Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
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}^{m - 1}... | 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 \( {\beta }_{0}{\varepsilon \Phi },\varepsilon \mathfrak{P}\left( \Phi \right) \) . 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 |
Corollary 2. 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\[ \n{\left( {\epsilon }^{{p}^{e}}\r... | 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 \( \mathrm{P} \) \( \otimes \Phi \mathbf{E} \) is a field. On the other hand, if \( \mathrm{P}/\Phi \) is not separable, then... | 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}\right. \cup... | 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 \( \mathfrak{o} \) and \( \alpha \... | 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 \alph... | 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 \( \mathfrak{o} \) such that \... | 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 linearly... | 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 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 8. Every ordered field \( \Phi \) has a real closure. If \( {\Phi }_{1} \) and \( {\Phi }_{2} \) are ordered fields with the real closures \( {\Delta }_{1} \) and \( {\Delta }_{2} \), respectively, then any order isomorphism of \( {\Phi }_{1} \) onto \( {\Phi }_{2} \) has a unique extension to an isomorphism of... | Proof. Let \( \Phi \) be an ordered field, \( \Omega \) an algebraic closure of \( \Phi \) . Let \( \mathrm{E} \) be the subfield of \( \Omega \) obtained by adjoining to \( \Phi \) the square roots of all the positive elements of \( \Phi \) . Then \( \mathrm{E} \) is formally real and \( \Omega \) is an algebraic clos... | 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 13. Let \( \Phi \) be a field of real numbers, \( \Phi \left( {x}_{i}\right) \equiv \Phi \left( {{x}_{1},\cdots }\right. \) , \( \left. {x}_{n}\right) \) the field of rational expressions in \( n \) indeterminates \( {x}_{i} \) with coefficients in \( \Phi \) and suppose an ordering has been given to \( \Phi \l... | We shall prove Theorem 13-after some necessary preliminaries -by induction on the number \( n \) of \( {x}_{i} \) . The result is clear if \( n = 0 \) since in this case \( \Phi \left( {x}_{i}\right) = \Phi \), so the functions are just constant functions. It remains to prove the inductive step, so we assume the result... | 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 |
Theorem 1.3. (The Fundamental Theorem of Arithmetic) Every integer \( n > 1 \) is a product of positive primes unique up to order, i.e., there exist unique primes\n\n(*) \n\n\[ \n1 < {p}_{1} < \cdots < {p}_{r} \n\] \n\n\[ \n\text{and unique integers}{e}_{1},\ldots ,{e}_{r} > 0\text{such that}n = {p}_{1}^{{e}_{1}}\cdots... | Assuming this theorem is true (Euclid knew its proof and we shall prove in Theorem 4.16) below, we show Statement 4. | No |
Theorem 1.8. (Cantor) The set \( \mathbb{R} \) of real numbers is not countable. | Proof. Suppose that \( \mathbb{R} \) is countable. As \( \mathbb{R} \) and the closed interval \( \left\lbrack {0,1}\right\rbrack \) have the same cardinality by Exercise \( {1.12}\left( 9\right) \), the interval \( \left\lbrack {0,1}\right\rbrack \) must also be countable by the first fact above. As \( {\mathbb{Z}}^{ ... | No |
