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Lemma 3. Let \( \mathfrak{Q} \) be primary, \( \mathfrak{P} \) its associated prime and let \( \mathfrak{C} \) be any ideal not contained in \( \mathfrak{P} \) ; then \( \mathfrak{Q} : \mathfrak{C} = \mathfrak{Q} \) . | Proof. An element \( u \) in \( \mathfrak{Q} : \mathfrak{C} \) satisfies the condition that \( {uc} \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{Q}}\right) \) for all \( c\mathrm{e}\mathfrak{C} \) . If we choose \( c ≢ 0\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \), then this implies that \( u \equiv 0\left(... | Yes |
Theorem 6. If \( \mathfrak{B} \) and \( \mathfrak{C} \) are ideals in a Noetherian ring, \( \mathfrak{B} : \mathfrak{C} = \) \( \mathfrak{B} \) if and only if \( \mathfrak{C} \) is not contained in any of the associated primes of \( \mathfrak{B} \) . | Proof. Let \( \mathfrak{B} = {\mathfrak{Q}}_{1} \cap {\mathfrak{Q}}_{2} \cap \cdots \cap {\mathfrak{Q}}_{r} \) be an irredundant decomposition of \( \mathfrak{B} \) into primary ideals. Let \( {\mathfrak{P}}_{i} = \mathfrak{R}\left( {\mathfrak{Q}}_{i}\right) \) and assume that \( \mathfrak{C} \nsubseteq {\mathfrak{P}}_... | Yes |
Theorem 7. If \( \mathfrak{g} \) is Noetherian, an element a \( \mathfrak{e}\mathfrak{A} \) is a \( \mathfrak{g} \) -integer if and only if there exists a finitely generated submodule of \( \mathfrak{A} \) that contains all the powers of a. | Proof. We have just seen that this condition is necessary. Now let \( \mathfrak{N} \) be a finitely generated \( \mathfrak{g} \) -module containing all the powers of \( a \) . Since \( \mathfrak{g} \) is Noetherian, \( \mathfrak{N} \) satisfies the ascending chain condition for submodules. Hence, there exists an intege... | Yes |
Theorem 8. The totality \( \mathfrak{G} \) of elements of \( \mathfrak{A} \) that are \( \mathfrak{g} \) -integral is a subring of \( \mathfrak{A} \) containing \( \mathfrak{g} \) . | Proof. Any element \( \gamma \) of \( \mathfrak{g} \) satisfies an equation \( x - \gamma = 0 \) . Hence, it belongs to \( \mathfrak{G} \) . Next let \( a \) and \( {b\varepsilon }\mathfrak{G} \) and let \( \left( {{u}_{1},{u}_{2},\cdots ,{u}_{s}}\right) \) and \( \left( {{v}_{1},{v}_{2},\cdots ,{v}_{t}}\right) \) be \... | Yes |
Theorem 9. The ring \( \mathfrak{G} \) of \( \mathfrak{g} \) -integral elements is integrally closed in \( \mathfrak{A} \) . | Proof. Let \( a \) be a \( \mathfrak{G} \) -integer and let\n\n\[ {a}^{n} = {g}_{0} + {g}_{1}a + \cdots + {g}_{n - 1}{a}^{n - 1} \]\n\nwhere the \( {g}_{i}\varepsilon \mathcal{B} \) . We can use this relation to show that every power of \( a \) is expressible as a linear combination of the powers 1, \( a,\cdots ,{a}^{n... | Yes |
Theorem 11. Let \( \mathfrak{g} \) be a Gaussian subring of a field \( \mathfrak{F} \) and let \( {\mathfrak{F}}_{0} \) be the subfield of \( \mathfrak{F} \) generated by \( \mathfrak{g} \) . Then an element \( {a\varepsilon }\mathfrak{F} \) is integrally dependent on \( \mathfrak{g} \) if and only if it is algebraic o... | This criterion is particularly useful if every element of \( \mathfrak{F} \) is algebraic over \( {\mathfrak{F}}_{0} \) ; for in this case it asserts that an element of \( \mathfrak{F} \) is \( \mathfrak{g} \) -integral if and only if its minimum polynomial is in \( \mathfrak{g}\left\lbrack x\right\rbrack \) . We note ... | Yes |
Theorem 1. A 1-1 mapping \( a \rightarrow {a}^{\prime } \) of a lattice \( L \) onto a lattice \( {L}^{\prime } \) is an isomorphism if and only if \( a \geq b \) in \( L \) implies and is implied by \( {a}^{\prime } \geq {b}^{\prime } \) in \( {L}^{\prime } \) . | Proof. A mapping \( a \rightarrow {a}^{\prime } \) of a lattice \( L \) into a lattice \( {L}^{\prime } \) is called order preserving if \( a \geq b \) implies that \( {a}^{\prime } \geq {b}^{\prime } \) . If \( a \rightarrow {a}^{\prime } \) is an isomorphism and \( a \geq b \), then \( a \cup b = a \) . Hence \( {a}^... | Yes |
Theorem 2. The lattice of invariant subgroups of any group is modular. | Proof. Let \( \mathfrak{G} \) be the given group and let \( {\mathfrak{H}}_{1},{\mathfrak{H}}_{2},{\mathfrak{H}}_{3} \) be invariant subgroups such that \( {\mathfrak{H}}_{1} \geq {\mathfrak{H}}_{2}\left( {{\mathfrak{H}}_{1} \supseteq {\mathfrak{H}}_{2}}\right) \) . Consider the intersection \( {\mathfrak{H}}_{1} \cap ... | Yes |
Theorem 3. A lattice \( L \) is modular if and only if \( a \geq b \) and \( a \cup c = b \cup c, a \cap c = b \cap c \) for any \( c \) imply that \( a = b \) . | Proof. Let \( L \) be modular and let \( a, b, c \) be elements of \( L \) such that \( a \geq b \) and \( a \cup c = b \cup c, a \cap c = b \cap c \) . Then\n\n\[ a = a \cap \left( {a \cup c}\right) = a \cap \left( {b \cup c}\right) = b \cup \left( {a \cap c}\right) \]\n\n\[ = b \cup \left( {b \cap c}\right) = b\text{... | No |
Theorem 4. If a and \( b \) are any two elements of a modular lattice, then the intervals \( I\left\lbrack {a \cup b, a}\right\rbrack \) and \( I\left\lbrack {b, a \cap b}\right\rbrack \) are isomorphic. | Proof. Let \( x \) be in the interval \( I\left\lbrack {a \cup b, a}\right\rbrack \), so that \( a \cup b \geq \) \( x \geq a \) . Then \( b \geq x \cap b \geq a \cap b \) and \( x \cap b \) is in the interval \( I\lbrack b \) , \( a \cap b\rbrack \) . Similarly, if \( y \) is in \( I\left\lbrack {b, a \cap b}\right\rb... | Yes |
Theorem 8. The number of terms in any two irredundant representations of an element as g.l.b. of irreducible elements is the same. | Proof. Applying Theorem 7 we can write\n\n(11)\n\n\[ a = {r}_{{1}^{\prime }} \cap {q}_{2} \cap \cdots \cap {q}_{m} = {r}_{{1}^{\prime }} \cap {r}_{{2}^{\prime }} \cap {q}_{3}\cdots \cap {q}_{m} \]\n\n\[ = \cdots = {r}_{{1}^{\prime }} \cap {r}_{{2}^{\prime }} \cap \cdots \cap {r}_{{m}^{\prime }}{.}^{ * } \]\n\nSince the... | Yes |
Theorem 9. If the elements \( {a}_{1},{a}_{2},\cdots ,{a}_{n} \) are independent, then\n\n\[ \left( {{a}_{1} \cup \cdots \cup {a}_{r} \cup {a}_{r + 1} \cup \cdots \cup {a}_{s}}\right) \cap \left( {{a}_{1} \cup \cdots \cup {a}_{r} \cup {a}_{s + 1} \cup \cdots \cup {a}_{t}}\right) = {a}_{1} \cup \cdots \cup {a}_{r}. \] | Proof. We prove first that\n\n\[ \left( {{a}_{1} \cup \cdots \cup {a}_{s}}\right) \cap \left( {{a}_{s + 1} \cup \cdots \cup {a}_{n}}\right) = 0. \]\n\nThis is true by assumption if \( s = 1 \) . Assume now that we have it for \( s - 1 \) . Then\n\n\[ \left( {{a}_{1} \cup \cdots \cup {a}_{s}}\right) \cap \left( {{a}_{s ... | Yes |
Theorem 10. If \( L \) is a complemented modular lattice that satisfies both chain conditions, then the element 1 of \( L \) is a l.u.b. of independent points. Conversely, if \( L \) is a modular lattice with 0 and 1 such that 1 is a l.u.b. of a finite number of points, then \( L \) is complemented and satisfies both c... | A cyclic subgroup of prime order is a point in the lattice \( \mathfrak{L} \) of subgroups of a group \( \mathcal{O} \) . Hence if \( \mathcal{O} \) is finite and commutative and every element of \( \otimes \) is of prime order, then \( \mathfrak{L} \) satisfies the chain conditions, is modular and 1 in \( \& \) is a l... | No |
