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Proposition 2.2. Let \( \sigma : X \rightarrow Y \) be a finite dominant morphism between irreducible affine varieties over an algebraically closed field \( F \) . Then \( \sigma \) is surjective. | Proof. Let \( y \) be a point of \( Y \), and view \( y \) as an \( F \) -algebra homomorphism from \( \mathcal{P}\left( Y\right) \) to \( F \) . Since \( \sigma \) is dominant, its transpose is injective from \( \mathcal{P}\left( Y\right) \) to \( \mathcal{P}\left( X\right) \) . Consequently, \( y \) defines an \( F \... | Yes |
Lemma 4.2. Let \( A \) be an integral domain, \( B \) a subring of \( A \) such \( A \) is finitely generated as a B-algebra. Then there is a finite subset \( \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) of \( A \) that is algebraically free over \( \left\lbrack B\right\rbrack \), and a non-zero element \( b \) of \( B \... | Proof. Let \( {A}^{\prime } \) be the sub \( \left\lbrack B\right\rbrack \) -algebra of \( \left\lbrack A\right\rbrack \) that is generated by \( A \) . Then \( {A}^{\prime } \) is a finitely generated \( \left\lbrack B\right\rbrack \) -algebra to which we can apply Theorem 1.2. This yields a finite subset \( \left( {{... | Yes |
Theorem 4.4. Let \( \sigma : X \rightarrow Y \) be a morphism between varieties over an algebraically closed field. If \( A \) is a constructible subset of \( X \) then \( \sigma \left( A\right) \) is a constructible subset of \( Y \) . | Proof. Since a constructible subset of \( X \) is a subvariety, it suffices to prove that \( \sigma \left( X\right) \) is a constructible subset of \( Y \) . Moreover, no generality is lost in assuming that \( X \) is irreducible. Hence, it suffices to prove that \( \sigma \left( X\right) \) is constructible in the cas... | Yes |
Theorem 4.5. Let \( \sigma : X \rightarrow Y \) be a dominant morphism between irreducible varieties over an algebraically closed field. Suppose that, for every closed irreducible subset \( W \) of \( Y \), each irreducible component of \( {\sigma }^{-1}\left( W\right) \) has dimension \( \dim \left( W\right) + \dim \l... | Proof. If we apply the assumption with \( W \) any 1-point subset of \( Y \), we find that \( \sigma \) is surjective. Now let \( W \) be any closed irreducible subset of \( Y \), and let \( {Z}_{1},\ldots ,{Z}_{k} \) be the irreducible components of \( {\sigma }^{-1}\left( W\right) \) . Put\n\n\[ r = \dim \left( X\rig... | Yes |
Proposition 5.2. Let \( \sigma : X \rightarrow Y \) be an injective and dominant morphism between irreducible varieties over an algebraically closed field \( F \) . Then the field of rational functions of \( X \) is a finite purely inseparable algebraic extension of the image, under the transpose of \( \sigma \), of th... | Proof. Applying Theorem 2.1 with a 1-point subset of \( Y \) in the place of \( W \) and using that \( \sigma \) is injective, we see that we must have \( \dim \left( X\right) = \dim \left( Y\right) \) . Therefore the field of rational functions of \( X \) is a finite algebraic extension of the image of the field of ra... | Yes |
Proposition 5.3. Let \( \rho : X \rightarrow Y \) be a morphism between irreducible varieties whose transpose is an isomorphism of the field of rational functions of \( Y \) onto the field of rational functions of \( X \) . Then there is an open non-empty subset \( U \) of \( Y \) such that \( \rho \) induces an isomor... | Proof. Without loss of generality, we assume that \( Y \) is affine. Let \( V \) be an affine patch of \( X \), and let \( W \) denote the closure of \( \rho \left( {X \smallsetminus V}\right) \) in \( Y \) . Since the irreducible components of \( X \smallsetminus V \) are of dimension \( < \dim \left( X\right) \), the... | Yes |
Lemma 1.1. Let \( R \) be a commutative Noetherian ring, \( M \) a finitely generated \( R \) -module, \( N \) a sub \( R \) -module of \( M \), and \( J \) an ideal of \( R \) . There is a non-negative integer \( k \) such that, for every \( n \geq k \), \[ \left( {{J}^{n} \cdot M}\right) \cap N = {J}^{n - k} \cdot \l... | Proof. Let \( \left( {{a}_{1},\ldots ,{a}_{r}}\right) \) be a system of \( R \) -module generators of \( J \), let \( t \) be an auxiliary variable over \( R \), and let \( S \) be the sub \( R \) -algebra \( R\left\lbrack {{a}_{1}t,\ldots ,{a}_{r}t}\right\rbrack \) of the polynomial algebra \( R\left\lbrack t\right\rb... | Yes |
Proposition 1.2. Let \( R \) be a Noetherian commutative ring, \( M \) a finitely generated R-module, J an ideal of R. Suppose that, for every a in J and every non-zero element \( x \) of \( M \), we have \( \left( {1 + a}\right) \cdot x \neq 0 \) . Then \( \mathop{\bigcap }\limits_{{n > 0}}{J}^{n} \cdot M = \left( 0\r... | Proof. Put \( N = \mathop{\bigcap }\limits_{{n > 0}}{J}^{n} \cdot M \) . Then we have \( \left( {{J}^{m} \cdot M}\right) \cap N = N \) for every non-negative integer \( m \), so that Lemma 1.1 gives \( N = {J}^{n - k} \cdot N \) for all \( n \geq k \) . Thus \( N = J \cdot N \) . Now let \( \left( {{x}_{1},\ldots ,{x}_... | Yes |
Theorem 1.3. Let \( A\left\lbrack {{x}_{1},\ldots ,{x}_{q}}\right\rbrack \) be as described just above, and let \( M \) be a finitely generated graded \( A\left\lbrack {{x}_{1},\ldots ,{x}_{q}}\right\rbrack \) -module. There is a polynomial \( {P}_{M} \) of degree strictly less than \( q \), with rational coefficients,... | Proof. If \( q = 0 \) then, since \( M \) is finitely generated, we have \( {M}_{n} = \left( 0\right) \) for all sufficiently large \( n \) ’s, and \( {P}_{M} \) is the zero polynomial. Now we suppose that \( q > 0 \), and that the theorem has been established in the cases of fewer than \( q \) variables.\n\nThe endomo... | Yes |
Theorem 2.2. Let \( R \) be a Noetherian commutative ring. Let \( P \) be a prime ideal of \( R \) that is minimal prime over an ideal \( J \) generated by \( n \) elements. Then every properly increasing chain of prime ideals ending at \( P \) has length at most \( n \) . | Proof. Let \( {P}_{0} \subset \cdots \subset {P}_{k} = P \) be such a chain. We must prove that \( k \leq n \) . Replacing \( R \) with \( R/{P}_{0} \), if necessary, we reduce the problem to the situation where \( R \) is an integral domain. Consider the local ring \( {R}_{P} = R\left\lbrack {\left( R \smallsetminus P... | Yes |
Theorem 3.2. Let \( R \) be a commutative ring. If \( R \) satisfies the maximal condition for prime ideals, so does the polynomial ring \( R\left\lbrack x\right\rbrack \) . If \( R \) has finite Krull dimension then the same holds for \( R\left\lbrack x\right\rbrack \), and \[ k\left( R\right) + 1 \leq k\left( {R\left... | Proof. Let \( {P}_{0} \subset {P}_{1} \subset \cdots \) be a properly ascending chain of prime ideals of \( R\left\lbrack x\right\rbrack \) . By Lemma 3.1, the chain \( {P}_{0} \cap R \subset {P}_{2} \cap R \subset \cdots \) formed with the even indices is properly ascending. Evidently, this gives the first part of The... | Yes |
