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Theorem 8. Let \( \Phi \) be a field containing \( m \) distinct \( m \) -th roots of 1 and let \( N \) be a subgroup of \( {\Phi }^{ * } \) containing \( {\Phi }^{*m} \) such that \( N/{\Phi }^{*m} \) is finite. Then there exists a Kummer \( {m}^{\prime } \) -extension \( \mathrm{P}/\Phi \) with \( {m}^{\prime } \mid ...
Proof. The foregoing analysis of Kummer extensions gives the clue to the definition of \( \mathrm{P}/\Phi \) . In view of this, we are led to choose \( {\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{r} \) in the given group \( N \) so that the cosets \( {\alpha }_{i}{\Phi }^{*m} \) generate \( N/{\Phi }^{*m} \) . Let \...
Yes
Lemma 1. Let \( \mu \geq 1,0 \leq k \leq m - 1, a = \left( {a}_{\nu }\right), b = \left( {b}_{\nu }\right) \) , \( 0 \leq \nu \leq m - 1,{a}_{\nu },{b}_{\nu }{\varepsilon I}\left\lbrack {{x}_{i},{y}_{j}}\right\rbrack \) . Write \( {a}^{\varphi } = \left( {a}^{\left( \nu \right) }\right) ,{b}^{\varphi } = \left( {b}^{\l...
Proof. We have \( {a}^{\left( 0\right) } = {a}_{0},{b}^{\left( 0\right) } = {b}_{0} \), so the result is clear for \( k = 0 \) . To prove the result by induction on \( k \) we may assume that both sets (16) and (17) hold for \( 0 \leq \nu \leq k - 1 \) and prove that under these conditions \( {a}_{k} \equiv {b}_{k}\lef...
Yes
Theorem 10. \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) is a commutative ring.
Proof. Let \( a = \left( {a}_{\nu }\right), b = \left( {b}_{\nu }\right), c = \left( {c}_{\nu }\right) \) be any three elements of \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) . Then we have just seen that we have a homomorphism of \( {\mathfrak{J}}_{m} \) into \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) s...
Yes
Theorem 11. \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) is a ring of characteristic \( {p}^{m} \) .
Proof. It suffices to show that the order of 1 in the additive group of \( {\mathfrak{W}}_{m}\left( \mathfrak{A}\right) \) is \( {p}^{m} \) . We have seen that \( {p1} = {1}^{VR} = (0,1,0 \) , \( \cdots ,0) \) and by iterating (27) we obtain \( {p}^{2}1 = \left( {0,0,1,0,\cdots }\right) \) etc. This shows that \( {p}^{...
Yes
Theorem 13. Let \( s \rightarrow {\mu }_{s} \) be a mapping of \( G \) into \( {\mathfrak{W}}_{m}\left( \mathrm{P}\right) \) such that \( {\mu }_{st} = {\mu }_{s}{}^{t} + {\mu }_{t}, s,{t\varepsilon G} \) . Then there exists an element \( {\sigma \varepsilon }{\mathfrak{W}}_{m}\left( \mathrm{P}\right) \) such that \( {...
Proof. The proof is identical with that of the special case of Galois extension fields treated in Theorem 1.20. We choose \( \rho \) in \( {\mathfrak{W}}_{m}\left( \mathrm{P}\right) \) so that \( T{\left( \rho \right) }^{-1} \) exists in \( {\mathfrak{W}}_{m}\left( \Phi \right) \) and we let \( \tau = \) \( T{\left( \r...
Yes
Theorem 14. Let \( \Phi \) be a field of characteristic \( p \neq 0,\mathrm{P}/\Phi \) an abelian p-extension whose Galois group \( G \) is of exponent \( {p}^{e} \) and let \( {\mathfrak{W}}_{m}\left( \mathrm{P}\right) \) be the ring of Witt vectors of length \( m \) over \( \mathrm{P} \) where \( m \geq e \) . Let \(...
The proof of the last statement is exactly like that of the corresponding statement of Theorem 7. We leave it to the reader to check the details.
No
Lemma 2. Let \( \beta = \left( {{\beta }_{0},{\beta }_{1},\cdots ,{\beta }_{m - 1}}\right) \varepsilon {\mathfrak{W}}_{m}\left( \Phi \right) \). Then there exists a finite dimensional separable extension field \( \mathrm{P} \) of \( \Phi \) such that \( \mathrm{P} = \Phi \left( \rho \right) \equiv \Phi \left( {{\rho }_...
Proof. If \( m = 1 \) we just have to construct a separable extension \( \mathrm{P} = \Phi \left( \rho \right) \) generated by a root \( \rho \) of an equation \( {x}^{p} - x = \beta ,\beta \) a given element in \( \Phi \). Since the derivative \( {\left( {x}^{p} - x - \beta \right) }^{\prime } = - 1 \) the given equat...
Yes
Lemma 3. If \( {\beta }_{0},\cdots ,{\beta }_{m - 1}\mathrm{e}\Phi \), then \( {\beta }_{0}\mathrm{e}\mathfrak{P}\left( \Phi \right) \) if and only if \( \beta = \left( {{\beta }_{0},{\beta }_{1},\cdots ,{\beta }_{m - 1}}\right) \) satisfies \( {p}^{m - 1}{\beta \varepsilon }\Re \left( {{\mathfrak{W}}_{m}\left( \Phi \r...
Proof. By (27), \( {p}^{m - 1}\beta = \left( {0,\cdots ,0,{\beta }_{0}^{{p}^{m - 1}}}\right) \) . We have \( (0,\cdots \) , \( \left. {0,{\beta }_{0}}\right) - \left( {0,\cdots ,0,{\beta }_{0}{p}^{m - 1}}\right) = \left( {0,\cdots ,0,{\beta }_{0}}\right) - \left( {0,\cdots ,0,{\beta }_{0}{}^{p}}\right) + \) \( \left( {...
Yes
Theorem 16. Let \( \Phi \) be a field of characteristic \( p \neq 0 \) . Then there exist cyclic extensions of \( {p}^{m} \) dimensions, \( m = 1,2,3,\cdots \) over \( \Phi \) if and only if there exist such extensions of \( p \) dimensions. The condition for this is \( \Phi \neq \mathfrak{P}\left( \Phi \right) \) .
Proof. We have seen that there exists a cyclic extension of \( p \) dimensions over \( \Phi \) if and only if \( \Phi \neq \mathfrak{P}\left( \Phi \right) \) . Suppose this condition holds and choose \( {\beta }_{0}{\varepsilon \Phi },\varepsilon \$ \varphi \left( \Phi \right) \) . Let \( \beta = \left( {{\beta }_{0},{...
Yes
Theorem 1. Any field has an algebraic closure.
Proof. If \( \Phi \) is a given field, then we can imbed \( \Phi \) in a set \( \Omega \) which is very large compared to \( \Phi \) in the following sense: if \( \Phi \) is finite, then \( \Omega \) is not countable and, if \( \Phi \) is infinite, then \( \left| \Omega \right| > \left| \Phi \right| \) . We now make ex...
Yes
Theorem 2. Let \( \alpha \rightarrow \bar{\alpha } \) be an isomorphism of a field \( \Phi \) onto a field \( \Phi \) and let \( \Omega \) be a set of polynomials of positive degree contained in \( \Phi \left\lbrack x\right\rbrack ,\bar{\Omega } \) the set of images of the \( {f\varepsilon \Omega } \) under the isomorp...
Proof. We consider the collection \( \Delta \) of isomorphisms \( s \) of subfields of \( \mathrm{P}/\Phi \) onto subfields of \( \overline{\mathrm{P}}/\Phi \) which coincide with the given isomorphism \( \alpha \rightarrow \bar{\alpha } \) of \( \Phi \) onto \( \Phi \) . We can partially order \( \Delta = \{ s\} \) by...
Yes
Theorem 3. Any field of characteristic 0 is perfect and a field \( \Phi \) of characteristic \( p \neq 0 \) is perfect if and only if \( \Phi = {\Phi }^{p} \), that is, every element of \( \Phi \) is a p-th power in \( \Phi \) .
Proof. The first statement is clear since inseparable polynomials exist only for characteristic \( p \neq 0 \) . Now let \( \Phi \) be of characteristic \( p \neq 0 \) and suppose \( {\Phi }^{p} \subset \Phi \) . Let \( \alpha \) be an element of \( \Phi \) which is not a \( p \) -th power in \( \Phi \) . Then we know ...
Yes
Lemma 1. Any finite subset of \( \mathrm{P} \) is contained in a subfield \( \mathrm{E}/\Phi \) which is finite dimensional Galois.
Proof. Let \( f \) be a polynomial which is a product of a finite number of polynomials contained in the set \( \dot{\Omega } \) . Then it is clear that \( \mathrm{P} \) contains a splitting field \( {\mathrm{P}}_{f}/\Phi \) of \( f \) . Moreover, we know that \( {\mathrm{P}}_{f} \) is finite dimensional Galois over \(...
Yes
Lemma 2. \( \Phi = I\left( G\right) \), that is, the only elements of \( \mathrm{P} \) which are \( G \) -invariant are the elements of \( \Phi \) .
