the above criterion that \( {A}^{c} \in \mathcal{U},{B}^{c} \in \mathcal{U} \) ; hence \( {\left( A \cup B\right) }^{c} = {A}^{c} \cap {B}^{c} \in \mathcal{U} \), and \(... | Yes |
Corollary 2. Let \( f : X \rightarrow Y \) and let \( \mathcal{U} \) be an ultrafilter on the set \( X \) . Then \( f\left( \mathcal{U}\right) \) is an ultrafilter on \( f\left( X\right) \) and a basis of an ultrafilter on \( Y \) . | Proof. It is enough to prove the assertion when \( f \) is surjective. For any \( A \subset Y \) , either \( {f}^{-1}\left( A\right) \) or \( {f}^{-1}{\left( A\right) }^{c} = {f}^{-1}\left( {A}^{c}\right) \) belongs to \( \mathcal{U} \) ; hence\n\n\[ \n\text{either}A = f\left( {{f}^{-1}\left( A\right) }\right) \text{or... | Yes |
Corollary 1. If a polynomial \( f \in \mathbf{Q}\left\lbrack x\right\rbrack \) takes integral values on \( \mathbf{N} \), it also takes integral values on \( \mathbf{Z} \) . | Proof. It is enough to check this property for the basis of \( L \) consisting of the binomial polynomials. If \( x = - m \) is a negative integer, then\n\n\[ \left( \begin{matrix} - m \\ i \end{matrix}\right) = - m\left( {-m - 1}\right) \cdots \left( {-m - i + 1}\right) /i! = {\left( -1\right) }^{i}\left( \begin{matri... | Yes |
Corollary 2. If a polynomial of degree \( d \geq 0 \) (with rational coefficients) takes integral values on \( d + 1 \) consecutive integers, then it takes integral values on all integers. | Proof. Let \( f \) take integral values on the integers \( a, a + 1,\ldots, a + d \) and consider its translate \( g\left( x\right) = f\left( {x - a}\right) \) which takes integral values on the first integers \( 0,1,\ldots, d \) . Hence the first iterated differences of \( g \) at the origin are also integers, and if ... | No |
Proposition 2. The indefinite-sum and finite-difference operators are linked by the formulas\n\n\[ \nabla \circ S = \mathrm{{id}},\;S \circ \nabla = \mathrm{{id}} - {P}_{0},\;\nabla \circ S - S \circ \nabla = {P}_{0}. \] | The identity \( S\left( {\nabla f}\right) = f - f\left( 0\right) \cdot 1 \) gives a first-order limited expansion of \( f \) if we only rewrite it \( f = f\left( 0\right) \cdot 1 + S\left( {\nabla f}\right) \) . This point of view has been generalized by van Hamme. | No |
Theorem 2. Let \( f : \mathbf{N} \rightarrow {\mathbf{C}}_{p} \) be any map and define \( {a}_{k} = {\nabla }^{k}f\left( 0\right) \) . Then the following properties are equivalent:\n\n(i) \( \left| {a}_{k}\right| \rightarrow 0 \) when \( k \rightarrow \infty \) .\n\n(ii) The Mahler series \( \mathop{\sum }\limits_{{k \... | Proof. Here is a complete scheme of implications.\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) We have\n\n\[ \left| {{a}_{k}\left( \begin{array}{l} x \\ k \end{array}\right) }\right| \leq \left| {a}_{k}\right| \begin{Vmatrix}\left( \begin{array}{l} \cdot \\ k \end{array}\right) \end{Vmatrix} = \left| {a}_{k}... | Yes |
Corollary 1. Any continuous \( f : {\mathbf{Z}}_{p} \rightarrow {\mathbf{C}}_{p} \) has limited Mahler expansions | \[ f = f\left( 0\right) + \nabla f\left( 0\right) \cdot \left( \begin{array}{l} . \\ 1 \end{array}\right) + {\nabla }^{2}f\left( 0\right) \cdot \left( \begin{array}{l} . \\ 2 \end{array}\right) + \cdots \] \[ + {\nabla }^{n}f\left( 0\right) \cdot \left( \begin{matrix} \cdot \\ n \end{matrix}\right) + {R}_{n + 1}f\;\lef... | Yes |