Lemma 3.3. Let \( P \) be a minimal non-zero prime ideal of the Noetherian commutative ring \( R \). Then \( {PR}\left\lbrack x\right\rbrack \) is a minimal non-zero prime ideal of \( R\left\lbrack x\right\rbrack \). | Proof. Let \( p \) be a non-zero element of \( P \). We show that \( {PR}\left\lbrack x\right\rbrack \) is minimal prime over \( {pR}\left\lbrack x\right\rbrack \). Suppose that \( Q \) is a prime ideal of \( R\left\lbrack x\right\rbrack \) containing \( p \) and contained in \( {PR}\left\lbrack x\right\rbrack \). Then... | Yes |
Theorem 3.4. Let \( R \) be a Noetherian commutative ring, and assume that \( R \) has finite Krull dimension \( k\left( R\right) \) . Then \( k\left( {R\left\lbrack x\right\rbrack }\right) = k\left( R\right) + 1 \) . | Proof. If \( k\left( R\right) = 0 \) then Theorem 3.2 gives \( k\left( {R\left\lbrack x\right\rbrack }\right) = 1 \) . Now assume that \( k\left( R\right) = n > 0 \), and that the theorem has been established in the lower cases. Let \( {P}_{0} \subset \cdots \subset {P}_{m} \) be a properly ascending chain of prime ide... | Yes |
Corollary 3.5. Let \( R \) be a Noetherian integral domain, \( S \) an integral domain containing \( R \) and algebraic over \( \left\lbrack R\right\rbrack \) . Then \( k\left( S\right) \leq k\left( R\right) \) . | Proof. We suppose, without loss of generality, that \( k\left( R\right) \) is finite; say \( k\left( R\right) = n \) . Assume the corollary is false, and consider a properly ascending chain \( \left( 0\right) \subset {Q}_{1} \subset \cdots \subset {Q}_{n + 1} \) of prime ideals of \( S \) . Choose an element \( {q}_{i}... | Yes |
Theorem 3.6. Let \( R \) and \( S \) be integral domains such that \( R \subset S \) and \( S \) is integral over \( R \) . Let \( Q \) be a prime ideal of \( S \), and let \( {P}_{1} \) be a prime ideal of \( R \) containing \( Q \cap R \) . There is a prime ideal \( {Q}_{1} \) of \( S \) such that \( {Q}_{1} \cap R =... | Proof. Let \( \gamma \) denote the canonical homomorphism \( R/\left( {Q \cap R}\right) \rightarrow R/{P}_{1} \) . The inclusion map \( R \rightarrow S \) clearly induces an injective ring homomorphism \( R/\left( {Q \cap R}\right) \rightarrow S/Q \), by means of which we identify \( R/\left( {Q \cap Q}\right) \) with ... | Yes |
Corollary 3.7. Let \( R \) and \( S \) be as in Theorem 3.6. Then \( S \) has finite Krull dimension if and only if \( R \) has finite \( K \) rull dimension, and \( k\left( S\right) = k\left( R\right) \) . | Proof. Let \( {P}_{0} \subset {P}_{1} \subset \cdots \) be a chain of prime ideals of \( R \) . It follows at once from Theorem 3.6 that there is a chain \( {Q}_{0} \subset {Q}_{1} \subset \cdots \) of prime ideals of \( S \) such that \( {Q}_{i} \cap R = {P}_{i} \) for each \( i \) . Therefore, if \( k\left( S\right) ... | Yes |
Theorem 3.8. Let \( F \) be a field, and let \( R \) be a finitely generated integral domain F-algebra. Then \( R \) has finite Krull dimension, and \( k\left( R\right) \) is equal to the degree of transcendence of \( \left\lbrack R\right\rbrack \) over \( F \) . | Proof. Let \( r \) denote the degree of transcendence of \( \left\lbrack R\right\rbrack \) over \( F \) . By Theorem X.1.2, there is a transcendence basis \( \left( {{z}_{1},\ldots ,{z}_{r}}\right) \) for \( \left\lbrack R\right\rbrack \) over \( F \) such that \( R \) is integral over the polynomial \( F \) -algebra \... | Yes |
In the notation introduced above, let \( \left( {{\gamma }_{1},\ldots ,{\gamma }_{q}}\right) \) be an \( R/m\left( R\right) \) - basis of \( m\left( R\right) /m{\left( R\right) }^{2} \), and let \( {x}_{i} \) be a representative of \( {\gamma }_{i} \) in \( m\left( R\right) \) . Then \( \left( {{x}_{1},\ldots ,{x}_{q}}... | Put \( P = m\left( R\right) \) and \( Q = R{x}_{1} + \cdots + R{x}_{q} \) . Then \( P = Q + {P}^{2} \) , and it follows inductively that \( P = Q + {P}^{n} \) for every positive exponent \( n \) . Now \( R/Q \) is a Noetherian local ring, with \( m\left( {R/Q}\right) = P/Q \) . From Proposition 1.2, we see that \( \mat... | Yes |
Corollary 4.2. If \( R \) is a regular Noetherian local ring then \( R \) is an integral domain, and integrally closed in [R]. | Proof. By Theorem 4.1, \( G\left( R\right) \) is an integral domain. Clearly, this implies that \( R \) is an integral domain. In order to proceed, we introduce the following notation. For a non-zero element \( x \) of \( R \), let \( \mu \left( x\right) \) be the largest exponent \( n \) such that \( x \) belongs to \... | Yes |
Theorem 5.1. For every point \( p \) of the irreducible variety \( X \), we have \( \dim \left( {X}_{p}\right) \geq \dim \left( X\right) \) . If the base field is algebraically closed then the set of points where the equality holds, i.e., the set of non-singular points, is non-empty and open in \( X \) . | It is clear from the definitions that a point \( p \) of \( X \) is non-singular if and only if the local ring at \( p \) is regular. From Corollary 4.2, we have that every non-singular point is a normal point. | No |
Proposition 6.3. Let \( \rho : X \rightarrow Y \) be a dominant morphism between irreducible varieties over an algebraically closed field. Suppose that \( \rho \) is separable. Then there is a non-empty open subset \( U \) of \( X \) such that, for every \( p \) in \( U \), the points \( p \) and \( \rho \left( p\right... | Proof. Since \( \rho \) is separable, the map \( {\rho }^{\prime } \) from \( {\operatorname{Der}}_{F}\left( {F\left( X\right), F\left( X\right) }\right) \) to \( {\operatorname{Der}}_{F}\left( {F\left( Y\right), F\left( X\right) }\right) \) used in the proof of Proposition 6.2 is surjective. It is clear from Theorem 5... | Yes |
Proposition 1.2. Let \( G \) be an irreducible algebraic group, \( H \) an algebraic subgroup of \( G \) . Let \( f \) be an element of \( {\left\lbrack \mathcal{P}\left( G\right) \right\rbrack }^{H} \) . There is a polynomial character \( g \) of \( H \) and elements \( s \) and \( t \) of \( \mathcal{P}\left( G\right... | Proof. We assume, without loss of generality, that \( f \neq 0 \), and we consider the polynomial \( H \) -module \( \left( {\mathcal{P}\left( G\right) f}\right) \cap \mathcal{P}\left( G\right) \) . This contains a simple sub \( H \) -module \( V \neq \left( 0\right) \) . Let \( {V}^{ \circ } \) denote the dual \( H \)... | Yes |
Theorem 2.2. Let \( F \) be an algebraically closed field, \( G \) an irreducible affine algebraic F-group, \( H \) an algebraic subgroup of \( G \) . Then \( G/H \) can be endowed with the structure of an \( F \) -variety, actually, a quasi projective variety, such that the following requirements are fulfilled:\n\n(1)... | Proof. We have already established (1), (2) and (3). Clearly, (5) follows from our above construction of the variety structure on \( G/H \) . It remains only to verify (4).\n\nLet \( U \) be an open subset of \( V \) . Then \( {\alpha }^{-1}\left( U\right) \) is an open subset of \( G \) . Since \( \pi \) is an open ma... | Yes |
Lemma 4.1. Let \( G \) be a unipotent algebraic group over an algebraically closed field, and let \( V \) be an affine strict G-variety. Then every G-orbit in \( V \) is closed. | Proof. Without loss of generality, we assume that \( G \) is irreducible. Let \( T \) be a \( G \) -orbit in \( V \) . In proving that \( T \) is closed, we assume without loss of generality that \( T \) is dense in \( V \) . Then we know from Theorem X.4.3 that \( T \) contains a non-empty open subset, \( U \) say, of... | Yes |
