Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Lemma 3.10. Assume that \( M \) is flat over \( R \) . Let \( {a}_{i} \in A,{x}_{i} \in M \) for \( i = 1 \) , \( \ldots, n \), and suppose that we have the relation\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{x}_{i} = 0 \]\n\nThen there exists an integer \( s \) and elements \( {b}_{ij} \in A \) and \( {y}_{j} \i... | Proof. We consider the exact sequence\n\n\[ 0 \rightarrow K \rightarrow {R}^{\left( n\right) } \rightarrow R \]\n\nwhere the map \( {R}^{\left( n\right) } \rightarrow R \) is given by\n\n\[ \left( {{b}_{1},\ldots ,{b}_{n}}\right) \mapsto \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{b}_{i} \]\n\nand \( K \) is its kernel.... | Yes |
Proposition 4.2. Let \( R \rightarrow A \) be an \( R \) -algebra, and assume \( A \) commutative.\n\n(i) Base change. If \( F \) is a flat \( R \) -module, then \( A{ \otimes }_{R}F \) is a flat \( A \) -module.\n\n(ii) Transitivity. If \( A \) is a flat commutative \( R \) -algebra and \( M \) is a flat \( A \) -modu... | The proofs are immediate, and will be left to the reader. | No |
Lemma 5.2. Let \( E \) and \( {E}_{i}\left( {i = 1,\ldots, m}\right) \) be modules over a ring. Let \( {\varphi }_{i} : {E}_{i} \rightarrow E \) and \( {\psi }_{i} : E \rightarrow {E}_{i} \) be homomorphisms having the following properties:\n\n\[ \n{\psi }_{i} \circ {\varphi }_{i} = \mathrm{{id}},\;{\psi }_{i} \circ {\... | Proof. The proof is routine, and is essentially the same as that of Proposition 3.1 of Chapter III. We shall leave it as an exercise to the reader. | No |
Corollary 5.3. Let \( {E}^{\prime }, E,{F}^{\prime }, F \) be free and finite dimensional over \( R \) . Then we have a functorial isomorphism\n\n\[ L\left( {{E}^{\prime }, E}\right) \otimes L\left( {{F}^{\prime }, F}\right) \rightarrow L\left( {{E}^{\prime } \otimes {F}^{\prime }, E \otimes F}\right) \]\n\n such that\... | Proof. Keep \( E,{F}^{\prime }, F \) fixed, and view \( L\left( {{E}^{\prime }, E}\right) \otimes L\left( {{F}^{\prime }, F}\right) \) as a functor in the variable \( {E}^{\prime } \) . Similarly, view\n\n\[ L\left( {{E}^{\prime } \otimes {F}^{\prime }, E \otimes F}\right) \]\n\n as a functor in \( {E}^{\prime } \) . T... | Yes |
Corollary 5.4. Let \( E, F \) be free and finite dimensional. Then we have a natural isomorphism\n\n\[{\operatorname{End}}_{R}\left( E\right) \otimes {\operatorname{End}}_{R}\left( F\right) \rightarrow {\operatorname{End}}_{R}\left( {E \otimes F}\right) . | Proof. Special case of Corollary 5.3. | No |
Corollary 5.5. Let \( E, F \) be free finite dimensional over \( R \) . There is a functorial isomorphism\n\n\[ \n{E}^{ \vee } \otimes F \rightarrow L\left( {E, F}\right)\n\]\n\ngiven for \( \lambda \in {E}^{ \vee } \) and \( y \in F \) by the map\n\n\[ \n\lambda \otimes y \mapsto {A}_{\lambda, y}\n\]\n\nwhere \( {A}_{... | To prove Corollary 5.5, justify that there is a well-defined homomorphism of \( {E}^{ \vee } \otimes F \) to \( L\left( {E, F}\right) \), by the formula written down. Verify that this homomorphism is both injective and surjective. We leave the details as exercises. | No |
Corollary 5.6. Let \( E, F \) be free and finite dimensional over \( R \) . There is a functorial isomorphism\n\n\[ \n{E}^{ \vee } \otimes {F}^{ \vee } \rightarrow {\left( E \otimes F\right) }^{ \vee }.\n\]\n\ngiven for \( \lambda \in {E}^{ \vee } \) and \( \mu \in {F}^{ \vee } \) by the map\n\n\[ \n\lambda \otimes \mu... | Proof. As before. | No |
Proposition 5.7. Let \( E \) be free and finite dimensional over \( R \) . The trace function on \( L\left( {E, E}\right) \) is equal to the composite of the two maps\n\n\[ L\left( {E, E}\right) \rightarrow {E}^{ \vee } \otimes E \rightarrow R, \]\n\nwhere the first map is the inverse of the isomorphism described in Co... | Of course, it is precisely in a situation involving the trace that the isomorphism of Corollary 5.5 becomes important, and that the finite dimensionality of \( E \) is used. In many applications, this finite dimensionality plays no role, and it is better to deal with \( L\left( {E, E}\right) \) directly. | No |
Finite coproducts exist in the category of commutative rings, and in the category of commutative algebras over a commutative ring. If \( R \rightarrow A \) and \( R \rightarrow B \) are two homomorphisms of commutative rings, then their coproduct over \( R \) is the homomorphism \( R \rightarrow A \otimes B \) given by... | Proof. We shall limit our proof to the case of the coproduct of two ring homomorphisms \( R \rightarrow A \) and \( R \rightarrow B \) . One can use induction.\n\nLet \( A, B \) be commutative rings, and assume given ring-homomorphisms into a commutative ring \( C \) ,\n\n\[ \varphi : A \rightarrow C\text{ and }\psi : ... | Yes |
Proposition 7.1. Let \( E \) be free of dimension n over \( R \) . Then \( T\left( E\right) \) is isomorphic to the non-commutative polynomial algebra on \( n \) variables over \( R \) . In other words, if \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is a basis of \( E \) over \( R \), then the elements\n\n\[ \n{M}_... | Proof. This follows at once from Proposition 2.3. | No |
Proposition 7.2. Let \( E \) be free, finite dimensional over \( R \) . Then we have an algebra-isomorphism\n\n\[ T\left( {L\left( E\right) }\right) = T\left( {{\operatorname{End}}_{R}\left( E\right) }\right) \rightarrow {LT}\left( E\right) = {\bigoplus }_{r = 0}^{\infty }{\operatorname{End}}_{R}\left( {{T}^{r}\left( E... | Proof. By Proposition 2.5, we have a linear isomorphism in each dimension, and it is clear that the map preserves multiplication. | No |
Proposition 8.1. Let \( E \) be free of dimension \( n \) over \( R \) . Let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( E \) over \( k \) . Viewed as elements of \( {S}^{1}\left( E\right) \) in \( S\left( E\right) \), these basis elements are algebraically independent over \( R \), and \( S\left( ... | Proof. Let \( {t}_{1},\ldots ,{t}_{n} \) be algebraically independent variables over \( R \), and form the polynomial algebra \( R\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . Let \( {P}_{r} \) be the \( R \) -module of homogeneous polynomials of degree \( r \) . We define a map of \( {E}^{\left( r\right) } ... | Yes |