For any continuous function \( f : {\mathbf{Z}}_{p} \rightarrow {\mathbf{C}}_{p} \), the indefinite sum \( {Sf} = f\underline{ * }1 \) of \( f \) extends continuously to \( {\mathbf{Z}}_{p} \) . More precisely, if \( f = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}\left( {}_{k}^{ \cdot }\right) \) is the Mahler expansion ... | \[ {Sf} = 1 * f = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}\left( \begin{matrix} \cdot \\ k + 1 \end{matrix}\right) ,\parallel {Sf}\parallel = \parallel f\parallel . \]\n\nProof. We have noticed that\n\n\[ S\left( \begin{array}{l} \cdot \\ k \end{array}\right) = 1\underline{ * }\left( \begin{array}{l} \cdot \\ k \end{a... | Yes |
Corollary 3. The only linear form \( \varphi : C\left( {{\mathbf{Z}}_{p};K}\right) \rightarrow K \) that is invariant under translation is the trivial one \( \varphi = 0 \) . | Proof. In fact, we prove that if \( \varphi \left( F\right) = \varphi \left( {F}_{1}\right) \) for all \( F \in C\left( {{\mathbf{Z}}_{p};K}\right) \), where \( {F}_{1}\left( x\right) = \) \( F\left( {x + 1}\right) \), then \( \varphi = 0 \) . Indeed, take any \( f \in C\left( {{\mathbf{Z}}_{p};K}\right) \) . There exi... | Yes |
Corollary 4. Let \( \sigma : {\mathbf{Z}}_{p} \rightarrow {\mathbf{Z}}_{p}, x \mapsto - 1 - x \), be the canonical involution (I.1.2). Then \( S\left( {f \circ \sigma }\right) \left( x\right) = - {Sf}\left( {-x}\right) \) . | Proof. For integers \( n, m \geq 1 \) we have\n\n\[ \n{Sf}\left( {n + m}\right) - {Sf}\left( n\right) = f\left( n\right) + \cdots + f\left( {n + m - 1}\right) .\n\] \n\nBy density of the integers \( n \geq 1 \) in \( {\mathbf{Z}}_{p} \) and continuity of both sides, we get more generally\n\n\[ \n{Sf}\left( {x + m}\righ... | Yes |
Corollary 2. Let \( E \) be a Banach space. Then the sum map \( {E}^{\left( I\right) } \rightarrow E \) has a unique continuous extension \( \sum : {c}_{0}\left( {I;E}\right) \rightarrow E \) . | Proof. The sum \( x = \left( {x}_{i}\right) \mapsto \mathop{\sum }\limits_{{i \in I}}{x}_{i} : {E}^{\left( I\right) } \rightarrow E \) is a contracting linear map\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{i \in I}}{x}_{i}}\end{Vmatrix} \leq \mathop{\sup }\limits_{i}\begin{Vmatrix}{x}_{i}\end{Vmatrix} = \parallel x\... | Yes |
Corollary 2. Let \( E \) be a Banach space. Then the sum map \( {E}^{\left( I\right) } \rightarrow E \) has a unique continuous extension \( \sum : {c}_{0}\left( {I;E}\right) \rightarrow E \) . | Proof. The sum \( x = \left( {x}_{i}\right) \mapsto \mathop{\sum }\limits_{{i \in I}}{x}_{i} : {E}^{\left( I\right) } \rightarrow E \) is a contracting linear map\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{i \in I}}{x}_{i}}\end{Vmatrix} \leq \mathop{\sup }\limits_{i}\begin{Vmatrix}{x}_{i}\end{Vmatrix} = \parallel x\... | Yes |
Corollary 3 (Universal Property of Direct Sums). Let \( {\varepsilon }_{j} \) denote the canonical injection of a factor into the direct sum \( {E}_{j} \rightarrow {\bigoplus }_{i \in I}{E}_{i} \subset \mathop{\bigoplus }\limits_{{i \in I}}{E}_{i} \). Then for each Banach space \( E \) and family \( \left( {f}_{j}\righ... | Proof. Under the assumptions made,\n\n\[ \left( {x}_{i}\right) \mapsto \left( {{f}_{i}{x}_{i}}\right) : {\widehat{\bigoplus }}_{i \in I}{E}_{i} \rightarrow {c}_{0}\left( {I;E}\right) \]\n\n is a linear contracting map, and composition with the sum \( \sum \) yields the unique solution to the factorization problem\n\n\[... | Yes |