Proof. We have to show that, if \( {\rho \varepsilon }\mathrm{P},\sharp \Phi \), then there exists an automorphism \( s \) of \( \mathrm{P} \) over \( \Phi \) such that \( {\rho }^{s} \neq \rho \) . By Lemma 1, \( \rho \) is contained in a subfield \( \mathrm{E}/\Phi \) which is finite dimensional Galois over \( \Phi \...
Yes
Theorem 6. Algebraic dependence in \( \mathrm{P}/\Phi \) is a dependence relation in the sense of \( I - {IV} \) .
Proof. I. This is evident. II. This was proved before. III. Let \( \xi \) be algebraic over \( \Phi \left( S\right) \) and suppose every \( {\eta \varepsilon S} \) is algebraic over \( \Phi \left( T\right) \) . Consider the subset \( \mathrm{A} \) of \( \mathrm{P} \) of elements which are algebraic over \( \Phi \left( ...
Yes
Theorem 9. If \( \mathrm{P} \) is an algebraic extension of \( \Phi \) (possibly infinite dimensional), then \( \mathrm{P} \) is separable over \( \Phi \) if and only if \( \mathrm{P} \) is linearly disjoint to \( {\Phi }^{{p}^{-1}} \) over \( \Phi \) .
Proof. We recall that an algebraic element \( \rho \) of \( \mathrm{P} \) over \( \Phi \) is separable if and only if \( {\rho \varepsilon \Phi }\left( {\rho }^{p}\right) \) (Lemma 2 of \( §{1.9} \) ). Suppose first that \( \mathrm{P} \) and \( {\Phi }^{{p}^{-1}} \) are linearly disjoint over \( \Phi \) and let \( {\rh...
Yes
Theorem 10. If \( \mathrm{P} \) is purely transcendental over \( \Phi \), then \( \mathrm{P} \) is linearly disjoint to \( {\Phi }^{{p}^{-1}} \) over \( \Phi \) .
Proof. Our assumption is that \( \mathrm{P} = \Phi \left( B\right) \) where \( B \) is an algebraically independent set. We have seen also that \( \mathrm{P} \) is linearly disjoint to \( {\Phi }^{{p}^{-1}} \) over \( \Phi \) if and only if the subalgebra \( \Phi \left\lbrack B\right\rbrack \) of polynomials in the ele...
Yes
Theorem 11. (1) If \( \mathrm{P} \) is separable over \( \Phi \) and \( \mathrm{E} \) is a subfield of \( \mathrm{P} \) over \( \Phi \), then \( \mathrm{E} \) is separable over \( \Phi \) . (2) If \( \mathrm{P} \) is separable over \( \mathrm{E} \) and \( \mathrm{E} \) is separable over \( \Phi \), then \( \mathrm{P} \...
Proof. We may assume the characteristic is \( p \neq 0 \) . (1) This is clear since the linear disjointness of \( \mathrm{P} \) and \( {\Phi }^{{p}^{-1}} \) implies the linear disjointness of \( \mathrm{E} \) and \( {\Phi }^{{p}^{-1}} \) . (2) We are assuming that \( {\Phi }^{{p}^{-1}} \) is linearly disjoint to \( \ma...
Yes
Theorem 12. If \( \mathfrak{A} \) is a subalgebra of \( \mathfrak{B} \) and \( D \) is a derivation of \( \mathfrak{A} \) into \( \mathfrak{B} \), then \( s : a \rightarrow a + \left( {aD}\right) t \) is an isomorphism of \( \mathfrak{A} \) into the algebra of dual numbers \( \mathfrak{B} \otimes \mathfrak{T} \) over \...
We shall now obtain some simple consequences of this connection between derivations and isomorphisms. First, let \( \mathfrak{X} \) be a set of generators of the subalgebra \( \mathfrak{A} \) of the algebra \( \mathfrak{B} \) and let \( {D}_{1} \) and \( {D}_{2} \) be derivations of \( \mathfrak{A} \) into \( \mathfrak...
Yes
Theorem 13. Let \( \mathrm{P} \) be a field over \( \Phi \) , \( \mathfrak{A} \) a subalgebra of \( \mathrm{P}/\Phi \) (containing 1), M a multiplicatively closed subset of non-zero elements of \( \mathfrak{A} \) containing 1, and let \( {\mathfrak{A}}_{M} \) be the subalgebra of \( \mathrm{P} \) of elements of the for...
Proof. Let \( s \) be the isomorphism \( a \rightarrow a + \left( {aD}\right) t \) of \( \mathfrak{A} \) into \( \mathrm{P} \otimes \mathfrak{T} \) . If \( a \neq 0 \), then \( {a}^{s} = a + \left( {aD}\right) t \) has the inverse \( {a}^{-1} - \) \( {a}^{-2}\left( {aD}\right) t \) since\n\n\[ \left( {a + \left( {aD}\r...
Yes
Theorem 14. Let \( \mathfrak{A} \) be a subalgebra over \( \Phi \) of the field \( \mathrm{P}/\Phi \) and let \( {\xi }_{1},{\xi }_{2},\cdots ,{\xi }_{m},{\eta }_{1},{\eta }_{2},\cdots ,{\xi }_{m} \) be elements of \( \mathrm{P}, D \) a derivation of \( \mathfrak{A} \) into \( \mathrm{P} \) . Let \( \mathfrak{K} \) be ...
(14)\n\n\[ {g}^{D}\left( {{\xi }_{1},\cdots ,{\xi }_{m}}\right) + \mathop{\sum }\limits_{{i = 1}}^{m}{\left( \frac{\partial g}{\partial {x}_{i}}\right) }_{{x}_{j} = {\xi }_{j}}{\eta }_{i} = 0 \] for every \( {g\varepsilon }\mathfrak{X} \) . If the extension exists, then it is unique.
Yes
Theorem 15. Let \( \mathrm{P} = \Phi \left( {{\xi }_{1},{\xi }_{2},\cdots ,{\xi }_{m}}\right) \) a field of algebraic functions over \( \Phi \) . Let \( \mathfrak{X} \) be a set of generators for the ideal \( \mathfrak{K} \) of polynomials \( f\left( {{x}_{1},{x}_{2},\cdots ,{x}_{m}}\right) \) such that \( f\left( {{\x...
If \( \mathfrak{X} = \left\{ {{g}_{1},{g}_{2},\cdots ,{g}_{r}}\right\} \), then it is clear from the definition of \( {d}_{f} \) and from the relation between dimensionality and determinantal rank (Vol. II, p. 22) that \( {\left\lbrack d\mathfrak{X} : \mathrm{P}\right\rbrack }_{R} \) is the rank of the matrix\n\n(22)\n...
Yes
Theorem 16. If \( \mathrm{P} = \Phi \left( {{\xi }_{1},{\xi }_{2},\cdots ,{\xi }_{m}}\right) \), then \( {\left\lbrack {\mathfrak{D}}_{\Phi }\left( \mathrm{P}\right) : \mathrm{P}\right\rbrack }_{R} \) is the smallest integer \( s \) such that there exists a subset \( \left\{ {{\xi }_{{i}_{1}},{\xi }_{{i}_{2}},\cdots ,{...
Proof. As before, we consider the mapping \( D \rightarrow \left( {{\xi }_{1}D,{\xi }_{2}D}\right. \) , \( \left. {\cdots ,{\xi }_{m}D}\right) \) of \( \mathfrak{D} = {\mathfrak{D}}_{\Phi }\left( \mathrm{P}\right) \) into \( {\mathrm{P}}^{\left( m\right) } \) . We know that this is a \( \mathrm{P} \) isomorphism into \...
Yes
Theorem 17. Let \( \mathrm{P} \) be an arbitrary field of characteristic \( \neq 0 \) , \( \Phi \) a subfield and \( \mathrm{E} \) an intermediate field. Let \( B \) be a p-basis of \( \mathrm{E} \) over \( \Phi \) . Let \( \delta \) be an arbitrary mapping of \( B \) into \( \mathrm{P} \) . Then there exists one and o...
Proof. As we indicated, there is no loss in generality in assuming \( \mathbf{E} \) is purely inseparable of exponent \( \leq 1 \) over \( \Phi \) . Also, we may suppose \( \mathrm{E} \supset \Phi \) which means that \( B \) is non-vacuous and the exponent of \( \mathrm{E}/\Phi \) is exactly one. Let \( \epsilon \) e \...
Yes
Corollary 1. \( {\left\lbrack {\mathfrak{D}}_{\Phi }\left( \mathrm{E},\mathrm{P}\right) : \mathrm{P}\right\rbrack }_{R} < \infty \) if and only if \( \mathrm{E}/\Phi \) has a finite p-basis. Then \( {\left\lbrack {\mathfrak{D}}_{\Phi }\left( \mathrm{E},\mathrm{P}\right) : \mathrm{P}\right\rbrack }_{R} = \left| B\right|...
Proof. Let \( B \) be a \( p \) -basis for \( \mathrm{E} \) over \( \Phi \) . Let \( \Delta \left( {B,\mathrm{P}}\right) \) be the set of mappings of \( B \) into \( \mathrm{P} \) which we consider as a right vector space over \( \mathrm{P} \) in the obvious way: \( \left( {{\delta }_{1} + {\delta }_{2}}\right) \left( ...