Corollary 2.13. Let \( n \in {\mathbb{Z}}^{ + } \) . Then there exist \( n \) consecutive composite (i.e., non-prime and not \( 0 \) or \( \pm 1 \) ) positive integers. | Proof. Let \( m \in {\mathbb{Z}}^{ + } \) and \( N = {\left( m\right) }_{n + 1} = m\left( {m + 1}\right) \cdots \left( {m + n}\right) \) . Then by the example, we have \( \left( {n + 1}\right) ! \mid N \) . If follows that for any integer \( s \) satisfying \( 2 \leq s \leq n + 1 \), we have \( s \mid N + s \) . So the... | Yes |
Corollary 2.14. Let \( m \) and \( n \) be positive integers with \( m \leq n \) . Define the binomial coefficients\n\n\[ \left( \begin{matrix} m \\ n \end{matrix}\right) \mathrel{\text{:=}} \frac{m!}{\left( {m - n}\right) !n!} = \frac{m\left( {m - 1}\right) \cdots \left( {m - n + 1}\right) }{n!} \]\n\nand\n\n\[ \left(... | Proof. We know that \( \left( \begin{matrix} m \\ n \end{matrix}\right) = \frac{{\left( m - n + 1\right) }_{n}}{n!} \) is an integer by the example. The second statement follows from the identity\n\n\[ \left( \begin{matrix} - m \\ n \end{matrix}\right) = {\left( -1\right) }^{n}\left( \begin{matrix} m + n - 1 \\ n \end{... | Yes |
Corollary 2.15. Let \( p > 1 \) be a prime, then\n\n\[ \left( \begin{array}{l} p \\ 1 \end{array}\right) ,\left( \begin{array}{l} p \\ 2 \end{array}\right) ,\ldots ,\left( \begin{matrix} p \\ p - 1 \end{matrix}\right) \]\n\nare all divisible by \( p \), i.e, \( p \mid \left( \begin{array}{l} p \\ n \end{array}\right) \... | Proof. If \( 1 \leq n \leq p - 1 \), then the example says that\n\n\[ n! \mid p\left( {p - 1}\right) \cdots \left( {p - n + 1}\right) = {\left( p - n + 1\right) }_{n} \]\n\nWe also know that \( s \) and \( p \) have no common non-trivial factors if \( 1 < \) \( s < p \) . (Proof?) We say that \( s \) and \( p \) are re... | No |
Theorem 3.3. Let \( n \) be a positive integer and \( p \) a positive prime. Suppose that \( {p}^{e}\parallel n! \), then\n\n\[ e = \mathop{\sum }\limits_{{i = 1}}^{\infty }\left\lbrack \frac{n}{{p}^{i}}\right\rbrack \] | Proof. If \( {p}^{i} > n \), then \( \left\lbrack \frac{n}{{p}^{i}}\right\rbrack = 0 \) ; so the sum is really a finite sum. We prove the result by induction of \( n \) .\n\n\( n = 1 \) . There is nothing to prove.\n\nWe can, therefore, make the following:\n\nInduction Hypothesis. Let \( {e}^{\prime } = \mathop{\sum }\... | Yes |
Using the properties of [ ], we compute \( e \) such that \( {7}^{e}\parallel {1000}! \) . | \[ \left\lbrack \frac{1000}{7}\right\rbrack = {142} \] \[ \left\lbrack \frac{1000}{{7}^{2}}\right\rbrack = \left\lbrack \frac{\left\lbrack \frac{1000}{7}\right\rbrack }{7}\right\rbrack = \left\lbrack \frac{142}{7}\right\rbrack = {20} \] \[ \left\lbrack \frac{1000}{{7}^{3}}\right\rbrack = \left\lbrack \frac{\left\lbrack... | Yes |
Corollary 3.6. Suppose that \( {a}_{1},\ldots ,{a}_{r} \) are non-negative integers sat-isfing \( {a}_{1} + \cdots + {a}_{r} = n \), then the multinomial coefficient \( \frac{n!}{{a}_{1}!\cdots {a}_{r}!} \) is an integer. | Proof. By the Fundamental Theorem of Arithmetic 4.16 and the theorem, it suffices to prove for each prime \( p \), we have\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{\infty }\left\lbrack \frac{n}{{p}^{i}}\right\rbrack \geq \mathop{\sum }\limits_{{i = 1}}^{\infty }\left\lbrack \frac{{a}_{1}}{{p}^{i}}\right\rbrack + \cdots +... | Yes |