Proposition 1. If \( F \) is complete, then \( L\left( {E;F}\right) \) is also complete. | Proof. Let \( \left( {T}_{n}\right) \) be a Cauchy sequence in \( L\left( {E;F}\right) \) . For each \( x \in E,\left( {{T}_{n}\left( x\right) }\right) \) is a Cauchy sequence in the complete space \( F \), and hence has a limit \( {Tx} \) which obviously depends linearly on \( x \in E \) . This defines a linear map \(... | Yes |
Proposition 2. The topological dual of the space \( {c}_{0}\left( {I;E}\right) \) is canonically isomorphic as a normed space to \( {l}^{\infty }\left( {I;{E}^{\prime }}\right) \) . | Proof. If \( \varphi \) is a continuous linear form on \( {c}_{0}\left( {I;E}\right) \), we let \( {\varphi }_{i} = \varphi \circ {\varepsilon }_{i} \) denote the restriction of \( \varphi \) to the \( i \) th factor \( E \) in \( {c}_{0}\left( {I;E}\right) \) (families having a zero component for all indices except \(... | Yes |
Proposition 1. Assume that \( E \) admits a normal basis and fix an isomorphism \( {c}_{0}\left( J\right) \cong E \) . Then the map\n\n\[ \nL\left( {E;F}\right) \rightarrow {l}^{\infty }\left( {J;F}\right)\n\]\n\ndefined above is an isometric isomorphism. | Proof. We have already seen that \( \begin{Vmatrix}{\left( {\mathbf{f}}_{j}\right) }_{J}\end{Vmatrix} \leq \parallel T\parallel \) . Conversely,\n\n\[ \n\mathbf{x} = \mathop{\sum }\limits_{j}{x}_{j}{\mathbf{e}}_{j} \Rightarrow T\left( \mathbf{x}\right) = \mathop{\sum }\limits_{j}{x}_{j}{\mathbf{f}}_{j}\text{ (this sum ... | Yes |
Proposition 3. When \( E = {c}_{0}\left( J\right) \) and \( F = {c}_{0}\left( I\right) \), we can make canonical identifications\n\n\[ L\left( {E;F}\right) = {l}^{\infty }\left( {J;{c}_{0}\left( I\right) }\right) \subset {l}^{\infty }\left( {I;{E}^{\prime }}\right) = {l}^{\infty }\left( {I;{l}^{\infty }\left( J\right) ... | More particularly, if \( T \) is continuous and of rank less than or equal to 1, we can write\n\n\[ T\left( \mathbf{x}\right) = \varphi \left( \mathbf{x}\right) \mathbf{a} = {\left( \varphi \left( \mathbf{x}\right) {a}_{i}\right) }_{I} \]\n\nfor some \( \varphi \in {E}^{\prime } \) . In this case, \( {\varphi }_{i}\lef... | Yes |
Let us consider the delta operator\n\n\\[ \nabla = {\nabla }_{ + } = \tau - 1 = {e}^{D} - 1 = \varphi \\left( D\\right) ,\\]\n\nfor which\n\n\\[ z = \varphi \\left( u\\right) = {e}^{u} - 1,\\;{e}^{u} = z + 1,\\;u = \\log \\left( {1 + z}\\right) = {\\varphi }^{-1}\\left( z\\right) .\\] | We have\n\n\\[ \exp \\left( {x{\\varphi }^{-1}\\left( z\\right) }\\right) = \exp \\left( {x\\log \\left( {1 + z}\\right) }\\right) = {\\left( 1 + z\\right) }^{x}\\]\n\n\\[ = \\mathop{\\sum }\\limits_{{n \\geq 0}}\\left( \\begin{array}{l} x \\\\ n \\end{array}\\right) {z}^{n} = \\mathop{\\sum }\\limits_{{n \\geq 0}}{\\l... | Yes |
Let \( f : {\mathbf{Z}}_{p} \rightarrow {\mathbf{Z}}_{p} \) be the continuous function defined by\n\n\[ \n x = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{p}^{n} \mapsto f\left( x\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{p}^{2n}.\n\]\n\nThen \( f \) is differentiable at all points \( x \in {\mathbf{Z}}_{p} \) w... | Again \( {f}^{\prime } = 0 \in {\mathcal{C}}^{1} \) , but \( f \) is injective, and hence far from being locally constant. | Yes |