Theorem 11. The complement \( {a}^{\prime } \) of any element \( a \) of a Boolean algebra \( B \) is uniquely determined. The mapping \( a \rightarrow {a}^{\prime } \) is \( 1 - 1 \) of \( B \) onto itself; it is of period two \( \left( {{a}^{\prime \prime } = a}\right) \) ; and it satisfies the conditions\n\n(15)\n\n... | Proof. Let \( a \) be any element of \( B \) and let \( {a}^{\prime } \) and \( {a}_{1} \) be elements such that \( a \cup {a}^{\prime } = 1, a \cap {a}_{1} = 0 \) . Then\n\n\[{a}_{1} = {a}_{1} \cap 1 = {a}_{1} \cap \left( {a \cup {a}^{\prime }}\right) = \left( {{a}_{1} \cap a}\right) \cup \left( {{a}_{1} \cap {a}^{\pr... | Yes |
Lemma 1. If \( {x}_{1},{x}_{2},\cdots ,{x}_{m} \) are linearly independent and \( {x}_{1},{x}_{2} \) , \( \cdots ,{x}_{m},{x}_{m + 1} \) are linearly dependent, then \( {x}_{m + 1} \) is linearly dependent on \( {x}_{1},\cdots ,{x}_{m} \) . | Proof. We have \( {\beta }_{1}{x}_{1} + {\beta }_{2}{x}_{2} + \cdots + {\beta }_{m}{x}_{m} + {\beta }_{m + 1}{x}_{m + 1} = 0 \) where some \( {\beta }_{k} \neq 0 \) . If \( {\beta }_{m + 1} = 0 \), this implies that \( {x}_{1},\cdots ,{x}_{m} \) are linearly dependent contrary to assumption. Hence \( {\beta }_{m + 1} \... | Yes |
Lemma 2. Let \( {x}_{1},{x}_{2},\cdots ,{x}_{m} \) be a set of \( m > 1 \) vectors and define \( {x}_{i}{}^{\prime } = {x}_{i} \) for \( i = 1,2,\cdots, m - 1 \) and \( {x}_{m}{}^{\prime } = {x}_{m} + \rho {x}_{1} \) . Then the \( {x}_{i} \) are linearly independent if and only if the \( {x}_{i}{}^{\prime } \) are line... | Proof. Suppose that the \( {x}_{i} \) are linearly independent and let \( {\beta }_{i} \) be elements of \( \Delta \) such that \( \sum {\beta }_{i}{x}_{i}{}^{\prime } = 0 \) . Then\n\n\[{\beta }_{1}{x}_{1} + {\beta }_{2}{x}_{2} + \cdots + {\beta }_{m - 1}{x}_{m - 1} + {\beta }_{m}\left( {{x}_{m} + \rho {x}_{1}}\right)... | Yes |
Theorem 2. If \( \Re \) has a basis of \( n \) vectors, then any \( n + 1 \) vectors in \( \Re \) are linearly dependent. | Proof. We prove the theorem by induction on \( n \) . Let \( {e}_{1},{e}_{2} \) , \( \cdots ,{e}_{n} \) be a basis and let \( {x}_{1},{x}_{2},\cdots ,{x}_{n + 1} \) be vectors in \( \Re \) . The theorem is clear for \( n = 1 \) ; for, in this case, \( {x}_{1} = {\alpha }_{1}{e}_{1},{x}_{2} = {\alpha }_{2}{e}_{1} \) and... | Yes |
Theorem 4. If \( {f}_{1},{f}_{2},\cdots ,{f}_{r} \) are linearly independent, then we can supplement these vectors with \( n - r \) vectors chosen from a basis \( {e}_{1},{e}_{2},\cdots ,{e}_{n} \) to obtain a basis. | Proof. We consider the set \( \left( {{f}_{1},{f}_{2},\cdots ,{f}_{r};{e}_{1},{e}_{2},\cdots ,{e}_{n}}\right) \), and we choose in this set a maximum linearly independent set \( \left( {f}_{1}\right. \) , \( \left. {{f}_{2},\cdots ,{f}_{r};{e}_{{i}_{1}},{e}_{{i}_{2}},\cdots ,{e}_{{i}_{h}}}\right) \) including the \( {f... | Yes |
Theorem 5. The matrix of any ordered basis \( \\left( {{f}_{1},{f}_{2},\\cdots ,{f}_{n}}\\right) \) relative to the ordered basis \( \\left( {{e}_{1},{e}_{2},\\cdots ,{e}_{n}}\\right) \) is non-singular. | Conversely, let \( \\left( \\alpha \\right) \) be an element of \( L\\left( {\\Delta, n}\\right) \) . Let \( \\left( \\beta \\right) = {\\left( \\alpha \\right) }^{-1} \) . Define \( {f}_{i} \) by \( {f}_{i} = \\sum {\\alpha }_{ij}{e}_{j} \) . Then we assert that \( \\left( {{f}_{1},{f}_{2},\\cdots ,{f}_{n}}\\right) \)... | No |
Theorem 6. If \( \left( \alpha \right) \) and \( \left( \beta \right) \varepsilon {\Delta }_{n} \) and \( \left( \beta \right) \left( \alpha \right) = 1 \), then also \( \left( \alpha \right) \left( \beta \right) = 1 \) so that \( \left( \alpha \right) \) and \( \left( \beta \right) \) e \( L\left( {\Delta, n}\right) \... | Proof. If \( \left( \beta \right) \left( \alpha \right) = 1 \), the equation (8) shows that if \( {f}_{i} = \) \( \sum {\alpha }_{ij}{e}_{j} \), then the \( {e}_{k} \) are dependent on the \( f \) ’s. The argument given above then shows that the \( f \) ’s form a basis. Hence the matrix \( \left( \alpha \right) \) of \... | Yes |
Theorem 7. If \( \left( \alpha \right) \) is not a right (left) zero divisor in \( {\Delta }_{n} \), then \( \left( \alpha \right) {\varepsilon L}\left( {\Delta, n}\right) \) . | Proof. We have to show that the vectors \( {f}_{i} = \sum {\alpha }_{ij}{e}_{j} \) form a basis. By Theorem 4 it suffices to show that the \( f \) ’s are linearly independent. Suppose therefore that \( \sum {\beta }_{i}{f}_{i} = 0 \) . Then \( \sum {\beta }_{i}{\alpha }_{ij}{e}_{j} \) \( = 0 \) and hence \( \mathop{\su... | No |
Theorem 8. Any matrix \( \left( \alpha \right) \) in \( L\left( {\Delta, n}\right) \) is a product of elementary matrices. | Proof. We note first that, if \( \left( {{f}_{1},{f}_{2},\cdots ,{f}_{n}}\right) \) is an ordered basis, then so are the following sets:\n\n\[ \left( {{f}_{1},{f}_{2},\cdots ,{f}_{p - 1},{f}_{p}^{\prime },{f}_{p + 1},\cdots ,{f}_{n}}\right) ,\;{f}_{p}^{\prime } = {f}_{p} + \beta {f}_{q},\;q \neq p \]\n\n\[ \left( {{f}_... | No |
Theorem 9. The vectors \( {x}_{i} = \sum {\alpha }_{ij}{e}_{j}, i = 1,2,\cdots, r \), are linearly independent if and only if \( \left( \alpha \right) \) is of determinantal rank \( r \) . | Proof. Evidently the determinantal rank \( \rho \leq n \) . Also the \( x \) ’s are linearly independent only if \( r \leq n \) . Hence we may assume that \( r \leq n \) . Suppose first that the \( x \) ’s are dependent, so that, say, \( {x}_{1} = {\beta }_{2}{x}_{2} + \cdots + {\beta }_{r}{x}_{r} \) . Then \( {\alpha ... | Yes |
Theorem 11. The spaces \( {\mathfrak{S}}_{i} \) are independent if and only if \( \dim \left( {{\mathfrak{S}}_{1} + {\mathfrak{S}}_{2} + \cdots + {\mathfrak{S}}_{r}}\right) = \sum \dim {\mathfrak{S}}_{i}. \) | Proof. Suppose first that the \( {\mathfrak{S}}_{i} \) are independent and let \( \left( {{f}_{1i},{f}_{2i},\cdots ,{f}_{{n}_{i}i}}\right) \) be a basis for \( {\mathfrak{S}}_{i} \) . Then if \( \sum {\beta }_{ji}{f}_{ji} = 0,\sum {y}_{i} = 0 \) where \( {y}_{i} = \mathop{\sum }\limits_{j}{\beta }_{ji}{f}_{ji}\mathrm{e... | Yes |
Theorem 5. If \( \left( \alpha \right) \) is an \( {n}_{1} \times {n}_{2} \) matrix with elements in a division ring \( \Delta \) and if \( \left( \alpha \right) \) has row rank \( \rho \), then \( \left( \alpha \right) \) is equivalent to the matrix given in equation (18). | If \( \left( \alpha \right) \) and \( \left( \widetilde{\alpha }\right) \) are equivalent matrices, then we know that these can be taken to be matrices of the same linear transformation \( A \) . The row ranks of \( \left( \alpha \right) \) and of \( \left( \widetilde{\alpha }\right) \) coincide with the rank of \( A \... | Yes |
Theorem 9. The row rank and the column rank of any matrix are the same. | A second, and somewhat more geometric, proof of this result will be given in \( §{12} \) . We note finally that the theory of right-handed systems of linear equations can be developed in a manner completely analogous to that of left-handed systems considered above. We have only to replace left vector spaces by right ve... | No |