Theorem 4.3. Let \( G \) be an irreducible solvable algebraic group over an algebraically closed field, \( H \) an algebraic subgroup of \( G \) . Then the variety \( G/H \) is affine. | Proof. Proposition 4.2 reduces the theorem to the case where \( H \) is irreducible. Then, by Theorem VI.3.2, we can write \( H = {H}_{u} \bowtie T \), where \( T \) is a toroid, and \( G = {G}_{u} \bowtie S \), where \( S \) is a toroid containing \( T \) . It follows immediately from Theorem V.5.3 that \( S \) is a d... | Yes |
Proposition 5.1. Let \( G \) be an algebraic group, \( H \) an algebraic subgroup of \( G \) . Suppose that, for every 1-dimensional polynomial H-module that is a sub H-module of a polynomial G-module, the dual H-module is also a sub H-module of a polynomial G-module. Then \( H \) is an observable subgroup of \( G \) . | Proof. Let \( N \) be a finite-dimensional polynomial \( H \) -module, and let \( M \) and \( g \) be as in Proposition 1.1. Let \( R \) be the 1-dimensional polynomial \( H \) -module determined by the polynomial character \( g \), so that the automorphism of \( R \) corresponding to an element \( x \) of \( H \) is t... | Yes |
Theorem 5.2. Let \( G \) be an irreducible algebraic group, \( H \) an algebraic subgroup of \( G \) . Then \( H \) is observable in \( G \) if and only if \( {\left\lbrack \mathcal{P}\left( G\right) \right\rbrack }^{H} = \left\lbrack {\mathcal{P}{\left( G\right) }^{H}}\right\rbrack \) . | Proof. First, suppose that \( H \) is observable in \( G \), and let \( f \) be a non-zero element of \( {\left\lbrack \mathcal{P}\left( G\right) \right\rbrack }^{H} \) . We must show that \( f \) belongs to \( \left\lbrack {\mathcal{P}{\left( G\right) }^{H}}\right\rbrack \) . For this, it evidently suffices to show th... | Yes |
Lemma 1.2. Let \( F \) be an algebraically closed field, and let \( \rho : X \rightarrow Y \) be a bijective morphism between irreducible algebraic F-varieties. Suppose that an arbitrary group \( G \) acts transitively by variety automorphisms on \( X \) and on \( Y \) , and that \( \rho \) commutes with the action of ... | Proof. By Proposition X.5.2, \( F\left( X\right) \) is a finite algebraic extension of \( F\left( Y\right) \circ \rho \) . As in the proof of Theorem X.4.3, with \( r = 0 \), we see from this that there are affine patches \( U \) of \( X \) and \( V \) of \( Y \) such that \( \rho \) restricts to a finite morphism from... | Yes |
Theorem 1.3. Let \( G \) be an irreducible solvable algebraic group over an algebraically closed field, and let \( X \) be a complete strict \( G \) -variety. Then the set \( {X}^{G} \) of \( G \) -fixed points of \( X \) is not empty. | Proof. Making an induction on the dimension of \( G \), we suppose that the theorem has been established in the lower cases. Then we know that \( {X}^{\left\lbrack G, G\right\rbrack } \) is not empty. Being closed in \( X \), this is a complete variety, and it is evidently stable under the action of \( G \) . This acti... | Yes |
Theorem 2.1. Let \( G \) be an irreducible algebraic group over an algebraically closed field. For every Borel subgroup \( B \) of \( G \), the algebraic variety \( G/B \) is a projective variety, and every Borel subgroup of \( G \) is a conjugate of \( B \) . | Proof. Let \( C \) be a Borel subgroup of the largest possible dimension. By Theorem II.2.1, there is an injective polynomial representation of \( G \) on a finite-dimensional vector space \( V \) having a 1-dimensional subspace \( {S}_{1} \) whose stabilizer in \( G \) coincides with \( C \) . Consider the induced rep... | Yes |
Theorem 2.2. Let \( G \) be an irreducible algebraic group over an algebraically closed field. An algebraic subgroup \( P \) of \( G \) contains a Borel subgroup of \( G \) if and only if \( G/P \) is complete. | Proof. If \( P \) contains a Borel subgroup \( B \) of \( G \), then the canonical morphism \( \pi \) from \( G \) to \( G/P \) is constant on the cosets \( {xB} \) and therefore induces a morphism \( {\pi }^{B} \) from \( G/B \) to \( G/P \) . Since \( G/B \) is complete and \( {\pi }^{B} \) is surjective, it follows ... | Yes |
Proposition 2.3. Let \( G \) be an irreducible algebraic group over an algebraically closed field. Let \( \alpha \) be an automorphism of \( G \) leaving the elements of some Borel subgroup B fixed. Then \( \alpha \) is the identity automorphism. | Proof. Consider the map \( \delta \) from \( G \) to \( G \) where \( \delta \left( x\right) = \alpha \left( x\right) {x}^{-1} \) . This is a morphism of varieties that is constant on each coset \( {xB} \) . Therefore, \( \delta \) defines a morphism of varieties \( {\delta }^{B} \) from \( G/B \) to \( G \) . Since \(... | Yes |
Proposition 2.4. Let \( G \) be as above. If a Borel subgroup of \( G \) is nilpotent then it coincides with \( G \) . | Proof. We make an induction on the dimension of \( G \), and suppose that the proposition has been established in the lower cases. Let \( B \) be a nilpotent Borel subgroup of \( G \) . If \( B \) is trivial then it follows from Theorem 2.1 that \( G \) is trivial. Therefore, we suppose that \( B \) is non-trivial. For... | Yes |
Proposition 2.5. Let \( G \) be as above, and let \( T \) be a maximal toroid in \( G \) . Let \( C = {\mathcal{C}}_{G}{\left( T\right) }_{1} \) . Then \( C \) is nilpotent and coincides with \( {\mathcal{N}}_{G}{\left( C\right) }_{1} \) . | Proof. There is a Borel subgroup \( S \) of \( C \) such that \( T \subset S \) . Evidently, \( T \) is a maximal toroid in \( S \) . By Theorem VI.3.2, we have \( S = {S}_{u} \sim T \) . Since \( T \) is central in \( S \), this means that \( S \) is the direct product of \( {S}_{u} \) and \( T \), so that \( S \) is ... | Yes |
Proposition 3.2. Let \( G \) be as above, and let \( S \) be any toroid in \( G \) . There is an element \( s \) in \( S \) such that every element of \( G \) that commutes with \( s \) belongs to \( {\mathcal{C}}_{G}\left( S\right) \) . | Proof. Let \( V \) be a finite-dimensional polynomial \( G \) -module such that the representation of \( G \) on \( V \) is injective. We may write \( V \) as a direct sum of \( S \) -stable subspaces \( {V}_{i} \) corresponding to mutually distinct polynomial characters \( {f}_{i} \) such that every element \( s \) of... | Yes |
Theorem 3.3. Let \( G \) be an irreducible algebraic group over an algebraically closed field, and let \( B \) be a Borel subgroup of \( G \). Then \( { \cup }_{x \in G}{xB}{x}^{-1} \) coincides with \( G \). | Proof. Choose a maximal toroid \( T \) in \( G \), and write \( C \) for \( {\mathcal{C}}_{G}{\left( T\right) }_{1} \). By Proposition 2.5, \( C \) is nilpotent, so that it follows from Theorem VI.3.1 and the maxi-mality of \( T \) that \( C = {C}_{u} \times T \). By Proposition 3.2, there is an element \( t \) in \( \... | Yes |