Proposition 1. Let \( f : X \rightarrow K \) be strictly differentiable at a point \( a \in X \) with \( {f}^{\prime }\left( a\right) \neq 0 \) . Then there is a neighborhood \( V \) of \( a \) in \( X \) such that the restriction of \( f/{f}^{\prime }\left( a\right) \) to \( V \) is isometric. | Proof. Since \( f \in {S}^{1}\left( a\right) \), for each \( \varepsilon > 0 \) there is a neighborhood \( {V}_{\varepsilon } \) of \( a \) for which\n\n\[ \left| {{\Phi f}\left( {x, y}\right) - {f}^{\prime }\left( a\right) }\right| < \varepsilon \text{ if }x \in {V}_{\varepsilon }\text{ and }y \in {V}_{\varepsilon }.\... | Yes |
Proposition 2. For \( f : X \rightarrow K \), the following properties are equivalent:\n\n(i) \( f \in {S}^{1}\left( a\right) \) for all \( a \in X \) .\n\n(ii) The function \( {\Phi f} \), initially defined only on \( X \times X - {\Delta }_{X} \), admits a continuous extension \( \widetilde{\Phi } \) to \( X \times X... | Proof. The implication \( \left( i\right) \Rightarrow \left( {ii}\right) \) is given by the double limit theorem, which we recall: Let \( {X}_{0} \) be a dense subset of a topological space \( X, Y \) a metric space, and \( f \) a continuous map \( {X}_{0} \rightarrow Y \) such that for each \( x \in X \)\n\n\[ z \in {... | Yes |
Proposition 1. If \( f \in {S}^{2}\left( a\right) \), then \( f \in {S}^{1}\left( a\right) \) . | Proof. Let us take two pairs \( \left( {x, y}\right) \) and \( \left( {z, t}\right) \in X \times X - {\Delta }_{X} \) in the vicinity of \( \left( {a, a}\right) \) and estimate the difference\n\n\[ \n{\Phi f}\left( {x, y}\right) - {\Phi f}\left( {z, t}\right) = {\Phi f}\left( {x, y}\right) - {\Phi f}\left( {z, y}\right... | Yes |
Proposition 2. If \( f \in {S}^{2} \), then \( {f}^{\prime } \in {S}^{1} \) . | Proof. We have to prove that the difference quotients\n\n\[ \Phi \left( {f}^{\prime }\right) \left( {x, y}\right) = \frac{{f}^{\prime }\left( x\right) - {f}^{\prime }\left( y\right) }{x - y} \]\n\nhave a continuous extension across the diagonal of \( X \times X \) . By assumption, there is a continuous function \( {\wi... | Yes |
Corollary 1. Let \( f \in \operatorname{Lip}\left( {\mathbf{Z}}_{p}\right) \) and \( f = \sum {c}_{n}\left( \begin{array}{l} \cdot \\ n \end{array}\right) \) its Mahler expansion. Then \[ \parallel {\Phi f}\parallel = \mathop{\sup }\limits_{{n \geq 1}}{\kappa }_{n}\left| {c}_{n}\right| < \infty . \] | The number \( \parallel {\Phi f}\parallel \) does not define a norm on the vector space \( \operatorname{Lip}\left( {\mathbf{Z}}_{p}\right) \) because \( {\Phi f} \) vanishes for constant functions: It is only a seminorm. In order to have a norm, we take \[ \parallel f{\parallel }_{1} = \sup \left( {\left| {f\left( 0\r... | Yes |
Corollary 2. Let \( f \in \operatorname{Lip}\left( {\mathbf{Z}}_{p}\right) \) and \( {Sf} \) its indefinite sum. Then \( {Sf} \in \operatorname{Lip}\left( {\mathbf{Z}}_{p}\right) \) and \[ \parallel f{\parallel }_{1} \leq \parallel {Sf}{\parallel }_{1} \leq p\parallel f{\parallel }_{1} \] | Proof. We have \[ \parallel f{\parallel }_{1} = \mathop{\sup }\limits_{{n \geq 0}}{\kappa }_{n}\left| {a}_{n}\right| \] \[ \parallel {Sf}{\parallel }_{1} = \mathop{\sup }\limits_{{n \geq 1}}{\kappa }_{n}\left| {a}_{n - 1}\right| \] by Corollary 2 in (IV.3.5). Now observe that \[ {\kappa }_{n - 1} \leq {\kappa }_{n} \le... | Yes |
Example 1. Let the Mahler coefficients \( {c}_{k} \) of a continuous function \( f \) be\n\n\[ \n{c}_{k} = \left\{ \begin{array}{ll} {p}^{j} & \text{ if }k = {p}^{j}, \\ 0 & \text{ if }k\text{ is not a power of }p, \end{array}\right. \n\]\n\nso that\n\n\[ \n\left| {{c}_{k}/k}\right| \text{takes alternatively values 0 a... | But \( {\Phi f} \) is bounded, since \( k\left| {c}_{k}\right| \) (taking values 0 and 1 only) is bounded. | No |