Theorem 10. If \( \\left( {{e}_{1}*,{e}_{2}*,\\cdots ,{e}_{n} * }\\right) \) is a basis for \( {\\Re }^{ * } \), then there exists a basis \( \\left( {{e}_{1},\\cdots ,{e}_{n}}\\right) \) for \( \\Re \) such that \( {e}_{i} * \\left( {e}_{j}\\right) = {\\delta }_{ij} \) . | Proof. We know that we can find in \( {\\Re }^{* * } \) a basis \( \\left( {{e}_{1}{}^{* * },{e}_{2}{}^{* * }}\\right. \), \( \\cdots ,{e}_{n} * * ) \) that is complementary to the given basis \( \\left( {{e}_{1}*,{e}_{2}*}\\right. \), \( \\left. {\\cdots ,{e}_{n}{}^{ * }}\\right) \), that is, \( {e}_{i}{}^{* * }\\left... | Yes |
Theorem 11. The null space of \( A \) is the subspace incident with the rank space of the transpose \( {A}^{ * } \) of \( A \) . The rank space of \( A \) is the subspace incident with the null space of \( {A}^{ * } \) . | Proof. It is sufficient to prove the first of these statements. Hence let \( z \) be a vector such that \( {zA} = 0 \) . Then \( f\left( {zA}\right) = 0 \) and \( \left( {f{A}^{ * }}\right) \left( z\right) = 0 \) for all \( f \) . Hence \( {zgj}\left( {{\Re }_{2} * {A}^{ * }}\right) \) . The converse follows by retraci... | No |
Theorem 1. There exists a vector in \( \Re \) whose order is the minimum polynomial of \( A \) . | We know that the degree of the order of any vector \( u \) is the dimensionality of the space \( \{ u\} \) . Hence \( \deg {\mu }_{u}\left( \lambda \right) \leq n \) . Thus we have the\n\nCorollary. The degree of the minimum polynomial of \( A \) is \( \leq n \) . | No |
Theorem 2. A linear transformation is cyclic if and only if its minimum polynomial has degree \( n \) . | Suppose now that \( \Re = \{ e\} \) and let\n\n\[ \mu \left( \lambda \right) = {\mu }_{e}\left( \lambda \right) = {\lambda }^{n} - {\mu }_{n - 1}{\lambda }^{n - 1} - \cdots - {\mu }_{0}. \]\n\nWe know that the vectors \( \left( {e,{eA},\cdots, e{A}^{n - 1}}\right) \) constitute a basis. We wish to determine the matrix ... | Yes |
Theorem 3. Let \( \Re = \{ e\} \) and \( \mu \left( \lambda \right) = {\mu }_{1}\left( \lambda \right) \cdots {\mu }_{s}\left( \lambda \right) \) where the \( {\mu }_{i}\left( \lambda \right) \) are relatively prime in pairs. Then \( \Re = \left\{ {e}_{1}\right\} \oplus \left\{ {e}_{2}\right\} \oplus \cdots \oplus \lef... | Let \( {\nu }_{i}\left( \lambda \right) = \mu \left( \lambda \right) {\left( {\mu }_{i}\left( \lambda \right) \right) }^{-1} \) and let \( {e}_{i} = e{\nu }_{i}\left( A\right) \) . Then \( {\mu }_{{e}_{i}}\left( \lambda \right) = {\mu }_{i}\left( \lambda \right) \) . We know also that the order of \( {e}^{\prime } = {e... | Yes |
Theorem 7. If \( \mathfrak{o} \) is a principal ideal domain, any finitely generated \( \mathfrak{o} \) -module is a direct sum of the submodule of elements of finite order and a free module. | We note finally that in the decomposition \( \Re = \left\{ {{e}_{1}{}^{\prime }}\right\} \oplus \left\{ {{e}_{2}{}^{\prime }}\right\} \) \( \oplus \cdots \oplus \left\{ {e}_{n}\right\} \) we may drop the terms \( \left\{ {e}_{i}\right\} \) whose order ideals are \( \left( {\delta }_{j}\right) = \left( 1\right) \) . If ... | Yes |
Theorem 9. Any finitely generated commutative group is a direct sum of a finite group and a free group. | The finite group is \( \mathfrak{S} = \left\{ {f}_{1}\right\} \oplus \left\{ {f}_{2}\right\} \oplus \cdots \oplus \left\{ {f}_{u}\right\} \) . Its order is \( {\delta }_{1}{\delta }_{2}\cdots {\delta }_{u} \) if \( {\delta }_{j} \) is normalized to be positive. | No |
Theorem 11. If diag \( \left\{ {{\delta }_{1},{\delta }_{2},\cdots ,{\delta }_{r},0,\cdots ,0}\right\} \) and diag \( \left\{ {{\delta }_{1}{}^{\prime }}\right. \) , \( \left. {{\delta }_{2}{}^{\prime },\cdots ,{\delta }_{{r}^{\prime }}{}^{\prime },0,\cdots ,0}\right\} \) are equivalent \( m \times n \) matrices with e... | Proof. Let \( \mathfrak{F} \) be a free module with basis \( \left( {{t}_{1},{t}_{2},\cdots ,{t}_{n}}\right) \), and let \( \mathfrak{N} \) be the submodule generated by \( {v}_{1} = {\delta }_{1}{t}_{1},\cdots ,{v}_{r} = {\delta }_{r}{t}_{r} \) , \( {v}_{r + 1} = 0,\cdots ,{v}_{m} = 0 \) . Then \( \mathfrak{F}/\mathfr... | Yes |
Theorem 13. Let \( \\left( \\alpha \\right) \) be a matrix in \( {\\Phi }_{n} \) and let \( {\\Delta }_{n}\\left( \\lambda \\right) = \\det (╏ - \\) \( \\left( \\alpha \\right) ) \) and \( \\mu \\left( \\lambda \\right) = {\\Delta }_{n}\\left( \\lambda \\right) {\\left\\lbrack {\\Delta }_{n - 1}\\left( \\lambda \\right... | The first two statements are clear from what we have proved about \( A \) . The last statement follows from (31) and the fact that all the \( {\\delta }_{i}\\left( \\lambda \\right) \) are factors of \( {\\delta }_{t}\\left( \\lambda \\right) = \\mu \\left( \\lambda \\right) \) . | Yes |
Theorem 14. (Hamilton-Cayley) If \( \left( \alpha \right) \) e \( {0}_{n},0 \) a commutative ring with an identity, and \( {\Delta }_{n}\left( \lambda \right) \) is the characteristic polynomial, then \( {\Delta }_{n}\left( \left( \alpha \right) \right) = 0 \) . | Proof. We recall the identity (Vol. I, p. 59)\n\n(37)\n\n\[ \left\lbrack {╏ - \left( \alpha \right) }\right\rbrack \operatorname{adj}\left\lbrack {╏ - \left( \alpha \right) }\right\rbrack = \det \left( {╏ - \left( \alpha \right) }\right) 1 = {\Delta }_{n}\left( \lambda \right) 1. \]\n\nThe matrix \( \operatorname{adj}\... | Yes |
Theorem 17. If \( \mathfrak{o} \) is a commutative ring with an identity and \( \Re \) is a cyclic \( \mathfrak{o} \) -module, then the only \( \mathfrak{o} \) -endomorphisms of \( \Re \) are the mappings \( x \rightarrow {\alpha x} \) . | Proof. Let \( \Re = \{ e\} \) and let \( B \) be an o-endomorphism of \( \Re \) . Suppose that \( {eB} = {\beta e} \) . Then if \( x = {\alpha e},{xB} = \left( {\alpha e}\right) B = \alpha \left( {eB}\right) \) \( = \beta \left( {\alpha e}\right) = {\beta x} \) . Thus \( B = {\beta }_{l} \) . | No |
Theorem 18. Let \( \Re = \left\{ {f}_{1}\right\} \oplus \left\{ {f}_{2}\right\} \oplus \cdots \oplus \left\{ {f}_{t}\right\} \) where the order ideal of \( {f}_{i} \) is \( \left( {\delta }_{i}\right) \) . Then the ring \( \mathfrak{B} \) of 0 -endomorphisms of \( \mathfrak{R} \) is isomorphic to the difference ring \(... | An explicit determination of the matrices of \( \mathfrak{M} \) can be made if use is made of the conditions (45) on the \( \delta \) ’s. We note the following cases of (47):\n\n1. \( i \geq j \) . Here \( {\delta }_{i} \equiv 0\left( {\;\operatorname{mod}\;{\delta }_{j}}\right) \) . Hence these \( {\beta }_{ij} \) are... | Yes |
Theorem 19. (Frobenius) Let \( \left( \alpha \right) \varepsilon {\Phi }_{n} \) and let \( {\delta }_{1}\left( \lambda \right) ,{\delta }_{2}\left( \lambda \right) ,\cdots \) , \( {\delta }_{t}\left( \lambda \right) \) be the invariant factors \( \neq 1 \) of \( ╏ - \left( \alpha \right) \) . Then if the degree of \( {... | Clearly, if \( t > 1 \), then \( N > \mathop{\sum }\limits_{{j = 1}}^{t}{n}_{j} = n > {n}_{t} \) . Since the dimensionality of \( \mathfrak{A} = {\Phi }_{l}\left\lbrack A\right\rbrack \) over \( {\Phi }_{l} \) is \( {n}_{t} \), this shows that, in this case, \( \mathfrak{B} \supset \mathfrak{A} \) . If we recall that \... | Yes |
