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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.
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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 \( M = \left\{ {\sum {x}_{i}{g}_{i} : \sum {x}_{i} = 0}\right\} \) . Now \( k\left\lbrack G\right\rbrack /M \approx k \) is a field, so it suffices to check that \( J\left( {k\left\lbrack G\right\rbrack }\right) = M \) . Since \( M \) is a maximal left ideal, \( M \supset J\left( {k\left\lbrack G\right\rbrack }\right) \), so to get \( M \subset J\left( {k\left\lbrack G\right\rbrack }\right) \) it suffices to check that \( M \) consists of quasiregular elements. But \( k\left\lbrack G\right\rbrack \) is also a \( k \) -algebra, and \( M \) has a vector-space basis \( \{ g - 1 : g \in G, g \neq 1\} \), each element of which is nilpotent: \( \left| G\right| = {p}^{n} \Rightarrow {\left( g - 1\right) }^{{p}^{n}} = {g}^{{p}^{n}} - 1 = 0 \) since \( k \) has characteristic \( p \) . A theorem of Wedderburn (see Herstein [32, pp. 56-58]) asserts that if an ideal in a finite-dimensional algebra over a field has a basis consisting of nilpotent elements, then the ideal must consist of nilpotent elements. But nilpotent elements are left quasiregular:\n\n\[ \left( {1 - x + {x}^{2} - {x}^{3} + \cdots }\right) \left( {1 + x}\right) = 1. \]
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Yes
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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\[ \n{\operatorname{Tor}}_{1}\left( {D, B}\right) \approx K/{MK} \n\]\n\nand\n\n\[ \n{\operatorname{Ext}}^{1}\left( {B, D}\right) \approx \operatorname{Hom}\left( {K, D}\right) \n\]
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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\[ \nD \otimes K = \left( {R/M}\right) \otimes K \approx K/{MK} \n\]\n\nand\n\n\[ \nD \otimes F = \left( {R/M}\right) \otimes F \approx F/{MF} \n\]\n\nby Proposition 2.2(b). But \( D \otimes K \rightarrow D \otimes F \) then becomes \( K/{MK} \rightarrow \) \( F/{MF} \), which is zero since \( K \subset {MF} \) (Proposition 9.33). Hence, \( {\operatorname{Tor}}_{1}\left( {D, B}\right) \) \( \approx K/{MK} \) .\n\nAs for \( {\mathrm{{Ext}}}^{1} \), we have from the long exact sequence for Ext:\n\n\[ \n0 \rightarrow \operatorname{Hom}\left( {B, D}\right) \rightarrow \operatorname{Hom}\left( {F, D}\right) \rightarrow \operatorname{Hom}\left( {K, D}\right) \rightarrow {\operatorname{Ext}}^{1}\left( {B, D}\right) \rightarrow 0.\n\]\n\nIf \( f \in \operatorname{Hom}\left( {F, D}\right) \), then for all \( m \in M, f\left( {mx}\right) = {mf}\left( x\right) = 0 \), since \( {mD} = 0 \) . That is, any \( f \in \operatorname{Hom}\left( {F, D}\right) \) restricts to zero on \( {MF} \supset K \), so \( \operatorname{Hom}\left( {F, D}\right) \rightarrow \) \( \operatorname{Hom}\left( {K, D}\right) \) is the zero map and \( {\operatorname{Ext}}^{1}\left( {B, D}\right) \approx \operatorname{Hom}\left( {K, D}\right) \) .
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Yes
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Lemma 9.36 Suppose \( R \) is quasilocal with maximal ideal \( M \) and quotient division ring \( D = R/M \) . Then for any \( B \in {}_{R}\mathbf{M},{\operatorname{Hom}}_{R}\left( {B, D}\right) \approx \) \( {\operatorname{Hom}}_{D}\left( {B/{MB}, D}\right) \) .
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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 \( \operatorname{Hom}\left( {B, D}\right) \) to \( \operatorname{Hom}\left( {{MB}, D}\right) \) is the zero map. Applying \( \operatorname{Hom}\left( {\bullet, D}\right) \) to \( 0 \rightarrow {MB} \rightarrow B \rightarrow B/{MB} \rightarrow 0 \) yields exactness of\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{R}\left( {B/{MB}, D}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {B, D}\right) \overset{0}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {{MB}, D}\right) ,\]\n\nso that \( {\operatorname{Hom}}_{R}\left( {B, D}\right) \approx {\operatorname{Hom}}_{R}\left( {B/{MB}, D}\right) = {\operatorname{Hom}}_{D}\left( {B/{MB}, D}\right) \), since \( B/{MB} \) and \( D \) are \
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Yes
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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 \( 0 \leq n \leq \mathrm{P} - \dim \left( B\right) \), and the dual of \( {\operatorname{Tor}}_{n}\left( {D, B}\right) \) as a left vector space over \( D \) is \( {\operatorname{Ext}}^{n}\left( {B, D}\right) \) . In particular, the left global dimension of \( R \) is equal both to the flat dimension of \( D \) as a right \( R \) -module and the injective dimension of \( D \) as a left \( R \) -module.
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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 \( n \) . The \( n = 0 \) case is just\n\n\[ \n{\operatorname{Tor}}_{0}\left( {D, B}\right) \approx \left( {R/M}\right) \otimes B \approx B/{MB} \neq 0 \n\]\n\nby Proposition 2.2(b) and Nakayama's lemma. Furthermore,\n\n\[ \n{\operatorname{Ext}}^{0}\left( {B, D}\right) \approx {\operatorname{Hom}}_{R}\left( {B, D}\right) \approx {\operatorname{Hom}}_{D}\left( {B/{MB}, D}\right) \n\]\n\nby Lemma 9.36.\n\nFor \( n > 0 \), let \( \pi : F \rightarrow B \) be a projective cover as described in Lemma 9.35. Then by Lemma 9.35, for P-dim \( B \geq 1 \) we get \( K \neq 0 \), so that \( K/{MK} \neq 0 \) by Nakayama’s lemma. But then \( {\operatorname{Tor}}_{1}\left( {D, B}\right) \neq 0 \) and \( {\operatorname{Ext}}^{1}\left( {B, D}\right) \) is its dual by Lemmas 9.35 and 9.36. As for the induction step \( n - 1 \mapsto n \), when \( n > 1 \) observe that\n\n\[ \n{\operatorname{Tor}}_{n}\left( {D, B}\right) \approx {\operatorname{Tor}}_{n - 1}\left( {D, K}\right) \neq 0\text{ for }n - 1 \leq \mathrm{P}\text{-dim }K \n\]\n\nby the long exact sequence for Tor and the induction hypothesis. But P-dim \( K = \) P-dim \( B - 1 \) by Exercise 2, Chapter 4. Furthermore, \( {\operatorname{Ext}}^{n}\left( {B, D}\right) \) \( \approx {\operatorname{Ext}}^{n - 1}\left( {K, D}\right) \) is the dual of \( {\operatorname{Tor}}_{n - 1}\left( {D, K}\right) \) by the long exact sequence for Ext and the induction hypothesis.\n\nFinally, since \( {\operatorname{Tor}}_{n}\left( {D, B}\right) \neq 0 \) for all \( n \leq \mathrm{P} \) -dim \( B \), the flat dimension of \( D \) as a right \( R \) -module must be \( \geq \mathrm{P} \) -dim \( B \) . Letting \( B \) float and using the global dimension theorem, the flat dimension of \( D \) as a right \( R \) -module is \( \geq \) LG-dim \( R = \mathrm{W} \) -dim \( R \) . The situation for the injective dimension of \( D \) as a left \( R \) -module is similar.
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Yes
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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 \( {\operatorname{Tor}}_{n}\left( {D, D}\right) \) and \( {\operatorname{Ext}}^{n}\left( {D, D}\right) \) are nonzero for \( 0 \leq n \leq \mathrm{{LG}} \) -dim \( R \) .
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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 from Theorem 9.37.
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No
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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.
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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 without loss of generality we may assume that \( \mathrm{P} \) -dim \( B = 1 \) . Let \( \pi : F \rightarrow B \) be a projective cover as described in Proposition 9.33 with \( K = \ker \pi \) . Then P-dim \( B \leq 1 \Rightarrow K \) is projective by the projective dimension theorem, while P-dim \( B > 0 \Rightarrow K \neq 0 \) . But \( K \subset {MF} \), so \( {xK} \subset x{MF} = 0 \) . But \( {xK} \neq 0 \) when \( K \) is free and nonzero since \( 0 \neq x \in {xR} \) , a contradiction since all projective left \( R \) -modules are free (Corollary 9.34).
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Yes
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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\[ \text{to a}{C}^{\infty }\text{function near 0}\} \text{.} \]
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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\[ f\left( {t, x, y}\right) = {y}^{n}\left( {g\left( {t, x}\right) + {y\phi }\left( {t, x, y}\right) }\right) \]\n\nwith \( g\left( {t, x}\right) \neq 0 \) . There is a lowest power \( m \) of \( x \) that appears in \( g\left( {t, x}\right) \) so that one may write\n\n\[ g\left( {t, x}\right) = {x}^{m}\left( {h\left( t\right) + {x\psi }\left( {t, x}\right) }\right) \]\n\nwith \( h\left( t\right) \neq 0 \) . Finally, \( h\left( t\right) = {t}^{l}\left( {a + {t\theta }\left( t\right) }\right) \) for some \( l \) and \( a \neq 0 \) . Thus,\n\n\[ f\left( {t, x, y}\right) = {y}^{n}\left( {g\left( {t, x}\right) + {y\phi }\left( {t, x, y}\right) }\right) \]\n\n\[ = {y}^{n}\left( {{x}^{m}\left( {h\left( t\right) + {x\phi }\left( {t, x}\right) }\right) + {y\phi }\left( {t, x, y}\right) }\right) \]\n\n\[ = {y}^{n}\left( {{x}^{m}\left( {{t}^{l}\left( {a + {t\theta }\left( t\right) }\right) + {x\psi }\left( {t, x}\right) }\right) + {y\phi }\left( {t, x, y}\right) }\right) \]\n\n\[ = {y}^{n}{x}^{m}\left( {{t}^{l}\left( {a + {t\theta }\left( t\right) }\right) + {x\psi }\left( {t, x}\right) + {x}^{-m}{y\phi }\left( {t, x, y}\right) }\right) \]\n\n\[ = {y}^{n}{x}^{m}{t}^{l}\left( {a + {t\theta }\left( t\right) + {t}^{-l}{x\psi }\left( {t, x}\right) + {t}^{-l}{x}^{-m}{y\phi }\left( {t, x, y}\right) }\right) \]\n\n\[ = {y}^{n}{x}^{m}{t}^{l} \cdot U\left( {t, x, y}\right) \]\n\nwhere \( U\left( {t, x, y}\right) \) is a unit in \( R \) . Hence, by dividing, we get that any nonzero \( f\left( {t, x, y}\right) \in \mathbb{R}\left( {t, x, y}\right) \) can be written as \( f\left( {t, x, y}\right) = {t}^{l}{x}^{m}{y}^{n} \cdot U\left( {t, x, y}\right) \) ; \( l, m, n \in \mathbb{Z} \) and \( U\left( {t, x, y}\right) \) a unit in \( R \) . Thus, \( f\left( {t, x, y}\right) \in R \) if and only if one of the following holds:\n\n\[ \text{i)}n > 0\text{, or} \]\n\nii) \( n = 0 \) and \( m > 0 \), or\n\niii) \( n = m = 0 \) and \( l \geq 0 \) .
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Yes
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Proposition. The ring \( {\mathbf{Z}}_{p} \) is a principal ideal domain. More precisely, its ideals are the principal ideals \( \{ 0\} \) and \( {p}^{k}{\mathbf{Z}}_{p}\left( {k \in \mathbf{N}}\right) \) .
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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 \( \left( {p}^{k}\right) = {p}^{k}{\mathbf{Z}}_{p} \subset I \) . Conversely, for any \( b \in I \) let \( w = v\left( b\right) \geq k \) and write\n\n\[ b = {p}^{w}{u}^{\prime } = {p}^{k} \cdot {p}^{w - k}{u}^{\prime } \in {p}^{k}{\mathbf{Z}}_{p}. \]\n\nThis shows that \( I \subset {p}^{k}{\mathbf{Z}}_{p} \) .
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Yes
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Proposition 1. The discrete subgroups of \( \mathbf{R} \) are the subgroups\n\n\[ a\mathbf{Z}\;\left( {0 \leq a \in \mathbf{R}}\right) . \]
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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, and there is a smallest positive element \( a \in H \) . Obviously, \( \mathbf{Z} \cdot a \subset H \) . In fact, this inclusion is an equality. Indeed, if we take any \( b \in H \) and assume (without loss of generality) \( b > 0 \), we can write\n\n\[ b = {ma} + r\;\left( {m \in \mathbf{N},0 \leq r < a}\right) \]\n\n(take for \( m \) the integral part of \( b/a \) ). Since \( r = b - {ma} \in H \) and \( 0 \leq r < a \) , we must have \( r = 0 \) by construction. This proves \( b = {ma} \in \mathbf{Z} \cdot a \), and hence the reverse inclusion \( H \subset \mathbf{Z} \cdot a \) .
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Yes
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Proposition 2. Any nondiscrete subgroup of \( \mathbf{R} \) is dense.
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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 subgroup, we must also have \( \mathbf{Z} \cdot {\varepsilon }_{n} \subset H \) (for all \( n \geq 0 \) ), and the subgroup \( H \) is dense in \( \mathbf{R} \) .
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No
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Corollary 1. The quotient of \( {\mathbf{Z}}_{p} \) by a closed subgroup \( H \neq \{ 0\} \) is discrete.
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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 \( H \supset \mathbf{N} \cdot h \) . In particular, \( H \ni {p}^{n}h \rightarrow 0\left( {n \rightarrow \infty }\right) \) . Since the elements \( {p}^{n}h \) are distinct, \( H \) is not discrete.
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No
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Corollary 2. The only discrete subgroup of the additive group \( {\mathbf{Z}}_{p} \) is the trivial subgroup \( \{ 0\} \) .
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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 \( H \supset \mathbf{N} \cdot h \) . In particular, \( H \ni {p}^{n}h \rightarrow 0\left( {n \rightarrow \infty }\right) \) . Since the elements \( {p}^{n}h \) are distinct, \( H \) is not discrete.