Proposition 1. The unit ball in \( K\{ X\} \) is uniformly equicontinuous. More precisely, \( K\{ X\} \subset \operatorname{Lip}\left( A\right) \) and \( \parallel {\Phi f}\parallel \leq \parallel f{\parallel }_{1} \leq \parallel f\parallel \) for \( f \in K\{ X\} \) . In particular, | Proof. Write \( f = \sum {a}_{n}{X}^{n} \), so that\n\n\[ f\left( x\right) - f\left( y\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}\left( {{x}^{n} - {y}^{n}}\right) = \left( {x - y}\right) \mathop{\sum }\limits_{{n \geq 1}}{a}_{n}\left( {{x}^{n - 1} + \cdots + {y}^{n - 1}}\right) .\n\]\n\nIf \( \left| x\right| \l... | Yes |
Proposition 2. If \( f \in K\{ X\} \), then\n\n\[ \left| {f\left( {x + y}\right) - f\left( x\right) - f\left( y\right) + f\left( 0\right) }\right| \leq \parallel f\parallel \cdot \left| {xy}\right| \;\left( {\left| x\right| \leq 1,\left| y\right| \leq 1}\right) . | Proof. With the same notation as before,\n\n\[ \left. {f\left( {x + y}\right) - f\left( x\right) - f\left( y\right) + f\left( 0\right) = \mathop{\sum }\limits_{{n \geq 2}}{a}_{n}\left( {{\left( x + y\right) }^{n} - {x}^{n} - {y}^{n}}\right) }\right) ,\n\nwhence the result, since each term \( {\left( x + y\right) }^{n} ... | Yes |
Theorem 1. Let \( f \in K\{ X\} \) . Then \( f \) defines a strictly differentiable function on the unit ball \( A \) of \( K : f \in {S}^{1}\left( A\right) \) . The derivative of \( f \) is given by the restricted formal power series\n\n\[ \n{f}^{\prime } = {\left. \Phi f\right| }_{\Delta } = \mathop{\sum }\limits_{{n... | It is easy to give more precise estimates for the convergence:\n\n\[ \n{\Phi f}\left( {x, y}\right) - \mathop{\sum }\limits_{{n \geq 1}}n{a}_{n}{\xi }^{n - 1} = \mathop{\sum }\limits_{{n \geq 1}}{a}_{n}\left( {{x}^{n} - {y}^{n}}\right) /\left( {x - y}\right) - \mathop{\sum }\limits_{{n \geq 1}}n{a}_{n}{\xi }^{n - 1}\n\... | Yes |
Theorem 2. A restricted formal power series \( f = \sum {a}_{n}{X}^{n} \) defines a twice strictly differentiable function on the unit ball \( A \) of \( K : f \in {S}^{2}\left( A\right) \) . | Proof. As we have seen in the proof of the preceding theorem,\n\n\[ \n{\Phi f}\left( {x, y}\right) = \mathop{\sum }\limits_{{n \geq 1}}{a}_{n}\mathop{\sum }\limits_{{i + j = n - 1}}{x}^{i}{y}^{j} = {a}_{1} + {a}_{2}\left( {x + y}\right) + \cdots , \n\] \n\nand hence, for distinct \( x, y, z \) , \n\n\[ \n{\Phi }_{2}f\l... | Yes |
Proposition 1. For \( \left| x\right| < {r}_{p} \) we have\n\n\[ \left| {\log \left( {1 + x}\right) }\right| = \left| x\right| ,\;\left| {\exp \left( x\right) }\right| = 1,\;\left| {1 - \exp \left( x\right) }\right| = \left| x\right| . \] | Proof. For \( k \geq 1 \) we have \( {S}_{p}\left( k\right) \geq 1 \) and hence \( {\operatorname{ord}}_{p}\left( {k!}\right) \leq \left( {k - 1}\right) /\left( {p - 1}\right) \) . We infer\n\n\[ \left| k\right| \geq \left| {k!}\right| \geq {\left| p\right| }^{\frac{k - 1}{p - 1}} = {r}_{p}^{k - 1}, \]\n\n\[ \left| {{x... | Yes |
Proposition 2. For two indeterminates \( X \) and \( Y \), we have the following formal identities:\n\n\[ \exp \left( {X + Y}\right) = \exp \left( X\right) \cdot \exp \left( Y\right) \] | Proof. The first identity is easily obtained if we observe that the product of two monomials \( {X}^{i}/i \) ! and \( {Y}^{j}/j \) ! is\n\n\[ \frac{{X}^{i}{Y}^{j}}{i!j!} = \left( \begin{matrix} i + j \\ i \end{matrix}\right) \frac{{X}^{i}{Y}^{j}}{\left( {i + j}\right) !}. \]\n\nGrouping the terms with \( i + j = n \) l... | No |