Theorem 4.2. (Division Algorithm) Let \( m \) and \( n \) be integers with \( m \) positive. Then there exist unique integers \( q \) and \( r \) satisfying:\n\n(i) \( n = {qm} + r \) .\n\n(ii) \( 0 \leq r < m \) | Proof. We have two things to show: existence and uniqueness. We first show\n\nUniqueness: Let \( \\left( {q, r}\\right) \) and \( \\left( {{q}^{\\prime },{r}^{\\prime }}\\right) \) be two pairs of integers satisfying the conclusion. We must show \( q = {q}^{\\prime } \) and \( r = {r}^{\\prime } \) . We have\n\n(*)\n\n... | Yes |
Theorem 4.5. Let \( n \neq 0 \) and \( m \) be integers. Then a gcd of \( m \) and \( n \) exists and is unique. | Proof. Uniqueness: If both \( d \) and \( {d}^{\prime } \) satisfy \( \left( i\right) ,\left( {ii}\right) \), and \( \left( {iii}\right) \) then \( d\left| {{d}^{\prime }\text{and}{d}^{\prime }}\right| d \) so \( d = \pm {d}^{\prime } \) by Property 4.1(4), hence \( d = \left| d\right| = \) \( \left| {d}^{\prime }\righ... | No |
Theorem 4.11. (Euclid’s Lemma) Let \( a, b \) be integers and \( p \) be a prime satisfying \( p \mid {ab} \). Then \( p \mid a \) or \( p \mid b \). | (proof of) Euclid’s Lemma. If \( p \) is a prime and \( p \mid {ab} \) but \( p/a \), then \( \left( {p, a}\right) = 1 \) as only \( \pm 1, \pm p \) divide \( p \). Since \( p \mid {ab} \), we conclude that \( p \mid b \) by (3). | Yes |
Theorem 4.16. (The Fundamental Theorem of Arithmetic) Every integer \( n > 1 \) is a product of positive primes unique up to order, i.e., there exist unique primes\n\n(*) \n\n\[ \n1 < {p}_{1} < \cdots < {p}_{r} \n\] \nand integers \( {e}_{1},\ldots ,{e}_{r} > 0 \) such that \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{... | Proof. Existence. Let\n\n\[ \nS = \{ n > 1\text{ in }\mathbb{Z} \mid n\text{ is not a product of primes }\} . \n\] \n\nWe must show \( S = \varnothing \) . Suppose this is false. By the Well-Ordering Principle, there exists a minimal element \( n \in S \) . Clearly, no prime lies in \( S \), so \( n \) is not a prime. ... | No |
Proposition 5.11. Let \( \sim \) be an equivalence relation on \( A \) . Then\n\n\[ \nA = \mathop{\bigvee }\limits_{\bar{A}}\bar{a} \n\]\n\nIn particular, if \( a, b \in A \), then\n\n\[ \n\text{either}\bar{a} = \bar{b}\text{or}\bar{a} \cap \bar{b} = \varnothing \text{,} \n\]\n\nhence\n\n\[ \n\bar{a} = \bar{b}\text{if ... | Proof. As \( a \sim a \) for all \( a \in A \) by Reflexitivity and \( a \in \bar{a} \) by definition, we have \( A = \mathop{\bigcup }\limits_{A}\bar{a} \) . By Symmetry, we have \( a \sim b \) if \( \bar{a} = \bar{b} \) . If \( c \in \bar{a} \cap \bar{b} \), then \( c \sim a \) and \( c \sim b \), so \( a \sim c \) b... | Yes |
Proposition 6.3. Let \( m > 1 \) in \( \mathbb{Z} \) . Then \( \equiv {\;\operatorname{mod}\;m} \) is an equivalence relation. In particular,\n\n\[ \mathbb{Z} = \overline{0} \vee \overline{1} \vee \cdots \vee \overline{m - 1} \]\n\nWrite \( \mathbb{Z}/m\mathbb{Z} \) for \( \mathbb{Z}/ \equiv {\;\operatorname{mod}\;m} \... | Proof. Exercise. | No |