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Yes
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Corollary 1. The topological group \( {\mathbf{Z}}_{p} \) is a completion of the additive group \( \mathbf{Z} \) equipped with the induced topology.
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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 \( x \) .
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No
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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.
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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 \( y \neq 0 \) is fixed, \( \left| k\right| < \left| y\right| /2 \), and \( c = \max \left( {\left| x\right| ,\left| y\right| }\right) \),\n\n\[ \left| {\frac{x + h}{y + k} - \frac{x}{y}}\right| < {2c}\frac{\left| h\right| + \left| k\right| }{\left| {y}^{2}\right| } \rightarrow 0\;\left( {\left| h\right| ,\left| k\right| \rightarrow 0}\right) . \]\n\nThis proves that \( K \) is a topological field.
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Yes
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Proposition 2. Let \( K \) be a valued field. Then the completion \( \widehat{K} \) of \( K \) is again a valued field.
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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 \( \widehat{K} \) . This means that the sequence \( \left| {x}_{n}\right| \) does not converge to zero. There is a positive \( \varepsilon > 0 \) together with an index \( N \) such that \( \left| {x}_{n}\right| > \varepsilon \) for all \( n \geq N \) . The sequence \( {\left( 1/{x}_{n}\right) }_{n \geq N} \) is also a Cauchy sequence\n\n\[ \left| {\frac{1}{{x}_{n}} - \frac{1}{{x}_{m}}}\right| = \left| \frac{{x}_{n} - {x}_{m}}{{x}_{n}{x}_{m}}\right| \leq {\varepsilon }^{-2}\left| {{x}_{n} - {x}_{m}}\right| \rightarrow 0\;\left( {n, m \rightarrow \infty }\right) .\n\]\n\nThe sequence \( {\left( 1/{x}_{n}\right) }_{n \geq N} \) (completed with 1 ’s for \( n < N \) ) defines an inverse of the original sequence \( \left( {x}_{n}\right) \) in the completion \( \widehat{K} \) .
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Yes
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Proposition 1. A projective limit of nonempty compact spaces is nonempty and compact.
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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\[ \n{K}_{n}^{\prime } = {\varphi }_{n}\left( {K}_{n + 1}\right) \supset {K}_{n}^{\prime \prime } = {\varphi }_{n}\left( {{\varphi }_{n + 1}\left( {K}_{n + 2}\right) }\right) \left( { = {\varphi }_{n}\left( {K}_{n + 1}^{\prime }\right) }\right) \supset \cdots .\n\] \n\nThese subsets are compact and nonempty. Their intersection \( {L}_{n} \) is not empty in the compact space \( {K}_{n} \) . Moreover, \( {\varphi }_{n}\left( {L}_{n + 1}\right) = {L}_{n} \), and the restriction of the maps \( {\varphi }_{n} \) to the subsets \( {L}_{n} \) leads to a projective system having surjective transition mappings. By the corollary in (4.3), this system has a nonempty limit (with surjective projections). Since \( \mathop{\lim }\limits_{ \leftarrow }{L}_{n} \subset \mathop{\lim }\limits_{ \leftarrow }{K}_{n} \), the proof is complete.
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Yes
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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 \( {E}_{n} \) .
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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 \( {x}_{n} \in {V}_{n} \) and \( {x}_{n - 1} \in {V}_{n - 1} \) means that\n\n\[ \n{\psi }_{n}\left( x\right) = {x}_{n} \in {V}_{n} \cap {\varphi }_{n - 1}^{-1}\left( {V}_{n - 1}\right) .\n\]\n\nCall \( {U}_{n} \) the open set \( {V}_{n} \cap {\varphi }_{n - 1}^{-1}\left( {V}_{n - 1}\right) \) of \( {E}_{n} \) . Then the preceding condition is still equivalent to \( x \in {\psi }_{n}^{-1}\left( {U}_{n}\right) \) . By induction, one can show that a basic open set in the product - say \( \mathop{\prod }\limits_{{n \leq N}}{V}_{n} \times \mathop{\prod }\limits_{{n > N}}{E}_{n} \) - has an intersection with the projective limit of the form \( {\psi }_{N}^{-1}\left( {U}_{N}\right) \) for some open set \( {U}_{N} \subset {E}_{N} \) .
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Yes
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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\[ \bar{A} = \mathop{\bigcap }\limits_{{n \geq 0}}{\psi }_{n}^{-1}\left( \overline{{\psi }_{n}\left( A\right) }\right) \]
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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 \( b \) . Without loss of generality, we can assume that \( V \) is of the form \( {\psi }_{n}^{-1}\left( {U}_{n}\right) \) for some open set \( {U}_{n} \subset {E}_{n} \) . Hence \( {\psi }_{n}\left( b\right) \subset {U}_{n} \) . Since by assumption \( b \in {\psi }_{n}^{-1}\left( \overline{{\psi }_{n}\left( A\right) }\right) \), we have \( {\psi }_{n}\left( b\right) \in \overline{{\psi }_{n}\left( A\right) } \), and the open set \( {U}_{n} \) containing \( b \) must meet \( {\psi }_{n}\left( A\right) \) : There is a point \( a \in A \) with \( {\psi }_{n}\left( a\right) \in {U}_{n} \) . This shows that\n\n\[ a \in A \cap {\psi }_{n}^{-1}\left( {U}_{n}\right) \]\n\nIn particular, this intersection is nonempty, and the given neighborhood of \( b \) indeed meets \( A \) .
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Yes
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Proposition 1. When \( p \) is an odd prime, the group of roots of unity in the field \( {\mathbf{Q}}_{p} \) is \( {\mu }_{p - 1} \) .
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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 unity, say \( \zeta \) has order \( n \geq 1 \) ,\n\n\[ \n{\zeta }^{n} = {\left( 1 + pt\right) }^{n} = 1 \n\]\n\nHence \( {npt} + \left( \begin{array}{l} n \\ 2 \end{array}\right) {p}^{2}{t}^{2} + \cdots + {p}^{n}{t}^{n} = 0 \), or\n\n\[ \nt\left( {n + \left( \begin{array}{l} n \\ 2 \end{array}\right) {pt} + \cdots + {p}^{n - 1}{t}^{n - 1}}\right) = 0. \n\]\n\nThis shows that \( t = 0 \) (when \( p \nmid n \) ) or \( p \mid n \) . In the second case, replace \( \zeta \) by \( {\zeta }^{p} \) and \( n \) by \( n/p \) : Starting the same computation, we see that \( t = 0 \) or \( {p}^{2} \mid n \) (original \( n) \), and so on. Finally, we are reduced to the case \( n = p \) . In this case, the above equation is simply\n\n\[ \nt\left( {p + \left( \begin{array}{l} p \\ 2 \end{array}\right) {pt} + \cdots + {p}^{p - 1}{t}^{p - 1}}\right) = 0, \n\]\n\nand since \( p \geq 3 \) ,\n\n\[ \np + \left( \begin{array}{l} p \\ 2 \end{array}\right) {pt} + \cdots + {p}^{p - 1}{t}^{p - 1} = p + {p}^{2}\left( \cdots \right) \neq 0. \n\]\n\nThis proves that \( t = 0 \) in all cases and \( \zeta = 1 \) .
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Yes
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Proposition 2. The group of roots of unity in the field \( {\mathbf{Q}}_{2} \) is \( {\mu }_{2} = \{ \pm 1\} \) .
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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 \( {\mathbf{Z}}_{2}^{ \times } \) have order a power of 2. But -1 is not a square of \( {\mathbf{Z}}_{2}^{ \times } \) (6.6), and there is no fourth root of 1 in \( {\mathbf{Q}}_{2} \) .
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Yes
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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 \( m \) .
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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\[ \n{\varphi }_{n} : \mathbf{R}/{p}^{n + 1}\mathbf{Z} \rightarrow \mathbf{R}/{p}^{n}\mathbf{Z} \n\]\n\nhave a kernel of order \( p \) prime to \( m \) (by assumption), they induce isomorphisms \( {C}_{m}^{n + 1} \rightarrow {C}_{m}^{n} \) . The projective limit of this constant sequence is the cyclic subgroup \( {C}_{m} \subset {\mathbf{S}}_{p} \) . To prove uniqueness, let us consider any homomorphism \( \sigma : \mathbf{Z}/m\mathbf{Z} \rightarrow \) \( {\mathbf{S}}_{p} \) . The composite\n\n\[ \n{\psi }_{n} \circ \sigma : \mathbf{Z}/m\mathbf{Z} \rightarrow {\mathbf{S}}_{p} \rightarrow \mathbf{R}/{p}^{n}\mathbf{Z} \n\]\n\nhas an image in the unique cyclic subgroup \( {C}_{m}^{n} \) of the circle \( \mathbf{R}/{p}^{n}\mathbf{Z} \) . Hence \( \sigma \) has an image in \( {C}_{m} \), and this concludes the proof.
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Yes
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Proposition 2. The p-adic solenoid \( {S}_{p} \) has no p-torsion.
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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\[ \n{\varphi }_{n} \circ {\psi }_{n + 1} \circ \sigma : \mathbf{Z}/p\mathbf{Z} \rightarrow {\mathbf{S}}_{p} \rightarrow \mathbf{R}/{p}^{n + 1}\mathbf{Z} \rightarrow \mathbf{R}/{p}^{n}\mathbf{Z}\n\]\n\nare trivial. Indeed, the composite\n\n\[ \n{\psi }_{n + 1} \circ \sigma : \mathbf{Z}/p\mathbf{Z} \rightarrow {\mathbf{S}}_{p} \rightarrow \mathbf{R}/{p}^{n + 1}\mathbf{Z}\n\]\n\nmust have an image in the unique cyclic subgroup of order \( p \) of the circle \( \mathbf{R}/{p}^{n + 1}\mathbf{Z} \) , and this subgroup is precisely the kernel of the connecting homomorphism \( {\varphi }_{n} \) and \( {\psi }_{n} \circ \sigma = {\varphi }_{n}\left( {{\psi }_{n + 1} \circ \sigma }\right) \) . Consequently, there is no element of order \( p \) in \( {\mathbf{S}}_{p} \) (and a fortiori no element of order \( {p}^{k} \) for \( k \geq 1 \) in \( {\mathbf{S}}_{p} \) ).
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Yes
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Corollary 1. The solenoid can also be viewed as a quotient of \( \\mathbf{R} \\times {\\mathbf{Z}}_{p} \) by the discrete subgroup \( {\\Delta }_{\\mathbf{Z}} = \\{ \\left( {m, - m}\\right) : m \\in \\mathbf{Z}\\} \)
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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\[ {f}^{\\prime } : \\left( {\\mathbf{R} \\times {\\mathbf{Z}}_{p}}\\right) /\\ker {f}^{\\prime } \\cong {\\mathbf{Z}}_{p}. \]\n\nBut\n\n\[ \\ker {f}^{\\prime } = \\left( {\\ker f}\\right) \\cap \\left( {\\mathbf{R} \\times {\\mathbf{Z}}_{p}}\\right) = {\\Delta }_{\\mathbf{Z}} = \\{ \\left( {m, - m}\\right) : m \\in \\mathbf{Z}\\} . \]
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Yes
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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) .
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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.
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Yes
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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
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\[ \left( {t, x}\right) = \left( {t - \left\lbrack t\right\rbrack, x + \left\lbrack t\right\rbrack }\right) \mapsto \left( {0, x + \left\lbrack t\right\rbrack }\right) \] furnishes by restriction a continuous retraction of \( {\psi }^{-1}\left( \left\lbrack {0,\eta }\right\rbrack \right) \subset {\mathbf{S}}_{p} \) onto the neutral fiber \( {\mathbf{Z}}_{p} \subset {\mathbf{S}}_{p}\left( {0 < \eta < 1}\right) \) .
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No
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Proposition 2. The solenoid \( {\mathbf{S}}_{p} \) is an indecomposable compact connected topological space.
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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\[ K = \mathop{\bigcap }\limits_{{n \geq 1}}{\psi }_{n}^{-1}\left( {{\psi }_{n}\left( K\right) }\right) \]\n\nfor every compact set \( K \), the assumption \( A \neq {\mathbf{S}}_{p} \) leads to \( {\psi }_{n}\left( A\right) \neq \mathbf{R}/{p}^{n}\mathbf{Z} \) for some integer \( n = {n}_{0} \) and hence also for all integers \( n \geq {n}_{0} \) (the transition maps \( {\varphi }_{m} \) are surjective). It will suffice to show \( {\psi }_{n}\left( B\right) = \mathbf{R}/{p}^{n}\mathbf{Z} \) for all \( n \geq {n}_{0} \) . Take such an \( n \) and an element \( b \in B \) . Then\n\n\[ {\varphi }_{n}^{-1}\left( b\right) \subset \mathbf{R}/{p}^{n + 1}\mathbf{Z} \]\n\nhas cardinality \( p \geq 2 \), and the restriction of \( {\varphi }_{n} \) to the connected set\n\n\[ C = {\varphi }_{n}^{-1}{\psi }_{n}\left( B\right) = {\psi }_{n + 1}\left( B\right) \]\n\nis not injective. The proof will be complete as soon as the following statement (in which the situation and notation are simplified) is established.\n\nLet \( a > 1 \) be any integer, \( \varphi : \mathbf{R}/a\mathbf{Z} \rightarrow \mathbf{R}/\mathbf{Z} \) the canonical projection, and \( {Ca} \) connected subset of \( \mathbf{R}/a\mathbf{Z} \) containing two distinct points \( s \neq t \) with \( \varphi \left( s\right) = \varphi \left( t\right) \) . Then \( \varphi \left( C\right) = \mathbf{R}/\mathbf{Z} \) .\n\nIn terms of the restriction \( {\left. \varphi \right| }_{C} \) of the map \( \varphi \) to \( C \), we have to prove\n\n\[ {\left. \varphi \right| }_{C}\text{ not injective } \Rightarrow {\left. \varphi \right| }_{C}\text{ surjective } \]\n\nunder the stated assumptions. It is obviously enough to do so when \( C \neq \mathbf{R}/a\mathbf{Z} \) . In this case, take a point \( P \notin C \subset \mathbf{R}/a\mathbf{Z} \) and consider a stereographic projection from the point \( P \) of the circle \( \mathbf{R}/a\mathbf{Z} \) onto a line \( \mathbf{R} \) . This is a homeomorphism\n\n\[ f : \mathbf{R}/a\mathbf{Z} - \{ P\} \overset{ \sim }{ \rightarrow }\mathbf{R}. \]\n\nThe image \( f\left( C\right) \) of the subset \( C \) is a connected subset of the real line containing the images of two different congruent points mod \( \mathbf{Z} \) . Since any connected set in the real line is an interval, this proves that \( f\left( C\right) \) contains the whole interval \( J \) linking these two different congruent points. Hence \( C \) contains a whole arc \( I \) of the circle having image \( \varphi \left( I\right) = \mathbf{R}/\mathbf{Z} \) .