For \( \left| x\right| < {r}_{p} \) and \( \left| y\right| < {r}_{p} \) we have\n\n\[ \exp \left( {x + y}\right) = \exp \left( x\right) \cdot \exp \left( y\right) \]\n\n\[ \log \exp \left( x\right) = x \]\n\n\[ \exp \log \left( {1 + x}\right) = 1 + x. \] | Proof. Observe first that if \( {a}_{n} \) and \( {b}_{n} \rightarrow 0 \), then the family \( {\left( {a}_{n}{b}_{m}\right) }_{n, m \geq 0} \) is summable. In particular, its sum is independent of the way terms are grouped before summing. Hence the first identity holds as soon as the variables \( x \) and \( y \) are ... | Yes |
Proposition 1. (a) For \( f \in {S}^{1}\left( {\mathbf{Z}}_{p}\right) \) we have\n\n\[ \left| {{\int }_{{\mathbf{Z}}_{p}}f\left( x\right) {dx}}\right| \leq p\parallel f{\parallel }_{1} \]\n\n(b) If \( {f}_{n} \rightarrow f \) in \( {S}^{1} \), namely \( {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{1} \rightarrow 0 \), th... | Proof. By definition,\n\n\[ \left| {{\int }_{{\mathbf{Z}}_{p}}f\left( x\right) {dx}}\right| = \left| {{\left( Sf\right) }^{\prime }\left( 0\right) }\right| \leq \parallel {Sf}{\parallel }_{1} = \sup \left( {\parallel {\Phi Sf}\parallel ,\left| {{Sf}\left( 0\right) }\right| }\right) ,\]\n\nso that \( \left( a\right) \) ... | Yes |
Proposition 2. For \( f \in {S}^{1}\left( {\mathbf{Z}}_{p}\right) \) we have\n\n\[{\int }_{{\mathbf{Z}}_{p}}\nabla f\left( x\right) {dx} = {f}^{\prime }\left( 0\right)\] | Proof. By definition,\n\n\[{\int }_{{\mathbf{Z}}_{p}}\nabla f\left( x\right) {dx} = {\left( S\nabla f\right) }^{\prime }\left( 0\right) = {\left( f - f\left( 0\right) \right) }^{\prime }\left( 0\right) = {f}^{\prime }\left( 0\right) ,\]\n\nsince \( S\nabla f = f - f\left( 0\right) \) (Proposition 2 of (IV.1.5)). | Yes |
Proposition 1. Let \( {P}_{0} : f \mapsto f\left( 0\right) \cdot 1 \) be the projection of \( {S}^{1}\left( {\mathbf{Z}}_{p}\right) \) onto constants. Then the following relations hold:\n\n(a) \( {S\tau } = {\tau S} - {P}_{0} \).\n\n(b) DS commutes with all translations \( {\tau }_{x} \).\n\n(c) \( {SD} = {DS} - {P}_{0... | Proof. By definition, for integers \( n \geq 1 \),\n\n\[ S\left( {\tau f}\right) \left( n\right) = \mathop{\sum }\limits_{{0 \leq j < n}}{\tau f}\left( j\right) = \mathop{\sum }\limits_{{0 \leq j < n}}f\left( {j + 1}\right) \]\n\n\[ = \mathop{\sum }\limits_{{0 < i \leq n}}f\left( i\right) = {Sf}\left( {n + 1}\right) - ... | Yes |
Proposition 3. Let \( f \in {S}^{2}\left( {\mathbf{Z}}_{p}\right) \subset {S}^{1}\left( {\mathbf{Z}}_{p}\right) \) and define \( F\left( x\right) = {\int }_{{\mathbf{Z}}_{p}}f\left( {x + t}\right) {dt} \). Then \( F \in {S}^{1}\left( {\mathbf{Z}}_{p}\right) \) and\n\n\[ \n{F}^{\prime }\left( x\right) = {\int }_{{\mathb... | Proof. By Proposition 2 of (1.3),\n\n\[ \nf \in {S}^{2}\left( {\mathbf{Z}}_{p}\right) \; \Rightarrow \;{f}^{\prime } \in {S}^{1}\left( {\mathbf{Z}}_{p}\right) \n\]\n\nso that\n\n\[ \nG\left( x\right) = {\int }_{{\mathbf{Z}}_{p}}{f}^{\prime }\left( {x + t}\right) {dt} \n\]\n\ndefines a function \( G \in {S}^{1}\left( {\... | Yes |
Proposition 4. Let \( \\sigma \) denote the involution (I.1.2) \( x \\mapsto - 1 - x \) of \( {\\mathbf{Z}}_{p} \) . Then\n\n\[ \n{\\int }_{{\\mathbf{Z}}_{p}}\\left( {f \\circ \\sigma }\\right) {dx} = {\\int }_{{\\mathbf{Z}}_{p}}{fdx} \n\] | Proof. We have seen that\n\n\[ \n{\\int }_{{\\mathbf{Z}}_{p}}{fdx} = {\\left( Sf\\right) }^{\\prime }\\left( 0\\right) = \\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\frac{{Sf}\\left( h\\right) }{h}. \n\]\n\nLet us take \( h = - {p}^{n}\\left( {n \\rightarrow \\infty }\\right) \) . Hence\n\n\[ \n{\\left( Sf\\right) }^... | Yes |