Yes
Every derivation of \( \mathrm{E}/\Phi \) into \( \mathrm{P}/\Phi \) can be extended to a derivation of \( \mathrm{P}/\Phi \) if and only if the elements of any p-basis \( B \) of \( \mathrm{E}/\Phi \) are p-independent in \( \mathrm{P}/\Phi \) .
Proof. If the condition holds, then \( B \) can be imbedded in a \( p \) - basis \( C \) of \( \mathrm{P} \) over \( \Phi \) . If \( D \) is a derivation of \( \mathrm{E}/\Phi \) into \( \mathrm{P}/\Phi \), then the restriction \( {\delta }_{B} \) of \( D \) to \( B \) can be extended to a mapping \( {\delta }_{C} \) o...
Yes
Theorem 19 (Jacobson). Let \( \mathrm{P} \) be a field of characteristic \( p \neq 0 \) and let \( \mathfrak{D} \) be a restricted \( \mathrm{P} \) -Lie algebra of derivations in \( \mathrm{P} \) such that \( {\left\lbrack \mathfrak{D} : \mathrm{P}\right\rbrack }_{R} = m < \infty \) . Then: (1) if \( \Phi \) is the sub...
Proof. The idea of the proof we shall give is basically the same as that we used for the Galois theory of automorphisms: we shall use the given set \( \mathfrak{D} \) to define a set of endomorphisms \( \mathfrak{A} \) satisfying the hypotheses of the Jacobson-Bourbaki theorem (Th. 1.2). In the present case we let \( \...
Yes
Theorem 20. Let \( \mathrm{P}/\Phi \) be a field of characteristic \( p \neq 0 \), E a subfield of \( \mathrm{P}/\Phi ,{D}^{\left( m\right) } \) a higher derivation of rank \( m \) and order \( q \) of \( \mathrm{E}/\Phi \) into \( \mathrm{P}/\Phi \) . Let \( \Gamma \) be the subfield of \( {D}^{\left( m\right) } \) -c...
Proof. We have to show that \( {\epsilon }^{{p}^{e}}{\varepsilon \Gamma } \) for every \( {\epsilon \varepsilon }\mathrm{E} \) and that there exists an \( \epsilon \mathrm{e}\mathrm{E} \) such that \( {\epsilon }^{{p}^{e - 1}} \notin \Gamma \) . The first is clear from (46) since\n\n\[{\left( {\epsilon }^{{p}^{e}}\righ...
Yes
Theorem 21. Let \( \mathrm{P}/\Phi \) and \( \mathrm{E}/\Phi \) be extension fields of \( \Phi \) . (1) If \( \mathrm{P}/\Phi \) is separable and \( \mathrm{E}/\Phi \) is purely inseparable, then \( \mathrm{P} \) \( \otimes \Phi \mathrm{E} \) is a field. On the other hand, if \( \mathrm{P}/\Phi \) is not separable, the...
Proof. In (1) and the first part of (3) we may assume the characteristic is \( p \neq 0 \) . In all cases we write \( \mathrm{P} \otimes \mathrm{E} \) for \( \mathrm{P} \otimes * \mathrm{E} \) and we identify \( \mathrm{P} \) and \( \mathrm{E} \) with subalgebras of \( \mathrm{P} \otimes \mathrm{E} = \mathrm{{PE}} \) ....
Yes
Theorem 22. Let \( \mathrm{P} \) be purely transcendental over \( \Phi \), say, \( \mathrm{P} = \) \( \Phi \left( B\right) \) where \( B \) is a transcendency basis and let \( \mathrm{E}/\Phi \) be arbitrary. Then \( \mathrm{P} \otimes * \mathrm{E} \) has no zero-divisors, and if \( \Omega \) is its field of fractions,...
Proof. As usual, we consider \( \mathrm{P} \) and \( \mathrm{E} \) as subalgebras of \( \mathrm{P} \otimes \Phi \mathrm{E} \) . Since \( B \) is an algebraically independent set, the set \( M \) of distinct monomials \( {\beta }_{1}{}^{{k}_{1}}{\beta }_{2}{}^{{k}_{2}}\cdots {\beta }_{r}{}^{{k}_{r}},{k}_{i} \geq 0 \) in...
Yes
Theorem 23. If \( \mathrm{P}/\Phi \) is separable and \( \mathrm{E}/\Phi \) is arbitrary, then \( \mathrm{P}{ \otimes }_{\Phi }\mathrm{E} \) has no non-zero nilpotent elements.
Proof. It is clear that it suffices to prove this result under the additional assumption that \( \mathrm{P} \) is finitely generated. Then \( \mathrm{P} \) is separably generated, so that \( \mathrm{P} \) has a transcendency basis \( B \) such that \( \mathrm{P} \) is separable algebraic over \( \Phi \left( B\right) \)...
Yes
Lemma 1. Let \( \left( {\Gamma, s, t}\right) \) be a field composite of the fields \( \mathbf{E} \) over \( \Phi \) and \( \mathrm{P} \) over \( \Phi \) . Suppose there exists a transcendency basis \( B \) for \( \mathrm{E} \) over \( \Phi \) and a transcendency basis \( {B}^{\prime } \) for \( \mathrm{P} \) over \( \P...
We remark also that if the condition of the lemma holds for \( B \) and \( {B}^{\prime } \), then \( {B}^{s} \cup {B}^{\prime t} \) is a transcendency basis for \( \Gamma \) . For, it is clear that the elements of \( {\mathbf{E}}^{s} \) and of \( {\mathbf{P}}^{t} \) are algebraic over \( \Phi \left( {{B}^{s} \cup }\rig...
Yes
Lemma 3. Let \( B \) and \( {B}^{\prime } \) be transcendency bases for \( \mathrm{E}/\Phi \) and \( \mathrm{P}/\Phi \) respectively. Then every element of \( \mathrm{E} \otimes \Phi \mathrm{P} \) is integral over \( \Phi \left( B\right) \Phi \left( {B}^{\prime }\right) \) .
Proof. Since \( \mathrm{E} \) and \( \mathrm{P} \) are algebraic over \( \Phi \left( B\right) \) and \( \Phi \left( {B}^{\prime }\right) \) respectively, it is clear that the elements of \( \mathbf{E} \) and of \( \mathbf{P} \) are integral over \( \Phi \left( B\right) \Phi \left( {B}^{\prime }\right) \) . Since \( \ma...
Yes
Theorem 2. If \( \varphi \) is a non-archimedean real valuation, then \( \varphi \left( {\alpha + \beta }\right) \leq \max \left( {\varphi \left( \alpha \right) ,\varphi \left( \beta \right) }\right) \) for every \( \alpha ,\beta \) in \( \Phi \) .
Proof. We have\n\n\[ \varphi {\left( \alpha + \beta \right) }^{n} = \varphi \left( {{\alpha }^{n} + \left( \begin{array}{l} n \\ 1 \end{array}\right) {\alpha }^{n - 1}\beta + \cdots + {\beta }^{n}}\right) \]\n\n\[ \leq \varphi {\left( \alpha \right) }^{n} + \varphi {\left( \alpha \right) }^{n - 1}\varphi \left( \beta \...
Yes
Theorem 3. Any archimedean real valuation of the rationals is equivalent to the absolute value valuation.
Proof (Artin). Let \( n \) and \( {n}^{\prime } \) be integers \( > 1 \) and write \( {n}^{\prime } = {a}_{0} \) \( + {a}_{1}n + \cdots + {a}_{k}{n}^{k},0 \leq {a}_{i} < n,{a}_{k} \neq 0 \) . Then,\n\n\[ \varphi \left( {n}^{\prime }\right) \leq \varphi \left( {a}_{0}\right) + \varphi \left( {a}_{1}\right) \varphi \left...
Yes
Theorem 4. Any non-trivial non-archimedean real valuation of the rationals is equivalent to a p-adic valuation for some prime p.
Proof. We have \( \varphi \left( n\right) \leq 1 \) for every integer \( n \) . If \( \varphi \left( n\right) = 1 \) for every integer, then \( \varphi \) is trivial. Hence there exist non-zero integers \( b \) such that \( \varphi \left( b\right) < 1 \) . Let \( \mathfrak{P} \) be the collection of integers \( b \) sa...
Yes
Theorem 6. Let \( {\Phi }_{i}, i = 1,2 \), be a complete field with a valuation \( {\varphi }_{i} \) and \( {\Phi }_{i} \) a dense subfeld of \( {\bar{\Phi }}_{i} \) . Let \( s \) be an isometric isomorphism of \( {\Phi }_{1} \) onto \( {\Phi }_{2} \) . Then \( s \) has a unique extension to an isometric isomorphism of...
This result implies, in particular, that, if \( {\Phi }_{1} \) and \( {\Phi }_{2} \) are completions of the same field \( \Phi \), then there exists an isometric isomorphism of \( {\Phi }_{1}/\Phi \) onto \( {\Phi }_{2}/\Phi \) . We just have to apply the theorem to the identity mapping in \( \Phi \) . In this sense th...