Corollary 3.4. If \( G \) and \( B \) are as in Theorem 3.3 then \( B = {\mathcal{N}}_{G}{\left( B\right) }_{1} \) . | Proof. Evidently, \( B \) is a Borel subgroup of \( {\mathcal{N}}_{G}{\left( B\right) }_{1} \) . Since \( B \) is normal in \( {\mathcal{N}}_{G}{\left( B\right) }_{1} \), the corollary follows at once from Theorem 3.3, with \( {\mathcal{N}}_{G}{\left( B\right) }_{1} \) in the place of \( G \) . | Yes |
Theorem 4.2. Let \( G \) be an irreducible algebraic group over an algebraically closed field, \( S \) a toroidal algebraic subgroup of \( G \) . Then \( {\mathcal{C}}_{G}\left( S\right) \) is irreducible. | Proof. Let \( x \) be any element of \( {\mathcal{C}}_{G}\left( S\right) \), and let \( B \) be a Borel subgroup of \( G \) . By Theorem 3.3, \( x \) belongs to some conjugate of \( B \) . This means that the fixed point set, \( P \) say, for \( x \) in \( G/B \) is non-empty. Being closed in the complete variety \( G/... | Yes |
Theorem 4.4. Let \( G \) be an irreducible algebraic group over an algebraically closed field. Let \( S \) be a toroidal subgroup of \( G \), and let \( B \) be a Borel subgroup of \( G \) containing \( S \) . Then \( {\mathcal{C}}_{G}\left( S\right) \cap B \) is a Borel subgroup of \( {\mathcal{C}}_{G}\left( S\right) ... | Proof. Applying Lemma 4.3, with the point \( B \) of \( G/B \) taking the place of \( p \) , we conclude that the canonical image of \( {\mathcal{C}}_{G}\left( S\right) \) in \( G/B \) is closed in \( G/B \), and therefore complete. The canonical map induces a bijective morphism from \( {\mathcal{C}}_{G}\left( S\right)... | Yes |
Corollary 5.2. With \( G \) as in Theorem 5.1, let \( P \) be an algebraic subgroup of \( G \) containing a Borel subgroup of \( G \) . Then \( P \) is irreducible, and \( P = {\mathcal{N}}_{G}\left( P\right) \) . | Proof. Let \( B \) be a Borel subgroup of \( G \) that is contained in \( P \), and let \( x \) be an element of \( {\mathcal{N}}_{G}\left( P\right) \) . Then \( B \) and \( {xB}{x}^{-1} \) are Borel subgroups of \( {P}_{1} \) . Hence, there is an element \( p \) in \( {P}_{1} \) such that \( {px} \) normalizes \( B \)... | Yes |
Corollary 5.3. Let \( G \) be as above, \( B \) a Borel subgroup of \( G \). Then \( B = {\mathcal{N}}_{G}\left( {B}_{u}\right) \) . | Proof. Write \( P \) for \( {\mathcal{N}}_{G}\left( {B}_{u}\right) \). Since \( P \) contains \( B \), we know from Corollary 5.2 that \( P \) is irreducible. From the conjugacy of Borel subgroups, it follows that \( {B}_{u} \) is maximal in the family of irreducible unipotent subgroups of \( G \). Hence, \( P/{B}_{u} ... | Yes |
Proposition 1.1. Let \( F \) be a Galois extension of a field \( K \), with Galois group \( S \) . Let \( G \) be an affine algebraic \( F \) -group that is defined over \( K \) by the \( K \) -form \( A \) of \( \mathcal{P}\left( G\right) \) . Then the S-stable algebraic subgroups of \( G \) are precisely the algebrai... | Proof. Let \( H \) be an algebraic subgroup of \( G \), and let \( J \) be its annihilator in \( \mathcal{P}\left( G\right) \) . First, suppose that \( H \) is associated with \( A \), so that \( J \) is generated as an ideal by \( A \cap J \) . Clearly, this implies that \( \left( {{i}_{A} \otimes \sigma }\right) \lef... | Yes |
Proposition 1.2. Let \( F, K, S, G \) and \( A \) be as in Proposition 1.1. Suppose that \( H \) is a properly normal algebraic subgroup of \( G \) that is associated with \( A \) . Then \( G/H \) is defined over \( K \), with \( {A}^{H} \) as the \( K \) -form of \( \mathcal{P}\left( {G/H}\right) \) . | Proof. Let \( f \) be an element of \( \mathcal{P}\left( {G/H}\right) = {\left( A \otimes F\right) }^{H} \) . For every element \( y \) of \( G \), every element \( x \) of \( H \) and every element \( \sigma \) of \( S \), we have\n\n\[ \left( {{i}_{A} \otimes \sigma }\right) \left( f\right) \left( {{\sigma }^{\prime ... | Yes |
Lemma 1.3. Let \( S \) be a finite group of automorphisms of a field \( F \). Then every multiplicative and every additive cocycle for \( S \) in \( F \) is a coboundary. | Proof. There is an element \( t \) in \( F \) such that \( \mathop{\sum }\limits_{{\sigma \in S}}\sigma \left( t\right) = 1 \). Let \( f \) be an additive cocycle for \( S \) in \( F \), so that\n\n\[ f\left( {\sigma \tau }\right) = \sigma \left( {f\left( \tau \right) }\right) + f\left( \sigma \right) \]\n\nfor all ele... | Yes |
Proposition 1.4. Let \( F \) be a Galois extension of a field \( K \) with Galois group \( S \) . Suppose that \( G \) is an affine algebraic \( F \) -group that is defined over \( K \) so as to be split solvable with respect to \( K \) . Then every Galois cocycle for \( S \) in \( G \) is a coboundary. | Proof. Let \( f \) be a Galois cocycle for \( S \) in \( G \), and let the \( {G}_{i} \) ’s be as in the above definition of \ | No |
Theorem 2.2. Let \( K \) be a perfect field, \( G \) an irreducible affine algebraic \( K \) - group, \( U \) an irreducible unipotent normal algebraic subgroup of \( G \) . Then \( U \) is properly normal in \( G \) . | Proof. Let \( F \) and \( S \) be as in Theorem 2.1, and let \( {G}^{F} \) and \( {U}^{F} \) denote the groups obtained from \( G \) and \( U \) by the canonical base field extension. Then every element \( h \) of \( \mathcal{G}\left( {\mathcal{P}{\left( G\right) }^{U}}\right) \) is the restriction to \( \mathcal{P}{\l... | Yes |
Theorem 4.2. Let \( H \) be an irreducible unipotent algebraic group over an algebraically closed field \( F \) . Then \( \mathcal{P}\left( H\right) = F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \), where the \( {t}_{i} \) ’s are algebraically independent over \( F \), and \( \delta \left( {t}_{i}\right) - {t}... | Proof. Clearly, the result holds when \( H \) is of dimension 1 . Therefore, we suppose that the dimension of \( H \) is greater than 1, and that the theorem has been established in the lower cases. There is a 1-dimensional irreducible central algebraic subgroup \( Z \) of \( H \) . Let \( \pi \) be the canonical morph... | Yes |
Proposition 2.2. Let \( G \) and \( K \) be as in Proposition 2.1. Suppose that \( L \) is another algebraic automorphism group of \( G \) that is normalized by \( K \) . Then \( {LK} \) is an algebraic automorphism group of \( G \) . | Proof. It is easy to see that \( \mathcal{P}\left( G\right) \) is locally finite as a right \( {LK} \) -module. Evidently, the restriction maps from \( {\mathcal{R}}_{G}\left( {LK}\right) \) to \( {\mathcal{R}}_{G}\left( L\right) \) and \( {\mathcal{R}}_{G}\left( K\right) \) are surjective morphisms of Hopf algebras. T... | Yes |
Proposition 2.3. Let \( F \) be a field, \( G \) an affine algebraic \( F \) -group, \( K \) an algebraic automorphism group of \( G \) . Then the restriction to \( K \) of the canonical group homomorphism from \( \mathcal{W}\left( G\right) \) to the affine algebraic group of all Lie algebra automorphisms of \( \mathca... | Proof. For every element \( \alpha \) of \( \mathcal{W}\left( G\right) \), let \( {\alpha }^{\prime } \) denote the corresponding Lie algebra automorphism of \( \mathcal{L}\left( G\right) \) . Then the transform by \( {\alpha }^{\prime } \) of an element \( \tau \) of \( \mathcal{L}\left( G\right) \) is given by\n\n\[{... | Yes |