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Yes
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Lemma 1. (a) Any point of a ball is a center of the ball.
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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 \( {B}_{ < r}\left( a\right) = {B}_{ < r}\left( b\right) \) . The case of a dressed ball is similar.
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Yes
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Lemma 3. (a) The spheres \( {S}_{r}\left( a\right) \left( {r > 0}\right) \) are both open and closed.
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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.
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No
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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.
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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 \( m \geq 0 \) .
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No
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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\[ f\left( X\right) = {X}^{n} + {a}_{n - 1}{X}^{n - 1} + \cdots + {a}_{0}, \]\n\n\[ \operatorname{ord}\left( {a}_{i}\right) \geq 1\left( {0 \leq i \leq n - 1}\right) ,\;\operatorname{ord}\left( {a}_{0}\right) = 1. \]\n\nThen \( f \) is irreducible in the rings \( {\mathbf{Z}}_{p}\left\lbrack X\right\rbrack \) and \( {\mathbf{Q}}_{p}\left\lbrack X\right\rbrack \) .
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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\[ g = {b}_{\ell }{X}^{\ell } + \cdots + {b}_{0},\;h = {c}_{m}{X}^{m} + \cdots + {c}_{0}. \]\n\nHence\n\n\[ \ell + m = n,\;{b}_{\ell }{c}_{m} = 1,\;{b}_{0}{c}_{0} = {a}_{0}. \]\n\nSince \( {a}_{0} \) is not divisible by \( {p}^{2}, p \) can divide only one of the two coefficients \( {b}_{0} \) and \( {c}_{0} \) . Without loss of generality we can assume that \( p \) divides \( {c}_{0} \) but \( p \) does not divide \( {b}_{0} \) . Consider now all these polynomials mod \( p \) . By assumption \( \widetilde{f} = {X}^{n} \) is a monomial, so that its factorization \( \widetilde{f} = \widetilde{g} \cdot \widetilde{h} \) must be a product of monomials and \( \widetilde{h} = \widetilde{{c}_{0}} \) is a constant. Considering that \( {b}_{\ell }{c}_{m} = 1 \), the only possibility now is \( m = 0 \) and a trivial factorization.\n\nThe preceding argument \( {\;\operatorname{mod}\;p} \) can be made directly on the coefficients. Let \( r \geq 1 \) be the smallest power of \( X \) in \( h \) having a coefficient not divisible by \( p \) :\n\n\[ p\text{does not divide}{c}_{r}\text{but}p\text{divides}{c}_{r - 1},{c}_{r - 2},\ldots ,{c}_{0}.\text{ }\]\n\nThe coefficient of \( {X}^{r} \) in the product of \( g \) and \( h \) is\n\n\[ {a}_{r} = {b}_{0}{c}_{r} + {b}_{1}{c}_{r - 1} + {b}_{2}{c}_{r - 2} + \cdots = {b}_{0}{c}_{r} + p\left( \cdots \right) . \]\n\nSince \( {b}_{0}{c}_{r} \) is not divisible by \( p \), the preceding equality shows that \( p \) does not divide \( {a}_{r} \) either. By assumption, this shows that \( r = n \) . Summing up,\n\n\[ n = m + \ell \geq m \geq r = n \]\n\nimplies \( m = n \) and \( \ell = 0 \) . The factorization \( g \cdot h \) of \( f \) is necessarily trivial.
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Yes
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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.
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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 \( {\left( {\pi }^{i}\right) }_{0 \leq i < e} \) generate \( K \) and \( K = {\mathbf{Q}}_{p}\left\lbrack \pi \right\rbrack \). The irreducible polynomial of this element can be factored (in a Galois extension of \( {\mathbf{Q}}_{p} \) containing \( K \) ) as\n\n\[ f\left( X\right) = \mathop{\prod }\limits_{\sigma }\left( {X - {\pi }^{\sigma }}\right) = {X}^{e} + \mathop{\sum }\limits_{{0 < i < e}}{a}_{i}{X}^{i} \pm \mathop{\prod }\limits_{\sigma }{\pi }^{\sigma }.\]\n\nThe constant term has absolute value \( \left| {{\Pi }_{\sigma }{\pi }^{\sigma }}\right| = {\left| \pi \right| }^{e} = \left| p\right| \) (by (3.3) all automor-phisms \( \sigma \) are isometric), whereas the intermediate coefficients \( {a}_{i} \) satisfy \( \left| {a}_{i}\right| < 1 \)\n\n(each is divisible by one \( {\pi }^{\sigma } \) at least, and \( {a}_{i} \in {\mathbf{Z}}_{p} \)). Hence these intermediate coefficients are in \( p{\mathbf{Z}}_{p} \) as required: \( f \) is an Eisenstein polynomial.
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Yes
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Proposition 1. Let \( K \) be any ultrametric extension of \( {\mathbf{Q}}_{p} \) . Then\n\n\[ \n{\mu }_{{p}^{\infty }}\left( K\right) = \mu \left( K\right) \cap \left( {1 + M}\right) .\n\]
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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\[ \n{\zeta }^{{p}^{f}} = 1 \Rightarrow {\widetilde{\zeta }}^{{p}^{f}} = \widetilde{1} \in k \Rightarrow \widetilde{\zeta } = \widetilde{1} \Rightarrow \zeta \in 1 + M,\n\]\n\nsince the field \( k \) has characteristic \( p \) . Conversely, if \( \zeta \in 1 + M \) has order \( n > 1 \) , write \( \zeta = 1 + \xi \) with \( 0 \neq \left| \xi \right| < 1 \) . Then\n\n\[ \n1 = {\left( 1 + \xi \right) }^{n} = 1 + {n\xi } + \cdots + {\xi }^{n} = 1 + \xi \left( {n + {\xi \alpha }}\right)\n\]\n\nimplies \( n + {\xi \alpha } = 0 \), and\n\n\[ \n\left| n\right| = \left| {\xi \alpha }\right| \leq \left| \xi \right| < 1\n\]\n\nimplies \( p \mid n \) . If \( n \neq p \), we can replace \( \zeta \) by \( {\zeta }^{p} \), which has order \( n/p > 1 \), and iterate the procedure. Eventually, we see that \( n \) is a power of \( p \) .
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Yes
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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\[\n\\left( 1\\right) \\rightarrow {\\mu }_{{p}^{\\infty }}\\left( K\\right) \\rightarrow \\mu \\left( K\\right) \\rightarrow {k}^{ \\times } \\rightarrow \\left( 1\\right) .\n\]\n\nIf the residue field is finite, say \( f = \\left\\lbrack {k : {\\mathbf{F}}_{p}}\\right\\rbrack < \\infty \), then the cyclic group \( {\\mu }_{\\left( p\\right) }\\left( K\\right) \) has order \( {p}^{f} - 1 \) .
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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 \( {\\mu }_{\\left( p\\right) }\\left( K\\right) \\cong {k}^{ \\times } \) . By the preceding proposition, the reduction map induces an isomorphism of \( {\\mu }_{\\left( p\\right) }\\left( K\\right) \) into \( {k}^{ \\times } \) . We have to prove that it is surjective. Let \( \\alpha \\in {k}^{ \\times } \) and replace \( k \) by the finite field \( {\\mathbf{F}}_{p}\\left( \\alpha \\right) \\cong {\\mathbf{F}}_{q} \) so that \( \\alpha \) is a root of unity of order prime to \( p \), dividing \( m = q - 1 = \\# \\left( {k}^{ \\times }\\right) \) . Choose an element \( a \\in A \) in the coset \( \\alpha \\left( {\\;\\operatorname{mod}\\;M}\\right) \) and consider the solutions \( x \) of the following problem:\n\n\[\n{X}^{m} - 1 = 0\\text{ with }x \\equiv a\\;\\left( {\\;\\operatorname{mod}\\;M}\\right) \\text{ (i.e.,}\\varepsilon \\left( x\\right) = \\alpha \\text{). }\n\]\n\nSince \( m \) is prime to \( p \), and \( K \) is complete, Hensel’s lemma (1.4) can be applied, and this furnishes an element \( x \) in \( {K}^{ \\times } \) with \( {x}^{m} = 1 \) ; hence \( x \\in {\\mu }_{\\left( p\\right) }\\left( K\\right) \) and \( \\varepsilon \\left( x\\right) = \\varepsilon \\left( a\\right) = \\alpha . \n\nThis proves that - when the residue field \( k \) is algebraic - the restriction of the reduction mod \( M \) is an isomorphism \( {\\mu }_{\\left( p\\right) }\\left( K\\right) \\overset{ \\sim }{ \\rightarrow }{k}^{ \\times } \) .
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Yes
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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\[ \n\# \left( {{\mu }_{{p}^{\infty }}\left( K\right) }\right) \leq \frac{e}{1 - 1/p}.\n\]
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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 \( {p}^{t} \leq {ep}/\left( {p - 1}\right) \), and\n\n\[ \n\# \left( {{\mu }_{{p}^{\infty }}\left( K\right) }\right) \leq \frac{ep}{p - 1}.\n\]\n\nObserve that the result of this corollary is valid for any valued field \( K \) of characteristic 0, provided that its absolute value extends the \( p \) -adic one on \( \mathbf{Q} \) .
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Yes
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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\[ \mu \left( {\mathbf{Q}}_{2}\right) = {\mu }_{2}\left( {\mathbf{Q}}_{2}\right) = \{ \pm 1\} \]
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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 determined by the previous theory, and a diagram summarizes the situation. \n\nThe element \( \pi = {\zeta }_{p} - 1 \) has absolute value \( \left| \pi \right| = {\left| p\right| }^{1/\left( {p - 1}\right) } \) generating the group of values \( \left| {K}^{ \times }\right| : P = {\pi R} \subset R \subset K = {\mathbf{Q}}_{p}\left( {\zeta }_{p}\right) \) . Similarly, the element \( {\pi }^{\prime } = {\zeta }_{{p}^{2}} - 1 \) has absolute value \( \left| {\pi }^{\prime }\right| = {\left| p\right| }^{1/p\left( {p - 1}\right) } \) generating the group of values \( \left| {K}^{\prime \times }\right| \) :\n\n\[ {P}^{\prime } = {\pi }^{\prime }{R}^{\prime } \subset {R}^{\prime } \subset {K}^{\prime } = {\mathbf{Q}}_{p}\left( {\zeta }_{{p}^{2}}\right) . \]
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No
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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\[ \n{a}^{n} \rightarrow 0\text{ in }K \Leftrightarrow m\left( a\right) < 1.\n\]
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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 \( {B}_{r} - V \) is positive; for \( 0 < {r}^{\prime \prime } < {r}^{\prime } \) it is clear that \( {B}_{{r}^{\prime \prime }} \subset V \) .
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Yes
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Corollary 2. Any discrete subfield of \( K \) is finite.
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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 \( F \subset {B}_{1} \) . But we know that \( F \) is closed (I.3.2). Thus \( F \) is compact and discrete, hence finite.
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Yes
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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.
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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\[ \n\forall \varepsilon > 0\exists U\text{ neighborhood of }0\text{ in }V\text{ such that }{av} \in U \Rightarrow \left| a\right| < \varepsilon .\n\]\n\nWe proceed as follows. If \( \varepsilon > 0 \) is chosen, we take \( b \in K \) with \( 0 < \left| b\right| \leq \varepsilon \) and a balanced neighborhood \( U \) of 0 in \( V \) such that \( U ∌ {bv} \neq 0 \) (this is possible, since we assume that \( V \) is Hausdorff). Now, if \( {av} \in U \), then\n\n\[ \n{bv} = \frac{b}{a} \cdot \underset{ \in U}{\underbrace{aa}} \notin U\underset{U\text{ balanced }}{ \Rightarrow }\left| \frac{b}{a}\right| > 1 \Rightarrow \left| a\right| < \left| b\right| \leq \varepsilon .\n\]
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Yes
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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\[ \left( {\lambda }_{i}\right) \mapsto \mathop{\sum }\limits_{i}{\lambda }_{i}{e}_{i} : {K}^{d} \rightarrow V \]\n\nis an isomorphism of topological vector spaces.
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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 the first \( d - 1{e}_{i} \) . By the induction assumption, the space \( W \) is isomorphic to \( {K}^{d - 1} \) and hence complete and closed in \( V \) . The linear form\n\n\[ \varphi : \mathop{\sum }\limits_{i}{\lambda }_{i}{e}_{i} \mapsto {\lambda }_{d}, V \rightarrow K \]\n\nis continuous, since its kernel \( \ker \left( \varphi \right) = W \) is closed. The one-to-one linear map\n\n\[ {K}^{d} = {K}^{d - 1} \times K\overset{ \approx }{ \rightarrow }W \times K{e}_{d}\overset{\text{ sum }}{ \rightarrow }V \]\n\nis continuous. Its inverse is\n\n\[ x \mapsto \left( {x - \varphi \left( x\right) {e}_{d},\varphi \left( x\right) {e}_{d}}\right) \]\n\nand hence is also continuous.
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Yes
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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 \( r = \mathop{\min }\limits_{{{a}^{\sigma } \neq a}}\left| {{a}^{\sigma } - a}\right| \) . Then every element \( b \in {B}_{ < r}\left( {a;{\mathbf{Q}}_{p}^{a}}\right) \) generates (over \( K \) ) an extension containing \( K\left( a\right) \) .