No
Lemma 1. Let \( \mathfrak{o} \) be a subring of a field \( \Phi \) and let \( \mathfrak{m} \) be a proper ideal in \( \mathfrak{o} \). If \( \alpha \) is a non-zero element of \( \Phi \) and \( \mathfrak{o}\left\lbrack \alpha \right\rbrack \) is the subring of \( \Phi \) generated by 0 and \( \alpha \), then either \( ...
Proof. Suppose the contrary: \( \mathfrak{m}\mathfrak{o}\left\lbrack \alpha \right\rbrack = \mathfrak{o}\left\lbrack \alpha \right\rbrack ,\mathfrak{m}\mathfrak{o}\left\lbrack {\alpha }^{-1}\right\rbrack = \mathfrak{o}\left\lbrack {\alpha }^{-1}\right\rbrack \). Then \( {1\varepsilon }\mathfrak{{mo}}\left\lbrack \alpha...
Yes
Lemma 2. Let \( \varphi \) be a valuation of a field \( \Phi ,{\Phi }_{0} \) a subfeld of finite co-dimension in \( \Phi \) . Then the value group of \( \Phi \) is order isomorphic to a subgroup of the value group of \( {\Phi }_{0} \) (relative to the restriction of \( \varphi \) ).
Proof. Let \( \xi \) e \( \Phi \) and let \( {\alpha }_{1}{\xi }^{{n}_{1}} + {\alpha }_{2}{\xi }^{{n}_{2}} + \cdots + {\alpha }_{k}{\xi }^{{n}_{k}} = 0 \) where the \( {\alpha }_{i} \neq 0 \) in \( {\Phi }_{0} \) and \( {n}_{1} > {n}_{2} > \cdots > {n}_{k} \) . As in the case of non-archimedean real valuations, if \( \...
Yes
Lemma 1. Let \( \\mathfrak{o} \) be a commutative ring, \( \\mathfrak{A} \) an ideal in \( \\mathfrak{o} \) and \( S \) a non-vacuous multiplicatively closed subset of \( \\mathfrak{o} \) such that \( \\mathfrak{A} \\cap S = \\varnothing \) . Then there exists a prime ideal \( \\mathfrak{P} \) in \( \\mathfrak{o} \) su...
Proof. Let \( U \) be the collection of ideals \( \\mathfrak{B} \) in \( \\mathfrak{o} \) such that: 1 . \( \\mathfrak{B} \\supseteq \\mathfrak{A},2.\\mathfrak{B} \\cap S = \\varnothing \) . Then \( U \) is non-vacuous since \( \\mathfrak{A} \) e \( U \) . We order the elements of \( U \) by inclusion. Let \( V \) be a...
Yes
Theorem 12. Let \( \mathfrak{A} \) be an ideal in the commutative ring \( \mathfrak{o} \) . Then the radical \( \Re \left( \mathfrak{A}\right) = \cap \mathfrak{P} \) the intersection of the prime ideals \( \mathfrak{B} \) containing \( \mathfrak{A} \) .
Proof. Let \( {a\varepsilon }\Re \left( \mathfrak{A}\right) \) and let \( \mathfrak{P} \) be a prime ideal containing \( \mathfrak{A} \) . A suitable power \( {a}^{n}\varepsilon \mathfrak{A} \) so \( {a}^{n}\varepsilon \mathfrak{P} \) . Since \( \mathfrak{P} \) is prime, this implies that \( {a\varepsilon }\widetilde{\...
Yes
Theorem 13. If the algebra \( \mathrm{P} = \Phi \left\lbrack {{\gamma }_{1},{\gamma }_{2},\cdots ,{\gamma }_{n}}\right\rbrack \) over \( \Phi \) generated by the \( {\gamma }_{i} \) is a field, then the \( {\gamma }_{i} \) are algebraic over \( \Phi \) .
Proof. Let \( \Phi \left\lbrack {{x}_{1},{x}_{2},\cdots ,{x}_{n}}\right\rbrack \) be the polynomial algebra over \( \Phi \) in indeterminates \( {x}_{i} \) and consider the homomorphism of this algebra onto \( \mathrm{P}/\Phi \) mapping \( {x}_{i} \rightarrow {\gamma }_{i},1 \leq i \leq n \) . Let \( \mathfrak{P} \) be...
Yes
Lemma 1. If \( \mathfrak{o} \) is a commutative ring (with an identity 1), any proper ideal \( \mathfrak{A} \) of \( \mathfrak{o} \) can be imbedded in a maximal ideal.
Proof. The proof is obtained as a special case of the argument in the proof of Lemma 1 of \( §{12} \) . We let \( S = \{ 1\} \), so \( S \) is multiplicatively closed and \( S \cap \mathfrak{A} = \varnothing \) . Let \( U \) be the set of ideals \( \mathfrak{B} \) such that \( \mathfrak{B} \supseteq \mathfrak{A} \) and...
Yes
Theorem 15. Let \( \mathrm{P} \) be a finite dimensional extension field of a field which is complete with respect to a non-trivial real valuation \( \varphi \) . Then if \( \varphi \) can be extended to a real valuation of \( \mathrm{P} \), this valuation is unique and is given by the formula\n\n(38)\n\n\[ \varphi \le...
Proof. Assume the extension \( \varphi \) exists and suppose there exists a \( {\rho \varepsilon }\mathrm{P} \) such that (38) does not hold. Then \( \varphi \left( {\rho }^{n}\right) \neq \varphi \left( {N\left( \rho \right) }\right) \), so \( \rho \neq 0 \) and either \( \varphi \left( {\rho }^{n}\right) < \varphi \l...
Yes
Lemma 3. Let \( \Phi \) be a field which is complete relative to a real valuation \( \varphi \) and let \( {x}^{2} - {ax} + b = 0 \) be an equation with coefficients \( a, b \) in \( \Phi \) such that \( \varphi {\left( a\right) }^{2} > {4\varphi }\left( b\right) \) . Then the equation has roots in \( \Phi \) .
Proof. A non-zero root \( \alpha \) of this equation will be a root of \( \alpha = a - b{\alpha }^{-1} \) . We shall obtain such a root as a limit of a sequence \( \left\{ {a}_{n}\right\} \) where \( {a}_{n} \) is defined recursively by \( {a}_{1} = \frac{1}{2}a,{a}_{n + 1} = \) \( a - b{a}_{n}{}^{-1} \) . We show firs...
Yes
Theorem 16 (Ostrowski). The only fields which are complete relative to a real archimedean valuation are the field of real numbers and the field of complex numbers.
Proof. Let \( \Phi \) be complete relative to the archimedean valuation \( \varphi \) . Then \( \Phi \) is of characteristic 0 and so it contains the rationals. Since any real archimedean valuation of the rationals is equivalent to the absolute value valuation and \( \Phi \) is complete, it is clear that \( \Phi \) con...
Yes
Theorem 17. If \( \Phi \) is complete relative to a real valuation \( \varphi \) and \( \mathrm{P} \) is a finite dimensional extension of \( \Phi \), then the valuation can be extended in one and only one way to \( \mathrm{P} \) . The extension is given by the formula (38). Moreover, \( \mathrm{P} \) is complete relat...
15. Extension of real valuations to finite dimensional extension fields. We now take up the problem of determining all the extensions of a real valuation defined in a field \( \Phi \) to a finite dimensional extension field \( \mathrm{P}/\Phi \) . The case in which \( \Phi \) is complete has been treated in the last se...
Yes
Theorem 19. Let \( \Phi \) be a field with a non-archimedean real valuation. Let \( \mathrm{P} \) be a finite dimensional extension field of \( \Phi ,{\psi }_{1},{\psi }_{2},\cdots \) , \( {\psi }_{h} \) the different valuations of \( \mathrm{P} \) which extend \( \varphi \) and let \( {e}_{i},{f}_{i} \) be the ramific...
Proof. Let \( {\mathrm{E}}_{i} \) be the completion of \( \mathrm{P} \) relative to \( {\psi }_{i} \) . Then for \( \Phi \) the completion of \( \Phi \) and \( {n}_{i} = \left\lbrack {{\mathrm{E}}_{i} : \bar{\Phi }}\right\rbrack \) we have \( \sum {n}_{i} \leq n \) and \( \sum {n}_{i} = \) \( n \) for \( \mathrm{P} \) ...
Yes
Theorem 1. If \( \Phi \) is real closed, then any element of \( \Phi \) is either a square or the negative of a square.
Proof. Let \( \alpha \) be an element of \( \Phi \) which is not a square. Then we can construct the proper algebraic extension \( \Omega = \Phi \left( \sqrt{\alpha }\right) \) . This field is not formally real, so there exist \( {\beta }_{i},{\gamma }_{i} \) not all 0 in \( \Phi \) such that \( \sum {\left( {\beta }_{...
Yes
Theorem 2. Any real closed field can be ordered in one and only one way. Any automorphism of such a field is an order isomorphism.
Proof. Let \( P \) be the subset of non-zero squares in the real closed field \( \Phi \) . Then \( 0 \notin P \) and, if \( \alpha \neq 0 \) and \( \alpha \notin P \), then \( - {\alpha \varepsilon P} \) by Theorem 1. If \( \alpha = {\beta }^{2} \) and \( \gamma = {\delta }^{2}{\varepsilon P} \), then \( \alpha + {\gam...