The radius of convergence of \( f = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{X}^{n} \) is\n\n\[ \n{r}_{f} = \frac{1}{\mathop{\lim }\limits_{{n \geq 0}}{\left| {a}_{n}\right| }^{1/n}} = \frac{1}{\mathop{\limsup }\limits_{{n \rightarrow \infty }}{\left| {a}_{n}\right| }^{1/n}}.\n\] | Proof. Define \( {r}_{f} \) by the Hadamard formula. If \( \left| x\right| > {r}_{f} \) (this can happen only if \( {r}_{f} < \infty \) !), we have\n\n\[ \n\mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sup }\limits_{{k \geq n}}\left| x\right| {\left| {a}_{k}\right| }^{1/k} = \left| x\right| \cdot \mathop{\lim... | Yes |
Proposition 2. Let \( f \) and \( g \) be two convergent power series. Their product \( {fg} \) (computed formally ) is a convergent power series, and more precisely, the radius of convergence of \( {fg} \) is greater than or equal to \( \min \left( {{r}_{f},{r}_{g}}\right) \) . Moreover, the numerical evaluation of th... | Proof. All statements are consequences of (V.2.2). | No |
For any polynomial \( f \), the radius of convergence of the composite \( f \circ g \) is \( \geq {r}_{g} \) and \[ \left( {f \circ g}\right) \left( x\right) = f\left( {g\left( x\right) }\right) \;\left( {\left| x\right| < {r}_{g}}\right) . | Proof. If \( \left| x\right| < {r}_{g} \), taking \( f = g \) in the preceding proposition, we obtain \( {g}^{2}\left( x\right) = \) \( g{\left( x\right) }^{2} \) and by induction \( {g}^{n}\left( x\right) = g{\left( x\right) }^{n}\left( {n \geq 0}\right) \) . Taking linear combinations of these equalities, we deduce \... | No |
Proposition 3. The radius of convergence of \( f = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{X}^{n} \) and of its derivative \( {Df} = \mathop{\sum }\limits_{{n \geq 1}}n{a}_{n}{X}^{n - 1} \) are the same: \( {r}_{f} = {r}_{Df} \) . | Proof. Let us prove this proposition when the field is either an extension of \( {\mathbf{Q}}_{p} \) or an extension of \( \mathbf{R} \) with the normalized absolute value. We know that\n\n\[ \frac{1}{n} \leq \left| n\right| \leq n\;\left( {n \in \mathbf{N}}\right) \]\n\nand also\n\n\[ {n}^{\pm 1/n} \rightarrow 1\;\lef... | Yes |
Theorem 1. Let \( f\left( X\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{X}^{n} \) be a formal power series. The following properties are equivalent:\n\n(i) \( \exists g \in K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) with \( g\left( 0\right) = 0 \) and \( \left( {f \circ g}\right) \left( X\right) ... | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) If \( g\left( X\right) = \mathop{\sum }\limits_{{m \geq 1}}{b}_{m}{X}^{m} \), then the identity \( \left( {f \circ g}\right) \left( X\right) = X \) can be written more explicitly as\n\n\[ \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}g{\left( X\right) }^{n} = {a}_{... | Yes |
Theorem 2 (Chain Rule). Let \( f \) and \( g \) be two formal power series with \( g\left( 0\right) = 0 \) . Then the formal derivative of \( f \circ g \) is given by\n\n\[ D\left( {f \circ g}\right) \left( Y\right) = {Df}\left( X\right) {Dg}\left( Y\right) = {Df}\left( {g\left( Y\right) }\right) {Dg}\left( Y\right) . ... | Proof. Fix the power series \( g \) and let \( f \) vary in \( K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) . Then\n\n\[ \omega \left( f\right) \geq k \Rightarrow \omega \left( {f \circ g}\right) \geq k \Rightarrow \omega \left( {D\left( {f \circ g}\right) }\right) \geq k - 1 \]\n\nas well as\n\n\[ \omega... | Yes |