Theorem 1.1. Let \( L \) be a Lie algebra over a field \( F \), and let \( X \) be a totally ordered \( F \) -basis of \( L \) . Let \( S\left( X\right) \) denote the set of all finite non-decreasing sequences of elements of \( X \) . For \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) in \( S\left( X\right) \), put\n\n\... | Proof. An evident \ | No |
Theorem 2.1. Let \( L \) be a Lie algebra over a field \( F \) of characteristic 0 . Then the space of primitive elements of \( \mathcal{U}\left( L\right) \) coincides with \( L \) . | Proof. Choose a totally ordered \( F \) -basis \( \left( {x}_{\alpha }\right) \) for \( L \) . By Theorem 1.1, the element 1 of \( F \) and the ordered monomials \( {x}_{{\alpha }_{1}}^{{e}_{1}}\cdots {x}_{{\alpha }_{n}}^{{e}_{n}} \), where the \( {e}_{i} \) ’s are strictly positive integers, and \( {x}_{{\alpha }_{1}}... | Yes |
Proposition 2.2. Let \( F, V, T, L \) be as above, and assume that \( F \) is of characteristic 0 . For every element \( t \) of \( T \), let \( {D}_{t} \) be the derivation effected by \( t \) in \( T \), so that\n\n\( {D}_{t}\left( u\right) = {tu} - {ut} \) . There is an \( F \) -linear projection \( \pi \) of \( T \... | Proof. Clearly, there is one and only one linear endomorphism \( \pi \) of \( T \) satisfying the equalities of the proposition and the condition \( \pi \left( v\right) = v \) for every element \( v \) of \( V \) . Moreover, it is evident from these properties that \( \pi \left( T\right) \) is contained in \( L \) . It... | Yes |
Proposition 1.1. For an element \( x \) of \( L \), let \( {L}^{x} \) denote the subspace of \( L \) consisting of all elements that are annihilated by some power of \( {D}_{x} \) . Then \( {L}^{x} \) is a sub Lie algebra of \( L \) and coincides with its stabilizer in \( L \) . | Proof. The formula expressing \( {D}_{x}^{n}\left( \left\lbrack {u, v}\right\rbrack \right) \) as a sum of terms \( \left\lbrack {{D}_{x}^{p}\left( u\right) ,{D}_{x}^{q}\left( v\right) }\right\rbrack \) , where \( p + q = n \), shows that \( \left\lbrack {{L}^{x},{L}^{x}}\right\rbrack \subset {L}^{x} \).\n\nNext, if \(... | Yes |
Proposition 1.2. In the notation of Proposition 1.1, \( {L}^{x} \) has an F-space complement \( {L}_{x} \) in \( L \) such that \( \left\lbrack {{L}^{x},{L}_{x}}\right\rbrack \subset {L}_{x} \) . | Proof. We let \( {L}_{x} \) be the subspace of \( L \) obtained from Fitting’s Lemma, so that \( {L}_{x} \) is the largest subspace of \( L \) on which \( {D}_{x} \) induces a linear automorphism. We have \( {L}_{x} = {D}_{x}^{p}\left( L\right) \) for some positive exponent \( p \), and it remains only to be shown that... | Yes |
Theorem 1.3. Let \( L \) be a finite-dimensional Lie algebra over an infinite field \( F \), and let \( x \) be an element of \( L \) such that \( {L}^{x} \) is of the smallest possible dimension. Then \( {L}^{x} \) is a Cartan subalgebra of \( L \) . | Proof. By Proposition 1.1, it suffices to prove that \( {L}^{x} \) is a nilpotent Lie algebra. By Theorem VII.1.5, it suffices to show that, for every element \( y \) of \( {L}^{x} \), the restriction of \( {D}_{y} \) to \( {L}^{x} \) is nilpotent. Let \( {f}_{y} \) denote the characteristic polynomial of \( {D}_{y} \)... | Yes |
Theorem 2.2. Let \( F \) be an algebraically closed field of characteristic 0, and let \( L \) be a finite-dimensional \( F \) -Lie algebra. Let \( H \) be a Cartan subalgebra of \( L \) . For all elements \( \alpha \) and \( \beta \) of \( {H}^{ \circ } \), one has \( \left\lbrack {{L}_{\alpha },{L}_{\beta }}\right\rb... | Proof. The first statement follows immediately from the formula\n\n\[ \n{\left( {D}_{z,\alpha + \beta }\right) }^{n}\left( \left\lbrack {x, y}\right\rbrack \right) = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) \left\lbrack {{\left( {D}_{z,\alpha }\right) }^{k}\left( x\right) ,{\... | Yes |
Theorem 2.3. Let \( F, L \) and \( H \) be as in Theorem 2.2, and let \( B \) denote the trace form of the adjoint representation of \( L \) . For \( \gamma \) in \( {H}^{ \circ } \), put \( {d}_{\gamma } = \dim \left( {L}_{\gamma }\right) \) . Then, for all elements \( x \) and \( y \) of \( H \), one has\n\n\[ B\left... | Proof. By Theorem 2.1, \( L \) is the direct sum of the family of sub \( H \) -modules \( {L}_{\gamma } \) . Hence, if \( T \) stands for trace and \( {D}_{x/\gamma } \) for the restriction of \( {D}_{x} \) to \( {L}_{\gamma } \), we have\n\n\[ B\left( {x, x}\right) = T\left( {D}_{x}^{2}\right) = \mathop{\sum }\limits_... | Yes |
Theorem 2.4. In the notation of Theorem 2.3, if \( \alpha \) and \( \beta \) are elements of \( {H}^{ \circ } \) such that \( \alpha + \beta \neq 0 \) then \( B\left( {{L}_{\alpha },{L}_{\beta }}\right) = \left( 0\right) \) . | Proof. By Theorem 2.2, if \( x \) belongs to \( {L}_{\alpha } \) and \( y \) to \( {L}_{\beta } \), then \( {D}_{x}{D}_{y} \) maps each \( {L}_{\gamma } \) into \( {L}_{\alpha + \beta + \gamma } \) . Now, it \( \gamma \) is a root then \( \alpha + \beta + \gamma \) is either no root or a root distinct from \( \gamma \)... | Yes |
Theorem 3.2. In the above notation, \( H \) is abelian, and the roots of \( L \) with respect to \( H \) span \( {H}^{ \circ } \) over \( F \) . | Proof. Let \( \alpha \) be a non-zero root. Since \( \dim \left( {L}_{\alpha }\right) = 1,{D}_{h} \) acts as the scalar multiplication by \( \alpha \left( h\right) \) on \( {L}_{\alpha } \), for every element \( h \) of \( H \) . Now let \( N \) be the subspace of \( H \) consisting of the elements that are annihilated... | Yes |
Proposition 3.3. In the above notation, let \( \left( {{\beta }_{1},\ldots ,{\beta }_{r}}\right) \) be any maximal \( F \) - linearly independent set of roots of \( L \) with respect to \( H \) . Then every root is a rational linear combination of the \( {\beta }_{i} \) ’s. | Proof. Let \( \beta \) be any root. Then \( \beta = \mathop{\sum }\limits_{{i = 1}}^{r}{b}_{i}{\beta }_{i} \), with each \( {b}_{i} \) in \( F \) . Applying this to \( {h}_{{\beta }_{j}} \), we obtain\n\n\[ B\left( {{h}_{\beta },{h}_{{\beta }_{j}}}\right) = \mathop{\sum }\limits_{{i = 1}}^{r}{b}_{i}B\left( {{h}_{{\beta... | Yes |
Lemma 2.1. Let \( L \) be a finite-dimensional Lie algebra over a field of characteristic 0 . There is one and only one direct Lie algebra decomposition\n\n\[ L = {L}_{0} + {L}_{1} \]\n\nwhere \( {L}_{0} \) is semisimple and \( {L}_{1} \) is regular. | Proof. Let \( K \) denote the centralizer of \( M \) in \( L \), and let \( T \) be the radical of \( K \) . Since \( K \) is an ideal of \( L \), its radical \( T \) is contained in the radical of \( L \) . An application of Theorem VII.3.2 shows that \( \left\lbrack {T, T}\right\rbrack \subset M \), whence\n\n\[ \lef... | Yes |
Corollary 2.3. Let \( F \) be an algebraically closed field of characteristic 0, and let \( G \) be an irreducible affine algebraic \( F \) -group whose center is unipotent. Then the group of all Lie algebra automorphisms of \( \mathcal{L}\left( G\right) \) is the semidirect product \( E \bowtie W \), where \( E \) is ... | Proof. By Theorem XV.4.3, \( \mathcal{W}\left( G\right) \) is an algebraic automorphism group of \( G \), and thus is an affine algebraic \( F \) -group in the canonical fashion of Chapter XV. By Proposition XV.2.3, the canonical map from \( \mathcal{W}\left( G\right) \) to Aut \( \left( {\mathcal{L}\left( G\right) }\r... | Yes |