Theorem 20. The center of the ring of 0-endomorphisms of \( \Re \) consists of the scalar multiplications. | Let \( \mathfrak{B} \) be the ring of \( \mathfrak{o} \) -endomorphisms, \( \mathfrak{C} \) its center and \( {\mathfrak{o}}_{l} \) the ring of scalar multiplications \( x \rightarrow {\alpha x} \) . We have seen that \( {\mathfrak{o}}_{l} \subseteq \mathfrak{C} \) . Now let \( C \) be any element of \( \mathfrak{C} \)... | Yes |
Corollary 2. The center of the ring of linear transformations of a vector space over a field \( \Phi \) is the set \( {\Phi }_{l} \) of scalar multiplications. | A slightly more direct proof of this result will be given later (Chapter VIII, p. 229). | No |
Theorem 2. A set \( \Omega \) of linear transformations is completely reducible if and only if \( \Re \) can be expressed as a direct sum of subspaces \( {\Re }_{i} \) that are invariant and irreducible relative to \( \Omega \) . | Sufficiency. Let \( \Re = {\Re }_{1} \oplus {\Re }_{2} \oplus \cdots \oplus {\Re }_{s} \) where the \( {\Re }_{i} \) are irreducible invariant subspaces. If \( \mathfrak{S} \) is any invariant subspace, either \( \mathfrak{S} = \mathfrak{R} \) or there exists an \( {\mathfrak{R}}_{i} \), say \( {\mathfrak{R}}_{1} \), s... | Yes |
Theorem 3. A linear transformation \( A \) in \( \Re \) over \( \Phi \) is irreducible if and only if it is cyclic and has prime minimum polynomial. | We consider next the question of decomposability of a single linear transformation. We know that \( \Re = \left\{ {f}_{1}\right\} \oplus \left\{ {f}_{2}\right\} \oplus \cdots \oplus \) \( \left\{ {f}_{t}\right\} \) . Hence a necessary condition for indecomposability is that \( t = 1 \), that is, \( \Re = \{ f\} \) is c... | Yes |
Theorem 6. Let \( \mu \left( \lambda \right) \) be the minimum polynomial of the linear transformation \( A \) in \( \Re \) over \( \Phi \) and let (13) be the factorization of \( \mu \left( \lambda \right) \) into prime powers. Then, if \( {\Re }_{i} \) is defined to be the subspace of vectors \( {x}_{i} \) such that ... | We shall call the spaces \( {\Re }_{i} \) the primary components of \( \Re \) relative to \( A \) . The projections \( {E}_{i} \) determined by the decomposition (15) will be called the principal idempotent elements of \( A \) . We specialize now by assuming that \( \Phi \) is algebraically closed. In this case the \( ... | Yes |
Theorem 1. If \( g\left( {x,{y}^{\prime }}\right) \) is a bilinear form connecting the vector spaces \( \Re \) and \( {\Re }^{\prime } \), then there exist bases \( \left( {{u}_{1},{u}_{2},\cdots ,{u}_{n}}\right) ,\left( {{v}_{1}{}^{\prime },{v}_{2}{}^{\prime }}\right. \) , \( \left. {\cdots ,{v}_{{n}^{\prime }}{}^{\pr... | Evidently the number \( r \) is the rank (row or column) of the matrix \( \left( \beta \right) \) of \( g\left( {x,{y}^{\prime }}\right) \) . An abstract characterization of this number will be given in the next section. | No |
Theorem 2. Necessary and sufficient conditions that a bilinear form \( g\left( {x,{y}^{\prime }}\right) \) connecting \( \Re \) and \( {\Re }^{\prime } \) be non-degenerate are: 1) \( \Re \) and \( {\Re }^{\prime } \) have the same dimensionality; 2) the matrix of the form relative to any pair of bases is non-singular.... | If two vector spaces \( \mathfrak{R} \) and \( {\mathfrak{R}}^{\prime } \) are connected by a non-degenerate bilinear form \( g \), then we shall say that these spaces are dual relative to \( g \) . Suppose that this is the case and let \( \left( {{e}_{1},{e}_{2},\cdots ,{e}_{n}}\right) \) be a given basis in \( \mathf... | Yes |
Theorem 3. If \( g\left( {x, y}\right) \) is a hermitian scalar product, then there exists a basis \( \left( {{u}_{1},{u}_{2},\cdots ,{u}_{r},{z}_{1},{z}_{2},\cdots ,{z}_{n - r}}\right) \) such that\n\n(31)\n\n\[ g\left( {{u}_{i},{u}_{i}}\right) = {\beta }_{i} \neq 0,\;i = 1,2,\cdots, r, \]\n\nand all other products ar... | Proof. The result is trivial if \( g = 0 \) ; for then any basis serves as a set of \( z \) ’s. If \( g \neq 0 \), we can take \( {u}_{1} \) to be any vector such that \( g\left( {{u}_{1},{u}_{1}}\right) = {\beta }_{1} \neq 0 \) . Such vectors exist by the Lemma. Now suppose that we have already determined linearly ind... | Yes |
Theorem 7. If \( g\left( {x, y}\right) \) is an alternate scalar product, there exists a basis for \( \mathfrak{R} \) relative to which the matrix has the form (38). | If \( g\left( {x, y}\right) \neq 0 \) is alternate, the anti-automorphism is the identity and the matrices \( \left( \beta \right) \) of \( g\left( {x, y}\right) \) are alternate in the sense that \( {\left( \beta \right) }^{\prime } = - \left( \beta \right) \) and \( {\beta }_{ii} = 0 \) for \( i = 1,2,\cdots, n \) . ... | No |
Theorem 8. Any g-equivalence of a subspace of \( \Re \) can be extended to a g-unitary transformation in \( {\Re }^{ * } \) | A hermitian scalar product is called totally regular if \( g\left( {x, x}\right) \) \( \neq 0 \) for every \( x \neq 0 \) in \( \Re \) . This is equivalent to saying that every non-zero subspace of \( \mathfrak{R} \) is not isotropic. Hence if \( g \) is totally regular and \( \otimes \) is any subspace, then \( \Re = ... | Yes |
Theorem 3. If \( \Omega \) is any set of linear transformations and \( \mathfrak{S} \) is invariant under \( \Omega \), then the orthogonal complement \( {\mathfrak{S}}^{ \bot } \) is invariant under \( {\Omega }^{\prime } \) the set of transposes of the linear transformations in \( \Omega \) . | Proof. Let \( {x\varepsilon }\mathfrak{S} \) and \( {y\varepsilon }{\mathfrak{S}}^{ \bot } \) . Then \( {xA\varepsilon }\mathfrak{S} \) for any \( A \) in \( \Omega \) . Hence \( \left( {{xA}, y}\right) = 0 \) and also \( \left( {x, y{A}^{\prime }}\right) = 0 \) . Since this holds for all \( x \) in \( \mathfrak{S}, y{... | Yes |
Theorem 6. If a Euclidean space \( \mathfrak{R} \) is irreducible relative to a commutative set of linear transformations which are either symmetric or skew, then \( \dim \Re \leq 2 \) . | Proof. Let \( \pi \left( \lambda \right) \) be an irreducible factor of the characteristic polynomial of any \( A \) e \( \Omega \) . Then the characteristic space \( {\Re }_{\pi \left( \lambda \right) } \) of \( A \) is invariant relative to \( \Omega \) . Hence \( {\Re }_{\pi \left( \lambda \right) } = \Re \) so that... | Yes |
Theorem 10. A symmetric transformation \( A \) is positive definite (semi-definite) if and only if its characteristic roots are all positive (non-negative). | Proof. We know that there exists a Cartesian basis \( \left( {{y}_{1},{y}_{2}}\right. \) , \( \left. {\cdots ,{y}_{n}}\right) \) for \( \Re \) such that each \( {y}_{i} \) is a characteristic vector in the sense that \( {y}_{i}A = {\rho }_{i}{y}_{i} \) . The \( {\rho }_{i} \) are the characteristic roots. Now \( \left(... | Yes |