Lemma 6.8. Let \( m, n,{a}_{i} \), and \( 1 \leq i \leq r \) be integers.\n\n(1) If \( \left( {{a}_{i}, m}\right) = 1 \) for \( i = 1,\ldots, r \), then \( \left( {{a}_{1}\cdots {a}_{r}, m}\right) = 1 \) . | Proof. (1). By induction, it suffices to do the case \( r = 2 \) . (Why?) By Key Observation (4.13), we have equations\n\n\[ \n{x}_{1}{a}_{1} + {y}_{1}m = 1 = {x}_{2}{a}_{2} + {y}_{2}m, \n\]\n\nfor some \( {x}_{1},{x}_{2},{y}_{1},{y}_{2} \in \mathbb{Z} \), so\n\n\[ \n1 = \left( {{x}_{1}{a}_{1} + {y}_{1}m}\right) \left(... | No |
Theorem 6.9. (Chinese Remainder Theorem) Let \( {m}_{i} \) be integers with \( \left( {{m}_{i},{m}_{j}}\right) = 1 \) for \( 1 \leq i, j \leq r \) and \( i \neq j \) . Set \( m = {m}_{1}\cdots {m}_{r} \) and suppose that \( {c}_{1},\ldots ,{c}_{r} \) are integers. Then there exists an integer \( x \) satisfying all of ... | Proof. Existence. Let \( {n}_{i} = \frac{m}{{m}_{i}} = {m}_{1}\cdots \widehat{{m}_{i}}\cdots {m}_{r} \) where \( \hat{} \) means omit. By Lemma 6.8 (1), we have \( \left( {{m}_{i},{n}_{i}}\right) = 1 \) for \( 1 \leq i \leq r \), so by Key Observation 4.13, there exist equations\n\n\[ 1 = {d}_{i}{m}_{i} + {e}_{i}{n}_{i... | Yes |
Proposition 9.2. Let \( G \) be a group and \( H \) be a non-empty subset of \( G \) . Then \( H \) is a subgroup of \( G \) if and only if the following two conditions hold:\n\n(i) If \( a \) and \( b \) are elements of \( H \), then \( {ab} \) is an element of \( H \) .\n\n[We say that \( H \) is closed under \( \cdo... | Proof. \( \left( \Rightarrow \right) \) follows from the definition of subgroup, since by definition, \( {e}_{G} = {e}_{H} \) .\n\n\( \left( \Leftarrow \right) \) : We first note that \( \left( i\right) \) and \( \left( {ii}\right) \) implies \( \left( *\right) \) for if \( a, b \in H \), then \( a,{b}^{-1} \in H \), h... | Yes |
Corollary 9.3. Let \( G \) be a group and \( H \subset G \) a non-empty subset. If \( H \) is a finite set, then \( H \) is a subgroup if and only if \( H \) is closed under \( \cdot \) . | Proof. We need only check:\n\n\( \left( \Leftarrow \right) \) : It suffices to show if \( a \in H \) then \( {a}^{-1} \in H \) . Our hypothesis implies that\n\n\[ S \mathrel{\text{:=}} \left\{ {{a}^{n} \mid n \in {\mathbb{Z}}^{ + }}\right\} \subset H\text{is also a finite set.} \]\n\nWe need the following, whose proof ... | No |
Theorem 9.9. (Classification of Cyclic Groups) Let \( G = \langle a\rangle \) be a cyclic group and \( \theta : \mathbb{Z} \rightarrow G \) the map given by \( m \rightarrow {a}^{m} \) . Then \( \theta \) is a group epimorphism. It is an isomorphism if and only if \( G \) is infinite. If \( G \) is finite, then \( \lef... | Proof. As \( \theta \left( {i + j}\right) = {a}^{i + j} = {a}^{i}{a}^{j} = \theta \left( i\right) \theta \left( j\right) \), the map is a group homomorphism. It is clearly an epimorphism.\n\nWe must show that \( \theta \) is an isomorphism if and only if \( \left| G\right| \) is infinite and if not then \( \ker \theta ... | Yes |
Theorem 10.4. (Lagrange’s Theorem) Let \( G \) be a finite group and \( H \) a subgroup of \( G \). Then \[ \left| G\right| = \left\lbrack {G : H}\right\rbrack \left| H\right| \text{. In particular,}\left| H\right| \left| \right| G\left| \right| \text{and}\left\lbrack {G : H}\right\rbrack \left| \right| G \mid \text{.}... | Proof. We begin with the following: Claim 10.5. If \( G \) is an arbitrary group (i.e., without assuming it is finite) and \( H \) a subgroup, then \( \left| {aH}\right| = \left| H\right| \) for all \( a \in G \). Define the left translation map \[ {\lambda }_{a} : H \rightarrow {aH}\text{ by }h \mapsto {ah}. \] By our... | Yes |