Proposition 3.1. Let \( G \) be a simply connected affine algebraic group over an algebraically closed field, and let \( N \) be a normal irreducible algebraic subgroup of \( G \) . Then \( G/N \) is simply connected. | Proof. Let \( \pi \) denote the canonical morphism from \( G \) to \( G/N \), and consider a group covering \( \eta : H \rightarrow G/N \) . We must show that the kernel, \( K \) say, of \( \eta \) is trivial. Consider the fibered product \( P = H{ \times }_{\left( \eta ,\pi \right) }G \), i.e., the algebraic subgroup ... | Yes |
Proposition 3.2. Let \( G \) be a simply connected affine algebraic group over an algebraically closed field of characteristic 0. Then the radical of \( G \) is unipotent. | Proof. Write \( P \) for \( G/{G}_{u} \), and note that \( P \) is an irreducible linearly reductive affine algebraic group. Appealing to Theorems VII.1.2 and VIII.3.3, we see that \( \mathcal{L}\left( \left\lbrack {P, P}\right\rbrack \right) \) is semisimple. By Proposition 3.1, \( P \), and hence also \( P/\left\lbra... | Yes |
Theorem 3.3. Let \( G \) be an irreducible affine algebraic group over an algebraically closed field of characteristic 0, and let \( R \) denote the radical of \( G \) . Then \( G \) is simply connected if and only if \( R \) is unipotent and \( G/R \) is simply connected. | Proof. The necessity of the conditions has already been established in Proposition 3.1 and 3.2. Now suppose that the conditions are satisfied, and consider a group covering \( \eta : H \rightarrow G \), with kernel \( K \) . Let \( S \) denote the irreducible component of the neutral element in \( {\eta }^{-1}\left( R\... | Yes |
Proposition 3.4. Let \( G \) be as in Theorem 3.3. If \( G \) is simply connected, so is every irreducible normal algebraic subgroup of \( G \) . | Proof. Let \( R \) denote the radical of \( G \), and let \( N \) be any irreducible normal algebraic subgroup of \( G \) . Clearly, \( N \cap R \) is a normal algebraic subgroup of \( N \), and \( N/\left( {N \cap R}\right) \) may be identified with its canonical image in \( G/R \) . Since \( \mathcal{L}\left( {G/R}\r... | Yes |
Theorem 5.1. Let \( F \) be an algebraically closed field of characteristic 0, and let \( L \) be a finite-dimensional \( F \)-Lie algebra with nilpotent radical \( R \). Then \( \mathcal{G}\left( L\right) \) is simply connected, its Lie algebra may be identified with L, and its algebra of polynomial functions is \( \m... | Proof. What remains to be proved is the last statement concerning \( \sigma \), and the fact that \( \mathcal{G}\left( L\right) \) is simply connected. However, if the last statement is applied to the inverse of the differential of a group covering of \( \mathcal{G}\left( L\right) \), it shows that the group covering i... | No |
Theorem 5.2. Let \( F \) be an algebraically closed field of characteristic 0, and let \( \eta : H \rightarrow G \) be a covering of irreducible affine algebraic \( F \) -groups. Suppose that \( T \) is a simply connected affine algebraic \( F \) -group, and \( \tau \) is a morphism of affine algebraic F-groups from \(... | Proof. By Theorem 5.1, we may identify \( T \) with \( \mathcal{G}\left( {\mathcal{L}\left( T\right) }\right) \) . Now \( \tau \left( T\right) \) is an irreducible algebraic subgroup of \( G \), and there is evidently an irreducible algebraic subgroup \( K \) of \( H \) such that the restriction of \( \eta \) to \( K \... | Yes |
If \( \left( {a, b}\right) = d \), then the equation\n\n\[ n = {ax} + {by} \]\n\nis solvable with integers \( x, y \) if and only if \( d \mid n \) . | Moreover it follows from this that every common divisor of \( a \) and \( b \) divides the GCD of \( a, b \) .\n\nTo ascertain the GCD one uses, as is well-known, a process which goes back to Euclid, the so-called Euclidean algorithm. The main point of this algorithm consists of reducing the calculation of \( \left( {a... | No |
Theorem 2. The numbers in a module \( S \) are identical with the multiples of certain number \( d.d \) is determined by \( S \) up to the factor \( \pm 1 \) . | For the proof we consider that \( S \) contains positive numbers in any case. Let \( d \) be the smallest positive number occurring in \( S \) . If \( n \) belongs to \( S \), then by what has gone before, \( n - {qd} \) also belongs to \( S \) for each integer \( q \), in particular so must the remainder of \( n{\;\op... | Yes |
Theorem 3. The range of values of an arbitrary linear form in \( n \) variables with integral coefficients, not all vanishing, is identical with the range of values of a certain form of one variable \( d \cdot x \) . Here \( d \) is the GCD of the coefficients of the original form. | In order that the equation (a so-called Diophantine equation)\n\n\[ k = {a}_{1}{x}_{1} + {a}_{2}{x}_{2} + \cdots + {a}_{n}{x}_{n} \]\n\nbe solvable in integers \( {x}_{1},\ldots ,{x}_{n} \), it is necessary and sufficient that the GCD of \( {a}_{1},\ldots ,{a}_{n} \) divides \( k \) . | No |
Theorem 4. For every three integers \( a, b, c \), where \( c > 0 \n\n\[ \left( {a, b}\right) c = \left( {{ac},{bc}}\right) . \]\n\n(2) | In fact if \( \left( {a, b}\right) = d \), then the equation \( {acx} + {bcy} = {cd} \) follows by Theorem 1 from the known solvable equation \( {ax} + {by} = d \) ; consequently \( {cd} \) is a multiple of \( \left( {{ac},{bc}}\right) \), again by Theorem 1 . On the other hand, however, \( {cd} \) is a common divisor ... | Yes |
Theorem 5. (Fundamental Theorem of Arithmetic). Each positive number \( > 1 \) can be represented in one-and except for the order of the factors-in only one way as a product of primes. | For this it is sufficient to show that a prime \( p \) can divide a product of two numbers \( a \cdot b \) only if it divides at least one factor. But this follows from Theorem 4. Namely, if the prime number does not divide \( a \), then as a prime it cannot have any factor at all in common with \( a \), hence \( \left... | Yes |
Theorem 7. If \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) forms a complete system of residues mod \( n\left( {n > 0}\right) \) , then \( a{x}_{1} + b,\ldots, a{x}_{n} + b \) is also such a system, as long as \( a \) and \( b \) are integers and \( \left( {a, n}\right) = 1 \) . | For by Theorem 6 the \( n \) numbers \( a{x}_{i} + b\left( {i = 1,2,\ldots, n}\right) \) are likewise incongruent numbers modulo \( n \) . | No |
Theorem 8. If \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) are pairwise relatively prime integers, then a complete residue system mod \( A \), where \( A = {a}_{1}{a}_{2}\cdots {a}_{n} \), is obtained in the form\n\n\[ L\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \frac{A}{{a}_{1}}{c}_{1}{x}_{1} + \frac{A}{{a}_{2}}{c}_{2}{x}_{2}... | The number of these \( L \) values is \( \left| A\right| \) and they are incongruent \( {\;\operatorname{mod}\;A} \) since from the congruence \( {\;\operatorname{mod}\;A} \)\n\n\[ L\left( {{x}_{1},\ldots ,{x}_{n}}\right) = L\left( {{x}_{1}^{\prime },\ldots ,{x}_{n}^{\prime }}\right) \left( {\;\operatorname{mod}\;A}\ri... | Yes |