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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 estimate the distance of \( a \) to \( b \) as follows:\n\n\[ \left| {b - {a}^{\sigma }}\right| = \left| {\left( b - a\right) }^{\sigma }\right| = \left| {b - a}\right| \]\n\n\[ \left| {a - {a}^{\sigma }}\right| \leq \max \left( {\left| {a - b}\right| ,\left| {b - {a}^{\sigma }}\right| }\right) = \left| {b - a}\right| . \]\n\nThis shows that\n\n\[ \left| {b - a}\right| \geq \left| {a - {a}^{\sigma }}\right| \geq r. \]\n\nHence if \( b \in {B}_{ < r}\left( a\right) \), namely \( \left| {b - a}\right| < r \), we have\n\n\[ a \in K\left( b\right) ,\;K\left( a\right) \subset K\left( b\right) . \]
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Yes
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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 \( n \) ). There is a positive \( \varepsilon \) such that any monic polynomial \( g \in K\left\lbrack X\right\rbrack \) of degree \( n \) with \( \parallel g - f\parallel < \varepsilon \) has a root \( b \in K\left( a\right) \) also generating this field: \( K\left( b\right) = K\left( a\right) \) .
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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\[ \n\prod \left( {a - {b}_{i}}\right) = g\left( a\right) = g\left( a\right) - f\left( a\right) .\n\]\n\nWith \( M = \mathop{\max }\limits_{{0 \leq i \leq n}}\left( {\left| a\right| }^{i}\right) = \max \left( {1,{\left| a\right| }^{n}}\right) \) we can estimate\n\n\[ \n\prod \left| {a - {b}_{i}}\right| = \left| {g\left( a\right) - f\left( a\right) }\right| \leq \parallel g - f\parallel \cdot M,\n\]\n\nhence for one index \( i \) at least,\n\n\[ \n\left| {a - {b}_{i}}\right| \leq \parallel g - f{\parallel }^{1/n} \cdot {M}^{1/n}.\n\]\n\nBy the preceding theorem, if \( \varepsilon > 0 \) is chosen small enough, then \( \parallel g - f\parallel < \varepsilon \) will imply \( K\left( {b}_{i}\right) \supset K\left( a\right) \) for some \( i \) . But the degree of \( {b}_{i} \) is less than or equal to \( n \), since it is a root of the \( n \) th degree polynomial \( g \in K\left\lbrack X\right\rbrack \) . This proves \( K\left( {b}_{i}\right) = \) \( K\left( a\right) \) .
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Yes
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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 \( {g}_{i} \rightarrow f \) (coefficientwise), there is a sequence \( \left( {x}_{i}\right) \) of roots of these polynomials such that \( {x}_{i} \in K\left( a\right) \) for large \( i \) and \( {x}_{i} \rightarrow a \) .
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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\[ \left| {a - {x}_{i}}\right| \leq {\begin{Vmatrix}{g}_{i} - f\end{Vmatrix}}^{1/n} \cdot {M}^{1/n} \]\n\nshows that \( \left| {a - {x}_{i}}\right| \rightarrow 0 \), and the convergence \( {x}_{i} \rightarrow a \) in \( K\left( a\right) \) follows.
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Yes
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Corollary 2. The algebraic closure \( {\mathbf{Q}}_{p}^{a} \) of \( {\mathbf{Q}}_{p} \) is a separable metric space.
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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 \( {g}_{n} \rightarrow f \), the continuity principle for the roots shows that \( a \) is a limit of roots \( {x}_{n} \) of the polynomials \( {g}_{n} \). This shows that the algebraic closure of \( \mathbf{Q} \) is dense in \( {\mathbf{Q}}_{p}^{a} \). But this algebraic closure is a countable field since the set of polynomials of fixed degree with coefficients in the countable field \( \mathbf{Q} \) is countable.
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Yes
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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 \( A \) is not Noetherian.
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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 \( M = {\\pi A} \) is principal. In the second case, for each \( x \\in M \), namely \( \\left| x\\right| < 1 \), we can find an element \( y \) such that \( \\left| x\\right| < \\left| y\\right| < 1 \), so that\n\n\[ x = y \\cdot \\left( {x/y}\\right) \\in {M}^{2}. \]\n\nSince \( y \) and \( x/y \) belong to \( M \), this shows that \( x \\in {M}^{2} \), and we have proved \( M = {M}^{2} \) . In this last case, the subgroup \( \\Gamma = \\left| {K}^{ \\times }\\right| \) is dense in \( {\\mathbf{R}}_{ > 0} \), and all the ideals\n\n\[ {I}_{r} = {B}_{ \\leq r} = {B}_{ \\leq r}\\left( {0;K}\\right) = \\{ x \\in K : \\left| x\\right| \\leq r\\} \]\n\nfor \( r \\in \\Gamma \\cap \\left( {0,1}\\right) \) are distinct: The ring \( A \) is not Noetherian.
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Yes
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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 \( K \) .
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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\[ {\left| \xi \right| }^{n} > \left| {\mathop{\sum }\limits_{{i < n}}{a}_{i}{\xi }^{i}}\right| \]\n\nand hence\n\n\[ \left| {{\xi }^{n} + \mathop{\sum }\limits_{{i < n}}{a}_{i}{\xi }^{i}}\right| = {\left| \xi \right| }^{n} > 1 \]\n\n\[ {\xi }^{n} + \mathop{\sum }\limits_{{i < n}}{a}_{i}{\xi }^{i} \neq 0 \]\n\nThis proves that any root of a monic polynomial having coefficients \( \left| {a}_{i}\right| \leq 1 \) belongs to the closed unit ball \( \left| x\right| \leq 1 \) .\n\n(a) Let \( {X}^{n} + \mathop{\sum }\limits_{{i < n}}{\alpha }_{i}{X}^{i} \in k\left\lbrack X\right\rbrack \) be a monic polynomial of degree \( n \geq 1 \) . Choose liftings \( {a}_{i} \in A \) of the coefficients, i.e., \( {\alpha }_{i} = {a}_{i}\left( {\;\operatorname{mod}\;M}\right) \), and consider the monic polynomial\n\n\[ {X}^{n} + \mathop{\sum }\limits_{{i < n}}{a}_{i}{X}^{i} \in A\left\lbrack X\right\rbrack \]\n\nSince the field \( K \) is algebraically closed, this polynomial has a root \( x \in K \) . By the preliminary observation, \( x \in A \) and \( x{\;\operatorname{mod}\;M} \) is a root of the reduced polynomial \( {X}^{n} + \mathop{\sum }\limits_{{i < n}}{\alpha }_{i}{X}^{i} \) . This proves that \( k \) is algebraically closed.\n\n(b) Let \( 0 \neq \xi \in {k}_{L} \) and choose a representative \( x \in {A}_{L} - {M}_{L} \) of the coset \( \xi \neq 0 : \left| x\right| = 1 \) . By assumption, this element is algebraic over \( K \), and hence \( x \) satisfies a nontrivial polynomial equation\n\n\[ \mathop{\sum }\limits_{{i \leq n}}{a}_{i}{x}^{n} = 0\;\left( {n \geq 1,{a}_{i} \in K}\right) .\n\nBy the principle of competitivity, at least two monomials have maximal competing absolute values\n\n\[ \left| {a}_{i}\right| = \left| {{a}_{i}{x}^{i}}\right| = \left| {{a}_{j}{x}^{j}}\right| = \left| {a}_{j}\right| \;\text{ for some }i < j.\n\nDividing by \( {a}_{i} \), we obtain a polynomial equation with coefficients \( \left| {a}_{k}^{\prime }\right| \leq 1,{a}_{k}^{\prime } \in A \) and at least two of them not in \( M \) . By reduction \( {\;\operatorname{mod}\;M} \) we get a nontrivial polynomial equation satisfied by \( \xi \) .
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Yes
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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 \( {\mathbf{Q}}_{p}^{a} \) .
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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 \( A \in \mathcal{U} \) such that \( r/2 < \left| {\alpha }_{i}\right| < {2r}\left( {i \in A}\right) \) . Define a sequence \( \beta = \left( {\beta }_{i}\right) \) by\n\n\[ \n{\beta }_{i} = \frac{1}{{\alpha }_{i}}\text{ for }i \in A\;\text{ and }\;{\beta }_{i} = 0\text{ for }i \notin A.\n\]\n\nSince \( \left| {\beta }_{i}\right| < 2/r\left( {i \in A}\right) \), the sequence \( \beta \) is bounded \( \beta \leq 2/r \) and \( \beta \in R \) . By construction \( 1 - {\alpha \beta } \) vanishes on the set \( A \), hence \( 1 - {\alpha \beta } \in \mathcal{J} \) . This shows that \( \alpha {\;\operatorname{mod}\;\mathcal{J}} \) is invertible in the quotient \( {\Omega }_{p} \) . Consequently, the quotient is a field, and \( \mathcal{J} \) a maximal ideal of \( R \) . Finally, constant sequences provide an embedding \( {\mathbf{Q}}_{p}^{a} \rightarrow {\Omega }_{p} \)
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Yes
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Proposition 2. The absolute value \( {\left. \mid .\right| }_{\Omega } \) coincides with the quotient norm of \( R/\mathcal{J} \), namely for \( a = \left( {\alpha {\;\operatorname{mod}\;\mathcal{J}}}\right) \), \[ {\left| a\right| }_{\Omega } = \parallel \alpha {\;\operatorname{mod}\;\mathcal{J}}{\parallel }_{R/\mathcal{J}} \mathrel{\text{:=}} \mathop{\inf }\limits_{{\beta \in \mathcal{J}}}\parallel \alpha - \beta \parallel . \]
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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 \[ \mathop{\lim }\limits_{\mathcal{U}}\left| {\alpha }_{i}\right| = \mathop{\lim }\limits_{\mathcal{U}}\left| {{\alpha }_{i} - {\beta }_{i}}\right| \leq \sup \left| {{\alpha }_{i} - {\beta }_{i}}\right| \;\left( {\beta \in \mathcal{J}}\right) , \] \[ {\left| a\right| }_{\Omega } \leq \parallel \alpha - \beta \parallel \;\left( {\beta \in \mathcal{J}}\right) . \] This proves \[ {\left| a\right| }_{\Omega } \leq \parallel a{\parallel }_{R/\mathcal{J}} \] Conversely, if \( a = \alpha {\;\operatorname{mod}\;\mathcal{J}} \), then for any subset \( A \in \mathcal{U} \) we can define the sequence \( \beta = \left( {\beta }_{i}\right) \) as \( {\beta }_{i} = 0\left( {i \in A}\right) \) and \( {\beta }_{i} = {\alpha }_{i}\left( {i/ \in A}\right) \) so that \( \beta \in \mathcal{J} \) and \( \parallel \alpha - \beta \parallel = \) \( \mathop{\sup }\limits_{{i \in A}}\left| {\alpha }_{i}\right| \) and \[ \parallel a{\parallel }_{R/\mathcal{J}} \leq \mathop{\inf }\limits_{{A \in \mathcal{U}}}\mathop{\sup }\limits_{{i \in A}}\left| {\alpha }_{i}\right| = \lim \sup \left| {\alpha }_{i}\right| = \mathop{\lim }\limits_{\mathcal{U}}\left| {\alpha }_{i}\right| = {\left| a\right| }_{\Omega }. \] From now on we shall simply write \( \left| a\right| = {\left| a\right| }_{\Omega } \) for either the absolute value on the field \( {\Omega }_{p} \) or the quotient norm in \( R/\mathcal{J} \) .
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Yes
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Proposition 3. We have\n\n\[ \left| {\Omega }_{p}^{ \times }\right| = {\mathbf{R}}_{ > 0} \]
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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 \( {r}_{n} = \left| {\alpha }_{n}\right| \left( {{\alpha }_{n} \in {\mathbf{Q}}_{p}^{a}}\right) \), so that the sequence \( \alpha \) is bounded and defines an element \( a \) in the quotient \( {\Omega }_{p} \) with \( \left| a\right| = r \) .
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Yes
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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 \( r = \mathop{\min }\limits_{{{a}^{\sigma } \neq a}}\left| {{a}^{\sigma } - a}\right| \) . Then every algebraic element \( b \) over \( K, b \in {B}_{ < r}\left( a\right) \), generates with \( K \) an extension containing \( K\left( a\right) \) .
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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 \( \sigma \) leaves all elements of \( K\left( b\right) \) fixed), and\n\n\[ \left| {b - {a}^{\sigma }}\right| = \left| {\left( b - a\right) }^{\sigma }\right| = \left| {b - a}\right| \]\n\n\[ \left| {a - {a}^{\sigma }}\right| \leq \max \left( {\left| {a - b}\right| ,\left| {b - {a}^{\sigma }}\right| }\right) = \left| {b - a}\right| . \]\n\nHence\n\n\[ \left| {b - a}\right| \geq \left| {a - {a}^{\sigma }}\right| \geq r. \]\n\nTaking the contrapositive, \( \left| {b - a}\right| < r \Rightarrow a \in K\left( b\right) \) and \( K\left( a\right) \subset K\left( b\right) \) .
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Yes
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Proposition 1. The group \( 1 + {\mathbf{M}}_{p} \) is divisible. For each \( m \geq 2 \) prime to \( p \), it is uniquely \( m \) -divisible.
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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, since this field is algebraically closed. Since \( {\left| x\right| }^{p} = \left| {x}^{p}\right| = \left| {1 + t}\right| = 1 \), we have \( \left| x\right| = 1 : x \in \mathbf{U}\left( 1\right) \) . Now\n\n\[{\left( x{\;\operatorname{mod}\;{\mathbf{M}}_{p}}\right) }^{p} = {x}^{p}{\;\operatorname{mod}\;{\mathbf{M}}_{p}} = 1 \in k\]\n\nimplies \( x{\;\operatorname{mod}\;{\mathbf{M}}_{p}} = 1 \), since \( k \) has characteristic \( p \) . This proves \( x = 1 + s \in \) \( 1 + {\mathbf{M}}_{p} \) .\n\n(2) Let \( 1 + t \in 1 + {\mathbf{M}}_{p} \) and select a positive integer \( m \) prime to \( p \) . We are looking for a root of the polynomial \( f\left( X\right) = {X}^{m} - \left( {1 + t}\right) \) . We already have an approximate root \( y = 1 \) for which the derivative \( m{X}^{m - 1} \) does not vanish \( {\;\operatorname{mod}\;{\mathbf{M}}_{p}} \) ( \( p \) does not divide \( m \) ):\n\n\[f\left( y\right) = 1 - \left( {1 + t}\right) = - t,{f}^{\prime }\left( y\right) = m,\left| {{f}^{\prime }\left( y\right) }\right| = 1.\]\n\nThus we have \( \left| {f\left( y\right) /{f}^{\prime }{\left( y\right) }^{2}}\right| = \left| {-t}\right| < 1 \), and Hensel’s lemma (II.1.5) is applicable: There is a unique root of \( f \) in the open ball of center 1 and radius 1 .\n\nIn fact, for each \( \zeta \in {\mu }_{m} \subset {\mu }_{\left( p\right) } \subset {\mathbf{F}}_{{p}^{\infty }}^{ \times } \), there is one root \( x \) of \( f \) with \( x \equiv \zeta \) (mod \( {\mathbf{M}}_{p} \) ). These \( m \) roots of \( f \) are all the roots of this polynomial, and for each given \( \zeta \in {\mu }_{m} \) there can be only one root of \( f \) congruent to this root of unity \( \zeta \) .