Corollary 10.9. Let \( G \) be a group and \( K \) and \( H \) two finite subgroups of \( G \) of relatively prime degree, then \( K \cap H = \left\{ {e}_{G}\right\} \) . | Proof. We leave this as an exercise. | No |
Corollary 10.12. Let \( G \) be a finite group and \( a \in G \) . Then \( {a}^{\left| G\right| } = e \) . | Proof. If \( n \) is the order of \( \langle a\rangle \), then \( \left| G\right| = {nm} \) for some \( m \in {\mathbb{Z}}^{ + } \) so \( e = {\left( {a}^{n}\right) }^{m} = {a}^{\left| G\right| } \) . | Yes |
Corollary 10.13. (Euler’s Theorem) Let \( m, n \) be relatively prime integers with \( m > 1 \) . Then \( {n}^{\varphi \left( m\right) } \equiv 1{\;\operatorname{mod}\;m} \), where \( \varphi \) is the Euler \( \varphi \) -function. | Proof. We have shown that the unit group \( {\left( \mathbb{Z}/mZ\right) }^{ \times } \) is given by\n\n\[ \n{\left( \mathbb{Z}/mZ\right) }^{ \times } = \{ \bar{a} \mid a \in \mathbb{Z},\left( {a, m}\right) = 1\}\n\]\n\nand by definition, its cardinality is \( \varphi \left( m\right) \) . | No |
Corollary 10.14. (Fermat's Little Theorem) Let \( p \) be a positive prime integer. Then \( {n}^{p} \equiv n{\;\operatorname{mod}\;p} \) for all integers \( n \) . If \( p \nmid n \), then \( {n}^{p - 1} \equiv 1 \) \( {\;\operatorname{mod}\;p} \) . | Proof. As \( \varphi \left( p\right) = p - 1 \), this follows from Euler’s Theorem together with the observation that \( {0}^{p} = 0 \) . | No |
Lemma 11.2. Let \( G \) be a group and \( x \in G \) . Then \( {\theta }_{x} : G \rightarrow G \) is an isomorphism. In particular, if \( H \subset G \) is a subgroup, so is \( {\theta }_{x}\left( H\right) \) and \( H \cong {\theta }_{x}\left( H\right) = {xH}{x}^{-1} \) . In particular, \( \left| H\right| = \left| {{xH... | Proof. Let \( g,{g}^{\prime } \in G \) . As \( {xg}{x}^{-1} = x{g}^{\prime }{x}^{-1} \) implies \( g = {g}^{\prime } \), the map \( {\theta }_{x} \) is one-to-one. The equation\n\n(11.3)\n\n\[ \n{\theta }_{x}\left( {g{g}^{\prime }}\right) = {xg}{g}^{\prime }{x}^{-1} = {xge}{g}^{\prime }{x}^{-1} = {xg}{x}^{-1}x{g}^{\pri... | Yes |
Example 12.2. Let \( G \) be a group. The map \( \theta : G \rightarrow \operatorname{Aut}\left( G\right) \) given by \( x \mapsto \left( {{\theta }_{x} : g \mapsto {xg}{x}^{-1}}\right) \) is a group homomorphism. | If \( {\theta }_{x} = {1}_{G} \) , the identity map on \( G \), then \( {xg}{x}^{-1} = g \), i.e., \( {xg} = {gx} \) for all \( g \in g \) . Therefore, \( \ker \theta = Z\left( G\right) \), the center of \( G \) . By the First Isomorphism Theorem, \( \theta \) induces an isomorphism \( \bar{\theta } : G/Z\left( G\right... | No |
Theorem 12.3. Let \( G \) be a group and \( H \) a normal subgroup of \( G \) . Then \( G/H \) is a group under the binary operation \( \cdot : G/H \times G/H \rightarrow \) \( G/H \) given by \( \left( {{aH},{bH}}\right) \mapsto {abH} \) . If this is the case then \( - : G \rightarrow G/H \) is a group epimorphism wit... | Proof. We first show that the map \( \cdot \) is well-defined. Suppose that \( {aH} = {a}^{\prime }H \) and \( {bH} = {b}^{\prime }H \) . We must show that \( {abH} = {a}^{\prime }{b}^{\prime }H \) . Equivalently, we must show that\n\n\[ \text{if}{a}^{\prime - 1}a,{b}^{\prime - 1}b \in H\text{then}x = {\left( {a}^{\pri... | Yes |