Theorem 10. If \( a \) is a root of the integral polynomial \( f\left( x\right) {\;\operatorname{mod}\;n} \), then \( f\left( x\right) \) is divisible by \( x - a{\;\operatorname{mod}\;n} \) and conversely. | Since \( f\left( a\right) \equiv 0\left( {\;\operatorname{mod}\;n}\right) \) we have\n\n\[ f\left( x\right) \equiv f\left( x\right) - f\left( a\right) \left( {\;\operatorname{mod}\;n}\right) \]\n\nHowever \( \left( {f\left( x\right) - f\left( a\right) }\right) /\left( {x - a}\right) \) is an integral polynomial, \( g\l... | Yes |
Theorem 11. If \( f\left( x\right) \equiv g\left( x\right) {g}_{1}\left( x\right) \left( {\;\operatorname{mod}\;p}\right) \), where \( p \) is a prime, then each root of \( f\left( x\right) {\;\operatorname{mod}\;p} \) is a root of at least one of the two polynomials \( g\left( x\right) ,{g}_{1}\left( x\right) {\;\oper... | If for the integer \( a, f\left( a\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) \), then\n\n\[ g\left( a\right) \cdot {g}_{1}\left( a\right) \equiv f\left( a\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) .\n\nIf the prime \( p \) divides the product \( g\left( a\right) \cdot {g}_{1}\left( a\right) \)... | Yes |
Theorem 12. An integral polynomial \( f\left( x\right) \) of degree \( k \) has no more than \( k \) incongruent roots modulo a prime \( p \), unless \( f\left( x\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) \), in which case all coefficients are divisible by \( p \) . | The theorem is true for the polynomials of degree 0 , the constants. For if \( f\left( x\right) = {c}_{0} \) is independent of \( x \), then \( f\left( x\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) has either 0 solutions-when \( p \) does not divide \( {c}_{0} \) -or it has more than 0 solutions-namely ev... | Yes |
Theorem 13. If for two integral polynomials \( f\left( x\right) \) and \( g\left( x\right) \)\n\n\[ f\left( x\right) \cdot g\left( x\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) ,\;p\text{ a prime,}\]\n\nthen either \( f\left( x\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) or \( g\left( x\right) ... | Suppose the theorem is false, i.e., neither \( f\left( x\right) \) nor \( g\left( x\right) \) is \( \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) . Then let all terms of \( f\left( x\right) \) and \( g\left( x\right) \) which are divisible by \( p \) be omitted and two nonvanishing polynomials \( {f}_{1}\left( x\r... | Yes |
Theorem 14. The congruence (6) has exactly one solution mod \( n \) if \( \left( {a, n}\right) = 1 \) . | For by Theorem 7, \( {ax} + b \) falls exactly once into the residue class 0 if \( x \) runs through a complete system of residues mod \( n \) . | No |
Theorem 15. The \( k \) congruences (7) have exactly one solution determined \( {\;\operatorname{mod}\;{n}_{1}}{n}_{2}\cdots {n}_{k} \) if the moduli are pairwise relatively prime. | For with Theorem 8 in mind let us set\n\n\[ x = \frac{v}{{n}_{1}}{x}_{1} + \frac{v}{{n}_{2}}{x}_{2} + \cdots + \frac{v}{{n}_{k}}{x}_{k}\;\left( {v = {n}_{1}{n}_{2}\cdots {n}_{k}}\right) \]\n\nand determine the \( {x}_{i} \) from the congruences\n\n\[ \frac{v}{{n}_{i}}{x}_{i} \equiv {a}_{i}\left( {\;\operatorname{mod}\;... | Yes |
Theorem 17. In every group there is exactly one element \( E \) such that\n\n\[ \n{AE} = {EA} = A \n\]\n\nfor every element of the group. \( E \) is called the unit (identity) element. | By (iv), to each \( A \) there is an \( E \) such that\n\n\[ \n{AE} = A\text{, thus also}{YAE} = {YA}\text{.} \n\]\n\n\nIf \( Y \) runs through all elements of the group, then, by (iv) this also holds for \( {YA} = B \), hence \( {BE} = B \) holds for each \( B \), and \( E \) is independent of \( B \) .\n\nMoreover th... | Yes |
Theorem 22. If a prime number \( p \) divides the order \( h \) of \( \mathfrak{G} \), then there is an element of order \( p \) in \( \mathfrak{G} \) . | Let \( {C}_{1},{C}_{2},\ldots ,{C}_{h} \) be the \( h \) elements of \( \mathfrak{G} \) and let \( {c}_{1},{c}_{2},\ldots ,{c}_{h} \) be their respective orders. We form all products\n\n\[ \n{C}_{1}^{{x}_{1}}{C}_{2}^{{x}_{2}}\cdots {C}_{h}^{{x}_{h}} \n\]\n\n(8)\n\nin which each \( {x}_{i} \) runs through a complete res... | Yes |
Theorem 23. Let \( h = {a}_{1} \cdot {a}_{2}\cdots {a}_{r} \) and suppose that the integers \( {a}_{1},\ldots ,{a}_{r} \) are pairwise relatively prime. Then each element \( C \) of \( \mathfrak{G} \) can be represented in one and only way in the form \[ C = {A}_{1} \cdot {A}_{2}\cdots {A}_{r} \] with the conditions \[... | For let \( r \) integers \( {n}_{1},\ldots ,{n}_{r} \) be determined so that \[ \frac{h}{{a}_{1}}{n}_{1} + \frac{h}{{a}_{2}}{n}_{2} + \cdots + \frac{h}{{a}_{r}}{n}_{r} = 1 \] which is always possible by Theorem 3 because of the assumption about the \( {a}_{i} \) . If we then set \[ {A}_{i} = {C}^{\left( {h/{a}_{i}}\rig... | Yes |
Theorem 27. If \( p \) is a prime, then the number of different elements of \( \mathfrak{G} \) with the property\n\n\[ \n{A}^{p} = 1 \n\]\n\nis equal to \( {p}^{e} \), where \( e \) is the basis number belonging to \( p \) . | If \( {B}_{1},{B}_{2},\ldots ,{B}_{e} \) are those basis elements whose orders are powers of \( p \) , then from\n\n\[ \nA = {B}_{1}^{{x}_{1}}{B}_{2}^{{x}_{2}}\cdots {B}_{e}^{{x}_{e}}{B}_{e + 1}^{{x}_{e + 1}}\cdots {B}_{r}^{{x}_{r}}\;\text{ and }\;{A}^{p} = 1 \n\]\n\nwe have the sequence of congruences\n\n\[ \np{x}_{i}... | Yes |
Theorem 28. An Abelian group \( \mathfrak{G} \) of order \( h \) is cyclic if and only if for each prime \( p \) dividing \( h \), the number of elements \( A \) with \( {A}^{p} = 1 \) is equal to \( p \) . | By the preceding theorem the condition is equivalent to: the basis number belonging to \( p \) should be \( = 1 \) . The condition is necessary. Namely if\n\n\[ C,{C}^{2},\ldots ,{C}^{h - 1},{C}^{h} = 1 \]\n\nare the \( h \) elements of \( \mathfrak{G} \), then from \( {A}^{p} = 1 \) it follows that for \( A = {C}^{x} ... | Yes |
Theorem 29. The order of \( \mathfrak{G}/{\mathfrak{U}}_{p} \) is \( {p}^{e} \) if \( e \) is the basis number of \( \mathfrak{G} \) belonging to p. The group \( \mathfrak{G}/{\mathfrak{U}}_{p} \) is isomorphic to the group of elements \( C \) of \( \mathfrak{G} \) for which \( {C}^{p} = 1 \) . | In fact we see from Theorem 26 that each element \( X \) of \( \mathfrak{G} \) can be represented in the form\n\n\[ X = {B}_{1}^{{x}_{1}}{B}_{2}^{{x}_{2}}\cdots {B}_{e}^{{x}_{e}}{A}^{p} \]\n\nwhere \( {B}_{1},\ldots ,{B}_{e} \) are the basis elements belonging to the prime \( p \) and the \( e \) numbers \( {x}_{1},\ld... | Yes |
\[ \mathop{\sum }\limits_{A}\chi \left( A\right) = \left\{ \begin{array}{ll} h & \text{ if }\chi \text{ is the principal character,} \\ 0 & \text{ if }\chi \text{ is not the principal character,} \end{array}\right. \] | The first half of each statement is trivial, as each summand \( = 1 \) . If \( B \) is an arbitrary element, then along with \( A,{AB} \) also runs through \( h \) all elements of \( \mathfrak{G} \), hence \[ \mathop{\sum }\limits_{A}\chi \left( A\right) = \mathop{\sum }\limits_{A}\chi \left( {AB}\right) = \chi \left( ... | Yes |