Proposition 8.2. Let \( E = {E}^{\prime } \oplus {E}^{\prime \prime } \) be a direct sum of finite free modules. Then there is a natural isomorphism\n\n\[ \n{S}^{n}\left( {{E}^{\prime } \oplus {E}^{\prime \prime }}\right) \approx {\bigoplus }_{p + q = n}{S}^{p}{E}^{\prime } \otimes {S}^{q}{E}^{\prime \prime }.\n\]\n\nI... | Proof. The isomorphism comes from the following maps. The inclusions of \( {E}^{\prime } \) and \( {E}^{\prime \prime } \) into their direct sum give rise to the functorial maps\n\n\[ \nS{E}^{\prime } \otimes S{E}^{\prime \prime } \rightarrow {SE}\n\]\n\nand the claim is that this is a graded isomorphism. Note that \( ... | No |
Proposition 1.1. Schur's Lemma. Let \( E, F \) be simple \( R \) -modules. Every non-zero homomorphism of \( E \) into \( F \) is an isomorphism. The ring \( {\operatorname{End}}_{R}\left( E\right) \) is a division ring. | Proof. Let \( f : E \rightarrow F \) be a non-zero homomorphism. Its image and kernel are submodules, hence \( \operatorname{Ker}f = 0 \) and \( \operatorname{Im}f = F \) . Hence \( f \) is an isomorphism. If \( E = F \), then \( f \) has an inverse, as desired. | Yes |
Proposition 1.2. Let \( E = {E}_{1}^{\left( {n}_{1}\right) } \oplus \cdots \oplus {E}_{r}^{\left( {n}_{r}\right) } \) be a direct sum of simple modules, the \( {E}_{i} \) being non-isomorphic, and each \( {E}_{i} \) being repeated \( {n}_{i} \) times in the sum. Then, up to a permutation, \( {E}_{1},\ldots ,{E}_{r} \) ... | Proof. The last statement follows from our previous considerations, taking into account Proposition 1.1. Suppose now that we have two \( R \) -modules, with direct sum decompositions into simple submodules, and an isomorphism \[ {E}_{1}^{\left( {n}_{1}\right) } \oplus \cdots \oplus {E}_{r}^{\left( {n}_{r}\right) } \rig... | Yes |
Lemma 2.1. Let \( E = \mathop{\sum }\limits_{{i \in I}}{E}_{i} \) be a sum (not necessarily direct) of simple submodules. Then there exists a subset \( J \subset I \) such that \( E \) is the direct sum \( \bigoplus {E}_{j} \) . \( j \in J \) | Proof. Let \( J \) be a maximal subset of \( I \) such that the sum \( \mathop{\sum }\limits_{{j \in J}}{E}_{j} \) is direct. We contend that this sum is in fact equal to \( E \) . It will suffice to prove that each \( {E}_{i} \) is contained in this sum. But the intersection of our sum with \( {E}_{i} \) is a submodul... | Yes |
Proposition 2.2. Every submodule and every factor module of a semisimple module is semisimple. | Proof. Let \( F \) be a submodule. Let \( {F}_{0} \) be the sum of all simple submodules of \( F \) . Write \( E = {F}_{0} \oplus {F}_{0}^{\prime } \) . Every element \( x \) of \( F \) has a unique expression \( x = {x}_{0} + {x}_{0}^{\prime } \) with \( {x}_{0} \in {F}_{0} \) and \( {x}_{0}^{\prime } \in {F}_{0}^{\pr... | Yes |
Lemma 3.1. Let \( E \) be semisimple over \( R \) . Let \( {R}^{\prime } = {\operatorname{End}}_{R}\left( E\right), f \in {\operatorname{End}}_{{R}^{\prime }}\left( E\right) \) as above. Let \( x \in R \) . There exists an element \( \alpha \in R \) such that \( {\alpha x} = f\left( x\right) \) . | Proof. Since \( E \) is semisimple, we can write an \( R \) -direct sum\n\n\[ E = {Rx} \oplus F \]\n\nwith some submodule \( F \) . Let \( \pi : E \rightarrow {Rx} \) be the projection. Then \( \pi \in {R}^{\prime } \), and hence\n\n\[ f\left( x\right) = f\left( {\pi x}\right) = {\pi f}\left( x\right) . \]\n\nThis show... | No |
Theorem 3.2. (Jacobson). Let \( E \) be semisimple over \( R \), and let \( {R}^{\prime } = {\operatorname{End}}_{R}\left( E\right) \) . Let \( f \in {\operatorname{End}}_{{R}^{\prime }}\left( E\right) \) . Let \( {x}_{1},\ldots ,{x}_{n} \in E \) . Then there exists an element \( \alpha \in R \) such that \[ \alpha {x}... | Proof. For clarity of notation, we shall first carry out the proof in case \( E \) is simple. Let \( {f}^{\left( n\right) } : {E}^{\left( n\right) } \rightarrow {E}^{\left( n\right) } \) be the product map, so that \[ {f}^{\left( n\right) }\left( {{y}_{1},\ldots ,{y}_{n}}\right) = \left( {f\left( {y}_{1}\right) ,\ldots... | Yes |
Corollary 3.3. (Burnside's Theorem). Let \( E \) be a finite-dimensional vector space over an algebraically closed field \( k \), and let \( R \) be a subalgebra of \( {\operatorname{End}}_{k}\left( E\right) \) . If \( E \) is a simple \( R \) -module, then \( R = {\operatorname{End}}_{{R}^{\prime }}\left( E\right) \) ... | Proof. We contend that \( {\operatorname{End}}_{R}\left( E\right) = k \) . At any rate, \( {\operatorname{End}}_{R}\left( E\right) \) is a division ring \( {R}^{\prime } \), containing \( k \) as a subring and every element of \( k \) commutes with every element of \( {R}^{\prime } \) . Let \( \alpha \in {R}^{\prime } ... | Yes |
Corollary 3.5. (Wedderburn’s Theorem). Let \( R \) be a ring, and \( E \) a simple, faithful module over \( R \). Let \( D = {\operatorname{End}}_{R}\left( E\right) \), and assume that \( E \) is finite dimensional over \( D \). Then \( R = {\operatorname{End}}_{D}\left( E\right) \). | Proof. Let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( E \) over \( D \). Given \( A \in {\operatorname{End}}_{D}\left( E\right) \), by Theorem 3.2 there exists \( \alpha \in R \) such that\n\n\[ \alpha {v}_{i} = A{v}_{i}\;\text{ for }\;i = 1,\ldots, n. \]\n\nHence the map \( R \rightarrow {\operat... | Yes |
Corollary 3.5. (Wedderburn’s Theorem). Let \( R \) be a ring, and \( E \) a simple, faithful module over \( R \) . Let \( D = {\operatorname{End}}_{R}\left( E\right) \), and assume that \( E \) is finite dimensional over \( D \) . Then \( R = {\operatorname{End}}_{D}\left( E\right) \). | Proof. Let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a basis of \( E \) over \( D \) . Given \( A \in {\operatorname{End}}_{D}\left( E\right) \), by Theorem 3.2 there exists \( \alpha \in R \) such that\n\n\[ \alpha {v}_{i} = A{v}_{i}\;\text{ for }\;i = 1,\ldots, n. \]\n\nHence the map \( R \rightarrow {\opera... | Yes |
Corollary 3.6. Let \( R \) be a ring, finite dimensional algebra over a field \( k \) which is algebraically closed. Let \( V \) be a finite dimensional vector space over \( k \), with a simple faithful representation \( \rho : R \rightarrow {\operatorname{End}}_{k}\left( V\right) \) . Then \( \rho \) is an isomorphism... | Proof. We apply Corollary 3.5, noting that \( D \) is finite dimensional over \( k \) . Given \( \alpha \in D \), we note that \( k\left( \alpha \right) \) is a commutative subfield of \( D \), whence \( k\left( \alpha \right) = k \) by assumption that \( k \) is algebraically closed, and the corollary follows. | No |