Yes
Theorem 3. Let \( \Phi \) be a formally real field and let \( \Omega \) be an algebraic closure of \( \Phi \) . Then \( \Omega \) contains a real closed field \( \Delta \) containing \( \Phi \) .
Proof. We consider the collection of formally real subfields of \( \Omega \) containing \( \Phi \) . This collection is not vacuous since it contains \( \Phi \) . Moreover, it is clear that the collection is inductive, so, by Zorn's lemma, it contains a maximal element \( \Delta \) . If \( \Delta \) is not real closed,...
Yes
Theorem 4. If \( \Phi \) is real closed, then every polynomial of odd degree with coefficients in \( \Phi \) has a root belonging to \( \Phi \) .
Proof. The result is clear for polynomials of degree 1 and we use induction on the degree \( n \) of \( f\left( x\right) \) . If \( f\left( x\right) \) is reducible, one of its factors is of odd degree so it has a root in \( \Phi \) . Hence we may assume \( f\left( x\right) \) is irreducible. Let \( \Delta = \Phi \left...
Yes
Theorem 6. If \( \Phi \) is a field such that \( \sqrt{-1}{\psi \Phi } \) and \( \Phi \left( \sqrt{-1}\right) \) is algebraically closed, then \( \Phi \) is real closed.
Proof. Suppose \( \Phi \) satisfies the conditions. We note first that the irreducible polynomials of positive degrees in \( \Phi \left\lbrack x\right\rbrack \) have degree 1 or 2. Let \( f\left( x\right) \) be such a polynomial and let \( \theta \) be a root of \( f\left( x\right) \) contained in \( \Omega = \Phi \lef...
Yes
Theorem 9. Let \( \Gamma \) be a finite dimensional extension of the field of rational numbers. Then the number of distinct orderings of \( \Gamma \) is the same as the number of isomorphisms of \( \Gamma /{R}_{0} \) into the field \( {\Delta }_{0}/{R}_{0} \) of real algebraic numbers.
In particular, this number cannot exceed \( \left\lbrack {\Gamma : {R}_{0}}\right\rbrack \) and there are no orderings of \( \Gamma = {R}_{0}\left( \theta \right) \) if and only if the minimum polynomial of \( \theta \) over \( {R}_{0} \) has no real roots, that is, no roots in \( {\Delta }_{0} \) .
No
Theorem 10. Let \( \Phi \) be a field of characteristic \( \neq 2 \) . Then an element \( \rho \neq 0 \) in \( \Phi \) is totally positive in \( \Phi \) if and only if \( \rho \) is a sum of squares of elements of \( \Phi \) .
Proof. If \( 0 \neq \rho = \sum {\alpha }_{i}{}^{2} \), then clearly \( \rho > 0 \) in every ordering of \( \Phi \) . Conversely, assume \( \rho \neq 0 \) is not a sum of squares in \( \Phi \) . Let \( \Omega \) be an algebraic closure of \( \Phi \) and consider the collection of subfields \( \mathrm{E} \) of \( \Omega...
Yes
Theorem 15. Let \( F\left( {{t}_{i};x, y}\right) \) e \( {R}_{0}\left\lbrack {{t}_{1},\cdots ,{t}_{r};x, y}\right\rbrack, G\left( {{t}_{i};x}\right) \) e \( {R}_{0}\left\lbrack {{t}_{1},\cdots ,{t}_{r};x}\right\rbrack ,{t}_{i}, x, y \) indeterminates, \( {R}_{0} \) the field of rational numbers. Then one can determine ...
The proof of this theorem is essentially a formalization of the decision method of the last section. We consider first some necessary preliminary notions.\n\nWe shall call the set \( {\Phi }^{\left( r\right) } \) of \( r \) -tuples \( \left( {{\tau }_{1},{\tau }_{2},\cdots ,{\tau }_{r}}\right) ,{\tau }_{i}{\varepsilon ...
Yes
Lemma 1. If \( f\left( z\right) \) is a polynomial of degree \( n \) or less, then \( \nabla f\left( z\right) \) is a polynomial of degree at most \( n - 2 \) (or zero if \( n = 0 \) or 1 ).
Proof: Since \( \nabla \) is linear it suffices to apply \( \nabla \) to simple polynomials of the form \( {x}^{j}{y}^{k} \) and to check that the result has no terms of degree \( j + k - 1 \) or higher.
Yes
Lemma 2. Let \( p\left( z\right) \) be a polynomial of degree \( n - 2 \) or less \( (p\left( z\right) = 0 \) if \( n = 0 \) or 1). Then there exists a polynomial \( q\left( z\right) \) of degree \( n \) or less such that \( \nabla q\left( z\right) = p\left( z\right) \) .
Proof: We may assume \( n > 2 \) ; otherwise the result is trivial. Let us write \( p\left( z\right) \) lexicographically, as\n\n\[ p\left( z\right) = {a}_{n - 2,0}{x}^{n - 2} \]\n\n\[ + {a}_{n - 3,1}{x}^{n - 3}y + {a}_{n - 3,0}{x}^{n - 3} \]\n\n\[ + \cdots \]\n\n\[ + {a}_{0, n - 2}{y}^{n - 2} + {a}_{0, n - 3}{y}^{n - ...
No
Proposition 2.2. Let \( B \) be a real Banach space and \( {B}^{ * } \) its dual space taken in the weak-* topology. Let \( K \) be a nonempty compact convex subset of \( {B}^{ * } \) . Then \( K \) has an extreme point.
Proof. Let \( \left\{ {L}_{n}\right\} \) be a countable dense subset of \( B \) . If \( y \in {B}^{ * } \), put\n\n\[ {L}_{n}\left( y\right) = y\left( {L}_{n}\right) \]\n\nDefine\n\n\[ {l}_{1} = \mathop{\sup }\limits_{{x \in K}}{L}_{1}\left( x\right) \]\n\nSince \( K \) is compact and \( {L}_{1} \) continuous, \( {l}_{...
Yes
Theorem 2.3 (Complex Stone-Weierstrass Theorem). A is a subalgebra of \( C\left( X\right) \) containing the constants and separating points. If\n\n(1)\n\n\[ f \in \mathfrak{A} \Rightarrow \bar{f} \in \mathfrak{A} \]\n\nthen \( \mathfrak{A} \) is dense in \( C\left( X\right) \) .
Proof. Let \( \mathcal{L} \) consists of all real-valued functions in \( \mathfrak{A} \) . Since by (1) \( \mathcal{L} \) contains Re \( f \) and \( \operatorname{Im}f \) for each \( f \in \mathfrak{A},\mathcal{L} \) separates points on \( X \) . Evidently \( \mathcal{L} \) is a subalgebra of \( {C}_{R}\left( X\right) ...
No
Corollary 1. Let \( X \) be a compact subset of \( {\sum }_{R} \) . Then \( P\left( X\right) = C\left( X\right) \) .
Proof. Let \( \mathfrak{A} \) be the algebra of all polynomials in \( {z}_{1},\ldots ,{z}_{n} \) restricted to \( X \) . \( \mathfrak{A} \) then satisfies the hypothesis of the last theorem, and so \( \mathfrak{A} \) is dense in \( C\left( X\right) \) ; i.e., \( P\left( X\right) = C\left( X\right) \) .
Yes
Lemma 2.4. The functions \[ \int \left| {\log \left| \frac{1}{z - \zeta }\right| }\right| d\left| \mu \right| \left( \zeta \right) \;\text{ and }\;\int \left| \frac{1}{\zeta - z}\right| d\left| \mu \right| \left( \zeta \right) \] are summable - dx dy over compact sets in \( \mathbb{C} \) . It follows that these functio...
Since \( 1/r \geq \left| {\log r}\right| \) for small \( r > 0 \), we need only consider the second integral. Fix \( R > 0 \) with supp \( \left| \mu \right| \subset \{ z\left| \right| z \mid < R\} \) . \[ \gamma = {\int }_{\left| z\right| \leq R}{dxdy}\left\{ {\int \left| \frac{1}{\zeta - z}\right| d\left| \mu \right|...
Yes
Lemma 2.5. Let \( F \in {C}_{0}^{1}\left( \mathbb{C}\right) \) . Then\n\n\[ F\left( \zeta \right) = - \frac{1}{\pi }{\int }_{\mathbb{C}}\int \frac{\partial F}{\partial \bar{z}}\frac{dxdy}{z - \zeta },\;\text{ all }\zeta \in C. \]
Proof. Fix \( \zeta \) and choose \( R > \left| \zeta \right| \) with supp \( F \subset \{ z\left| \right| z \mid < R\} \) . Fix \( \varepsilon > 0 \) and small. Put \( {\Omega }_{\varepsilon } = \{ \left| \right| z\left| { < R\text{and}}\right| z - \zeta \mid > \varepsilon \} \).\n\nThe 1 -form \( {Fdz}/z - \zeta \) i...