Theorem 12.4. (General Cayley Theorem) Let \( H \) be an arbitrary subgroup of a group \( G \). Let \( S = G/H \), the set of left cosets of \( H \) in \( G \). For each \( x \in G \), let \( {\lambda }_{x} : S \rightarrow S \) be defined by \( {gH} \mapsto {xgH} \). Then \( {\lambda }_{x} \) is a permutation and the m... | Proof. We first must show that the map \( {\lambda }_{x} : S \rightarrow S \), called left multiplication by \( x \), is a permutation. But this is clear since \( {\lambda }_{{x}^{-1}} \) is easily checked to be its inverse.\n\nFor all \( x, y \in G \), we have \( {\lambda }_{xy}\left( {gH}\right) = {xygH} = {\lambda }... | Yes |
Corollary 12.6. Let \( H \) be a subgroup of a group \( G \) with \( H < G \) . If there exists no normal subgroup \( N \) of \( G \) satisfying \( 1 < N \subset H \) then \( \lambda : G \rightarrow \sum \left( {G/H}\right) \) by \( x \mapsto \left( {{\lambda }_{x} : {aH} \mapsto {xaH}}\right) \) is a monomorphism. | Proof. \( \ker \lambda \) is the maximal such normal subgroup. | No |
Corollary 12.7. (Useful Counting Result) Let \( G \) be a finite group, \( H \) a subgroup of \( G \) satisfying \( \left| G\right| /\left\lbrack {G : H}\right\rbrack \) !. Then there exists a normal subgroup \( N \) of \( G \) satisfying \( 1 < N \subset H \) . In particular, \( G \) is not a simple group. | Proof. Exercise. | No |
Theorem 13.2. (Correspondence Principle) Let \( \varphi : G \rightarrow {G}^{\prime } \) be a group epimorphism.. Then\n\n(1) If \( A \) is a subgroup of \( G \) (respectively, a normal subgroup), then \( \varphi \left( A\right) \) is a subgroup of \( {G}^{\prime } \) (respectively, a normal subgroup). In particular, g... | Proof. Let \( K = \ker \varphi \) .\n\n(1). Let \( A \subset G \) be a subgroup, then \( {\left. \varphi \right| }_{A} : A \rightarrow {G}^{\prime } \) is a group homomorphism (why?), so \( \varphi \left( A\right) = {\left. \operatorname{im}\varphi \right| }_{A} \subset {G}^{\prime } \) is a subgroup. Next suppose that... | No |
Theorem 13.4. (Third Isomorphism Theorem) Let \( G \) be a group with normal subgroups \( K \) and \( H \) satisfying \( K \subset H \) . Then the map\n\n\[ \varphi : G/K \rightarrow G/H\text{defined by}{xK} \rightarrow {xH} \]\n\n is a group epimorphism with kernel \( H/K \) and induces an isomorphism\n\n\[ \bar{\varp... | Proof. As \( K \) and \( H \) are normal subgroups of \( G \), we know that \( G/H \) and \( G/K \) are groups. If \( {xK} = {yK} \) then \( {y}^{-1}x \in K \subset H \), hence \( {xH} = {yH} \) . Therefore, \( \varphi \) is well-defined and clearly surjective. As \( \varphi \left( {xKyK}\right) = \varphi \left( {xyK}\... | Yes |
Theorem 13.5. (Second Isomorphism Theorem) Let \( G \) be a group and \( H \) and \( N \) be subgroups with \( N \) normal in \( G \) . Then\n\n(1) \( H \cap N \vartriangleleft H \) .\n\n(2) \( {HN} = {NH} \) is a subgroup of \( G \) .\n\n(3) \( N \vartriangleleft {HN} \) .\n\n(4) \( H/H \cap N \cong {HN}/N \) . | Proof. (1). We know that \( H \cap N \) is a subgroup of \( H \) and it is a normal subgroup in \( H \), since \( N \) is normal in \( G \) and \( H \) is normal in \( H \) . (2). As \( {aN} = {Na} \) for all \( a \in G \) the sets \( {HN} \) and \( {NH} \) are equal, so we need only show it is a group. Let \( {h}_{1},... | Yes |