Theorem 3.7. Existence of projection operators. Let \( k \) be a field, \( {Ra} \) \( k \) -algebra, and \( {V}_{1},\ldots ,{V}_{m} \) finite dimensional \( k \) -spaces which are also simple \( R \) -modules, and such that \( {V}_{i} \) is not \( R \) -isomorphic to \( {V}_{j} \) for \( i \neq j \) . Then there exist ... | Proof. We observe that the projection \( {f}_{i} \) from the direct sum \( E \) to the \( i \) -th factor is in \( {\operatorname{End}}_{{R}^{\prime }}\left( E\right) \), because if \( \varphi \in {R}^{\prime } \) then \( \varphi \left( {V}_{j}\right) \subset {V}_{j} \) for all \( j \) . We may therefore apply the dens... | No |
Corollary 3.8. (Bourbaki). Let \( k \) be a field of characteristic 0 . Let \( R \) be a \( k \) -algebra, and let \( E, F \) be semisimple \( R \) -modules, finite dimensional over \( k \) . For each \( \alpha \in R \), let \( {\alpha }_{E},{\alpha }_{F} \) be the corresponding \( k \) -endomorphisms on \( E \) and \(... | Proof. Each of \( E \) and \( F \) is isomorphic to a finite direct sum of simple \( R \) - modules, with certain multiplicities. Let \( V \) be a simple \( R \) -module, and suppose\n\n\( E = {V}^{\left( n\right) } \oplus \) direct summands not isomorphic to \( V \)\n\n\( F = {V}^{\left( m\right) } \oplus \) direct su... | Yes |
Proposition 4.1. If \( R \) is semisimple, then every \( R \) -module is semisimple. | Proof. An R-module is a factor module of a free module, and a free module is a direct sum of \( R \) with itself a certain number of times. We can apply Proposition 2.2 to conclude the proof. | No |
Lemma 4.2. Let \( L \) be a simple left ideal, and let \( E \) be a simple \( R \) -module. If \( L \) is not isomorphic to \( E \), then \( {LE} = 0 \) . | Proof. We have \( {RLE} = {LE} \), and \( {LE} \) is a submodule of \( E \), hence equal to 0 or \( E \) . Suppose \( {LE} = E \) . Let \( y \in E \) be such that \[ {Ly} \neq 0\text{.} \] Since \( {Ly} \) is a submodule of \( E \), it follows that \( {Ly} = E \) . The map \( \alpha \mapsto {\alpha y} \) of \( L \) int... | Yes |
Theorem 4.4. Let \( R \) be semisimple, and let \( E \) be an \( R \) -module \( \neq 0 \) . Then\n\n\[ E = {\bigoplus }_{i = 1}^{s}{R}_{i}E = {\bigoplus }_{i = 1}^{s}{e}_{i}E \]\n\nand \( {R}_{i}E \) is the submodule of \( E \) consisting of the sum of all simple submodules isomorphic to \( {L}_{i} \) . | Proof. Let \( {E}_{i} \) be the sum of all simple submodules of \( E \) isomorphic to \( {L}_{i} \) . If \( V \) is a simple submodule of \( E \), then \( {RV} = V \), and hence \( {L}_{i}V = V \) for some \( i \) . By a previous lemma, we have \( {L}_{i} \approx V \) . Hence \( E \) is the direct sum of \( {E}_{1},\ld... | Yes |
Proposition 4.7. Let \( k \) be a field and \( E \) a finite dimensional vector space over \( k \) . Let \( S \) be a subset of \( {\operatorname{End}}_{k}\left( E\right) \) . Let \( R \) be the \( k \) -algebra generated by the elements of \( S \) . Then \( R \) is semisimple if and only if \( E \) is a semisimple \( ... | Proof. If \( R \) is semisimple, then \( E \) is semisimple by Proposition 4.1. Conversely, assume \( E \) semisimple as \( S \) -module. Then \( E \) is semisimple as \( R \) -module, and so is a direct sum\n\n\[ E = {\bigoplus }_{i = 1}^{n}{E}_{i} \]\n\nwhere each \( {E}_{i} \) is simple. Then for each \( i \) there ... | Yes |
Corollary 4.6. A simple ring has exactly one simple module, up to isomorphism. | Both these corollaries are immediate consequences of Theorems 4.3 and 4.4. | No |
Proposition 4.7. Let \( k \) be a field and \( E \) a finite dimensional vector space over \( k \) . Let \( S \) be a subset of \( {\operatorname{End}}_{k}\left( E\right) \) . Let \( R \) be the \( k \) -algebra generated by the elements of \( S \) . Then \( R \) is semisimple if and only if \( E \) is a semisimple \( ... | Proof. If \( R \) is semisimple, then \( E \) is semisimple by Proposition 4.1. Conversely, assume \( E \) semisimple as \( S \) -module. Then \( E \) is semisimple as \( R \) -module, and so is a direct sum\n\n\[ E = {\bigoplus }_{i = 1}^{n}{E}_{i} \]\n\nwhere each \( {E}_{i} \) is simple. Then for each \( i \) there ... | Yes |
Lemma 5.1. Let \( R \) be a ring, and \( \psi \in {\operatorname{End}}_{R}\left( R\right) \) a homomorphism of \( R \) into itself, viewed as \( R \) -module. Then there exists \( \alpha \in R \) such that \( \psi \left( x\right) = {x\alpha } \) for all \( x \in R \) . | Proof. We have \( \psi \left( x\right) = \psi \left( {x \cdot 1}\right) = {x\psi }\left( 1\right) \) . Let \( \alpha = \psi \left( 1\right) \) . | Yes |
Theorem 5.2. Let \( R \) be a simple ring. Then \( R \) is a finite direct sum of simple left ideals. There are no two-sided ideals except 0 and \( R \) . If \( L, M \) are simple left ideals, then there exists \( \alpha \in R \) such that \( {L\alpha } = M \) . We have \( {LR} = R \) . | Proof. Since \( R \) is by definition also semisimple, it is a direct sum of simple left ideals, say \( {\bigoplus }_{j \in J}{L}_{j} \) . We can write 1 as a finite sum \( 1 = \mathop{\sum }\limits_{{j = 1}}^{m}{\beta }_{j} \), with \( {\beta }_{j} \in {L}_{j} \) .\n\nThen\n\[ R = {\bigoplus }_{j = 1}^{m}R{\beta }_{j}... | Yes |
Corollary 5.3. Let \( R \) be a simple ring. Let \( E \) be a simple \( R \) -module, and \( L \) a simple left ideal of \( R \). Then \( {LE} = E \) and \( E \) is faithful. | Proof. We have \( {LE} = L\left( {RE}\right) = \left( {LR}\right) E = {RE} = E \). Suppose \( {\alpha E} = 0 \) for some \( \alpha \in R \). Then \( {R\alpha RE} = {R\alpha E} = 0 \). But \( {R\alpha R} \) is a two-sided ideal. Hence \( {R\alpha R} = 0 \), and \( \alpha = 0 \). This proves that \( E \) is faithful. | Yes |
Theorem 5.4. (Rieffel). Let \( R \) be a ring without two-sided ideals except 0 and \( R \) . Let \( L \) be a nonzero left ideal, \( {R}^{\prime } = {\operatorname{End}}_{R}\left( L\right) \) and \( {R}^{\prime \prime } = {\operatorname{End}}_{{R}^{\prime }}\left( L\right) \) . Then the natural map \( \lambda : R \rig... | Proof. The kernel of \( \lambda \) is a two-sided ideal, so \( \lambda \) is injective. Since \( {LR} \) is a two-sided ideal, we have \( {LR} = R \) and \( \lambda \left( L\right) \lambda \left( R\right) = \lambda \left( R\right) \) . For any \( x, y \in L \) , and \( f \in {R}^{\prime \prime } \), we have \( f\left( ... | Yes |