Proposition 1. When \( \left| {K}^{ \times }\right| \) is dense in \( {\mathbf{R}}_{ > 0} \) and \( f \in K\{ X\} \), the Gauss norm of \( f \) and the sup norm of the function defined by \( f \) on the unit ball \( A \) of \( K \) coincide. | Proof. The Gauss norm of \( f \) is \( {M}_{1}f \), and the inequality\n\n\[ \mathop{\sup }\limits_{{\left| x\right| \leq 1}}\left| {f\left( x\right) }\right| \leq {M}_{1}f \]\n\nholds in general. With our assumption, we can choose a sequence \( {x}_{n} \in K \) with \( \left| {x}_{n}\right| \) regular and \( \left| {x... | Yes |
Proposition 2. When \( r > 0 \) is fixed, \( f \mapsto {M}_{r}\left( f\right) \) is an ultrametric norm on the subspace consisting of formal power series \( f\left( X\right) = \sum {a}_{n}{X}^{n} \) such that \( \left| {a}_{n}\right| {r}^{n} \rightarrow 0\left( {n \rightarrow \infty }\right) \) . This norm is multiplic... | Proof. If \( f \neq 0 \), then one \( {a}_{n} \) at least is nonzero, and \( {M}_{r}\left( f\right) \geq \left| {a}_{n}\right| {r}^{n} > 0 \), since \( r > 0 \) . Hence \( {M}_{r} \) is a norm on the subspace considered. Moreover, the equality\n\n\[ \n{M}_{r}\left( {fg}\right) = {M}_{r}\left( f\right) {M}_{r}\left( g\r... | Yes |
Example 2. Let us treat the case of the power series\n\n\\[ \nf\left( X\right) = \log \left( {1 + X}\right) = \mathop{\sum }\limits_{{n \geq 1}}\frac{{\left( -1\right) }^{n - 1}}{n}{X}^{n}. \n\\]\n\nWe have\n\n\\[ \n{\operatorname{ord}}_{p}{a}_{n} = 0\text{ if }1 \leq n < p,\;{\operatorname{ord}}_{p}{a}_{p} = - 1, \n\\... | The Newton polygon of the logarithm\n\nThe vertices of the Newton polygon are the points\n\n\\[ \n{P}_{1} = \left( {1,0}\right) ,\;{P}_{p} = \left( {p, - 1}\right) ,\;{P}_{{p}^{2}} = \left( {{p}^{2}, - 2}\right) ,\;{P}_{{p}^{3}} = \left( {{p}^{3}, - 3}\right) ,\ldots . \n\\]\n\nThe successive slopes of the sides are\n\... | Yes |
Theorem 1. Let \( K \) be a complete and algebraically closed extension of \( {\mathbf{Q}}_{p} \) and \( f = \sum {a}_{n}{X}^{n} \in K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) a nonzero convergent power series. If \( f \) has a critical radius \( r < {r}_{f} \), then \( f \) has a zero on the critical s... | Proof. The result is trivial if \( r = 0 \), so we assume \( r > 0 \) from now on. Recall that \( \left| {a}_{\mu }\right| {r}^{\mu } = \left| {a}_{v}\right| {r}^{v},{r}^{v - \mu } = \left| {{a}_{\mu }/{a}_{v}}\right| \in \left| {K}^{ \times }\right| \) . Since \( K \) is algebraically closed, there is an element \( a ... | Yes |
Theorem 2. Let \( K \) be a complete extension of \( {\mathbf{Q}}_{p} \) in \( {\Omega }_{p} \) and \( f \in K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack a \) nonzero convergent power series.\n\n(a) If \( f\left( a\right) = 0 \) for some \( a \in {\Omega }_{p},\left| a\right| < {r}_{f} \), then \( a \) is al... | Proof. (a) If \( f \) has a zero \( a \) on the sphere \( \left| x\right| = r \) in \( {\mathbf{C}}_{p} \) (or \( {\Omega }_{p} \) ), then \( r \) is a critical radius, and the preceding theorem shows that \( f \) has \( v - \mu \) roots in \( {\Omega }_{p} \) (counting multiplicities). If \( \sigma \) is a \( K \) -au... | Yes |
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