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Yes
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Proposition 2. For \( x \in {\mathbf{C}}_{p} \) we have\n\n\[ x \in 1 + {\mathbf{M}}_{p} \Leftrightarrow {x}^{{p}^{n}} \rightarrow 1\;\left( {n \rightarrow \infty }\right) . \]
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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 \( n \) such that \( {x}^{{p}^{n}} \) belongs to the open neighborhood \( 1 + {\mathbf{M}}_{p} \) of 1 in \( {\mathbf{C}}_{p} \) . Since we have proved in (4.1) that there is a torsion-free subgroup \( \Gamma \left( { \cong {p}^{\mathbf{Q}}}\right) \) of \( {\mathbf{C}}_{p}^{ \times } \) and a direct-product decomposition\n\n\[ {\mathbf{C}}_{p}^{ \times } = \Gamma \cdot {\mu }_{\left( p\right) } \cdot \left( {1 + {\mathbf{M}}_{p}}\right) \]\n\nwe see that \( x \in {\mu }_{\left( p\right) } \cdot \left( {1 + {\mathbf{M}}_{p}}\right) \) . The first component \( \zeta \) of \( x \) is trivial simply because it has an order prime to \( p \) :\n\n\[ {x}^{{p}^{n}} \in 1 + {\mathbf{M}}_{p} \Rightarrow {\zeta }^{{p}^{n}} = 1 \Rightarrow \zeta = 1. \]\n\nObserve that the convergent sequence \( {\left( {x}^{{p}^{n}}\right) }_{n \geq 0} \) is eventually constant precisely when \( x \) is a \( p \) th power root of unity\n\n\[ x \in {\mu }_{{p}^{\infty }} \subset 1 + {\mathbf{M}}_{p} \]
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Yes
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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 \( {A}_{i} \in \mathcal{U} \) .
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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 \( A \cup B \notin \mathcal{U} \) .
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Yes
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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 \( Y \) .
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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\[ \n\text{either}A = f\left( {{f}^{-1}\left( A\right) }\right) \text{or}{A}^{c} = f\left( {{f}^{-1}\left( {A}^{c}\right) }\right) \text{belongs to}\mathcal{U}\text{.} \n\]\n\nBy the criterion, \( f\left( \mathcal{U}\right) \) is an ultrafilter on \( Y \) .
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Yes
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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 \( \mathbf{Z} \) .
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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\[ \left( \begin{matrix} - m \\ i \end{matrix}\right) = - m\left( {-m - 1}\right) \cdots \left( {-m - i + 1}\right) /i! = {\left( -1\right) }^{i}\left( \begin{matrix} m + i - 1 \\ i \end{matrix}\right) \in \mathbf{Z}. \]\n\nHence the \( \left( \begin{array}{l} \cdot \\ i \end{array}\right) \) and all \( f \in L \) define functions \( \mathbf{Z} \rightarrow \mathbf{Z} \) .
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Yes
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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.
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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 \( f \) is a polynomial of degree \( d \), so is \( g \) . The expansion \( g = \mathop{\sum }\limits_{{i \leq d}}{\nabla }^{i}g\left( 0\right) \left( \begin{matrix} . \\ i \end{matrix}\right) \) shows that \( g \in L \) .
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No
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Proposition 2. The indefinite-sum and finite-difference operators are linked by the formulas\n\n\[ \nabla \circ S = \mathrm{{id}},\;S \circ \nabla = \mathrm{{id}} - {P}_{0},\;\nabla \circ S - S \circ \nabla = {P}_{0}. \]
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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.
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No
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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 \( \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}\left( \begin{array}{l} \cdot \\ k \end{array}\right) \) converges uniformly.\n\n(iii) \( f \) admits a continuous extension to \( {\mathbf{Z}}_{p} \rightarrow {\mathbf{C}}_{p} \) .\n\n(iv) \( f \) is uniformly continuous (for the p-adic topology on \( \mathbf{N} \) ).\n\n(v) \( \begin{Vmatrix}{{\nabla }^{k}f}\end{Vmatrix} \rightarrow 0 \) when \( k \rightarrow \infty \) .
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Proof. Here is a complete scheme of implications.\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) We have\n\n\[ \left| {{a}_{k}\left( \begin{array}{l} x \\ k \end{array}\right) }\right| \leq \left| {a}_{k}\right| \begin{Vmatrix}\left( \begin{array}{l} \cdot \\ k \end{array}\right) \end{Vmatrix} = \left| {a}_{k}\right| \;\left( {x \in {\mathbf{Z}}_{p}}\right) ,\]\n\nhence the uniform convergence if \( \left| {a}_{k}\right| \rightarrow 0 \) .\n\n(ii) \( \Rightarrow \) (iii) This is the basic property of uniform convergence reviewed in (3.1).\n\n(iii) \( \Leftrightarrow \) (iv) On a compact metric space, any continuous function is uniformly continuous.\n\n(iii) \( \Rightarrow \left( v\right) \) Apply the Mahler theorem to the continuous extension of \( f \) to \( {\mathbf{Z}}_{p} \) (still denoted by \( f \) ):\n\n\[ f = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}\left( \begin{array}{l} \cdot \\ k \end{array}\right) \;\left( {{a}_{k} = {\nabla }^{k}f\left( 0\right) }\right) .\n\nSince \( \nabla \left( \begin{array}{l} \cdot \\ k \end{array}\right) = \left( \begin{matrix} \cdot \\ k - 1 \end{matrix}\right) \), we have\n\n\[ \nabla f = \mathop{\sum }\limits_{{k \geq 1}}{a}_{k}\left( \begin{matrix} \cdot \\ k - 1 \end{matrix}\right) \]\n\nand by induction\n\n\[ {\nabla }^{j}f = \mathop{\sum }\limits_{{k \geq j}}{a}_{k}\left( \begin{matrix} \cdot \\ k - j \end{matrix}\right) \]\n\nBy the same theorem\n\n\[ \begin{Vmatrix}{{\nabla }^{j}f}\end{Vmatrix} = \mathop{\sup }\limits_{{k \geq j}}\left| {a}_{k}\right| \rightarrow 0. \]\n\nIn particular, \( \left| {a}_{j}\right| = \left| {{\nabla }^{j}f\left( 0\right) }\right| \leq \begin{Vmatrix}{{\nabla }^{j}f}\end{Vmatrix} \rightarrow 0 \) ; hence \( \left( v\right) \Rightarrow \left( i\right) \) .
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Yes
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Corollary 1. Any continuous \( f : {\mathbf{Z}}_{p} \rightarrow {\mathbf{C}}_{p} \) has limited Mahler expansions
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\[ f = f\left( 0\right) + \nabla f\left( 0\right) \cdot \left( \begin{array}{l} . \\ 1 \end{array}\right) + {\nabla }^{2}f\left( 0\right) \cdot \left( \begin{array}{l} . \\ 2 \end{array}\right) + \cdots \] \[ + {\nabla }^{n}f\left( 0\right) \cdot \left( \begin{matrix} \cdot \\ n \end{matrix}\right) + {R}_{n + 1}f\;\left( {n \geq 1}\right) \] with the van Hamme form of the remainder \[ {R}_{n + 1}f = {\nabla }^{n + 1}f\underline{ * }\left( \begin{array}{l} \cdot \\ n \end{array}\right) ,\;\begin{Vmatrix}{{R}_{n + 1}f}\end{Vmatrix} \leq \begin{Vmatrix}{{\nabla }^{n + 1}f}\end{Vmatrix} \rightarrow 0\;\left( {n \rightarrow \infty }\right) . \] Proof. The announced formulas hold on \( \mathbf{N} \) by the preceding section. Taking \( g = \) \( \left( \cdot \right) \) in \( \parallel f * g\parallel \leq \parallel f\parallel \parallel g\parallel \), we see that they extend continuously to \( {\mathbf{Z}}_{p} \) by the proposition.
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Yes
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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 of \( f \), then
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\[ {Sf} = 1 * f = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}\left( \begin{matrix} \cdot \\ k + 1 \end{matrix}\right) ,\parallel {Sf}\parallel = \parallel f\parallel . \]\n\nProof. We have noticed that\n\n\[ S\left( \begin{array}{l} \cdot \\ k \end{array}\right) = 1\underline{ * }\left( \begin{array}{l} \cdot \\ k \end{array}\right) = \left( \begin{matrix} \cdot \\ k + 1 \end{matrix}\right) \]\n\nwhence the result.
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Yes
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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 \( \varphi = 0 \) .
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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 exists an \( F \in \) \( C\left( {{\mathbf{Z}}_{p};K}\right) \) with \( f = \nabla F = {F}_{1} - F \) (take \( F = {Sf} \) ), and thus\n\n\[ \varphi \left( f\right) = \varphi \left( {{F}_{1} - F}\right) = \varphi \left( {F}_{1}\right) - \varphi \left( F\right) = 0. \]
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Yes
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Corollary 4. Let \( \sigma : {\mathbf{Z}}_{p} \rightarrow {\mathbf{Z}}_{p}, x \mapsto - 1 - x \), be the canonical involution (I.1.2). Then \( S\left( {f \circ \sigma }\right) \left( x\right) = - {Sf}\left( {-x}\right) \) .
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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\[ \n{Sf}\left( {x + m}\right) - {Sf}\left( x\right) = f\left( x\right) + \cdots + f\left( {x + m - 1}\right) \;\left( {x \in {\mathbf{Z}}_{p}}\right) .\n\] \n\nTake now \( x = - m \) in this equality:\n\n\[ \n{Sf}\left( 0\right) - {Sf}\left( {-m}\right) = f\left( {-m}\right) + \cdots + f\left( {-1}\right)\n\] \n\n\[ \n= f\left( {\sigma \left( {m - 1}\right) }\right) + \cdots + f\left( {\sigma \left( 0\right) }\right)\n\] \n\n\[ \n= S\left( {f \circ \sigma }\right) \left( m\right) \text{.}\n\] \n\nSince \( {Sf}\left( 0\right) = 0 \), the result follows.
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Yes
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Corollary 2. Let \( E \) be a Banach space. Then the sum map \( {E}^{\left( I\right) } \rightarrow E \) has a unique continuous extension \( \sum : {c}_{0}\left( {I;E}\right) \rightarrow E \) .
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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\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{i \in I}}{x}_{i}}\end{Vmatrix} \leq \mathop{\sup }\limits_{i}\begin{Vmatrix}{x}_{i}\end{Vmatrix} = \parallel x\parallel \;\left( {x \in {E}^{\left( I\right) }}\right) .\n\]\n\nIt has a unique continuous extension \( \sum \) . This extension is also a contracting linear map by density and continuity. Hence we have more generally\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{i \in I}}{x}_{i}}\end{Vmatrix} \leq \mathop{\sup }\limits_{i}\begin{Vmatrix}{x}_{i}\end{Vmatrix} = \parallel x\parallel \;\left( {x \in {c}_{0}\left( {I;E}\right) }\right) .\n\]\n\nThis sum \( \sum \) can be computed using any ordering of the index set \( I \) and any grouping \( I = \mathop{\coprod }\limits_{i}{I}_{j} \) : The equality for families with finite support extends by continuity to the completion \( {c}_{0}\left( {I;E}\right) \) (cf. (II.1.2)).
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Yes
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Corollary 2. Let \( E \) be a Banach space. Then the sum map \( {E}^{\left( I\right) } \rightarrow E \) has a unique continuous extension \( \sum : {c}_{0}\left( {I;E}\right) \rightarrow E \) .
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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\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{i \in I}}{x}_{i}}\end{Vmatrix} \leq \mathop{\sup }\limits_{i}\begin{Vmatrix}{x}_{i}\end{Vmatrix} = \parallel x\parallel \;\left( {x \in {E}^{\left( I\right) }}\right) .\n\]\n\nIt has a unique continuous extension \( \sum \) . This extension is also a contracting linear map by density and continuity. Hence we have more generally\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{i \in I}}{x}_{i}}\end{Vmatrix} \leq \mathop{\sup }\limits_{i}\begin{Vmatrix}{x}_{i}\end{Vmatrix} = \parallel x\parallel \;\left( {x \in {c}_{0}\left( {I;E}\right) }\right) .\n\]\n\nThis sum \( \sum \) can be computed using any ordering of the index set \( I \) and any grouping \( I = \mathop{\coprod }\limits_{i}{I}_{j} \) : The equality for families with finite support extends by continuity to the completion \( {c}_{0}\left( {I;E}\right) \) (cf. (II.1.2)).
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Yes
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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 \( \left( {f}_{j}\right) \) consisting of linear contractions \( {f}_{j} : {E}_{j} \rightarrow E \), there is a unique linear contraction \( f \) such that the following diagram is commutative:\n\n\[ {E}_{j}\;\overset{{\varepsilon }_{j}}{ \rightarrow }\;{\bigoplus }_{i \in I}{E}_{i}\; \subset \;{\widehat{\bigoplus }}_{i \in I}{E}_{i} \]\n\n\[ {f}_{j} \searrow \; \downarrow \oplus {f}_{i}\; \swarrow f \]\n\n\( E \)
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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\[ f = \sum \circ \left( {f}_{i}\right) ,\;{fx} = \mathop{\sum }\limits_{i}{f}_{i}{x}_{i}. \]
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Yes
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Proposition 1. If \( F \) is complete, then \( L\left( {E;F}\right) \) is also complete.