Theorem 5.5. Let \( D \) be a division ring, and \( E \) a finite-dimensional vector space over \( D \) . Let \( R = {\operatorname{End}}_{D}\left( E\right) \) . Then \( R \) is simple and \( E \) is a simple \( R \) -module. Furthermore, \( D = {\operatorname{End}}_{R}\left( E\right) \) . | Proof. We first show that \( E \) is a simple \( R \) -module. Let \( v \in E, v \neq 0 \) . Then \( v \) can be completed to a basis of \( E \) over \( D \), and hence, given \( w \in E \), there exists \( \alpha \in R \) such that \( {\alpha v} = w \) . Hence \( E \) cannot have any invariant subspaces other than 0 o... | Yes |
Theorem 5.6. Let \( k \) be a field and \( E \) a finite-dimensional vector space of dimension \( m \) over \( k \) . Let \( R = {\operatorname{End}}_{k}\left( E\right) \) . Then \( R \) is a \( k \) -space, and\n\n\[ \n{\dim }_{k}R = {m}^{2}.\n\]\n\nFurthermore, \( m \) is the number of simple left ideals appearing in... | Proof. The \( k \) -space of \( k \) -endomorphisms of \( E \) is represented by the space of \( m \times m \) matrices in \( k \), so the dimension of \( R \) as a \( k \) -space is \( {m}^{2} \) . On the other hand, the proof of Theorem 5.5 showed that \( R \) is \( R \) -isomorphic as an \( R \) -module to the direc... | Yes |
Theorem 6.2. Let \( A \) be a semisimple algebra, finite dimensional over a field \( k \) . Let \( K \) be a finite separable extension of \( k \) . Then \( K{ \otimes }_{k}A \) is a semisimple over \( K \) . | Proof. In light of the radical criterion for semisimplicity, it suffices to prove that \( K{ \otimes }_{k}A \) has zero radical, and it suffices to do so for an even larger extension than \( K \), so that we may assume \( K \) is Galois over \( k \), say with Galois group \( G \) . Then \( G \) operates on \( K \otimes... | Yes |
Theorem 6.3. Let \( A, B \) be simple algebras, finite dimensional over a field \( k \) which is algebraically closed. Then \( A{ \otimes }_{k}B \) is also simple. We have \( A \approx {\operatorname{End}}_{k}\left( V\right) \) and \( B \approx {\operatorname{End}}_{k}\left( W\right) \) where \( V, W \) are finite dime... | Proof. The formula is a special case of Theorem 2.5 of Chapter XVI, and the isomorphisms \( A \approx {\operatorname{End}}_{k}\left( V\right), B \approx {\operatorname{End}}_{k}\left( W\right) \) exist by Wedderburn’s theorem or its corollaries. | No |
Theorem 6.4. Let \( A, B \) be absolutely semisimple algebras finite dimensional over a field \( k \) . Then \( A{ \otimes }_{k}B \) is absolutely semisimple. | Proof. Let \( F = {k}^{\mathrm{a}} \) . Then \( {A}_{F} \) is semisimple by hypothesis, so it is a direct product of simple algebras, which are matrix algebras, and in particular we can apply Theorem 6.3 to see that \( {A}_{F}{ \otimes }_{F}{B}_{F} \) has no radical. Hence \( A{ \otimes }_{k}B \) has no radical (becaus... | Yes |
Theorem 7.1. (Morita). Let \( E \) be an \( R \) -module. Then \( E \) is a generator if and only if \( E \) is balanced and finitely generated projective over \( {R}^{\prime }\left( E\right) \) . | Proof. We shall prove half of the theorem, leaving the other half to the reader, using similar ideas (see Exercise 12). So we assume that \( E \) is a generator, and we prove that it satisfies the other properties by arguments due to Faith.\n\nWe first prove that for any module \( F, R \oplus F \) is balanced. We ident... | No |
Proposition 1.1. Let \( G \) be a finite group and let \( {E}^{\prime }, E, F,{F}^{\prime } \) be \( G \) -modules. Let\n\n\[ \n{E}^{\prime }\overset{\varphi }{ \rightarrow }E\overset{f}{ \rightarrow }F\overset{\psi }{ \rightarrow }{F}^{\prime } \]\n\nbe R-homomorphisms, and assume that \( \varphi ,\psi \) are \( G \) ... | Proof. We have\n\n\[ \n{\operatorname{Tr}}_{G}\left( {\psi \circ f \circ \varphi }\right) = \mathop{\sum }\limits_{{\sigma \in G}}\sigma \left( {\psi \circ f \circ \varphi }\right) = \mathop{\sum }\limits_{{\sigma \in G}}\left( {\sigma \psi }\right) \circ \left( {\sigma f}\right) \circ \left( {\sigma \varphi }\right)\n... | Yes |
Theorem 1.2. (Maschke). Let \( G \) be a finite group of order \( n \), and let \( k \) be a field whose characteristic does not divide \( n \) . Then the group ring \( k\left\lbrack G\right\rbrack \) is semisimple. | Proof. Let \( E \) be a \( G \) -module, and \( F \) a \( G \) -submodule. Since \( k \) is a field, there exists a \( k \) -subspace \( {F}^{\prime } \) such that \( E \) is the \( k \) -direct sum of \( F \) and \( {F}^{\prime } \) . We let the \( k \) -linear map \( \pi : E \rightarrow F \) be the projection on \( F... | Yes |
Proposition 2.1. If \( E, F \) are \( G \) -spaces, then\n\n\[{\chi }_{E} + {\chi }_{F} = {\chi }_{E \oplus F}\;\text{ and }\;{\chi }_{E}{\chi }_{F} = {\chi }_{E \otimes F}.\]\n\nIf \( {\chi }^{ \vee } \) denotes the character of the dual representation on \( {E}^{ \vee } \), then\n\n\[{\chi }^{ \vee }\left( \sigma \ri... | Proof. The first relation holds because the matrix of an element \( \sigma \) in the representation \( E \oplus F \) decomposes into blocks corresponding to the representation in \( E \) and the representation in \( F \) . As to the second, if \( \left\{ {v}_{i}\right\} \) is a basis of \( E \) and \( \left\{ {w}_{j}\r... | Yes |
Theorem 2.2. There are only a finite number of simple characters of \( G \) (over \( k) \) . The characters of representations of \( G \) are the linear combinations of the simple characters with integer coefficients \( \geqq 0 \) . | We shall use the direct product decomposition of a semisimple ring. We have\n\n\[ \nk\left\lbrack G\right\rbrack = \mathop{\prod }\limits_{{i = 1}}^{s}{R}_{i} \]\n\nwhere each \( {R}_{i} \) is simple, and we have a corresponding decomposition of the unit element of \( k\left\lbrack G\right\rbrack \) :\n\n\[ 1 = {e}_{1}... | Yes |