Yes
Lemma 2.6. Let \( G \in {C}_{0}^{2}\left( \mathbb{C}\right) \) . Then\n\n\[ G\left( \zeta \right) = - \frac{1}{2\pi }{\int }_{C}\int {\Delta G}\left( z\right) \log \frac{1}{\left| z - \zeta \right| }{dxdy},\;\text{ all }\zeta \in \mathbb{C}. \]
Proof. The proof is very much like that of Lemma 2.5. With \( {\Omega }_{\varepsilon } \) as in that proof, start with Green's formula\n\n\[ {\int }_{{\Omega }_{\varepsilon }}\int \left( {{u\Delta v} - {v\Delta u}}\right) {dxdy} = {\int }_{\partial {\Omega }_{\varepsilon }}\left( {u\frac{\partial v}{\partial n} - v\fra...
No
Lemma 2.7. If \( \mu \) is a measure with compact support in \( C \), and if \( \widehat{\mu }\left( z\right) = 0 \) a.e. - \( {dxdy} \), then \( \mu = 0 \) .
Proof. Fix \( g \in {C}_{0}^{1}\left( \mathbb{C}\right) \) . By (4)\n\n\[ \n\int g\left( \zeta \right) {d\mu }\left( \zeta \right) = \int {d\mu }\left( \zeta \right) \left\lbrack {-\frac{1}{\pi }\int \frac{\partial g}{\partial \bar{Z}}\left( z\right) \frac{dxdy}{z - \zeta }}\right\rbrack .\n\]\n\nFubini's theorem now g...
Yes
Theorem 2.8 (Hartogs-Rosenthal). Assume that \( X \) has Lebesgue two-dimensional measure 0 . Then rational functions whose poles lie off \( X \) are uniformly dense in \( C\left( X\right) \) .
Proof. Let \( W \) be the linear space consisting of all rational functions holomorphic on \( X \) . \( W \) is a subspace of \( C\left( X\right) \) . To show \( W \) dense, we consider a measure \( \mu \) on \( X \) with \( \mu \bot W \) . Then \( \widehat{\mu }\left( z\right) = \int {d\mu }\left( \zeta \right) /\zeta...
Yes
Theorem 2.9 (Runge). If \( F \) is a holomorphic function defined on \( \Omega \), there exists a sequence \( \left\{ {R}_{n}\right\} \) of rational functions holomorphic in \( \Omega \) with\n\n\[ \n{R}_{n} \rightarrow F\text{ uniformly on }K\text{. } \n\]
Proof. Let \( {\Omega }_{1},{\Omega }_{2},\ldots \) be the components of \( \mathbb{C} \smallsetminus K \) . It is no loss of generality to assume that each \( {\Omega }_{j} \) meets the complement of \( \Omega \) . (Why?) Fix \( {p}_{i} \in {\Omega }_{j} \smallsetminus \Omega \) .\n\nLet \( W \) be the space of all ra...
Yes
Theorem 2.10 (Runge). Let \( \Omega \) be a simply connected region and fix \( G \) holomorphic in \( \Omega \). If \( K \) is a compact subset of \( \Omega \), then \( \exists \) a sequence \( \left\{ {P}_{n}\right\} \) of polynomials converging uniformly to \( G \) on \( K \).
Proof. Without loss of generality we may assume that \( \mathbb{C} \smallsetminus K \) is connected.\n\nFix a point \( p \) in \( \mathbb{C} \) lying outside a disk \( \{ z\left| \right| z \mid \leq R\} \) which contains \( K \). The proof of the last theorem shows that \( \exists \) rational functions \( {R}_{n} \) wi...
Yes
Lemma 2.12 (Carleson). Let \( E \) be a compact plane set with \( \mathbb{C} \smallsetminus E \) connected and fix \( {z}_{0} \in \partial E \) . Then \( \exists \) probability measures \( {\sigma }_{t} \) for each \( t > 0 \) with \( {\sigma }_{t} \) carried on \( \mathbb{C} \smallsetminus E \) such that:\n\nLet \( \a...
Proof. We may assume that \( {z}_{0} = 0 \) . Fix \( t > 0 \) . Since \( 0 \in \partial E \) and \( \mathbb{C} \smallsetminus E \) is connected, \( \exists \) a probability measure \( {\sigma }_{t} \) carried on \( \mathbb{C} \smallsetminus E \) such that\n\n\[ \n{\sigma }_{t}\left\{ {z\left| {{r}_{1} < }\right| z \mid...
Yes
Lemma 3.2. \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \) exist in \( \mathfrak{A} \) and depends only on \( x \) and \( \Phi \), not on the choice of \( \left\{ {f}_{n}\right\} \) .
Proof of Lemma 3.2. Choose \( \gamma \) as in Exercise 3.2. Then\n\n\[ \begin{Vmatrix}{{f}_{n}\left( x\right) - \frac{1}{2\pi i}{\int }_{\gamma }\frac{\Phi \left( t\right) {dt}}{t - z}}\end{Vmatrix} = \begin{Vmatrix}{\frac{1}{2\pi i}{\int }_{\gamma }\frac{{f}_{n}\left( t\right) - \Phi \left( t\right) }{t - x}{dt}}\end{...
No
Theorem 3.3. Let \( \\mathfrak{A} \) be a Banach algebra, \( x \\in \\mathfrak{A} \), and let \( \\Omega \) be an open set containing \( \\sigma \\left( x\\right) \) . Then there exists a map \( \\tau : H\\left( \\Omega \\right) \\rightarrow \\mathfrak{A} \) such that the following holds. We write \( F\\left( x\\right)...
Proof. Fix \( F \\in H\\left( \\Omega \\right) \) . Choose a sequence of rational functions \( \\left\{ {f}_{n}\\right\} \\in H\\left( \\Omega \\right) \) with \( {f}_{n} \\rightarrow F \) in \( H\\left( \\Omega \\right) \) . By Lemma 3.2\n\n(6)\n\n\[ \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{f}_{n}\\left( x...
No
Lemma 4.3. \( \tau \land \sigma \) as defined is \( \left( {k + l}\right) \) -linear and alternating and so \( \in { \land }^{k + l}\left( V\right) \) .
The operation \( \land \) (wedge) defines a product for pairs of elements, one in \( { \land }^{k}\left( V\right) \) and one in \( { \land }^{l}\left( V\right) \), the value lying in \( { \land }^{k + l}\left( V\right) \), hence in \( \mathcal{G}\left( V\right) \) . By linearity, \( \land \) extends to a product on arb...
No
Lemma 4.8. \( {d}^{2} = 0 \) for every \( k \) ; i.e., if \( {\omega }^{k} \in { \land }^{k}\left( \Omega \right), k \) arbitrary, then \( d\left( {d{\omega }^{k}}\right) = \) 0.
To prove Lemma 4.8, it is useful to prove first
No
Lemma 5.4. Assume that \( \partial f/\partial {\bar{z}}_{j} = 0, j = 1,\ldots, n \), in \( \Omega \) . then there exist constants \( {A}_{v} \) in \( \mathbb{C} \) for each tuple \( v = \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) of nonnegative integers such that\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{v}{A}_{v}...
For a proof of this result, see, e.g., [Hö, Th. 2.2.6].
No
Lemma 6.2. Let \( \phi \in {C}^{1}\left( {\mathbb{R}}^{2}\right) \) and assume that \( \phi \) has compact support. Put\n\n\[ \Phi \left( \zeta \right) = - \frac{1}{\pi }{\int }_{{\mathbb{R}}^{2}}\phi \left( z\right) \frac{dxdy}{z - \zeta }.\]\n\nThen \( \Phi \in {C}^{1}\left( {\mathbb{R}}^{2}\right) \) and \( \partial...
Proof. Choose \( \mathrm{R} \) with \( \operatorname{supp}\phi \subset \{ z\left| \right| z \mid \leq R\} \) .\n\n\[ {\pi \Phi }\left( \zeta \right) = {\int }_{\left| z\right| \leq R}\phi \left( z\right) \frac{1}{\zeta - z}{dxdy} = {\int }_{\left| {{z}^{\prime } - \zeta }\right| \leq R}\phi \left( {\zeta - {z}^{\prime ...
Yes
Lemma 6.3. Let \( \Omega \) be a neighborhood of \( {\Delta }^{n} \) and fix \( f \) in \( {C}^{\infty }\left( \Omega \right) \) . Fix \( j,1 \leq j \) \( \leq n \) . Assume that\n\n\[ \frac{\partial f}{\partial {\bar{z}}_{k}} = 0\text{ in }\Omega, k = {k}_{1},\ldots ,{k}_{s}\text{, each }{k}_{i} \neq j. \]\n\nThen we ...
Proof. Choose \( \varepsilon > 0 \) so that if \( z = \left( {{z}_{1},\ldots ,{z}_{n}}\right) \in {\mathbb{C}}^{n} \) and \( \left| {z}_{v}\right| < 1 + {2\varepsilon } \) for all \( v \), then \( z \in \Omega \) .\n\nChoose \( \psi \in {C}^{\infty }\left( {\mathbb{R}}^{2}\right) \), having support contained in \( \{ z...
Yes
Theorem 7.1. Assume that \( \mathbb{C} \smallsetminus K \) is connected. Let \( F \) be holomorphic in some neighborhood \( \Omega \) of \( K \) . Then \( {\left. F\right| }_{K} \) is in \( P\left( K\right) \) .