Theorem 14. If \( \left( \alpha \right) \) is a matrix with complex elements, then there exists a unitary matrix \( \left( \sigma \right) \) such that \( \left( \sigma \right) \left( \alpha \right) {\left( \sigma \right) }^{-1} \) is triangular. | Proof. In order to prove this we let \( A \) be the linear transformation whose matrix relative to some unitary basis \( \left( {{v}_{1},{v}_{2},\cdots }\right. \) , \( \left. {v}_{n}\right) \) is the given matrix \( \left( \alpha \right) \) . If \( {\rho }_{1} \) is a characteristic root, there exists a vector \( {y}_... | Yes |
Theorem 1. Let \( \mathfrak{P} \) be a direct product of \( {\mathfrak{R}}^{\prime } \) and \( \mathfrak{S} \) relative to \( \times \) and let \( {\mathfrak{P}}_{1} \) be any product of these same spaces relative to the multiplication \( {\mathrm{X}}_{1} \) . Then the mapping \( \sum {x}_{i}{}^{\prime } \times {y}_{i}... | Proof. Suppose we have two ways of writing an element \( {z\varepsilon }\mathfrak{P} \) as a sum of products. We can suppose that these are \( z = \mathop{\sum }\limits_{1}^{m}{x}_{i}{}^{\prime } \times {y}_{i} = \mathop{\sum }\limits_{{m + 1}}^{q}\left( {-{x}_{j}{}^{\prime }}\right) \times {y}_{j} \) . Then \( \mathop... | Yes |
Theorem 3. Let \( \Re ,\varnothing \) and \( \mathfrak{P} \) be vector spaces over a field \( \Phi \) and suppose that there is defined a product \( x \times y,{x\varepsilon }\Re ,{y\varepsilon }(\varepsilon, x \times {y\varepsilon }\$ \) , such that \( {\mathbf{1}}^{\prime } \cdot ,{\mathbf{2}}^{\prime } \) . and \( {... | Proof. The necessity of this condition has already been proved. Conversely, let \( \dim \mathfrak{P} = \dim \mathfrak{R}\dim \mathfrak{S} \) . Let \( \left( {{e}_{1},{e}_{2},\cdots ,{e}_{n}}\right) \) be a basis for \( \Re \) and let \( \left( {{f}_{1},{f}_{2},\cdots ,{f}_{m}}\right) \) be a basis for \( \mathfrak{S} \... | Yes |
Theorem 4. If \( \mathfrak{R} \) and \( \mathfrak{S} \) are (finite dimensional) vector spaces, the space \( \mathfrak{L}\left( {\Re \times \mathfrak{S},\Re \times \mathfrak{S}}\right) = \mathfrak{L}\left( {\Re ,\Re }\right) \times \mathfrak{L}\left( {\mathfrak{S},\mathfrak{S}}\right) \) relative to Kronecker multiplic... | Assume now that\n\n(14)\n\n\[ \n{e}_{i}A = \sum {\alpha }_{ij}{e}_{j},\;{f}_{k}B = \sum {\beta }_{kl}{f}_{l} \n\]\n\nso that \( \left( \alpha \right) \) and \( \left( \beta \right) \) are the matrices of \( A \) and \( B \), respectively, relative to the chosen bases. Then we have\n\n(15)\n\n\[ \n\left( {{e}_{i} \times... | Yes |
Theorem 6. The algebra \( \mathfrak{L}\left( {\Re \times \varnothing ,\Re \times \varnothing }\right) = \mathfrak{L}\left( {\Re ,\Re }\right) \times \) \( \mathfrak{L}\left( {\mathfrak{S},\mathfrak{S}}\right) \) . | Proof. We have already seen that this relation holds in the vector space sense. Also we have the relation \( \left( {A \times B}\right) \left( {C \times D}\right) \) \( = {AC} \times {BD} \), and this shows that the ordinary product in \( \mathfrak{L}\left( {\Re \times \mathfrak{S},\Re \times \mathfrak{S}}\right) \) co... | No |
Theorem 1. The ring \( \mathfrak{L} \) of linear transformations in a finite dimensional vector space is simple. | ## EXERCISE\n\n1. Prove that \( \bar{\Delta } \) is the totality of linear transformations that commute with the \( {E}_{ij} \) .\n\n2. Operator methods. We shall now give a second proof of the simplicity of \( \mathfrak{L} \) using operator methods. Let \( \mathfrak{B} \) be a two-sided ideal \( \neq 0 \) in \( \mathf... | No |
Theorem 2. If \( \mathfrak{L} \) is the ring of linear transformations in \( \mathfrak{R} \) over \( \Delta \), then reciprocally \( \Delta \) is the complete set of endomorphisms in \( \Re \) which commute with all the transformations contained in \( \mathfrak{L} \) . | Proof. Let \( C \) be an endomorphism in \( \Re \) which commutes with every element of \( \mathfrak{L} \) . If \( x \) is any vector in \( \mathfrak{R} \), then \( x \) and \( {xC} \) are linearly dependent. Otherwise there exists an \( A \) e \( \Omega \) such that \( {xA} = 0 \), but \( \left( {xC}\right) A \neq 0 \... | Yes |
Theorem 3. Every left ideal \( {\mathfrak{J}}^{\prime } \) in \( \mathfrak{L} \) has the form \( {\mathfrak{R}}^{\prime } \times \mathfrak{S} \) where \( \mathfrak{S} \) is a subspace of \( \mathfrak{R} \) . The subspace \( \mathfrak{S} \) is in fact the join of all the rank spaces \( \Re B, B \) in \( {\Im }^{\prime }... | This result can also be formulated in another way. Let \( \otimes \) be any subspace of \( \mathfrak{R} \) . Define \( {\mathfrak{J}}^{\prime }\left( \mathfrak{S}\right) \) to be the totality of \( {B\varepsilon }\mathfrak{L} \) such that \( \Re B \subseteq \mathfrak{S} \) . Clearly \( {\Im }^{\prime }\left( \mathfrak{... | Yes |
Theorem 5. Every right ideal \( \mathfrak{F} \) in \( \mathfrak{L} \) has the form \( {\mathfrak{S}}^{\prime } \times \mathfrak{R} \) where \( {\mathfrak{S}}^{\prime } \) is a subspace of \( {\mathfrak{R}}^{\prime } \) . | We can obtain a correspondence also between subspaces of \( \Re \) and right ideals of \( \mathfrak{L} \) . For this purpose we consider the subspace \( \mathfrak{S} = j\left( {\mathfrak{S}}^{\prime }\right) \) of vectors \( y \) in \( \mathfrak{R} \) such that \( g\left( {y,{y}^{\prime }}\right) = 0 \) for all \( {y}^... | Yes |
Theorem 6. Every right ideal, \( \Im \) in \( \mathfrak{L} \) has the form \( \mathcal{Z}\left( \mathfrak{S}\right) \), the set of linear transformations which annihilate a subspace \( \mathfrak{S} \) . The subspace \( \mathfrak{S} \) is the totality of vectors annihilated by every \( B \) e \( \Im \) . | Since \( {\mathfrak{S}}^{\prime } \) is arbitrary in \( {\mathfrak{R}}^{\prime } \) , \( \mathfrak{S} \) is arbitrary in \( \mathfrak{R} \) . If \( \mathfrak{S} \) is any subspace of \( \mathfrak{R} \), it is clear at the start that the totality \( \mathfrak{J} = \mathfrak{Z}\left( \mathfrak{S}\right) \) of linear tran... | Yes |
Lemma 2. If \( \Im \) is a minimal right ideal and \( x\Im \neq 0 \), then the homomorphism \( \chi : B \rightarrow {xB} \) is an isomorphism of \( \Im \) onto \( x\Im = \Re \) . | If \( {\chi }^{-1} \) denotes the inverse mapping of \( \Re \) onto \( \Im \) and \( {\Lambda }_{r} \) denotes the right multiplication \( B \rightarrow {BA} \) in \( \Im \), then, by (13), \( {A}_{r}\chi = {\chi A} \) so that\n\n\( \left( {13}^{\prime }\right) \)\n\n\[ \n{A}_{r} = {\chi A}{\chi }^{-1} \n\] | Yes |
Theorem 1. Let \( \mathfrak{R} \) have a basis \( B = \left( {e}_{i}\right) \) . Then if \( \mathfrak{S} \) is any subspace of \( \Re \), we can divide \( B \) into two non-overlapping subsets \( C = \left( {e}_{j}\right), D = \left( {e}_{k}\right) \) such that \( \mathfrak{S} \) has a basis of the form \( {f}_{j} = {e... | Proof. There exists a complement \( {\mathfrak{S}}^{\prime } \) of \( \mathfrak{S} \) spanned by a subset \( D = \left( {e}_{k}\right) \) of \( B \) . Let \( C = \left( {e}_{j}\right) \) be the complement of \( D \) in \( B \) . Then each \( {e}_{j} = {f}_{j} - {u}_{j} \) where \( {f}_{j}\mathrm{e}\mathfrak{S} \) and \... | Yes |