Assume that \( k \) has characteristic 0 . Then every effective character has a unique expression as a linear combination\n\n\[ \chi = \mathop{\sum }\limits_{{i = 1}}^{s}{n}_{i}{\chi }_{i},\;{n}_{i} \in \mathbf{Z},{n}_{i} \geqq 0, \]\n\nwhere \( {\chi }_{1},\ldots ,{\chi }_{s} \) are the simple characters of \( G \) ov... | Let \( E \) be the representation space of \( \chi \) . Then by Theorem 4.4 of Chapter XVII,\n\n\[ E \approx {\bigoplus }_{i = 1}^{s}{n}_{i}{L}_{i} \]\n\nThe sum is finite because we assume throughout that \( E \) is finite dimensional. Since \( {e}_{i} \) acts as a unit element on \( {L}_{i} \), we find\n\n\[ {\chi }_... | Yes |
As functions of \( G \) into \( k \), the simple characters \[ {\chi }_{1},\ldots ,{\chi }_{s} \] are linearly independent over \( k \) . | Proof. Suppose that \( \sum {a}_{i}{\chi }_{i} = 0 \) with \( {a}_{i} \in k \) . We apply this expression to \( {e}_{j} \) and get \[ 0 = \left( {\sum {a}_{i}{\chi }_{i}}\right) \left( {e}_{j}\right) = {a}_{j}{\dim }_{k}{L}_{j} \] Hence \( {a}_{j} = 0 \) for all \( j \) . | Yes |
Theorem 3.1. Let \( G \) be a finite abelian group, and assume that \( k \) is algebraically closed. Then every simple representation of \( G \) is 1-dimensional. The simple characters of \( G \) are the homomorphisms of \( G \) into \( {k}^{ * } \) . | Proof. The group ring \( k\left\lbrack G\right\rbrack \) is semisimple, commutative, and is a direct product of simple rings. Each simple ring is a ring of matrices over \( k \) (by Corollary 3.6 Chapter XVII), and can be commutative if and only if it is equal to \( k \) . | Yes |
Corollary 3.2. Let \( k \) be algebraically closed. Let \( G \) be a finite group. For any character \( \chi \) and \( \sigma \in G \), the value \( \chi \left( \sigma \right) \) is equal to a sum of roots of unity with integer coefficients (i.e. coefficients in \( \mathbf{Z} \) or \( \mathbf{Z}/p\mathbf{Z} \) dependin... | Proof. Let \( H \) be the subgroup generated by \( \sigma \) . Then \( H \) is a cyclic subgroup. A representation of \( G \) having character \( \chi \) can be viewed as a representation for \( H \) by restriction, having the same character. Thus our assertion follows from Theorem 3.1. | No |
An element of \( k\left\lbrack G\right\rbrack \) commutes with every element of \( G \) if and only if it is a linear combination of conjugacy classes with coefficients in \( k \) . | Let \( \alpha = \mathop{\sum }\limits_{{\sigma \in G}}{a}_{\sigma }\sigma \) and assume \( {\alpha \tau } = {\tau \alpha } \) for all \( \tau \in G \) . Then\n\n\[\n\mathop{\sum }\limits_{{\sigma \in G}}{a}_{\sigma }{\tau \sigma }{\tau }^{-1} = \mathop{\sum }\limits_{{\sigma \in G}}{a}_{\sigma }\sigma\n\]\n\nHence \( {... | Yes |
Proposition 4.3. Let \( {\chi }_{\mathrm{{reg}}} \) be the regular character. Then\n\n\[ \n{\chi }_{\mathrm{{reg}}}\left( \sigma \right) = 0\;\text{ if }\;\sigma \in G,\sigma \neq 1 \n\]\n\n\[ \n{\chi }_{\text{reg }}\left( 1\right) = n\text{.} \n\] | Proof. Let \( 1 = {\sigma }_{1},\ldots ,{\sigma }_{n} \) be the elements of \( G \) . They form a basis of \( k\left\lbrack G\right\rbrack \) over \( k \) . The matrix of 1 is the unit \( n \times n \) matrix. Thus our second assertion follows. If \( \sigma \neq 1 \), then multiplication by \( \sigma \) permutes \( {\s... | Yes |
Assume again that \( k \) is algebraically closed. Let\n\n\[ \n{e}_{i} = \mathop{\sum }\limits_{{\tau \in G}}{a}_{\tau }\tau ,\;{a}_{\tau } \in k.\n\]\n\nThen\n\n\[ \n{a}_{\tau } = \frac{1}{n}{\chi }_{\mathrm{{reg}}}\left( {{e}_{i}{\tau }^{-1}}\right) = \frac{{d}_{i}}{n}{\chi }_{i}\left( {\tau }^{-1}\right) .\n\] | Proof. We have for all \( \tau \in G \) :\n\n\[ \n{\chi }_{\mathrm{{reg}}}\left( {{e}_{i}{\tau }^{-1}}\right) = {\chi }_{\mathrm{{reg}}}\left( {\mathop{\sum }\limits_{{\sigma \in G}}{a}_{\sigma }\sigma {\tau }^{-1}}\right) = \mathop{\sum }\limits_{{\sigma \in G}}{a}_{\sigma }{\chi }_{\mathrm{{reg}}}\left( {\sigma {\tau... | Yes |
Corollary 4.6. The dimensions \( {d}_{i} \) are not divisible by the characteristic of \( k \) . | Proof. Otherwise, \( {e}_{i} = 0 \), which is impossible. | Yes |
Corollary 4.6. The dimensions \( {d}_{i} \) are not divisible by the characteristic of \( k \) . | Proof. Otherwise, \( {e}_{i} = 0 \), which is impossible. | Yes |
Corollary 4.7. The simple characters \( {\chi }_{1},\ldots ,{\chi }_{s} \) are linearly independent over \( k \) . | Proof. The proof in Corollary 2.4 applies, since we now know that the characteristic does not divide \( {d}_{i} \) . | No |
Corollary 4.8. Assume in addition that \( k \) has characteristic 0 . Then \( {d}_{i} \mid n \) for each \( i \) . | Proof. Multiplying our expression for \( {e}_{i} \) by \( n/{d}_{i} \), and also by \( {e}_{i} \), we find\n\n\[\n\frac{n}{{d}_{i}}{e}_{i} = \mathop{\sum }\limits_{{\sigma \in G}}{\chi }_{i}\left( {\sigma }^{-1}\right) \sigma {e}_{i}\n\]\n\nLet \( \zeta \) be a primitive \( m \) -th root of unity, and let \( M \) be th... | Yes |
Theorem 4.9. Let \( k \) be algebraically closed. Let \( {Z}_{k}\left( G\right) \) be the center of \( k\left\lbrack G\right\rbrack \), and let \( {X}_{k}\left( G\right) \) be the \( k \) -space of class functions on \( G \) . Then \( {Z}_{k}\left( G\right) \) and \( {X}_{k}\left( G\right) \) are the dual spaces of eac... | Proof. The formula has been proved in the proof of Theorem 2.3. The two spaces involved here both have dimension \( s \), and \( {d}_{i} \neq 0 \) in \( k \) . Our proposition is then clear. | No |
Theorem 5.1. The symbol \( \langle f, g\rangle \) for \( f, g \in X\left( G\right) \) takes on values in the prime ring. The simple characters form an orthonormal basis for \( X\left( G\right) \), in other words\n\n\[ \left\langle {{\chi }_{i},{\chi }_{j}}\right\rangle = {\delta }_{ij} \] | Proof. By Proposition 4.4, we find\n\n\[ {\chi }_{j}\left( {e}_{i}\right) = \frac{{d}_{i}}{n}\mathop{\sum }\limits_{{\sigma \in G}}{\chi }_{i}\left( {\sigma }^{-1}\right) {\chi }_{j}\left( \sigma \right) \]\n\nIf \( i \neq j \) we get 0 on the left-hand side, so that \( {\chi }_{i} \) and \( {\chi }_{j} \) are orthogon... | Yes |
Theorem 5.2. Let \( k \) have characteristic 0, and let \( R \) be a subring containing the \( m \) -th roots of unity, and having a conjugation. Then the bilinear form on \( X\left( G\right) \) has a unique extension to a hermitian form\n\n\[ \n{X}_{R}\left( G\right) \times {X}_{R}\left( G\right) \rightarrow R \n\]\n\... | Proof. The formula given in the statement of the theorem gives the same value as before for the symbol \( \langle f, g\rangle \) when \( f, g \) lie in \( X\left( G\right) \) . Thus the extension exists, and is obviously unique. | Yes |