Proof. Let \( \mathcal{L} \) denote the space of all finite linear combinations of functions \( 1/(z - \) \( a{)}^{p} \), where \( a \in \mathbb{C} \smallsetminus \Omega, p \) an integer \( \geq 0 \) . By Runge’s theorem (Theorem 2.9), \( {\left. F\right| }_{K} \) lies in the uniform closure of \( \mathcal{L} \) on \( ...
Yes
Lemma 7.2. Let \( K \) be a compact set in \( \mathbb{C}.\mathbb{C} \smallsetminus K \) is connected if and only if for each \( {x}_{0} \in \mathbb{C} \smallsetminus K \) we can find a polynomial \( P \) such that\n\n(1)\n\n\[ \left| {P\left( {x}^{0}\right) }\right| > \mathop{\max }\limits_{K}\left| P\right| \]
Proof. If \( \mathbb{C} \smallsetminus K \) fails to be connected, we can choose \( {x}^{0} \) in a bounded component of \( \mathbb{C} \smallsetminus K \) and note that (1) violates the maximum principle.\n\nAssume that \( \mathbb{C} \smallsetminus K \) is connected. Fix \( {x}^{0} \in \mathbb{C} \smallsetminus K \) . ...
Yes
Lemma 7.4. Let \( X \) be a compact polynomially convex subset of \( {\Delta }^{n} \). Let \( \mathcal{O} \) be an open set containing \( X \). Then there exists a p-polyhedron \( \Pi \) with \( X \subset \Pi \subset \mathcal{O} \).
Proof. For each \( x \in {\Delta }^{n} \smallsetminus \mathcal{O} \) there exists a polynomial \( {P}_{x} \) with \( \left| {{P}_{x}\left( x\right) }\right| > 1 \) and \( \left| {P}_{x}\right| \leq 1 \) on \( X \). Then \( \left| {P}_{x}\right| > 1 \) in some neighborhood \( {\mathcal{N}}_{x} \) of \( x \). By compactn...
Yes
Theorem 7.5 (Oka Extension Theorem). Given f holomorphic in some neighborhood of \( \Pi \) ; then there exists \( F \) holomorphic in a neighborhood of \( {\Delta }^{n + r} \) such that \[ F\left( {z,{P}_{1}\left( z\right) ,\ldots ,{P}_{r}\left( z\right) }\right) = f\left( z\right) \text{, all}z \in \Pi \text{.} \]
Proof of Theorem 7.3. Without loss of generality we may assume that \( X \subset {\Delta }^{n} \) . (Why?) \( f \) is holomorphic in a neighborhood \( \mathcal{O} \) of \( X \) . By Lemma 7.4 there exists a \( p \) -polyhedron \( \Pi \) with \( X \subset \Pi \subset \mathcal{O} \) . Then \( f \) is holomorphic in a nei...
Yes
Theorem 7.6. Let \( \Pi \) be a p-polyhedron in \( {\mathbb{C}}^{n} \) and \( \Omega \) a neighborhood of \( \Pi \) . Given that \( \phi \in { \land }^{p, q}\left( \Omega \right), q > 0 \), with \( \bar{\partial }\phi = 0 \), then there exists a neighborhood \( {\Omega }_{1} \) of \( \Pi \) and \( \psi \in { \land }^{p...
Proof of Theorem 7.6. We denote\n\n\[ {P}^{k}\left( {{q}_{1},\ldots ,{q}_{r}}\right) = \left\{ {z \in {\Delta }^{k}\left| \right| {q}_{j}\left( z\right) \mid \leq 1, j = 1,\ldots, r}\right\} ,\]\n\nthe \( {q}_{j} \) being polynomials in \( {z}_{1},\ldots ,{z}_{k} \) . Every \( p \) -polyhedron is of this form.\n\nWe sh...
Yes
Fix \( k \) and polynomials \( {q}_{1},\ldots ,{q}_{r} \) in \( z = \left( {{z}_{1},\ldots ,{z}_{k}}\right) \) . Let \( f \) be holomorphic in a neighborhood \( W \) of \( \Pi = {P}^{k}\left( {{q}_{1},\ldots ,{q}_{r}}\right) \) . The \( \exists F \) holomorphic in a neighborhood of \( {\Pi }^{\prime } = {P}^{k + 1}\lef...
Proof. Let \( \sum \) be the subset of \( {\Pi }^{\prime } \) defined by \( {z}_{k + 1} - {q}_{1}\left( z\right) = 0 \) . Choose \( \phi \in \) \( {C}_{0}^{\infty }\left( {{\pi }^{-1}\left( W\right) }\right) \) with \( \phi = 1 \) in a neighborhood of \( \sum \) .\n\nWe seek a function \( G \) defined in a neighborhood...
Yes
Lemma 8.3. Assume (3). Then \( \sigma \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is a polynomially convex subset of \( {\mathbb{C}}^{n} \) .
Proof. Fix \( {z}^{0} = \left( {{z}_{1}^{0},\ldots ,{z}_{n}^{0}}\right) \) with\n\n\[ \left| {Q\left( {z}^{0}\right) }\right| \leq \mathop{\max }\limits_{\sigma }\left| Q\right| ,\;\text{ all polynomials }Q, \]\n\nwhere \( \sigma = \sigma \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) .\n\n\[ \mathop{\max }\limits_{\sigma ...
Yes
Theorem 8.4. Let \( \mathfrak{A} \) be a Banach algebra, \( a \in \mathfrak{A} \) . and assume that \( \exists h \in C\left( \mathcal{M}\right) \) with \( {h}^{2} = \widehat{a} \) . Assume also that \( \widehat{a} \) never vanishes on \( \mathcal{M} \) . Then a has a square root in \( \mathfrak{A} \) .
We approach the proof as follows: First find \( {a}_{2},\ldots ,{a}_{n} \in \mathfrak{A} \) such that \( \exists F \) holomorphic in a neighborhood of \( \sigma \left( {a,{a}_{2},\ldots ,{a}_{n}}\right) \) in \( {\mathbb{C}}^{n} \) with \( {F}^{2} = {z}_{1} \) . By Theorem \( {8.2},\exists y \in \mathfrak{A} \), with \...
No
Lemma 8.5. Given a as in Theorem 8.4, \( \exists {a}_{2},\ldots ,{a}_{n} \in \mathfrak{A} \) such that if \( K = \) \( \sigma \left( {a,{a}_{2},\ldots ,{a}_{n}}\right) \subset {\mathbb{C}}^{n} \), then we can find \( H \in C\left( K\right) \) with \( {H}^{2} = {z}_{1} \) on \( K \) .
Proof. In the topological product \( \mathcal{M} \times \mathcal{M} \) put\n\n\[ S = \left\{ {\left( {M,{M}^{\prime }}\right) \mid h\left( M\right) + h\left( {M}^{\prime }\right) = 0}\right\} ,\]\n\nwhere \( h \) is as in Theorem 8.4. \( S \) is compact and disjoint from the diagonal. (Why?) Let \( x = \left( {{M}_{1},...
No
Corollary 2. Let \( \mathfrak{A} \) be a uniform algebra on a compact space \( X \) . Assume that \( \mathcal{M} \) is totally disconnected. Then \( \mathfrak{A} = C\left( X\right) \) .
Proof of Corollary 2. If \( {x}_{1},{x}_{2} \in X,{x}_{1} \neq {x}_{2} \), choose an open and closed set \( {\mathcal{M}}_{1} \) in \( \mathcal{M} \) with \( {x}_{1} \in {\mathcal{M}}_{1},{x}_{2} \notin {\mathcal{M}}_{1} \) . Put \( {\mathcal{M}}_{2} = \mathcal{M} \smallsetminus {\mathcal{M}}_{1} \) . By Theorem 8.6, \...
Yes
Corollary 3. Let \( X \) be a compact subset of \( {\mathbb{C}}^{n} \). Assume that \( X \) is polynomially convex and totally disconnected. Then \( P\left( X\right) = C\left( X\right) \).
Proof. The result follows from Corollary 2, together with the fact that \( \mathcal{M}\left( {P\left( X\right) }\right) = X \) .
No
Lemma 9.2. Fix \( x \in X \smallsetminus S \) . \( \exists \) a neighborhood \( U \) of \( x \) with the following property: If \( \beta \) is a boundary, then \( \beta \smallsetminus U \) is also a boundary.
Proof. \( x \notin S \) and so \( \exists \) boundary \( {S}_{0} \) with \( x \notin {S}_{0} \) . For each \( y \in {S}_{0} \), choose \( {f}_{y} \in \mathcal{F} \) with \( {f}_{y}\left( x\right) = 0,{f}_{y}\left( y\right) = 2 \) . \( {\mathcal{N}}_{y} = \left\{ {\left| {f}_{y}\right| > 1}\right\} \) is a neighborhood ...
Yes
Lemma 9.4. Let \( X \) be a compact, polynomially convex set in \( {\mathbb{C}}^{n} \) and \( {U}_{1} \) and \( {U}_{2} \) be open sets in \( {\mathbb{C}}^{n} \) with \( X \subset {U}_{1} \cup {U}_{2} \) . If \( h \in H\left( {{U}_{1} \cap {U}_{2}}\right) \), then \( \exists \) a neighborhood \( W \) of \( X \) and \( ...