Lemma 1. If \( \dim \Re = b \) is infinite and the cardinal number of \( \Delta \) is \( d \), then the cardinal number of \( \Re \) is bd. | Proof. If \( \left( {e}_{i}\right) \) is a basis for \( \Re \), every non-zero vector \( x \) in \( \Re \) has a unique representation as\n\n\[ x = \mathop{\sum }\limits_{{j = 1}}^{N}{\xi }_{{i}_{j}}{e}_{{i}_{j}},\;{\xi }_{{i}_{j}} \neq 0. \]\n\nThus with each \( x \neq 0 \) we can associate a uniquely determined subse... | Yes |
Lemma 2. There exists a strongly independent collection of sequences with cardinal number \( \geq d \) . | Proof. We partially order the strongly independent collections \( F \) by inclusion. If a set \( \{ F\} \) of these collections is linearly ordered, clearly \( \cup F \) is strongly independent. By Zorn’s lemma, there exists a maximal strongly independent collection \( M \) . We shall prove, by induction, that if the c... | Yes |
Theorem 1. Let \( \mathrm{E}/\Phi ,\mathrm{P}/\Phi \) be fields over \( \Phi \) and let \( {\mathfrak{L}}_{\Phi }\left( {\mathrm{E},\mathrm{P}}\right) \) be the right vector space over \( \mathrm{P} \) of linear mappings of \( \mathrm{E}/\Phi \) into \( \mathrm{P}/\Phi \) . Then \( \left\lbrack {\mathrm{E} : \Phi }\rig... | Proof. Let \( {\eta }_{1},{\eta }_{2},\cdots ,{\eta }_{n} \) be elements of \( \mathrm{E} \) which are linearly independent over \( \Phi \) . Then we may imbed this set in a basis \( \left\{ {\eta }_{\alpha }\right\} \) for \( \widehat{\mathrm{E}} \) over \( \Phi \) (Vol. II, p. 239). If we choose a correspondent \( {\... | Yes |
Theorem 2 (Jacobson-Bourbaki). Let \( \mathrm{P} \) be a field and \( \mathfrak{A} \) a set of endomorphisms of \( \left( {\mathrm{P}, + }\right) \) such that:\n\n(i) \( \mathfrak{A} \) is a subring of \( \mathfrak{L}\left( {\mathrm{P},\mathrm{P}}\right) \) the ring of endomorphisms of \( \left( {\mathrm{P}, + }\right)... | Proof (Hochschild). The verification that \( \Phi \) is a subfield is immediate and will be omitted. Next we apply the lemma of \( §1 \) to obtain elements \( {\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{n} \) in \( \mathrm{P} \) and a right basis \( \left( {{E}_{1},{E}_{2}}\right. \) , \( \left. {\cdots ,{E}_{n}}\right) \... | No |
Theorem 3 (Dedekind). Let \( \mathrm{E} \) and \( \mathrm{P} \) be fields and let \( {s}_{1},{s}_{2},\cdots \) , \( {s}_{n} \) be distinct isomorphisms of \( \mathbf{E} \) into \( \mathrm{P} \) . Then the \( {s}_{i} \) are right linearly independent over \( \mathrm{P} : \sum {s}_{i}{\rho }_{i} = 0,{\rho }_{i}\varepsilo... | Proof. If the assertion is false, then we have a shortest relation, which by suitable ordering reads:\n\n(3)\n\n\[ \n{s}_{1}{\rho }_{1} + {s}_{2}{\rho }_{2} + \cdots + {s}_{r}{\rho }_{r} = 0, \n\] \n\nwhere every \( {\rho }_{i} \neq 0 \) . Suppose \( r > 1 \) . Since \( {s}_{1} \neq {s}_{2} \) there exists \( {\eta \va... | Yes |
Theorem 4. Let \( \mathrm{E} \) and \( \mathrm{P} \) be fields, \( {s}_{1},{s}_{2},\cdots ,{s}_{n} \) isomorphisms of \( \mathrm{E} \) into \( \mathrm{P} \), and let \( \mathfrak{A} \) be the right \( \mathrm{P} \) -subspace of \( \mathfrak{L}\left( {\mathrm{E},\mathrm{P}}\right) \) of endomorphisms \( \sum {s}_{i}{\rh... | Proof. It is clear that \( \left\{ {\mathop{\sum }\limits_{{j = 1}}^{r}{s}_{ij}{\rho }_{ij} \mid {\rho }_{ij}\text{ e }\mathrm{P}}\right\} \subseteq \mathfrak{B} \) . To prove the opposite inclusion it suffices to show that, if \( \mathop{\sum }\limits_{1}^{n}{s}_{i}{\rho }_{i}\varepsilon \mathfrak{B} \), then the \( {... | Yes |
Theorem 5. Let \( \mathrm{P} \) be a field and let \( \mathcal{A} \) be the collection of finite groups of automorphisms in \( \mathrm{P},\mathcal{I} \) the collection of subfields of \( \mathrm{P} \) which are Galois and of finite co-dimension in \( \mathrm{P} \) . If \( {\Phi \varepsilon }\mathcal{I} \), let \( A\lef... | Proof. (i)-(ii). If \( {G\varepsilon }\mathcal{A} \) and \( \mathfrak{A} = \left\{ {\sum {s}_{i}{\rho }_{i} \mid {s}_{i}{\varepsilon G},{\rho }_{i}\varepsilon \mathrm{P}}\right\} \), then \( \left\lbrack {\mathrm{P} : I\left( G\right) }\right\rbrack = {\left\lbrack \mathfrak{A} : \mathrm{P}\right\rbrack }_{R} = \left( ... | Yes |
Lemma 1. (1) If \( \mathrm{P}/\Phi \) is a splitting field of \( f\left( x\right) {\varepsilon \Phi }\left\lbrack x\right\rbrack \) and \( \sum /\Phi \) is a subfield of \( \mathrm{P}/\Phi \), then \( \mathrm{P}/\sum \) is a splitting field of \( f\left( x\right) \) . (2) If \( \mathrm{P}/\sum \) is a splitting field f... | Proof. (1) This is an immediate consequence of the definition. (2) By assumption we have \( \mathrm{P} = \mathbf{\sum }\left( {{\rho }_{1},\cdots ,{\rho }_{n}}\right) \) where (5) holds in \( \mathrm{P}\left\lbrack x\right\rbrack \) . Also \( \sum = \Phi \left( {{\sigma }_{1},\cdots ,{\sigma }_{r}}\right) \) and \( f\l... | Yes |
Theorem 6. Any polynomial \( f\left( x\right) {\varepsilon \Phi }\left\lbrack x\right\rbrack \) of positive degree has a splitting field \( \mathrm{P}/\Phi \) . | Proof. Let \( f\left( x\right) = {f}_{1}\left( x\right) {f}_{2}\left( x\right) \cdots {f}_{k}\left( x\right) \) be the factorization of \( f\left( x\right) \) into irreducible factors (with leading coefficients 1). Evidently \( k \leq n = \deg f\left( x\right) \) . We use induction on \( n - k \) . If \( n - k \) \( = ... | Yes |
Lemma 2. Let \( \mathrm{P} = \Phi \left( {{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{m}}\right) \) and assume that \( {\rho }_{i} \) is algebraic over \( \Phi \left( {{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{i - 1}}\right), i = 1,2,\cdots, m \) . Then \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack < \infty \) and \( \... | Proof. We have seen that this holds for \( m = 1 \) . Suppose \( m \) \( > 1 \) and assume the result holds for \( r < m \) . Then \( \Phi \left( {{\rho }_{1},\cdots ,{\rho }_{r}}\right) \) \( = \Phi \left\lbrack {{\rho }_{1},\cdots ,{\rho }_{r}}\right\rbrack \) and this is finite dimensional over \( \Phi \) . Since \(... | Yes |
Theorem 7. Let \( \alpha \rightarrow \bar{\alpha } \) be an isomorphism of a field \( \Phi \) onto the field \( \Phi \) and let \( f\left( x\right) \) be a polynomial of positive degree with leading coefficient \( 1, f\left( x\right) \) in \( \Phi \left\lbrack x\right\rbrack \), and let \( \bar{f}\left( x\right) \) be ... | Proof. Both assertions will be proved by induction on \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack \) . If \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack = 1,\mathrm{P} = \Phi \) and \( f\left( x\right) = \Pi \left( {x - {\rho }_{i}}\right) \) in \( \Phi \left\lbrack x\right\rbrack \) . Applying the isomorphism... | Yes |
Theorem 8. If \( f\left( x\right) {\varepsilon \Phi }\left\lbrack x\right\rbrack \) and \( \deg f > 0 \), then all the roots of \( f \) (in its splitting field) are simple if and only if \( \left( {f,{f}^{\prime }}\right) = 1 \) (that is, 1 is the highest common factor of \( f \) and \( {f}^{\prime } \) ). | Proof. Let \( d\left( x\right) \) be the highest common factor \( \left( {f,{f}^{\prime }}\right) \) of \( f \) and \( {f}^{\prime } \) in \( \Phi \left\lbrack x\right\rbrack \) (cf. Vol. I, p. 100, p. 122). Suppose \( f\left( x\right) \) has a multiple root in \( \mathrm{P}\left\lbrack x\right\rbrack \), so \( f\left(... | Yes |