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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 \( T : E \rightarrow F \) . Let \( \varepsilon > 0 \) be given. There exists an integer \( {N}_{\varepsilon } \) such that \( \begin{Vmatrix}{{T}_{n} - {T}_{m}}\end{Vmatrix} \leq \varepsilon \) for all\n\n\( n, m \geq {N}_{\varepsilon } \) . Letting \( m \rightarrow \infty \) we deduce \( \begin{Vmatrix}{{Tx} - {T}_{m}x}\end{Vmatrix} \leq \varepsilon \parallel x\parallel \) for all \( n, m \geq {N}_{\varepsilon } \) . This proves that the operator \( T - {T}_{m} \) is continuous (bounded); hence \( T = {T}_{m} + \) \( \left( {T - {T}_{m}}\right) \) is continuous. Moreover, \( \begin{Vmatrix}{T - {T}_{m}}\end{Vmatrix} \leq \varepsilon \) when \( m \geq {N}_{\varepsilon } \) . This shows that \( \begin{Vmatrix}{T - {T}_{m}}\end{Vmatrix} \rightarrow 0,{T}_{m} \rightarrow T\left( {m \rightarrow \infty }\right) \), and everything is proved.
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Yes
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Proposition 2. The topological dual of the space \( {c}_{0}\left( {I;E}\right) \) is canonically isomorphic as a normed space to \( {l}^{\infty }\left( {I;{E}^{\prime }}\right) \) .
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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 \( i \) ). Since \( \begin{Vmatrix}{\varphi }_{i}\end{Vmatrix} \leq \parallel \varphi \parallel \), we get a bounded family \( \left( {\varphi }_{i}\right) \in {l}^{\infty }\left( {I;{E}^{\prime }}\right) \) . Conversely, if \( \left( {\varphi }_{i}\right) \in {l}^{\infty }\left( {I;{E}^{\prime }}\right) \), we can define a linear form \( \varphi = \sum {\varphi }_{i} \) on \( {c}_{0}\left( {I;E}\right) \) by the formula \( \left( {a}_{i}\right) \mapsto \mathop{\sum }\limits_{i}{\varphi }_{i}\left( {a}_{i}\right) \) (a summable series, since the sequence \( {\varphi }_{i} \) is bounded and \( \left. {\begin{Vmatrix}{a}_{i}\end{Vmatrix} \rightarrow 0}\right) \) . Both maps\n\n\[ \n\varphi \mapsto \left( {\varphi \circ {\varepsilon }_{i}}\right) ,\;\left( {\varphi }_{i}\right) \mapsto \sum {\varphi }_{i} \n\]\n\nare linear and decrease norms. Hence they are inverse isometries.\n\nIn other words, the bilinear map\n\n\[ \n\left( {\left( {a}_{i}\right) ,\left( {\varphi }_{i}\right) }\right) \mapsto \mathop{\sum }\limits_{i}{\varphi }_{i}\left( {a}_{i}\right) ,\;{c}_{0}\left( {I;E}\right) \times {l}^{\infty }\left( {I;{E}^{\prime }}\right) \rightarrow K \n\]\n\nis a duality pairing that proves the proposition.
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Yes
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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.
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Proof. We have already seen that \( \begin{Vmatrix}{\left( {\mathbf{f}}_{j}\right) }_{J}\end{Vmatrix} \leq \parallel T\parallel \) . Conversely,\n\n\[ \n\mathbf{x} = \mathop{\sum }\limits_{j}{x}_{j}{\mathbf{e}}_{j} \Rightarrow T\left( \mathbf{x}\right) = \mathop{\sum }\limits_{j}{x}_{j}{\mathbf{f}}_{j}\text{ (this sum converges!),}\n\]\n\n\[ \n\parallel T\left( \mathbf{x}\right) \parallel \leq \mathop{\sup }\limits_{j}\begin{Vmatrix}{{x}_{j}{\mathbf{f}}_{j}}\end{Vmatrix} \leq \mathop{\sup }\limits_{j}\left| {x}_{j}\right| \mathop{\sup }\limits_{j}\begin{Vmatrix}{\mathbf{f}}_{j}\end{Vmatrix} = \parallel x\parallel \mathop{\sup }\limits_{j}\begin{Vmatrix}{\mathbf{f}}_{j}\end{Vmatrix}\n\]\n\nwhence \( \parallel T\parallel \leq \mathop{\sup }\limits_{j}\begin{Vmatrix}{\mathbf{f}}_{j}\end{Vmatrix} = \begin{Vmatrix}{\left( {\mathbf{f}}_{j}\right) }_{J}\end{Vmatrix} \) . Observe that for any choice of bounded family \( {\mathbf{f}}_{j} \in F \), there is a \( T \in L\left( {E;F}\right) \) with \( T{e}_{j} = {\mathbf{f}}_{j}\left( {j \in J}\right) \), so that the map \( L\left( {E;F}\right) \rightarrow {l}^{\infty }\left( {J;F}\right) \) is surjective.
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Yes
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Proposition 3. When \( E = {c}_{0}\left( J\right) \) and \( F = {c}_{0}\left( I\right) \), we can make canonical identifications\n\n\[ L\left( {E;F}\right) = {l}^{\infty }\left( {J;{c}_{0}\left( I\right) }\right) \subset {l}^{\infty }\left( {I;{E}^{\prime }}\right) = {l}^{\infty }\left( {I;{l}^{\infty }\left( J\right) }\right) \cong {l}^{\infty }\left( {I \times J}\right) . \]\n\nIn other words, when normal bases are chosen, continuous linear maps \( E \rightarrow F \) are represented by bounded matrices with columns in \( {c}_{0}\left( I\right) \cong F \) .
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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, \( {\varphi }_{i}\left( \mathbf{x}\right) = \varphi \left( \mathbf{x}\right) {a}_{i},\begin{Vmatrix}{\varphi }_{i}\end{Vmatrix} = \left| {a}_{i}\right| \parallel \varphi \parallel \rightarrow 0 \) . This proves that the image of \( T \) belongs to the closed subspace \( {c}_{0}\left( {I;{E}^{\prime }}\right) \) . By linearity, the same property will hold for any continuous linear map \( T \) of finite rank:\n\n\[ {L}_{\mathrm{{fr}}}\left( {E;{c}_{0}\left( I\right) }\right) \rightarrow {c}_{0}\left( {I;{E}^{\prime }}\right) : T \mapsto \left( {\varphi }_{i}\right) . \]
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Yes
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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) .\\]
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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}}{\\left( x\\right) }_{n}\\frac{{z}^{n}}{n!}.\\]
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Yes
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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 \( x \in {\mathbf{Z}}_{p} \) with \( {f}^{\prime }\left( x\right) = 0 \) .
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Again \( {f}^{\prime } = 0 \in {\mathcal{C}}^{1} \) , but \( f \) is injective, and hence far from being locally constant.
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Yes
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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.
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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\[ \left| {{\Phi f}\left( {x, y}\right) - {f}^{\prime }\left( a\right) }\right| < \varepsilon \text{ if }x \in {V}_{\varepsilon }\text{ and }y \in {V}_{\varepsilon }.\]\n\nLet us take \( \varepsilon = \left| {{f}^{\prime }\left( a\right) }\right| \left( { \neq 0\text{by assumption}}\right) \) and \( V \) the corresponding neighborhood. Then\n\n\[ \left| {{\Phi f}\left( {x, y}\right) - {f}^{\prime }\left( a\right) }\right| < \left| {{f}^{\prime }\left( a\right) }\right| \neq 0\text{, if }\left( {x, y}\right) \in V \times V \]\n\nand there is a competition between the terms \( {\Phi f}\left( {x, y}\right) \) and \( {f}^{\prime }\left( a\right) \)\n\n\[ \left| {{\Phi f}\left( {x, y}\right) }\right| = \left| {{f}^{\prime }\left( a\right) }\right| \text{ for }\left( {x, y}\right) \in V \times V.\]\n\nHence \( \left| {f\left( x\right) - f\left( y\right) }\right| = \left| {{f}^{\prime }\left( a\right) }\right| \cdot \left| {x - y}\right| \) for \( \left( {x, y}\right) \in V \times V \) .
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Yes
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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 \( X \times X \) .\n\n(iii) \( f \) is differentiable at each point \( a \in X \) and there is a continuous function \( \alpha \) on \( X \times X \) vanishing on \( {\Delta }_{X} \) with\n\n\[ f\left( y\right) = f\left( x\right) + \left( {y - x}\right) {f}^{\prime }\left( x\right) + \left( {y - x}\right) \alpha \left( {x, y}\right) \;\left( {x, y \in X}\right) . \]
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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\[ z \in {X}_{0}\text{and}z \rightarrow x\text{implies}f\left( z\right) \text{has a limit}g\left( x\right) \in Y\text{.} \]\n\nThen the extension \( g : X \rightarrow Y \) is continuous. (More generally, the conclusion is valid when the target space \( Y \) is a regular space, i.e., a topological space in which every point has a fundamental system of neighborhoods consisting of closed sets.)\n\nThe implication \( \left( {ii}\right) \Rightarrow \left( i\right) \) is obvious.\n\nFinally, if \( {\Phi f} \) has a continuous extension \( \widetilde{\Phi } \), it has a unique one by the density of \( X \times X - {\Delta }_{X} \) in \( X \times X \) . Since we can write\n\n\[ f\left( y\right) = f\left( x\right) + \left( {y - x}\right) {\Phi f}\left( {x, y}\right) \]\n\n\[ = f\left( x\right) + \left( {y - x}\right) {f}^{\prime }\left( x\right) + \left( {y - x}\right) \left\lbrack \underset{\alpha \left( {x, y}\right) }{\underbrace{{\Phi f}\left( {x, y}\right) - {f}^{\prime }\left( x\right) }}\right\rbrack \]\n\nit is obvious that \( \left( {ii}\right) \Leftrightarrow \left( {iii}\right) \) .
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Yes
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Proposition 1. If \( f \in {S}^{2}\left( a\right) \), then \( f \in {S}^{1}\left( a\right) \) .
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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\[ \n{\Phi f}\left( {x, y}\right) - {\Phi f}\left( {z, t}\right) = {\Phi f}\left( {x, y}\right) - {\Phi f}\left( {z, y}\right) + {\Phi f}\left( {z, y}\right) - {\Phi f}\left( {z, t}\right) \n\]\n\n\[ \n= \left( {x - z}\right) {\Phi }_{2}f\left( {x, z, y}\right) + \left( {y - t}\right) {\Phi }_{2}f\left( {y, t, z}\right) . \n\]\n\nIf we assume \( f \in {S}^{2}\left( a\right) \), then \( {\Phi }_{2}f \) will remain bounded in a neighborhood of \( \left( {a, a, a}\right) \), say \( \left| {{\Phi }_{2}f}\right| \leq M \), when the three variables of \( {\Phi }_{2}f \) are close enough to \( a \) . In particular if \( x, y, z \), and \( t \) are near enough to \( a \), we have\n\n\[ \n\left| {{\Phi f}\left( {x, y}\right) - {\Phi f}\left( {z, t}\right) }\right| \leq M\max \left( {\left| {x - z}\right| ,\left| {y - t}\right| }\right) , \n\]\n\na quantity that tends to zero when \( \left( {x, y}\right) \) and \( \left( {z, t}\right) \) tend to \( \left( {a, a}\right) \) . Since the target of \( {\Phi f} \) is a complete space, the Cauchy criterion is valid and shows that this function \( {\Phi f} \) has a limit as \( \left( {x, y}\right) \rightarrow \left( {a, a}\right) \) .
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Yes
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Proposition 2. If \( f \in {S}^{2} \), then \( {f}^{\prime } \in {S}^{1} \) .
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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 \( {\widetilde{\Phi }}_{2} \) that extends \( {\Phi }_{2}f \) to \( X \times X \times X \), and we have\n\n\[ {\Phi f}\left( {x, z}\right) - {\Phi f}\left( {y, z}\right) = \left( {x - y}\right) \cdot {\widetilde{\Phi }}_{2}\left( {x, y, z}\right) . \]\n\nIn this expression we let \( z \rightarrow x \) . We know that \( {\Phi f}\left( {x, z}\right) \) tends to \( {f}^{\prime }\left( x\right) \) and\n\n\[ {f}^{\prime }\left( x\right) - {\Phi f}\left( {y, x}\right) = \left( {x - y}\right) \cdot {\widetilde{\Phi }}_{2}\left( {x, y, x}\right) . \]\nSince the order of the variables in \( {\Phi f},{\Phi }_{2}f \), and \( {\widetilde{\Phi }}_{2} \) is irrelevant, we can write\n\n\[ {f}^{\prime }\left( x\right) = {\Phi f}\left( {x, y}\right) + \left( {x - y}\right) \cdot {\widetilde{\Phi }}_{2}\left( {x, x, y}\right) ,\]\n\nand interchanging \( x \) and \( y \) ,\n\n\[ {f}^{\prime }\left( y\right) = {\Phi f}\left( {y, x}\right) + \left( {y - x}\right) \cdot {\widetilde{\Phi }}_{2}\left( {y, y, x}\right) . \]\n\nSubtracting these expressions, we obtain\n\n\[ {f}^{\prime }\left( x\right) - {f}^{\prime }\left( y\right) = \left( {x - y}\right) \left\lbrack {{\widetilde{\Phi }}_{2}\left( {x, x, y}\right) + {\widetilde{\Phi }}_{2}\left( {x, y, y}\right) }\right\rbrack \]\n\n\[ \Phi {f}^{\prime }\left( {x, y}\right) = {\widetilde{\Phi }}_{2}\left( {x, x, y}\right) + {\widetilde{\Phi }}_{2}\left( {x, y, y}\right) . \]\n\nThis shows that \( \Phi {f}^{\prime } \) admits a continuous extension to \( X \times X : {f}^{\prime } \in {S}^{1} \) . Moreover,\n\n\[ {f}^{\prime \prime }\left( a\right) = {\left( {f}^{\prime }\right) }^{\prime }\left( a\right) = \Phi {f}^{\prime }\left( {a, a}\right) = 2{\widetilde{\Phi }}_{2}\left( {a, a, a}\right) . \]
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Yes
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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 \[ \parallel {\Phi f}\parallel = \mathop{\sup }\limits_{{n \geq 1}}{\kappa }_{n}\left| {c}_{n}\right| < \infty . \]
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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 \[ \parallel f{\parallel }_{1} = \sup \left( {\left| {f\left( 0\right) }\right| ,\parallel {\Phi f}\parallel }\right) . \] Since \( f\left( 0\right) = {c}_{0} \), we define in an ad hoc way the value \( {\kappa }_{0} = 1 \) in order to have \[ \parallel f{\parallel }_{1} = \mathop{\sup }\limits_{{n \geq 0}}{\kappa }_{n}\left| {c}_{n}\right| \]
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Yes
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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 \[ \parallel f{\parallel }_{1} \leq \parallel {Sf}{\parallel }_{1} \leq p\parallel f{\parallel }_{1} \]
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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 \[ {\kappa }_{n - 1} \leq {\kappa }_{n} \leq p{\kappa }_{n - 1} \] whence the assertion.