Proposition 5.3. For \( \alpha ,\beta \in Z\left( G\right) \), we can define a symbol \( \langle \alpha ,\beta \rangle \) by either one of the following expressions, which are equal:\n\n\[ \langle \alpha ,\beta \rangle = \frac{1}{n}{\chi }_{\text{reg }}\left( {\alpha {\beta }^{ - }}\right) = \frac{1}{n}\mathop{\sum }\l... | Proof. Each expression is linear in its first and second variable. Hence to prove their equality, it will suffice to prove that the two expressions are equal when we replace \( \alpha \) by \( {e}_{i} \) and \( \beta \) by an element \( \tau \) of \( G \) . But then, our equality is equivalent to\n\n\[ {\chi }_{\text{r... | Yes |
For each ring \( R \) contained in \( k \), the pairing of Proposition 5.3 has a unique extension to a map\n\n\[ \n{Z}_{R}\left( G\right) \times Z\left( G\right) \rightarrow R \]\n\nwhich is \( R \) -linear in its first variable. If \( R \) contains the \( m \) -th roots of unity, where \( m \) is an exponent for \( G ... | Proof. We note that \( {h}_{i} \) is not divisible by the characteristic because it is the index of a subgroup of \( G \) (the isotropy group of an element in \( {\gamma }_{i} \) when \( G \) operates by conjugation), and hence \( {h}_{i} \) divides \( n \) . The extension of our pairing as stated is obvious, since \( ... | Yes |
Theorem 5.5. The conjugacy classes \( {\gamma }_{1},\ldots ,{\gamma }_{s} \) constitute an orthogonal basis for \( Z\left( G\right) \) . We have \( \left\langle {{\gamma }_{i},{\gamma }_{i}}\right\rangle = {h}_{i} \) . For each ring \( R \) contained in \( k \), the bilinear map of Proposition 5.3 has a unique extensio... | Proof. We use the lemma. By linearity, the formula in the lemma remains valid when we replace \( R \) by \( k \), and when we replace \( {e}_{i} \) by any element of \( {Z}_{k}\left( G\right) \), in particular when we replace \( {e}_{i} \) by \( {\gamma }_{i} \) . But \( \left\{ {{\gamma }_{1},\ldots ,{\gamma }_{s}}\ri... | Yes |
Corollary 5.6. If \( G \) is commutative, then\n\n\[ \n\frac{1}{n}\mathop{\sum }\limits_{{v = 1}}^{n}{\chi }_{v}\left( \sigma \right) {\chi }_{v}\left( {\tau }^{-1}\right) = \left\{ \begin{array}{ll} 0 & \text{ if }\sigma \text{ is not equal to }\tau \\ 1 & \text{ if }\sigma \text{ is equal to }\tau . \end{array}\right... | Proof. When \( G \) is commutative, each conjugacy class has exactly one element, and the number of simple characters is equal to the order of the group. | No |
Theorem 5.7. Let \( k \) have characteristic 0, and let \( R \) be a subring of \( k \), containing the \( m \) -th roots of unity, and having a conjugation. Then the pairing of Proposition 5.3 has a unique extension to a hermitian form \[ {Z}_{R}\left( G\right) \times {Z}_{R}\left( G\right) \rightarrow R \] given by t... | Proof. The formula given in the statement of the theorem gives the same value as the symbol \( \langle \alpha ,\beta \rangle \) of Proposition 5.3 when \( \alpha ,\beta \) lie in \( Z\left( G\right) \) . Thus the extension exists, and is obviously unique. Using the second formula in Proposition 5.3, defining the scalar... | Yes |
Theorem 5.8. Let \( E, F \) be simple \( \left( {G, k}\right) \) -spaces. Let \( \lambda \) be a \( k \) -linear functional on \( E \), let \( x \in E \) and \( y \in F \) . If \( E, F \) are not isomorphic, then\n\n\[ \mathop{\sum }\limits_{{\sigma \in G}}\lambda \left( {\sigma x}\right) {\sigma }^{-1}y = 0 \]\n\nIf \... | Proof. The map \( x \mapsto \sum \lambda \left( {\sigma x}\right) {\sigma }^{-1}y \) is a \( G \) -homomorphism of \( E \) into \( F \), so Schur's lemma concludes the proof of the first statement. The second comes by applying the functional \( \mu \) . | Yes |
Lemma 5.9. Let \( E \) be a simple \( \left( {G, k}\right) \) -space. Then any \( G \) -endomorphism of \( E \) is equal to a scalar multiple of the identity. | Proof. The algebra \( {\operatorname{End}}_{G, k}\left( E\right) \) is a division algebra by Schur’s lemma, and is finite dimensional over \( k \) . Since \( k \) is assumed algebraically closed, it must be equal to \( k \) because any element generates a commutative subfield over \( k \) . This proves the lemma. | Yes |
Lemma 5.10. Let \( E \) be a representation space for \( G \) of dimension \( d \) . Let \( \lambda \) be a functional on \( E \), and let \( x \in E \) . Let \( {\varphi }_{\lambda, x} \in {\operatorname{End}}_{k}\left( E\right) \) be the endomorphism such that\n\n\[ \n{\varphi }_{\lambda, x}\left( y\right) = \lambda ... | Proof. If \( x = 0 \) the statement is obvious. Let \( x \neq 0 \) . If \( \lambda \left( x\right) \neq 0 \) we pick a basis of \( E \) consisting of \( x \) and a basis of the kernel of \( \lambda \) . If \( \lambda \left( x\right) = 0 \), we pick a basis of \( E \) consisting of a basis for the kernel of \( \lambda \... | Yes |
Theorem 5.11. Let \( \rho : G \rightarrow {\operatorname{Aut}}_{k}\left( E\right) \) be a simple representation of \( G \), of dimension \( d \) . Then the characteristic of \( k \) does not divide \( d \) . Let \( x, y \in E \) . Then for any functionals \( \lambda ,\mu \) on \( E \) ,\n\n\[ \mathop{\sum }\limits_{{\s... | Proof. It suffices to prove that\n\n\[ \mathop{\sum }\limits_{{\sigma \in G}}\lambda \left( {\sigma x}\right) {\sigma }^{-1}y = \frac{n}{d}\lambda \left( y\right) x. \]\n\nFor fixed \( y \) the map\n\n\[ x \mapsto \mathop{\sum }\limits_{{\sigma \in G}}\lambda \left( {\sigma x}\right) {\sigma }^{-1}y \]\n\nis immediatel... | Yes |
Corollary 5.12. Let \( \chi \) be the character of the representation of \( G \) on the simple space E. Then\n\n\[ \langle \chi ,\chi \rangle = 1\text{.} \] | Proof. This follows immediately from the theorem, and the expression of \( \chi \) as\n\n\[ \chi = {\rho }_{11} + \cdots + {\rho }_{dd} \] | Yes |
Corollary 5.13. \( {\chi }_{i}\left( {e}_{j}\right) = {\delta }_{ij}{d}_{i} \) and \( {\chi }_{\text{reg }} = \mathop{\sum }\limits_{{i = 1}}^{s}{d}_{i}{\chi }_{i} \) . | Proof. The first formula is a direct application of the orthonormality of the characters. The second formula concerning the regular character is obtained by writing\n\n\[ \n{\chi }_{\text{reg }} = \mathop{\sum }\limits_{j}{m}_{j}{\chi }_{j} \n\]\n\nwith unknown coefficients. We know the values \( {\chi }_{\mathrm{{reg}... | Yes |