Theorem 32. If \( A \) is an element of order \( f \), then \( {\chi }_{n}\left( A\right) \) is an \( f \) th root of unity. Among the \( h \) numbers \( {\chi }_{n}\left( A\right), n = 1,\ldots, h \), all \( f \) th roots of unity occur equally often, namely \( h/f \) times. | To begin with, since \( {A}^{f} = 1 : {\chi }_{n}{\left( A\right) }^{f} = {\chi }_{n}\left( {A}^{f}\right) = {\chi }_{n}\left( 1\right) = 1 \) . Thus the first part of the theorem is true. Now if \( \zeta \) is an arbitrary \( f \) th root of unity, let us consider the sum\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{h}\left... | Yes |
Theorem 36. The index of \( \mathfrak{U} \) in \( \mathfrak{G} \) is \( j = \left| {{r}_{11} \cdot {r}_{22}\cdots {r}_{nn}}\right| \) . | For the proof we must determine the maximum number of elements which can exist in \( \mathfrak{G} \) such that no two differ by a factor in \( \mathfrak{U} \) . We first show that an element\n\n\[ \n{B}_{1}^{{x}_{1}}{B}_{2}^{{x}_{2}}\cdots {B}_{n}^{{x}_{n}} \]\n\nwhere all \( \left| {x}_{i}\right| < {r}_{ii} \), belong... | Yes |
Theorem 37. If a torsion-free Abelian group \( \mathfrak{G} \) has a finite basis of \( n \) elements \( {B}_{1},\ldots ,{B}_{n} \), then \( n \) is the maximal number of independent elements of \( \mathfrak{G} \) , independent of the choice of basis. | Since the \( {B}_{1},\ldots ,{B}_{n} \) are independent in any case, there are \( n \) independent elements in \( \mathfrak{G} \) and thus we need only show that \( n + 1 \) elements in \( \mathfrak{G} \) are not independent. In fact, between \( n + 1 \) arbitrary elements\n\n\[ \n{A}_{i} = {B}_{1}^{{c}_{i1}} \cdot {B}... | Yes |
Theorem 38. From a basis \( {B}_{1},\ldots ,{B}_{n} \) of a torsion-free Abelian group ( \( \mathfrak{H} \) one can obtain all systems of bases \( {B}_{1}^{\prime },\ldots ,{B}_{n}^{\prime } \) of \( \mathfrak{G} \) in the form\n\n\[ \n{B}_{i}^{\prime } = {B}_{1}^{{a}_{i1}}{B}_{2}^{{a}_{i2}}\cdots {B}_{n}^{{a}_{in}},\;... | To begin with, the \( {B}_{i}^{\prime } \) always form a basis. To see this we need only show that the \( {B}_{i} \) can be represented through the \( {B}_{i}^{\prime } \) . The equation\n\n\[ \n{B}_{m} = {B}_{1}^{\prime {x}_{1}} \cdot {B}_{2}^{\prime {x}_{2}}\cdots {B}_{n}^{\prime {x}_{n}} \n\]\n\nis satisfied if the ... | Yes |
Theorem 39. If \( \mathfrak{G} \) is a torsion-free Abelian group with a finite basis \( {B}_{1},\ldots ,{B}_{n} \) , \( \mathfrak{U} \) a subgroup of finite index \( j \), then \( \mathfrak{U} \) also has a finite basis \( {U}_{1},\ldots ,{U}_{n} \), and the determinant \( \left| {a}_{ik}\right| \) in the \( n \) equa... | The last assertion holds for the special basis mentioned in Theorem 36. The passage from the special basis \( {U}^{\prime } \) to an arbitrary basis \( U \) is done by Theorem 38 using an array of exponents with determinant \( \pm 1 \) . However, in the passage from \( B \) to \( U \) we obviously obtain an array of ex... | Yes |
Theorem 40. If \( \mathfrak{G} \) is a group with a finite basis \( {B}_{1},\ldots ,{B}_{m} \), then a subgroup \( \mathfrak{U} \) is of finite index if and only if a power of each element of \( \mathfrak{G} \) belongs to \( \mathfrak{U} \) . | If the \( {N}_{h} \) th power \( \left( {{N}_{h} > 0}\right) \) of \( {B}_{h} \) belongs to \( \mathfrak{U} \) and if we set\n\n\[ N = {N}_{1}{N}_{2}\cdots {N}_{m} \]\n\nthen \( {B}_{h}^{N} \) also belongs to \( \mathfrak{U} \) and consequently the \( N \) th power of each element likewise belongs to \( \mathfrak{U} \)... | Yes |
Theorem 42. Suppose \( \left( {{n}_{1},{n}_{2}}\right) = 1, n = {n}_{1} \cdot {n}_{2} \) . Then\n\n\[ \Re \left( n\right) = \Re \left( {n}_{1}\right) \cdot \Re \left( {n}_{2}\right) \] | To prove this we assign to each element \( A \) of \( \Re \left( n\right) \) a pair of elements \( {C}_{1} \) from \( \mathfrak{R}\left( {n}_{1}\right) \) and \( {C}_{2} \) from \( \mathfrak{R}\left( {n}_{2}\right) \) as follows: If \( a \) is a number in \( A \), then choose any two numbers \( {c}_{1},{c}_{2} \) accor... | Yes |
Theorem 43. If \( p \) is a prime, then the group \( \mathfrak{R}\left( p\right) \) of residue classes mod \( p \) is a cyclic group of order \( p - 1 \) . | By Theorem 27, we need only show that if \( q \) is a prime dividing \( p - 1 \) , then the number of classes \( A \) with \( {A}^{q} = 1 \) is equal to \( q \) (by Theorem 22 it must be at least \( q \) ). However, the number of these classes \( \mathrm{A} \) is identical with the number of integers \( a \) which are ... | Yes |
Theorem 45. The groups \( \Re \left( 2\right) \) and \( \Re \left( 4\right) \) are cyclic. If \( \alpha \geq 3 \) then the group \( \Re \left( {2}^{\alpha }\right) \) of order \( h = \varphi \left( {2}^{\alpha }\right) = {2}^{\alpha - 1} \) has exactly two basis classes. One is of order 2, the other of order \( h/2 = {... | The statements are trivial for the moduli 2 and 4 . Thus suppose \( \alpha \geq 3 \) . The group of classes \( {\;\operatorname{mod}\;{2}^{\alpha }} \) has order \( h = \varphi \left( {2}^{\alpha }\right) = {2}^{\alpha - 1} \) . The number of incongruent solutions of \( {x}^{2} \equiv 1\left( {\;\operatorname{mod}\;{2}... | Yes |
Theorem 47. With odd \( q \) and \( a \), the congruence \( {x}^{q} \equiv a\left( {\;\operatorname{mod}\;{2}^{\alpha }}\right) \) always has exactly one solution. | The first part \( \left( {q\text{odd}}\right) \) is proved exactly as above in Case 1. Since, by Theorem 45, the classes \( {\;\operatorname{mod}\;{2}^{\alpha }}\left( {\alpha \geq 3}\right) \) can be represented in the form \( {B}_{1}^{{a}_{1}}{B}_{2}^{{a}_{2}} \) , where \( {B}_{1}^{2} = {B}_{2}^{{2}^{\alpha - 2}} = ... | No |
Theorem 48. Two arbitrary nonzero polynomials \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) over \( k \) have a uniquely determined greatest common divisor \( d\left( x\right) \), that is, there is a polynomial \( d\left( x\right) \) with leading coefficient 1, such that\n\n\[ d\left( x\right) \left| {{... | The proof is well known from elementary algebra, yet no importance is attached there to the nature of the numerical coefficients which appear. For this reason we reproduce quite briefly a proof based on the proof of the analogous fact for rational numbers (Theorems 1 and 2). Among the polynomials\n\n\[ L\left( x\right)... | Yes |
Theorem 49. If a polynomial \( f\left( x\right) \), irreducible over \( k \), has a common zero \( x = \alpha \) with a polynomial \( g\left( x\right) \) over \( k \), then \( f\left( x\right) \) is a divisor of \( g\left( x\right) \) and hence all zeros of \( f\left( x\right) \) are zeros of \( g\left( x\right) \) . | For \( \left( {f\left( x\right), g\left( x\right) }\right) \) is at least divisible by \( x - \alpha \) and thus not \( = 1 \) . On the other hand \( f\left( x\right) \) has no factors over \( k \) other than \( {cf}\left( x\right) \) . Consequently \( \left( {f\left( x\right), g\left( x\right) }\right) = {cf}\left( x\... | Yes |
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