Lemma 1. Let \( \mathrm{P} \supseteq \mathrm{E} \supseteq \Phi \) where \( \mathrm{E} \) and \( \Phi \) are subfields of \( \mathrm{P} \) and \( \mathrm{E}/\Phi \) is finite dimensional Galois. Then any element \( \theta \mathrm{e}\mathrm{P} \) which is separable algebraic over \( \mathrm{E} \) is separable algebraic o... | Proof. Let \( g\left( x\right) \) be the minimum polynomial of \( \theta \) over \( \mathbf{E} \) . If \( {s\varepsilon G} \) the Galois group of \( \mathbf{E}/\Phi \), then \( s \) has a unique extension to \( \mathrm{E}\left\lbrack x\right\rbrack \) satisfying \( {x}^{s} = x \) . Let \( {g}^{{s}_{1}}\left( x\right) ,... | Yes |
Theorem 11. If \( \mathrm{A}/\Phi \) is algebraic, then the set \( \sum \) of elements of \( \mathrm{A} \) which are separable over \( \Phi \) is a subfield containing \( \Phi \) . Moreover, \( \sum \) contains every element of \( \mathrm{A} \) which is separable algebraic over \( \sum \) . | Proof. Let \( \rho ,\sigma \) e \( \sum \) and let \( g\left( x\right) \) and \( h\left( x\right) \) be the minimum polynomials over \( \Phi \) of \( \rho \) and \( \sigma \) respectively. Then \( f\left( x\right) = g\left( x\right) h\left( x\right) \) is separable. If \( \Delta \) is a splitting field over \( \Phi \le... | Yes |
Theorem 14. Let \( \Phi \) be an infinite field and let \( \mathrm{P} = \Phi \left( {\xi ,\eta }\right) \) be a field generated over \( \Phi \) by a separable algebraic element \( \xi \) and an algebraic element \( \eta \) . Then \( \mathrm{P}/\Phi \) has a primitive element. | Proof. Let \( f\left( x\right) \) and \( g\left( x\right) \) be the minimum polynomial over \( \Phi \) of \( \xi \) and \( \eta \) respectively and let \( \Delta /\mathrm{P} \) be a splitting field of \( f\left( x\right) g\left( x\right) \) . Then \( \Delta /\Phi \) is a splitting field of \( f\left( x\right) g\left( x... | Yes |
Theorem 15 (Artin). Let \( \Phi \) be an infinite field and \( \mathrm{P} \) a finite dimensional extension field of \( \Phi \) . Then \( \mathrm{P}/\Phi \) is a simple extension if and only if there are only a finite number of intermediate fields between \( \mathrm{P} \) and \( \Phi \) . | Proof. Suppose first that \( \mathrm{P} = \Phi \left( \theta \right) \) and let \( \mathrm{E} \) be an intermediate field. Let \( g\left( x\right) \) be the minimum polynomial of \( \theta \) over \( \mathrm{E} \) and let \( {\mathrm{E}}^{\prime }/\Phi \) be the field generated by the coefficients of \( g\left( x\right... | Yes |
Theorem 16. Let \( \mathrm{P} \) be finite dimensional Galois over an infinite field \( \Phi ,\mathrm{E} \) a subfield of \( \mathrm{P}/\Phi \), and \( \Omega \) an arbitrary extension field of P. Let \( {s}_{1},{s}_{2},\cdots ,{s}_{m} \) be the different isomorphisms of \( \mathbf{E} \) over \( \Phi \) into P over \( ... | Proof. We recall that the number \( m \) of isomorphisms is \( \left\lbrack {\mathrm{E} : \Phi }\right\rbrack \) (§ 7). We note next that, if \( \left( {{\epsilon }_{1},{\epsilon }_{2},\cdots ,{\epsilon }_{m}}\right) \) is a basis for \( \mathrm{E}/\Phi \) , then the determinant det \( \left( {{\epsilon }_{i}{}^{{s}_{j... | Yes |
Theorem 17. Let \( \mathrm{P} \) be finite dimensional Galois over an infinite \( \Phi \) . Then \( \mathrm{P}/\Phi \) has a normal basis. | Proof. Let \( G = \left\{ {{s}_{1},\cdots ,{s}_{n}}\right\} \) be the Galois group of \( \mathrm{P}/\Phi \) . We have just seen that, if \( \left( {{\rho }_{1},\cdots ,{\rho }_{n}}\right) \) is a basis of \( \mathrm{P} \) over \( \Phi \), then det \( \left( {{\rho }_{i}{}^{{s}_{i}}}\right) \neq 0 \) . Conversely, this ... | Yes |
Lemma 1. Any finite subgroup \( A \) of the multiplicative group of a field is cyclic. | Proof. Let \( m \) be the order of \( A \) and let \( {m}^{\prime } \) be the highest order for the elements of \( A \) . It is known that, if \( a \) and \( b \) are two elements of a finite commutative group, then there exists a \( c \) in the group whose order is the least common multiple of the orders of \( a \) an... | Yes |
Lemma 2. Any cyclic extension \( \mathrm{P}/\Phi \) has a normal basis over \( \Phi \) . | Proof. Let \( s \) be a generator of the Galois \( G \) group of \( \mathrm{P}/\Phi \) . We consider \( s \) as a linear transformation in \( \mathrm{P} \) over \( \Phi \) and let \( \mu \left( x\right) \varepsilon \) \( \Phi \left\lbrack x\right\rbrack \) be its minimum polynomial. Now Dedekind’s independence theorem ... | Yes |
Theorem 19. Let \( s \rightarrow {\mu }_{s} \) be a mapping of \( G \) into \( {\mathrm{P}}^{ * } \) such that \( {\mu }_{st} = {\mu }_{s}{}^{t}{\mu }_{t}, s, t \) e \( G \) . Then there exists a non-zero element \( \gamma \) in \( \mathrm{P} \) such that \( {\mu }_{s} = \gamma {\left( {\gamma }^{s}\right) }^{-1} \) . | Proof. Since the \( {\mu }_{s} \) are \( \neq 0 \) and the automorphisms are right linearly independent over \( \mathrm{P} \), we see that the operator \( {\sum s}{\mu }_{s}( \equiv \)\n\n\( \left. {{\sum s}{\mu }_{sR}}\right) \) is \( \neq 0 \) . Thus we can find a \( {\beta \varepsilon }\mathrm{P} \) such that \( \ga... | Yes |
Theorem 20. Let \( {\delta }_{\varepsilon } \) , \( {s\varepsilon G} \), be elements of \( \mathrm{P} \) satisfying (63). Then there exists a \( \gamma \) e \( \mathrm{P} \) such that \( {\delta }_{s} = \gamma - {\gamma }^{s} \) . | Proof. We choose an element \( {\rho \varepsilon }\mathrm{P} \) such that \( {T}_{\mathrm{P} \mid \Phi }\left( \rho \right) = \sum {\rho }^{s} \neq \) 0. This can be done since \( \mathop{\sum }\limits_{{s \neq G}}s \neq 0 \) by the Dedekind independence theorem. Set \( \gamma = \mathop{\sum }\limits_{{s \in G}}T{\left... | Yes |
Theorem 21. Let \( \mathrm{E}/\Phi \) and \( \mathrm{P}/\Phi \) be fields such that \( \left\lbrack {\mathrm{P} : \Phi }\right\rbrack < \infty \) and let \( \mathfrak{J} \) be a maximal ideal in \( \mathrm{E}{ \otimes }_{\Phi }\mathrm{P} \) . Let \( s \) be the mapping \( \epsilon \rightarrow \) \( \epsilon \otimes 1 +... | Proof. If \( \Im \) is a maximal ideal in \( \mathrm{E} \otimes \mathrm{P} \), then \( \epsilon \rightarrow \epsilon \otimes 1 \) is a homomorphism into \( \mathrm{E} \otimes \mathrm{P} \) so \( s : \epsilon \rightarrow \epsilon \otimes 1 + \Im \) is a homomorphism into \( \Gamma = \left( {\mathrm{E} \otimes \mathrm{P}... | Yes |
Theorem 1. Let \( \Phi \) be a field of characteristic \( \neq 2 \) and \( f\left( x\right) \) a nonzero polynomial \( \mathrm{e}\Phi \left\lbrack x\right\rbrack \) without multiple roots. Let \( \mathrm{P}/\Phi \) be a splitting field of \( f\left( x\right) ,{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{m} \) its roots, \(... | Proof. We recall a standard characterization of the alternating group. For this one considers the ring \( \Phi \left\lbrack {{x}_{1},{x}_{2},\cdots ,{x}_{m}}\right\rbrack ,{x}_{i} \) in-determinates. If \( i \rightarrow {i}^{\sigma } \) is a permutation of \( 1,2,\cdots, m \), then we have the automorphism \( A\left( \... | Yes |