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Yes
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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\[ \n\left| {{c}_{k}/k}\right| \text{takes alternatively values 0 and 1 .} \n\]\n\nHence \( \left| {{c}_{k}/k}\right| \) does not tend to 0, thereby proving that \( f \) is not differentiable at the origin.
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But \( {\Phi f} \) is bounded, since \( k\left| {c}_{k}\right| \) (taking values 0 and 1 only) is bounded.
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No
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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,
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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 \( \left| x\right| \leq 1 \) and \( \left| y\right| \leq 1 \), the ultrametric inequality gives \( \left| {{x}^{n - 1} + \cdots + {y}^{n - 1}}\right| \leq 1 \) , and the result follows.
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Yes
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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) .
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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 \( {\left( x + y\right) }^{n} - {x}^{n} - {y}^{n} \) is divisible by \( {xy} \) .
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Yes
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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\[ \n{f}^{\prime } = {\left. \Phi f\right| }_{\Delta } = \mathop{\sum }\limits_{{n \geq 1}}n{a}_{n}{X}^{n - 1} \in K\{ X\} .\n\]
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It is easy to give more precise estimates for the convergence:\n\n\[ \n{\Phi f}\left( {x, y}\right) - \mathop{\sum }\limits_{{n \geq 1}}n{a}_{n}{\xi }^{n - 1} = \mathop{\sum }\limits_{{n \geq 1}}{a}_{n}\left( {{x}^{n} - {y}^{n}}\right) /\left( {x - y}\right) - \mathop{\sum }\limits_{{n \geq 1}}n{a}_{n}{\xi }^{n - 1}\n\]\n\n\[ \n= \mathop{\sum }\limits_{{n \geq 1}}{a}_{n}\left( {{x}^{n - 1} + \cdots + {y}^{n - 1} - n{\xi }^{n - 1}}\right) \rightarrow 0\n\]\n\nwhen \( \left( {x, y}\right) \rightarrow \left( {\xi ,\xi }\right) \) . In fact,\n\n\[ \n\left( {{x}^{n - 1} + \cdots + {y}^{n - 1} - n{\xi }^{n - 1}}\right) = \mathop{\sum }\limits_{{i + j = n - 1}}\left( {{x}^{i}{y}^{j} - {\xi }^{n - 1}}\right) ,\n\]\n\nand by (2.3),\n\n\[ \n\left| {{x}^{i}{y}^{j} - {\xi }^{n - 1}}\right| = \left| {{x}^{i}{y}^{j} - {\xi }^{i}{\xi }^{j}}\right|\n\]\n\n\[ \n\leq \max \left( {\left| {x}^{i}\right| \left| {{y}^{j} - {\xi }^{j}}\right| ,\left| {{x}^{i} - {\xi }^{i}}\right| \left| {\xi }^{j}\right| }\right)\n\]\n\n\[ \n\leq \max \left( {\left| {y - \xi }\right| ,\left| {x - \xi }\right| }\right) \;\left( {x, y,\xi \in A}\right) .\n\]\n\nWe have obtained\n\n\[ \n\left| {{\Phi f}\left( {x, y}\right) - {f}^{\prime }\left( \xi \right) }\right| \leq \parallel f\parallel \cdot \max \left( {\left| {y - \xi }\right| ,\left| {x - \xi }\right| }\right) .\n\]
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Yes
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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 \( K : f \in {S}^{2}\left( A\right) \) .
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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\[ \n{\Phi }_{2}f\left( {x, y, z}\right) = \frac{{\Phi f}\left( {x, z}\right) - {\Phi f}\left( {y, z}\right) }{x - y} \n\] \n\n\[ \n= \mathop{\sum }\limits_{{n \geq 2}}{a}_{n}\mathop{\sum }\limits_{{i + j = n - 1}}\frac{{x}^{i} - {y}^{i}}{x - y}{z}^{j} \n\] \n\n\[ \n= \mathop{\sum }\limits_{{n \geq 2}}{a}_{n}\mathop{\sum }\limits_{{k + \ell + m = n - 2}}{x}^{k}{y}^{\ell }{z}^{m} \n\] \n\nSince \( \left| {a}_{n}\right| \rightarrow 0 \), this series converges uniformly on \( A \times A \times A \) and represents a continuous extension of \( {\Phi }_{2}f \) .
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Yes
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Proposition 1. For \( \left| x\right| < {r}_{p} \) we have\n\n\[ \left| {\log \left( {1 + x}\right) }\right| = \left| x\right| ,\;\left| {\exp \left( x\right) }\right| = 1,\;\left| {1 - \exp \left( x\right) }\right| = \left| x\right| . \]
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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\[ \left| {{x}^{k}/k}\right| \leq \left| {{x}^{k}/k!}\right| \leq {\left( \left| x\right| /{r}_{p}\right) }^{k - 1} \cdot \left| x\right| < \left| x\right| < 1 \]\n\nfor \( k \geq 2 \) and \( 0 < \left| x\right| < {r}_{p} \) . Hence the absolute values of the terms in the series\n\n\[ 1 + x + \mathop{\sum }\limits_{{k \geq 2}}\frac{{x}^{k}}{k!} = \exp \left( x\right) \]\n\n\[ x + \mathop{\sum }\limits_{{k \geq 2}}{\left( -1\right) }^{k - 1}\frac{{x}^{k}}{k} = \log \left( {1 + x}\right) \]\n\nare strictly smaller than the first ones. By the ultrametric character of the absolute value, the strongest (we underline it!) wins:\n\n\[ \exp \left( x\right) = \underline{1} + x + \mathop{\sum }\limits_{{k \geq 2}}\cdots \Rightarrow \left| {\exp \left( x\right) }\right| = 1, \]\n\n\[ \exp \left( x\right) - 1 = \underline{x} + \mathop{\sum }\limits_{{k \geq 2}}\cdots \Rightarrow \left| {\exp \left( x\right) - 1}\right| = \left| x\right| , \]\n\n\[ \log \left( {1 + x}\right) = \underline{x} + \mathop{\sum }\limits_{{k \geq 2}}\cdots \Rightarrow \left| {\log \left( {1 + x}\right) }\right| = \left| x\right| \]\n\nif \( \left| x\right| < {r}_{p} \) .
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Yes
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Proposition 2. For two indeterminates \( X \) and \( Y \), we have the following formal identities:\n\n\[ \exp \left( {X + Y}\right) = \exp \left( X\right) \cdot \exp \left( Y\right) \]
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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 \( i + j = n \) leads to a sum \( {\left( X + Y\right) }^{n}/n \) !.
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No
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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\[ \exp \log \left( {1 + x}\right) = 1 + x. \]
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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 in the domain of convergence of the \( p \) -adic exponential\n\n\[ \exp \left( x\right) \cdot \exp \left( y\right) = \exp \left( {x + y}\right) .\n\nLet us check the second identity: We have to show that it is legitimate to substitute a value \( x \in {\mathbf{C}}_{p},\left| x\right| < {r}_{p} \) in the formal identity\n\n\[ X = \log {e}^{X} = \log \left( {1 + e\left( X\right) }\right) \]\n\nwhere\n\n\[ e\left( X\right) = \mathop{\sum }\limits_{{n \geq 1}}\frac{{X}^{n}}{n!} = {e}^{X} - 1 \]\n\nThe substitution in the sum can be made by addition of two contributions:\n\n\[ x = {\left\lbrack \mathop{\sum }\limits_{{n \leq N}}\frac{{\left( -1\right) }^{n - 1}}{n}e{\left( X\right) }^{n}\right\rbrack }_{X = x} + {\left\lbrack \mathop{\sum }\limits_{{m > N}}\frac{{\left( -1\right) }^{m - 1}}{m}e{\left( X\right) }^{m}\right\rbrack }_{X = x}. \]\n\nIn the first finite sum, the substitution can obviously be made in each term according to\n\n\[ {\left. e{\left( X\right) }^{n}\right| }_{X = x} = {\left( x + \frac{{x}^{2}}{2!} + \cdots \right) }^{n} = e{\left( x\right) }^{n}\;\left( {\left| x\right| < {r}_{p}}\right) .\n\nSince \( \left| {e\left( x\right) }\right| = \left| x\right| < {r}_{p} < 1 \), we have\n\n\[ \mathop{\sum }\limits_{{n \leq N}}\frac{{\left( -1\right) }^{n - 1}}{n}e{\left( x\right) }^{n} \rightarrow \log \left( {1 + e\left( x\right) }\right) = \log {e}^{x}\;\left( {N \rightarrow \infty }\right) .\n\nThe proof of the second identity will be completed if we show that the second contribution is arbitrarily small (for large \( N \) ). But when \( \left| x\right| < {r}_{p} \), each monomial appearing in the computation of \( e\left( x\right) \) satisfies \( \left| {{x}^{i}/i!}\right| < {r}_{p} \) (because \( i \geq 1 \) ), and each monomial appearing in the computation of \( e{\left( x\right) }^{m} \) has an absolute value less than \( {r}_{p}^{m} \) . All individual monomials appearing in the evaluation of the second contribution \( \mathop{\sum }\limits_{{m > N}}\cdots \) have an absolute value smaller than\n\n\[ \mathop{\sup }\limits_{{m > N}}\left| \frac{{\left( -1\right) }^{m - 1}}{m}\right| {r}_{p}^{m} \]\n\nSince the power series for the logarithm converges, it is possible to choose \( N \) large enough to ensure that all \( \left| {1/m}\right| {r}_{p}^{m}\left( {m > N}\right) \) are arbitrarily small and that the same holds for their sum (independently of the groupings made to compute it). Again, the verification of the third identity is similar.
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Yes
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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 \( {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{1} \rightarrow 0 \), then\n\n\[ {\int }_{{\mathbf{Z}}_{p}}{f}_{n}\left( x\right) {dx} \rightarrow {\int }_{{\mathbf{Z}}_{p}}f\left( x\right) {dx} \]
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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 \( \left( a\right) \) follows from Corollary 2 in (1.5):\n\n\[ \parallel {Sf}{\parallel }_{1} \leq p\parallel f{\parallel }_{1}. \]\n\n(b) is a consequence of \( \left( a\right) \) .
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Yes
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Proposition 2. For \( f \in {S}^{1}\left( {\mathbf{Z}}_{p}\right) \) we have\n\n\[{\int }_{{\mathbf{Z}}_{p}}\nabla f\left( x\right) {dx} = {f}^{\prime }\left( 0\right)\]
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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)).
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Yes
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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) \( {SD} = {DS} - {P}_{0}{DS} \).
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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\[ = \mathop{\sum }\limits_{{0 < i \leq n}}f\left( i\right) = {Sf}\left( {n + 1}\right) - f\left( 0\right) = {\tau Sf}\left( n\right) - f\left( 0\right) ,\]\n\nwhich proves \( {S\tau } = {\tau S} - {P}_{0} \) (by density of the integers \( n \geq 1 \) in \( {\mathbf{Z}}_{p} \) and continuity of the functions in question). On the other hand, differentiation of the function \( {S\tau f} = {\tau Sf} - f\left( 0\right) \) leads to \( {DS\tau f} = {D\tau Sf} = {\tau DSf} \). Moreover, recall that \( \nabla {Sf} = f \) but \( S\nabla f = f - f\left( 0\right) \) (IV.1.5). In other words,\n\n\[ \nabla S = \mathrm{{id}},\;S\nabla = \mathrm{{id}} - {P}_{0} \]\n\nWe infer\n\n\[ {SD} = {SD}\nabla S = S\nabla {DS} = {DS} - {P}_{0}{DS}. \]\n\nThe proposition is proved.
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Yes
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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\[ \n{F}^{\prime }\left( x\right) = {\int }_{{\mathbf{Z}}_{p}}{f}^{\prime }\left( {x + t}\right) {dt} \n\]
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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 \( G \in {S}^{1}\left( {\mathbf{Z}}_{p}\right) \). Moreover, by Proposition 2 (a),\n\n\[ \n{\int }_{{\mathbf{Z}}_{p}}{f}^{\prime }\left( {x + t}\right) {dt} = {\left( S{f}^{\prime }\right) }^{\prime }\left( x\right) = \left( {DSDf}\right) \left( x\right) , \n\]\n\nwhich proves \( G = {DSDf} \). Now by the first proposition \( {SD} = {DS} - {P}_{0}{DS} \) and\n\n\[ \nG = D\left( {{DS} - {P}_{0}{DS}}\right) f = {DDSf} = {\left( Sf\right) }^{\prime \prime } = {F}^{\prime }, \n\]\n\nbecause \( F = {\left( Sf\right) }^{\prime } \).
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Yes
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Proposition 4. Let \( \\sigma \) denote the involution (I.1.2) \( x \\mapsto - 1 - x \) of \( {\\mathbf{Z}}_{p} \) . Then\n\n\[ \n{\\int }_{{\\mathbf{Z}}_{p}}\\left( {f \\circ \\sigma }\\right) {dx} = {\\int }_{{\\mathbf{Z}}_{p}}{fdx} \n\]
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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\[ \n{\\left( Sf\\right) }^{\\prime }\\left( 0\\right) = \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{{Sf}\\left( {-{p}^{n}}\\right) }{-{p}^{n}} = \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{-{Sf}\\left( {-{p}^{n}}\\right) }{{p}^{n}}. \n\]\n\nBut by the Corollary 4 in (IV.3.5),\n\n\[ \n- {Sf}\\left( {-{p}^{n}}\\right) = S\\left( {f \\circ \\sigma }\\right) \\left( {p}^{n}\\right) \n\]\n\nwhence the result \( {\\left( Sf\\right) }^{\\prime }\\left( 0\\right) = {\\left( S\\left( f \\circ \\sigma \\right) \\right) }^{\\prime }\\left( 0\\right) \) .