Proposition 14.2. Let \( G \) be a finite abelian group and \( p > 0 \) a prime dividing the order of \( G \) . Then there exists an element of \( G \) of order \( p \) . | Proof. We prove this by induction on \( \left| G\right| \) . We may assume that \( G \neq 1 \) . As \( G \) has no nontrivial subgroups if and only if \( G \cong \mathbb{Z}/p\mathbb{Z} \) for some prime \( p \), we may assume that \( \left| G\right| \) is not a prime. In particular, \( G \) has a subgroup \( 1 < H < G ... | Yes |
Theorem 14.3. Let \( G \) be a finite abelian group and \( p > 0 \) a prime dividing \( \left| G\right| \), say \( \left| G\right| = {p}^{n}m \) with \( p \) and \( m \) relatively prime. Then \[ G\left( p\right) \mathrel{\text{:=}} \left\{ {x \in G \mid {x}^{{p}^{r}} = e\text{ some integer }r}\right\} \vartriangleleft... | Proof. The set \( G\left( p\right) \) is a subgroup of \( G \) (why?) and normal as \( G \) is abelian. As every element in \( G\left( p\right) \) has order a power of \( p \), it follows by the previous proposition that \( \left| {G\left( p\right) }\right| = {p}^{r} \) some \( r \geq 1 \) and by Lagrange’s Theorem tha... | Yes |
Corollary 14.5. Let \( G \) be a finite abelian group of order \( n = {p}_{1}^{{m}_{1}}\cdots {p}_{r}^{{m}_{r}} \) with positive primes \( {p}_{1} < \cdots < {p}_{r} \) and positive integers \( {m}_{1},\ldots ,{m}_{r} \). Then \( G = G\left( {p}_{1}\right) \cdots G\left( {p}_{r}\right) \). Moreover, \( G \cong G\left( ... | Proof. As \( G\left( {p}_{1}\right) \cdots G\left( {p}_{r}\right) \) is a group and we know that\n\n\[ \left| {G\left( {p}_{1}\right) \cdots G\left( {p}_{r}\right) }\right| = \left| {G\left( {p}_{1}\right) }\right| \cdots \left| {G\left( {p}_{r}\right) }\right| \]\n\nby Lemma 14.1, the first statement follows by the th... | Yes |
Lemma 14.6. Let \( G \) be a finite additive p-group and suppose that the element \( x \) in \( G \) has maximal order. Then there exists a subgroup \( H \) of \( G \) satisfying \( G = \langle x\rangle \oplus H \) . | Proof. Let \( {p}^{n} \) be the order of \( x \) . By the Well-ordering Principle, there exists a maximal subgroup \( H \) of \( G \) satisfying \( H \cap \langle x\rangle = \{ 0\} \) . Therefore, \( \langle x\rangle + H = \langle x\rangle \oplus H \), and we are done if \( \langle x\rangle + H = G \) . So suppose not.... | Yes |
Corollary 14.7. Let \( G \) be a finite p-group. Then \( G \) is a product of cyclic p-groups. | Proof. We may assume that \( G \) is additive and not cyclic. In particular \( \left| G\right| > p \) . By the lemma, \( G = \langle x\rangle \oplus H \) for some \( x \) in \( G \) and subgroup \( H \) of \( G \) . As \( G \) is not cyclic, \( H \neq G \), and the result follows by induction on \( \left| G\right| \) . | No |
Proposition 14.9. Every finite abelian group is a product of cyclic groups. | Proof. By 14.5 every finite abelian group is a product of finite abelian \( p \) -groups. By Corollary 14.5, every finite abelian \( p \) -group is a product of cyclic \( p \) -groups. | Yes |