Theorem 2. Let \( f\left( x\right) \) e \( \Phi \left\lbrack x\right\rbrack \) have no multiple roots in its splitting field \( \mathrm{P} \) . Then \( f\left( x\right) \) is irreducible in \( \Phi \left\lbrack x\right\rbrack \) if and only if the Galois group \( {G}_{f} \) of \( f\left( x\right) = 0 \) over \( \Phi \)... | Proof. We recall that a transformation group of a set \( M \) is called transitive if given any pair \( \left( {x, y}\right), x, y \) e \( M \) there exists a \( \sigma \) in the group such that \( {x}^{\sigma } = y \) . Suppose first that \( f\left( x\right) \) is irreducible in \( \Phi \left\lbrack x\right\rbrack \) ... | Yes |
Theorem 3. If the characteristic of \( \Phi \) is not a divisor of \( n \) ( \( 0 \) included), then the Galois group \( G \) of the cyclotomic field \( \mathrm{P}/\Phi \) of order \( n \) is isomorphic to a subgroup of the multiplicative group \( U\left( n\right) \) of units in \( I/\left( n\right), I \) the ring of i... | Proof. As in \( §1 \) let \( {G}_{f} \) denote the group of permutations of the set \( Z\left( n\right) \) of roots induced by \( G \) . Since the elements of \( {G}_{f} \) are restrictions of automorphisms, it is clear that they are automor-phisms of the multiplicative group of \( Z\left( n\right) \) . Hence \( {G}_{f... | Yes |
Theorem 4. If \( \Phi \) contains \( n \) distinct \( n \) -th roots of 1 then the Galois group of the equation \( {x}^{n} = \alpha \) over \( \Phi \) is cyclic of order a divisor of \( n \) . | Proof. Let \( \mathrm{P}/\Phi \) be a splitting field over \( \Phi \) of \( {x}^{n} - \alpha, G \) its Galois group. We have to show that \( G \) is cyclic. If \( \alpha = 0 \), we have \( \mathrm{P} = \Phi, G = 1 \) . Hence we assume \( \alpha \neq 0 \) . Let \( \rho \) be one of the roots of \( {x}^{n} - \alpha \) in... | Yes |
Theorem 5. Assume \( \Phi \) has \( n \) distinct \( n \)-th roots of 1 and let \( \mathrm{P}/\Phi \) be a cyclic \( n \) dimensional extension field. Then \( \mathrm{P} = \Phi \left( \xi \right) \) where \( {\xi }^{n} = {\alpha \varepsilon \Phi } \) | Proof. The hypothesis on \( \mathrm{P} \) is that \( \mathrm{P}/\Phi \) is Galois with Galois group \( G \) which is cyclic of order \( n \). Since \( \mathrm{P} \) is separable over \( \Phi \) it has a primitive element so \( \mathrm{P} = \Phi \left( \theta \right) \). Let \( s \) be a generator of \( G \) and let \( ... | No |
Theorem 6. Let \( \mathrm{P}/\Phi \) be finite dimensional Galois over \( \Phi \) and let \( {\mathrm{P}}^{\prime } \) be an extension field of \( \mathrm{P} \) such that \( {\mathrm{P}}^{\prime } \) is generated by \( \mathrm{P} \) and a second subfield \( {\Phi }^{\prime } \supseteq \Phi \) . Then \( {\mathrm{P}}^{\p... | Proof. We know that \( \mathrm{P} = \Phi \left( {{\xi }_{1},\cdots ,{\xi }_{n}}\right) \) where the \( {\xi }_{i} \) are the roots of a separable polynomial \( f\left( x\right) \) e \( \Phi \left\lbrack x\right\rbrack \) . Since \( {\mathrm{P}}^{\prime } \) is generated by \( {\Phi }^{\prime } \supseteq \Phi \) and \( ... | Yes |
Theorem 7. The general equation of the \( n \) -th degree (13) is irreducible in \( \sum = \Phi \left( {{t}_{1},{t}_{2},\cdots ,{t}_{n}}\right) \) and has distinct roots. The Galois group of \( f\left( x\right) = 0 \) is the symmetric group \( {S}_{n} \) . | Since \( {S}_{n} \) is not solvable if \( n > 4 \) this implies the\n\nTheorem of Abel-Ruffini. The general equation of the n-th degree is not solvable by radicals if \( n > 4 \) (characteristic 0 ). | Yes |
Theorem 8. Let \( f\left( x\right) \) be a polynomial of prime degree with rational coefficients which is irreducible in the rational field. Suppose \( f\left( x\right) = 0 \) has exactly two non-real roots in the field \( C \) of complex numbers. Then the group \( {G}_{f} \) of \( f\left( x\right) = 0 \) over the rati... | Proof. The fundamental theorem of algebra asserts that \( f\left( x\right) = \left( {x - {\rho }_{1}}\right) \left( {x - {\rho }_{2}}\right) \cdots \left( {x - {\rho }_{p}}\right) \) in \( C\left\lbrack x\right\rbrack \) . Then the subfield \( \mathrm{P} = {R}_{0}\left( {{\rho }_{1},{\rho }_{2},\cdots ,{\rho }_{p}}\rig... | Yes |
Theorem 4. \( U\left( 2\right) \) and \( U\left( 4\right) \) are cyclic and, if \( e \geq 3 \), then \( U\left( {2}^{e}\right) \) is a direct product of a cyclic group of order 2 and one of order \( {2}^{e - 2} \) . | Proof. The order of \( U\left( {2}^{e}\right) \) is \( \varphi \left( {2}^{e}\right) = {2}^{e - 1} \) . If \( e = 1,\left( {U\left( 2\right) : 1}\right) \) \( = 1 \) and if \( e = 2, U\left( {2}^{e}\right) = U\left( 4\right) \) has only two elements and so is cyclic. Suppose \( e \geq 3 \) . We show first that there ar... | Yes |
Corollary 1. If \( a \neq 1 \) in \( A \), then there exists a character \( {\chi \varepsilon } \) Hom \( \left( {A, Z}\right) \) such that \( {a}^{x} \neq 1 \) . | Proof. Let \( B \) be the subgroup of \( A \) of elements \( b \) such that \( {b}^{x} = \) 1 for all \( {\chi \varepsilon } \) Hom \( \left( {A, Z}\right) \) . Then we see immediately that our assertion will follow if we can show that \( B = 1 \) . Now let \( \chi \mathbf{e} \) Hom \( \left( {A, Z}\right) \) . Since \... | Yes |
For a \( \mathrm{e}A \) define a mapping \( {\eta }_{a} \) of Hom \( \left( {A, Z}\right) \) into \( Z \) by \( {\chi }^{{\eta }_{a}} = {a}^{\chi } \) . Then \( {\eta }_{a}\varepsilon \operatorname{Hom}\left( {\operatorname{Hom}\left( {A, Z}\right), Z}\right) \) and the mapping \( a \rightarrow {\eta }_{a} \) is an iso... | Proof. Observe first that \( a \rightarrow {\eta }_{a} \) is a homomorphism since \( {\chi }^{{\eta }_{ab}} = {\left( ab\right) }^{\chi } = {a}^{\chi }{b}^{\chi } = {\chi }^{{\eta }_{a}}{\chi }^{{\eta }_{b}} = {\chi }^{{\eta }_{a}{\eta }_{b}} \) (the last equation by the definition of the product in a character group).... | Yes |
Corollary 3. A set \( \left\{ {{\chi }_{1},{\chi }_{2},\cdots ,{\chi }_{r}}\right\} \) of characters generate the character group \( \operatorname{Hom}\left( {A, Z}\right) \) if and only if the only a \( {\varepsilon A} \) satisfying \( {a}^{{\chi }_{i}} = 1, i = 1,2,\cdots, r \) is \( a = 1 \) . | Proof. This is equivalent to the dual statement \( \left\{ {{a}_{1},{a}_{2},\cdots ,{a}_{r}}\right\} \) generate \( A \) if and only if \( {a}_{i}^{x} = 1 \), for \( i = 1,2,\cdots, r \) holds only for the character 1. This is easy; for, if \( {a}_{1},\cdots ,{a}_{r} \) generate \( A \) and \( {a}_{i}{}^{x} = 1 \) hold... | Yes |
Theorem 7. Let \( \Phi \) be a field containing \( m \) distinct \( m \)-th roots of 1 and let \( \mathrm{P}/\Phi \) be a Kummer \( {m}^{\prime } \)-extension where \( {m}^{\prime } \mid m \). Let \( M\left( \mathrm{P}\right) \) be defined by (7) where \( {\mathrm{P}}^{ * } \) is the multiplicative group of \( \mathrm{... | Proof. The first statement on the exactness of the displayed sequence means that \( {\Phi }^{ * } \) is the kernel of the mapping \( \rho \rightarrow {\chi }_{\rho } \) and this mapping is surjective on Hom \( \left( {G, Z}\right) \). Both of these facts were established above. Consequently, we have Hom \( \left( {G, Z... | Yes |
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