Proof. Write \( X = {X}_{1} \cup {X}_{2} \), where \( {X}_{j} \) is compact and \( {X}_{j} \subset {U}_{j}, j = 1,2 \) . Choose \( {f}_{1} \in {C}_{0}^{\infty }\left( {U}_{1}\right) \) with \( 0 \leq {f}_{1} \leq 1 \) and \( {f}_{1} = 1 \) on \( {X}_{1} \) . Similarly, choose \( {f}_{2} \in {C}_{0}^{\infty }\left( {U}_...
Yes
Lemma 9.5. Let \( K \) be a compact set in \( {\mathbb{C}}^{N} \) and \( {U}_{1} \) and \( {U}_{2} \) open sets with\n\n(4)\n\n\[ \n{U}_{1} \cup {U}_{2} \supset K \n\]\n\n(5) \( {U}_{1} \cap {U}_{2} \subset \left\{ {\operatorname{Re}{z}_{1} < 0}\right\} \; \) and \( \;\exists {h}_{1} \in H\left( {U}_{1}\right) ,{h}_{2}...
Proof. By (5) we have in \( {U}_{1} \cap {U}_{2} \) ,\n\n\[ \n{z}_{1}{h}_{1} - {z}_{1}{h}_{2} = \log {z}_{1}\;\text{ so }\;{e}^{{z}_{1}{h}_{1}} = {z}_{1}{e}^{{z}_{1}{h}_{2}}.\n\]\n\nIt follows that if we define\n\n\[ \nf = \left\{ \begin{array}{ll} {e}^{{z}_{1}{h}_{1}} & \text{ in }{U}_{1}, \\ {z}_{1}{e}^{{z}_{1}{h}_{2...
Yes
Theorem 9.7. Let \( \mathfrak{A} \) be a uniform algebra on a space \( X \) . Let \( {U}_{1},{U}_{2},\ldots ,{U}_{s} \) be an open covering of \( \mathcal{M} \) . Denote by \( \mathcal{L} \) the set of all \( f \) in \( C\left( \mathcal{M}\right) \) such that for \( j = 1,\ldots, s,{\left. f\right| }_{{U}_{j}} \) lies ...
Proof. The proof is a corollary of Theorem 9.3. We leave it to the reader as *Exercise 9.9.
No
Lemma 10.1 (Paul Cohen). Let \( a, b \in \mathfrak{A} \). Assume that\n\n\[ \parallel 1 + a + \bar{b}\parallel < 1\text{.}\]\n\nThen \( a + b \) is invertible in \( \mathfrak{A} \).
Proof. Put \( f = a + b \). We have\n\n\[ \parallel 1 + a + \bar{b}\parallel < 1,\;\text{ hence }\parallel 1 + \bar{a} + b\parallel < 1,\]\n\nwhence\n\n\[ \parallel 1 + a + b + 1 + \bar{a} + b\parallel < 2\text{ or }k = \parallel 1 + \operatorname{Re}f\parallel < 1.\]\n\nFor all \( x \in X \), then\n\n\[ \left| {1 + \o...
Yes
Theorem 10.2 (Maximality Of \( {\mathfrak{A}}_{0} \) ). Let \( B \) be a uniform algebra on \( \Gamma \) with\n\n\[ \n{\mathfrak{A}}_{0} \subseteq B \subseteq C\left( \Gamma \right) \n\]\n\nThen either \( {\mathfrak{A}}_{0} = B \) or \( B = C\left( \Gamma \right) \) .
We shall deduce this result by means of Lemma 10.1 as follows. Assuming \( B \neq {\mathfrak{A}}_{0} \), we construct elements \( u, v \in B \) with\n\n(1)\n\n\[ \n\parallel 1 + z \cdot u + \bar{z}\bar{v}\parallel < 1 \n\]\n\nwhere \( z = {e}^{i\theta } \) . Then we conclude that \( {zu} + {zv} \) is invertible in \( B...
Yes
Theorem 10.3 (Rudin). Let \( \mathcal{L} \) be an algebra of continuous functions on \( D \) such that\n\n(a) The function \( z \) is in \( \mathcal{L} \).\n\n(b) \( \mathcal{L} \) satisfies a maximum principle relative to \( \Gamma \):\n\n\[ \left| {G\left( x\right) }\right| \leq \mathop{\max }\limits_{\Gamma }\left| ...
Proof. The uniform closure of \( \mathcal{L} \) on \( D \), written \( \mathfrak{A} \), still satisfies (a) and (b).\n\nPut \( B = {\left. \mathfrak{A}\right| }_{\Gamma } \) . Because of (b), \( B \) is closed under uniform convergence on \( \Gamma \) and by (a), \( {\mathfrak{A}}_{0} \subseteq B \) . So Theorem 10.2 a...
Yes
Lemma 10.4 (Glicksberg). Let \( E \) be a subset of \( {X}_{0} \) and let \( f \in \mathcal{L} \) and \( f = 0 \) on \( E \) . Then for each \( x \in X \) either\n\n(a) \( f\left( x\right) = 0 \), or\n\n(b) \( \left| {g\left( x\right) }\right| \leq \mathop{\sup }\limits_{{{X}_{0} \smallsetminus E}}\left| g\right| \), a...
Proof. Fix \( g \in \mathcal{L} \) . Then \( f \cdot g \in \mathcal{L} \) . Fix \( x \in X \) with \( f\left( x\right) \neq 0 \) . We have\n\n\[ \left| {\left( {fg}\right) \left( x\right) }\right| \leq \mathop{\max }\limits_{{X}_{0}}\left| {fg}\right| = \mathop{\sup }\limits_{{{X}_{0} \smallsetminus E}}\left| {fg}\righ...
Yes
Theorem 10.5. Let \( f \in A\left( \Omega \right) \) and assume that \( f = 0 \) on \( \partial \Omega \cap U \) . Then \( f \equiv 0 \) in \( \Omega \) .
If we assume that\n\n(4)\n\n\[ \exists \text{a sequence}\left\{ {z}_{n}\right\} \text{in}\mathbb{C} \smallsetminus \bar{\Omega }\text{with}{z}_{n} \rightarrow {z}_{0}\text{,} \]\n\nthen Lemma 10.4 gives a direct proof, as follows.\n\nPut \( X = \bar{\Omega },\mathcal{L} = A\left( \Omega \right) \) . Then \( \partial \O...
Yes
Theorem 10.6 (Radó’s Theorem). Let \( h \) be a continuous function on the disk D. Let \( Z \) denote the set of zeros of \( h \) . If \( h \) is analytic on \( \overset{ \circ }{D} \smallsetminus Z \), then \( h \) is analytic on D.
Proof. We assume that \( Z \) has an empty interior. The case \( \mathring{Z} \neq \varnothing \) is treated similarly.\n\nLet \( \mathcal{L} \) consist of all sums\n\n\[ \mathop{\sum }\limits_{{v = 0}}^{N}{a}_{v}{h}^{v},\;{a}_{v} \in A\left( D\right) \]\n\nIf \( f \in \mathcal{L}, f \) is analytic in \( \left| z\right...
Yes
Theorem 10.3 (which clearly holds if \( D \) is replaced by an arbitrary disk) now applies to the algebra \( \mathcal{L} \) on \( \bar{U} \) . We conclude that \( \mathcal{L} \subseteq A\left( \bar{U}\right) \), and so \( G = g\left( {f}^{-1}\right) \) is analytic in \( U \) for every \( g \in \mathfrak{A} \) .
Thus \( G \) is analytic in \( D \), whence (8) holds.
No
Theorem 11.2. Let \( \\left( {A, X,\\Omega, p}\\right) \) be a maximum modulus algebra on \( \\Omega \) and fix \( F \\in A \) . Choose a closed disk \( \\Delta \) contained in \( \\Omega \), with center \( {\\lambda }_{0} \), and fix a point \( {x}^{0} \) in \( {p}^{-1}\\left( {\\lambda }_{0}\\right) \) . Then there e...
Proof. Let \( \\chi \\left( \\lambda \\right) = {a\\lambda } + b, a, b \\in \\mathbb{C} \), be a conformal map of \( \\Delta \) onto the unit disk \( \\left\\{ {\\left| z\\right| \\leq 1}\\right\\} ,\\chi \\left( {\\lambda }_{0}\\right) = 0 \) . Put \( \\Pi = \\chi \\circ p \) . Then \( \\Pi \\in A \) and \( \\chi \) m...
Yes
Corollary 11.3. Let \( \\left( {A, X,\\Omega, f}\\right) \) be a maximum modulus algebra and fix a closed disk \( \\Delta \\subseteq \\Omega \) . Assume that \( X \) lies one sheeted over \( \\Delta \), in the sense that \( {f}^{-1}\\left( \\lambda \\right) \) consists of a single point for each \( \\lambda \\in \\Delt...
Proof. Fix \( g \\in A \) . By Theorem 11.2, there exists a bounded analytic function \( G \) on int \( \\left( \\Delta \\right) \) such that\n\n\[ G\\left( \\lambda \\right) \\in \\operatorname{co}\\left( {g\\left( {{f}^{-1}\\left( \\lambda \\right) }\\right) }\\right. \\text{for a.a.}\\lambda \\in \\partial \\Delta \...
Yes