Theorem 14.10. (Fundamental Theorem of Finite Abelian Groups) Let \( G \) be a finite additive group and for each prime \( p > 0 \) dividing \( G \) , let \( G\left( p\right) \) be the unique p-subgroup of \( G \) of maximal order. Then\n\n\[ G = {\bigoplus }_{p\parallel G\parallel }G\left( p\right) \]\n\nMoreover, if ... | Proof. Let \( p\left| \right| G \mid \) . By Corollary 14.5 and Proposition 14.9, it suffices to show \( G\left( p\right) \cong \mathop{\sum }\limits_{{i = 1}}^{r}\mathbb{Z}/{p}^{{n}_{i}}\mathbb{Z} \) and uniquely up to isomorphism. As every abelian \( p \) -group is isomorphic to a product of cyclic \( p \) -groups by... | Yes |
Proposition 15.1. Let \( G \) be a finitely generated group and \( n \) a positive integer. Then there exist finitely many subgroups (if any) of \( G \) of index \( n \) . | Proof. Let \( G = \left\langle {{a}_{1},\ldots ,{a}_{r}}\right\rangle \) and\n\n(*) \n\n\[ \varphi : G \rightarrow {S}_{n} \] \n\nbe a group homomorphism. Then the map \( \varphi \) is completely determined by \n\n\[ \varphi \left( {a}_{1}\right) ,\ldots ,\varphi \left( {a}_{r}\right) \] \n\nin the finite set \( {S}_{n... | Yes |
Corollary 15.2. Let \( G \) be a finitely generated group. Suppose \( G \) contains a subgroup \( H \) of finite index. Then there exists a characteristic subgroup \( K \) of \( H \) of finite index in \( G \) . | Proof. By the proposition, there exist finitely many subgroups\n\n\[ \nH = {H}_{1},\ldots ,{H}_{m} \n\]\n\nof finite index \( \left\lbrack {G : H}\right\rbrack \) . It follows, given an automorphism \( \varphi \) of \( G \), that \( \varphi \left( H\right) = {H}_{i} \) for some \( i \), i.e., \( \varphi \) permutes the... | No |
Theorem 15.3. Let \( G \) be a finitely generated group and \( H \) be a subgroup of finite index. Then \( H \) is a finitely generated group. | Proof. Let \( G = \left\langle {{a}_{1},\ldots ,{a}_{r}}\right\rangle \) and\n\n\[ \n{y}_{1} = e,\ldots ,{y}_{n} \n\] \n\nbe a transversal (i.e., a system of representatives for the cosets) of \( H \) in \( G \) . Let \n\n\[ \n\lambda : G \rightarrow \sum \left( {G/H}\right) \text{be given by}a \mapsto {\lambda }_{x} :... | Yes |
Theorem 16.3. The following are true:\n\n(1) A subgroup of a solvable group is solvable.\n\n(2) The homomorphic image (i.e., the image of a group under a group homomorphism) of a solvable group is solvable.\n\n(3) If \( N \vartriangleleft G \) and both \( N \) and \( G/N \) are solvable then so is \( G \) . | Proof. We leave this as an exercise. It is important that you do this exercise, as it will teach you how to use the isomorphism theorems. | No |
Proposition 16.5. Let \( G \) be a group. Then \( {G}^{\left( n\right) } \) is a characteristic subgroup of \( {G}^{\left( n - 1\right) } \) for all \( n \) . In particular, \( {G}^{\left( n\right) } \) is characteristic in \( G \) , and\n\n\[ \n{G}^{\left( n\right) } \subset {G}^{\left( n - 1\right) } \subset \cdots \... | Proof. Being characteristic is transitive, so this follows immediately, using Properties 16.4 | Yes |
Theorem 16.6. Let \( G \) be a group. Then \( G \) is solvable if and only if there exist an integer \( n \) such that \( {G}^{\left( n\right) } = 1 \) . | Proof. By Exercise 11.9(17), each factor group \( {G}^{\left( i\right) }/{G}^{\left( i + 1\right) } \) is abelian, hence if \( {G}^{\left( n\right) } = 1 \) for some \( n \), then \( G \) is solvable. Conversely, suppose that\n\n\[ 1 = {N}_{n} \subset {N}_{n - 1} \subset {N}_{n - 2} \subset \cdots \subset {N}_{1} \subs... | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.