Proposition 5.14. We have\n\n\\[ \n{\\rho }_{i}\\left( {e}_{i}\\right) = \\mathrm{{id}}\\;\\text{ and }\\;{\rho }_{i}\\left( {e}_{j}\\right) = 0\\;\\text{ if }i \\neq j.\n\\]\n | Proof. The map \\( x \\mapsto {e}_{i}x \\) is a \\( G \\) -homomorphism of \\( {E}_{i} \\) into itself since \\( {e}_{i} \\) is in the center of \\( k\\left\\lbrack G\\right\\rbrack \\) . Hence by Lemma 5.9 this homomorphism is a scalar multiple of the identity. Taking the trace and using the orthogonality relations be... | No |
Theorem 5.15. Let \( f \) be a class function on \( G \) . Then\n\n\[ f = \mathop{\sum }\limits_{{i = 1}}^{s}\left\langle {f,{\chi }_{i}}\right\rangle {\chi }_{i} \] | Proof. The number of conjugacy class is equal to the number of distinct characters, and these are linearly independent, so they form a basis for the class functions. The coefficients are given by the stated formula, as one sees by taking the scalar product of \( f \) with any character \( {\chi }_{j} \) and using the o... | Yes |
Theorem 5.16. Let \( {\rho }^{\left( i\right) } \) be a matrix representation of \( G \) on \( {E}_{i} \) relative to a choice of basis, and let \( {\rho }_{v,\mu }^{\left( i\right) } \) be the coefficient functions of this matrix, \( i = 1,\ldots, s \) and \( v,\mu = 1,\ldots ,{d}_{i} \) . Then the functions \( {\rho ... | Proof. That the coefficient functions form an orthogonal basis follows from Theorems 5.8 and 5.11. The expression of \( f \) in terms of this basis is then merely the standard Fourier expansion relative to any scalar product. This concludes the proof. | No |
Theorem 5.17. (a) Let \( \chi \) be an effective character in \( X\left( G\right) \). Then \( \chi \) is simple over \( \mathbf{C} \) if and only if \( \parallel \chi {\parallel }^{2} = 1 \), or alternatively,\n\n\[ \mathop{\sum }\limits_{{\sigma \in G}}{\left| \chi \left( \sigma \right) \right| }^{2} = \# \left( G\rig... | Proof. The first part has been proved, and for (b), let \( \chi ,\psi \) be effective characters in \( X\left( G\right) \), and let \( E, F \) be their representation spaces over \( \mathbf{C} \). Then\n\n\[ \langle \chi ,\psi {\rangle }_{G} = \dim {\operatorname{Hom}}_{G}\left( {E, F}\right) . \]\n\nThen by orthonorma... | No |
Corollary 5.18 With the above notation and \( k = \mathbf{C} \) for simplicity, we have:\n\n(a) The multiplicity of \( {1}_{G} \) in \( {E}^{ \vee } \otimes F \) is \( {\dim }_{k}{\operatorname{inv}}_{G}\left( {{E}^{ \vee } \otimes F}\right) \) .\n\n(b) The \( \left( {G, k}\right) \) -space \( E \) is simple if and onl... | Proof. Immediate from Theorem 5.17 and formula (3) of §1. | No |
Proposition 7.1. Let \( \varphi : F \rightarrow {M}_{G}^{S}\left( F\right) \) be such that \( \varphi \left( x\right) = {\varphi }_{x} \) is the map\n\n\[{\varphi }_{x}\left( \tau \right) = \left\{ \begin{array}{lll} 0 & \text{ if } & \tau \notin S \\ {\tau x} & \text{ if } & \tau \in S \end{array}\right.\]\n\nThen \( ... | Proof. Let \( \sigma \in S \) and \( x \in F \) . Let \( \tau \in G \) . Then\n\n\[ \left( {\sigma {\varphi }_{x}}\right) \left( \tau \right) = {\varphi }_{x}\left( {\tau \sigma }\right) \]\n\nIf \( \tau \in S \), then this last expression is equal to \( {\varphi }_{\sigma x}\left( \tau \right) \) . If \( \tau \notin S... | Yes |
Proposition 7.2. Let \( G = \mathop{\bigcup }\limits_{{i = 1}}^{r}S{\bar{c}}_{i} \) be a decomposition of \( G \) into right cosets.\n\nLet \( {F}_{1} \) be the additive group of functions in \( {M}_{G}^{S}\left( F\right) \) having value 0 at elements \( \xi \in G,\xi \notin S \) . Then\n\n\[ \n{M}_{G}^{S}\left( F\righ... | Proof. For each \( f \in {M}_{G}^{S}\left( F\right) \), let \( {f}_{i} \) be the function such that\n\n\[ \n{f}_{i}\left( \xi \right) = \left\{ \begin{array}{lll} 0 & \text{ if } & \xi \notin S{\bar{c}}_{i} \\ f\left( \xi \right) & \text{ if } & \xi \in S{\bar{c}}_{i} \end{array}\right. \n\]\n\nFor all \( \sigma \in S ... | Yes |
Theorem 7.3. Let \( \left\{ {{\lambda }_{1},\ldots ,{\lambda }_{r}}\right\} \) be a system of left coset representatives of \( S \) in G. There exists a G-module \( E \) containing \( F \) as an S-submodule, such that \[ E = {\bigoplus }_{i = 1}^{r}{\lambda }_{i}F \] is a direct sum (as R-modules). Let \( \varphi : F \... | Proof. By the usual set-theoretic procedure of replacing \( {F}_{1} \) by \( F \) in \( {M}_{G}^{S}\left( F\right) \) , obtain a \( G \) -module \( E \) containing \( F \) as a \( S \) -submodule, and having the desired direct sum decomposition. Let \( {\varphi }^{\prime } : F \rightarrow {E}^{\prime } \) be an \( S \)... | Yes |
Proposition 7.4. Let \( \psi \) be the character of the representation of \( S \) on the \( k \) -space \( F \) . Let \( E \) be the space of an induced representation. Then the character \( \chi \) of \( E \) is equal to the induced character \( {\psi }^{G} \), i.e. is given by the formula\n\n\[ \chi \left( \xi \right... | Proof. Let \( \left\{ {{w}_{1},\ldots ,{w}_{m}}\right\} \) be a basis for \( F \) over \( k \) . We know that\n\n\[ E = \bigoplus {\bar{c}}^{-1}F \]\n\nLet \( \sigma \) be an element of \( G \) . The elements \( {\left\{ {\overline{c\sigma }}^{-1}{w}_{j}\right\} }_{c, j} \) form a basis for \( E \) over \( k \) .\n\nWe... | Yes |
Lemma 7.5. The elements \( \left\{ {{\tau }_{\gamma }\gamma }\right\} \) form a family of left coset representatives for \( S \) in \( G \) ; that is, we have a disjoint union\n\n\[ G = \mathop{\bigcup }\limits_{{\gamma ,{\tau }_{\gamma }}}{\tau }_{\gamma }{\gamma S} \] | Proof. First we have by hypothesis\n\n\[ G = \mathop{\bigcup }\limits_{\gamma }\mathop{\bigcup }\limits_{{\tau }_{\gamma }}{\tau }_{\gamma }\left( {H \cap \left\lbrack \gamma \right\rbrack S}\right) {\gamma S} \]\n\nand so every element of \( G \) can be written in the form\n\n\[ {\tau }_{\gamma }\gamma {s}_{1}{\gamma ... | Yes |
Theorem 7.6. Applied to the S-module \( F \), we have an isomorphism of \( H \) - modules\n\n\[ \n{\operatorname{res}}_{H}^{G} \circ {\operatorname{ind}}_{S}^{G} \approx {\bigoplus }_{\gamma }{\operatorname{ind}}_{H \cap \left\lbrack \gamma \right\rbrack S}^{H} \circ {\operatorname{res}}_{H \cap \left\lbrack \gamma \ri... | Proof. The induced module \( {\operatorname{ind}}_{S}^{G}\left( F\right) \) is simply the direct sum\n\n\[ \n{\operatorname{ind}}_{S}^{G}\left( F\right) = {\bigoplus }_{\gamma ,{\tau }_{\gamma }}{\tau }_{\gamma }{\gamma F} \n\]\n\nby Lemma 7.5, which gives us coset representatives of \( S \) in \( G \), and Theorem 7.3... | Yes |