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Yes
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The radius of convergence of \( f = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{X}^{n} \) is\n\n\[ \n{r}_{f} = \frac{1}{\mathop{\lim }\limits_{{n \geq 0}}{\left| {a}_{n}\right| }^{1/n}} = \frac{1}{\mathop{\limsup }\limits_{{n \rightarrow \infty }}{\left| {a}_{n}\right| }^{1/n}}.\n\]
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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\[ \n\mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sup }\limits_{{k \geq n}}\left| x\right| {\left| {a}_{k}\right| }^{1/k} = \left| x\right| \cdot \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sup }\limits_{{k \geq n}}{\left| {a}_{k}\right| }^{1/k} = \left| x\right| \cdot \frac{1}{{r}_{f}} > 1.\n\]\n\nHence the decreasing sequence \( \mathop{\sup }\limits_{{k \geq n}}\left| x\right| {\left| {a}_{k}\right| }^{1/k} \) is greater than 1, and for infinitely many values of \( k \geq 0 \) we have \( \left| {a}_{k}\right| {\left| x\right| }^{k} > 1 \), namely, the general term \( {a}_{k}{x}^{k} \) of the series does not tend to zero: The series \( \sum {a}_{k}{x}^{k} \) diverges. Conversely, if \( \left| x\right| < {r}_{f} \) (this can happen only if \( {r}_{f} > 0 \) !) we can choose \( \left| x\right| < r < {r}_{f} \), and from\n\n\[ \n\mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sup }\limits_{{k \geq n}}r{\left| {a}_{k}\right| }^{1/k} = r \cdot \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sup }\limits_{{k \geq n}}{\left| {a}_{k}\right| }^{1/k} < 1\n\]\n\nwe infer that for some large \( N \)\n\n\[ \n\mathop{\sup }\limits_{{k \geq N}}r{\left| {a}_{k}\right| }^{1/k} < 1\n\]\n\nHence \( \left| {a}_{k}\right| {r}^{k} < 1 \) for all \( k \geq N \) and\n\n\[ \n\left| {{a}_{k}{x}^{k}}\right| = \left| {a}_{k}\right| {r}^{k}{\left( \frac{\left| x\right| }{r}\right) }^{k} < \frac{{\left| x\right| }^{k}}{{r}^{k}} \searrow 0\;\left( {k \rightarrow \infty }\right) .\n\]\n\nThis shows that the general term of the series \( \sum {a}_{k}{x}^{k} \) tends to zero, and the series converges.
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Yes
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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 the power series \( {fg} \) can be made according to the usual rule\n\n\[ \left( {fg}\right) \left( x\right) = f\left( x\right) g\left( x\right) \;\left( {\left| x\right| < \min \left( {{r}_{f},{r}_{g}}\right) }\right) . \]
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Proof. All statements are consequences of (V.2.2).
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No
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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) .
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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 \[ \left( {f \circ g}\right) \left( x\right) = f\left( {g\left( x\right) }\right) \;\left( {\left| x\right| < {r}_{g}}\right) \] for any polynomial \( f \) .
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No
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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: \( {r}_{f} = {r}_{Df} \) .
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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\[ {n}^{\pm 1/n} \rightarrow 1\;\left( {n \rightarrow \infty }\right) . \]\n\nThis proves\n\n\[ {\overline{\lim }}_{n \rightarrow \infty }{\left| n{a}_{n}\right| }^{1/\left( {n - 1}\right) } = {\overline{\lim }}_{n \rightarrow \infty }{\left| n{a}_{n}\right| }^{1/n} = {\overline{\lim }}_{n \rightarrow \infty }{\left| {a}_{n}\right| }^{1/n}, \]\n\nwhich concludes the proof.
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Yes
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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 \( \left( {f \circ g}\right) \left( X\right) = X \) .\n\n(ii) \( {a}_{0} = f\left( 0\right) = 0 \) and \( {a}_{1} = {f}^{\prime }\left( 0\right) \neq 0 \) .\n\nWhen they are satisfied, there is a unique formal power series \( g \) as required by (i), and this formal power series also satisfies \( \left( {g \circ f}\right) \left( X\right) = X \) .
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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\[ \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}g{\left( X\right) }^{n} = {a}_{0} + {a}_{1}{b}_{1}X + {X}^{2}\left( \cdots \right) = X. \]\n\nIn particular, \( {a}_{0} = 0 \) and \( {a}_{1}{b}_{1} = 1 \) ; hence \( {a}_{1} \neq 0 \).\n\n(ii) \( \Rightarrow \) (i) The equality \( \left( {f \circ g}\right) \left( X\right) = X \) requires that \( {a}_{1}{b}_{1} = 1 \) and that the coefficient of \( {X}^{n} \) in \( {a}_{1}g\left( X\right) + \cdots + {a}_{n}g{\left( X\right) }^{n} \) vanishes (for \( n \geq 2 \) ) (indeed, the coefficient of \( {X}^{n} \) in \( {a}_{m}g{\left( X\right) }^{m} \), whenever \( m > n \), vanishes). This coefficient of \( {X}^{n} \) is determined by an expression\n\n\[ {a}_{1}{b}_{n} + {P}_{n}\left( {{a}_{2},\ldots ,{a}_{n};{b}_{1},\ldots ,{b}_{n - 1}}\right) \]\n\nwith known polynomials \( {P}_{n} \) having integral coefficients (not that it matters, but these polynomials are linear in the first variables \( {a}_{i} \) ; cf. (V.4.2)). The hypothesis \( {a}_{1} \neq 0 \in K \) makes it possible to choose iteratively the coefficients \( {b}_{n} \) according to\n\n\[ {b}_{n} = - {a}_{1}^{-1}{P}_{n}\left( {{a}_{2},\ldots ,{a}_{n};{b}_{1},\ldots ,{b}_{n - 1}}\right) \;\left( {n \geq 2}\right) . \]\n\nThese choices furnish the required inverse formal power series \( g \).\n\nFinally, if \( f \) satisfies (ii) and \( g \) is chosen as in (i), then \( {b}_{0} = 0 \) and \( {b}_{1} = \) \( 1/{a}_{1} \neq 0 \), so that we may apply \( \left( i\right) \) to \( g \) and choose a formal power series \( h \) with \( \left( {g \circ h}\right) \left( X\right) = X \) . The associativity of composition shows that\n\n\[ h\left( X\right) = \left( \underset{\text{id }}{\underbrace{f \circ g}}\right) \circ h\left( X\right) = f \circ \left( \underset{\text{id }}{\underbrace{g \circ h}}\right) \left( X\right) = f\left( X\right) . \]\n\nThis proves \( g \circ f\left( X\right) = g \circ h\left( X\right) = X \) .
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Yes
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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\[ D\left( {f \circ g}\right) \left( Y\right) = {Df}\left( X\right) {Dg}\left( Y\right) = {Df}\left( {g\left( Y\right) }\right) {Dg}\left( Y\right) . \]
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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\[ \omega \left( f\right) \geq k \Rightarrow \omega \left( {Df}\right) \geq k - 1 \Rightarrow \omega \left\lbrack {{Df}\left( {g\left( Y\right) }\right) {Dg}\left( Y\right) }\right\rbrack \geq k - 1. \]\n\nThe identity \( D\left( {f \circ g}\right) \left( Y\right) = {Df}\left( {g\left( Y\right) }\right) {Dg}\left( Y\right) \), valid on the dense subspace of polynomials \( f \in K\left\lbrack X\right\rbrack \), extends by continuity to \( f \in K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) .
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Yes
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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.
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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 \( \left| {x}_{n}\right| \nearrow 1 \) ; hence we have\n\n\[ {M}_{1}f = \mathop{\sup }\limits_{{r \nearrow 1}}{M}_{r}f\overset{!}{ \leq }\mathop{\sup }\limits_{{x \in A}}\left| {f\left( x\right) }\right| \]
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Yes
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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 multiplicative, i.e., \( {M}_{r}\left( {f\bar{g}}\right) = {M}_{r}\left( f\right) {M}_{r}\left( g\right) \) when \( f \) and \( g \) belong to this subspace.
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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\[ \n{M}_{r}\left( {fg}\right) = {M}_{r}\left( f\right) {M}_{r}\left( g\right) \n\]\n\nis true if \( r \) is a regular radius for \( f, g \), and \( {fg} \), since it is the common value (V.2.2)\n\n\[ \n\left| {{fg}\left( x\right) }\right| = \left| {f\left( x\right) }\right| \left| {g\left( x\right) }\right| \;\left( {\left| x\right| = r, x \in {\Omega }_{p}}\right) . \n\]\n\nThe general result follows by density of regular values and continuity of the maps\n\n\[ \nr \mapsto {M}_{r}\left( f\right) ,\;r \mapsto {M}_{r}\left( g\right) ,\;r \mapsto {M}_{r}\left( {fg}\right) . \n\]
|
Yes
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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\\]\n\nand\n\n\\[ \n- 1 \leq {\operatorname{ord}}_{p}{a}_{n} \leq 0\text{ if }p < n < {p}^{2},\;\ldots . \n\\]
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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\n\\[ \n\frac{-1}{p - 1} > \frac{-1}{{p}^{2} - p} > \frac{-1}{{p}^{3} - {p}^{2}} > \cdots \;\left( { \rightarrow 0}\right) . \n\\]\n\nThey correspond to critical radii\n\n\\[ \n{p}^{-\frac{1}{p - 1}} < {p}^{-\frac{1}{{p}^{2} - p}} < {p}^{-\frac{1}{{p}^{3} - {p}^{2}}} < \cdots \;\left( { \rightarrow 1}\right) , \n\\]\n\nand we recognize the sequence\n\n\\[ \n{r}_{p} = {\left| p\right| }^{\frac{1}{p - 1}} < {r}_{p}^{\prime } = {r}_{p}^{1/p} < {r}_{p}^{\prime \prime } = {r}_{p}^{1/{p}^{2}} < \cdots \;\left( { \rightarrow 1}\right) . \n\\]\n\nBetween two consecutive critical radii, the absolute value of the logarithm coincides with the absolute value of the dominant monomial. We already know that\n\n\\[ \n\left| {\log \left( {1 + x}\right) }\right| = \left| x\right| \;\left( {0 \leq \left| x\right| < {r}_{p}}\right) , \n\\]\n\n- the isometry domain of log (inverted by exp) - and see further that\n\n\\[ \n{r}_{p} < \left| {\log \left( {1 + x}\right) }\right| = \left| \frac{{x}^{p}}{p}\right| = p{\left| x\right| }^{p} < p{r}_{p}\;\left( {{r}_{p} < \left| x\right| < {r}_{p}^{\prime }}\right) , \n\\]\n\nwhere \\( {r}_{p}^{\prime } = {r}_{p}^{1/p} \\) is the next critical radius. Quite generally,\n\n\\[ \n{p}^{j - 1}{r}_{p} < \left| {\log \left( {1 + x}\right) }\right| = \left| \frac{{x}^{{p}^{j}}}{{p}^{j}}\right| = {p}^{j}{\left| x\right| }^{{p}^{j}} < {p}^{j}{r}_{p} \n\\]\n\nfor\n\n\\[ \n{r}_{p}^{1/{p}^{j - 1}} < \left| x\right| < {r}_{p}^{1/{p}^{j}} \n\\]\n\nHere we see how \\( \left| {\log \left( {1 + x}\right) }\right| \\) increases: We already knew by (V.4.4) that it can be arbitrarily large, since \\( \log : 1 + {\mathbf{M}}_{p} \rightarrow {\mathbf{C}}_{p} \\) is surjective. On the other hand, the zeros of \\( \log \left( {1 + x}\right) \\) can occur only when \\( \left| x\right| \\) is equal to a critical radius. This gives an independent proof of (II.4.4) for the estimates of \\( \left| {\zeta - 1}\right| \\) when \\( \zeta \in {\mu }_{{p}^{\infty }} \\) .
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Yes
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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 sphere of radius \( r \) in \( K \) . More precisely, if \( \mu < v \) are the extreme indices for which \( \left| {a}_{n}\right| {r}^{n} = {M}_{r}f \), then \( f \) has exactly \( v - \mu \) zeros (counting multiplicities) on the critical sphere \( \left| x\right| = r \) of \( K \) : There is a polynomial \( P \in K\left\lbrack X\right\rbrack \) of degree \( v - \mu \) and a convergent power series \( g \in K\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) with
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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 \( a \in K \) with \( \left| a\right| = r \) . Replace \( f \) by \( {f}_{a}\left( X\right) = f\left( {aX}\right) \) having \( r = 1 \) as critical radius. This converts \( f \) into a series having a radius of convergence \( {r}_{{f}_{a}} = {r}_{f}/\left| a\right| > 1 \), and in particular, \( {f}_{a} \in K\{ X\} \) . We can similarly replace \( f \) by the multiple \( f/{a}_{v} \) and assume \( \left| {a}_{\mu }\right| = \left| {a}_{v}\right| = {M}_{r}f = 1 \) (and \( {a}_{v} = 1 \) ). To sum up, it is sufficient to study the normalized situation
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Yes
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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 algebraic over \( K \) .
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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 \( K \) -automorphism of \( {\Omega }_{p} \), it is continuous and isometric (III.3.2):\n\n\[ f\left( {a}^{\sigma }\right) = f{\left( a\right) }^{\sigma } = 0,\;\left| {a}^{\sigma }\right| = \left| a\right| = r. \]\n\nHence \( a \) has a finite number of conjugates contained in the finite set of roots of \( f \) on the sphere \( \left| x\right| = r \) of \( {\Omega }_{p} \) . By Galois theory, this proves that \( a \) is algebraic over \( K \) . The same argument shows that the product \( P = \mathop{\prod }\limits_{\xi }\left( {X - \xi }\right) \in {K}^{a}\left\lbrack X\right\rbrack \) extended over all roots of \( f \) having absolute value \( r \) (all multiplicities counted) has coefficients fixed by all \( K \) -automorphisms of \( {K}^{a} \) and hence coefficients in \( K \) . This is a monic polynomial \( P \in K\left\lbrack X\right\rbrack \) of degree \( v - \mu \) .
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Yes
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