Theorem 7.7. Let \( H, S \) be subgroups of finite index in \( G \) . Let \( {F}_{1},{F}_{2} \) be \( \left( {H, R}\right) \) and \( \left( {S, R}\right) \) -modules respectively. Then we have an isomorphism of \( R \) - modules\n\n\[ \n{\operatorname{Hom}}_{G}\left( {{\operatorname{ind}}_{H}^{G}\left( {F}_{1}\right) ,... | Proof. We have the isomorphisms:\n\n\( {\operatorname{Hom}}_{G}\left( {{\operatorname{ind}}_{H}^{G}\left( {F}_{1}\right) ,{\operatorname{ind}}_{S}^{G}\left( {F}_{2}\right) }\right) \approx {\operatorname{Hom}}_{H}\left( {{F}_{1},{\operatorname{res}}_{H}^{G} \circ {\operatorname{ind}}_{S}^{G}\left( {F}_{2}\right) }\righ... | Yes |
Corollary 7.8. Let \( R = k = \mathbf{C} \) . Let \( S, H \) be subgroups of the finite group G. Let \( D = {H\gamma S} \) range over the double cosets, with representatives \( \gamma \) . Let \( \chi \) be an effective character of \( H \) and \( \psi \) an effective character of \( S \) . Then\n\n\[ \n{\left\langle {... | Proof. Immediate from Theorem 5.17(b) and Theorem 7.7, taking dimensions on the left-hand side and on the right-hand side. | No |
Corollary 7.9. (Irreducibility of the induced character). Let \( S \) be a subgroup of the finite group \( G \) . Let \( R = k = \mathbf{C} \) . Let \( \psi \) be an effective character of \( S \) . Then \( {\operatorname{ind}}_{S}^{G}\left( \psi \right) \) is irreducible if and only if \( \psi \) is irreducible and\n\... | Proof. Immediate from Corollary 7.8 and Theorem 5.17(a). It is of course trivial that if \( \psi \) is reducible, then so is the induced character. | No |
Theorem 7.10. Let \( S \) be a subgroup of \( G \) and let \( F \) be a finite free \( R \) -module.\n\nThen there is a G-isomorphism\n\n\[ \n{\operatorname{ind}}_{S}^{G}\left( {F}^{ \vee }\right) \approx {\left( {\operatorname{ind}}_{S}^{G}\left( F\right) \right) }^{ \vee }.\n\] | Proof. Let \( G = \bigcup {\lambda }_{i}S \) be a left coset decomposition. Then, as in Theorem 7.3, we can express the representation space for \( {\operatorname{ind}}_{S}^{G}\left( F\right) \) as\n\n\[ \n{\operatorname{ind}}_{S}^{G}\left( F\right) = \bigoplus {\lambda }_{i}F \n\]\n\nWe may select \( {\lambda }_{1} = ... | Yes |
Theorem 7.11. Let \( S \) be a subgroup of finite index in \( G \) . Let \( F \) be an \( S \) - module, and \( E \) a \( G \) -module (over the commutative ring \( R \) ). Then there is an isomorphism\n\n\[ \n{\operatorname{ind}}_{S}^{G}\left( {{\operatorname{res}}_{S}\left( E\right) \otimes F}\right) \approx E \otime... | Proof. The \( G \) -module \( {\operatorname{ind}}_{S}^{G}\left( F\right) \) contains \( F \) as a summand, because it is the direct sum \( \bigoplus {\lambda }_{i}F \) with left coset representatives \( {\lambda }_{i} \) as in Theorem 7.3. Hence we have a natural \( S \) -isomorphism\n\n\[ \nf : {\operatorname{res}}_{... | Yes |
Theorem 7.12. There is a G-isomorphism\n\n\[ \n{\operatorname{ind}}_{H}^{G}\left( {F}_{1}\right) \otimes {\operatorname{ind}}_{S}^{G}\left( {F}_{2}\right) \approx {\bigoplus }_{\gamma }{\operatorname{ind}}_{H \cap \left\lbrack \gamma \right\rbrack S}^{G}\left( {{F}_{1} \otimes \left\lbrack \gamma \right\rbrack {F}_{2}}... | Proof. We have:\n\n\( {\operatorname{ind}}_{H}^{G}\left( {F}_{1}\right) \otimes {\operatorname{ind}}_{S}^{G}\left( {F}_{2}\right) \approx {\operatorname{ind}}_{H}^{G}\left( {{F}_{1} \otimes {\operatorname{res}}_{H}{\operatorname{ind}}_{S}^{G}\left( {F}_{2}\right) }\right) \; \) by Theorem 7.11\n\n\( \approx {\bigoplus ... | Yes |
Proposition 8.1. Let \( H \) be a subgroup of \( G \), and let \( \psi \) be a character of \( H \) . Let \( {\psi }^{G} \) be the induced character. Then the multiplicity of \( {1}_{H} \) in \( \psi \) is the same as the multiplicity of \( {1}_{G} \) in \( {\psi }^{G} \) . | Proof. By Theorem 6.1 (i), we have\n\n\[ \n{\left\langle \psi ,{1}_{H}\right\rangle }_{H} = {\left\langle {\psi }^{G},{1}_{G}\right\rangle }_{G} \n\]\n\nThese scalar products are precisely the multiplicities in question. | Yes |
Proposition 8.2. The regular representation is the representation induced by the trivial character on the trivial subgroup of \( G \) . | Proof. This follows at once from the definition of the induced character\n\n\[ \n{\psi }^{G}\left( \tau \right) = \mathop{\sum }\limits_{{\sigma \in G}}{\psi }_{H}\left( {{\sigma \tau }{\sigma }^{-1}}\right) \n\]\n\ntaking \( \psi = 1 \) on the trivial subgroup. | Yes |
Proposition 8.5. Let \( G \) be a finite group of order \( n \) . Then\n\n\[ n{u}_{G} = \sum {\lambda }_{A}^{G} \]\n\nthe sum being taken over all cyclic subgroups of \( G \) . | Proof. Given two class functions \( \chi ,\psi \) on \( G \), we have the usual scalar product:\n\n\[ \langle \psi ,\chi {\rangle }_{G} = \frac{1}{n}\mathop{\sum }\limits_{{\sigma \in G}}\psi \left( \sigma \right) \overline{\chi \left( \sigma \right) } \]\n\nLet \( \psi \) be any class function on \( G \) . Then:\n\n\[... | Yes |
Proposition 8.6. If \( A \neq \{ 1\} \), the function \( {\lambda }_{A} \) is a linear combination of irreducible nontrivial characters of \( A \) with positive integral coefficients. | Proof. If \( A \) is cyclic of prime order, then by Proposition 8.5, we know that \( {\lambda }_{A} = n{u}_{A} \), and our assertion follows from the standard structure of the regular representation.\n\nIn order to prove the assertion in general, it suffices to prove that the Fourier coefficients of \( {\lambda }_{A} \... | Yes |
Proposition 9.1. Every subgroup and every factor group of a super-solvable group is supersolvable. | Proof. Obvious, using the standard homomorphism theorems. | No |
Proposition 9.2. Let \( G \) be a non-abelian supersolvable group. Then there exists a normal abelian subgroup which contains the center properly. | Proof. Let \( C \) be the center of \( G \), and let \( \bar{G} = G/C \) . Let \( \bar{H} \) be a normal subgroup of prime order in \( \bar{G} \) and let \( H \) be its inverse image in \( G \) under the canonical map \( G \rightarrow G/C \) . If \( \sigma \) is a generator of \( \bar{H} \), then an inverse image \( \s... | Yes |
Theorem 9.3. (Blichfeldt). Let \( G \) be a supersolvable group, let \( k \) be algebraically closed. Let \( E \) be a simple \( \left( {G, k}\right) \) -space. If \( {\dim }_{k}E > 1 \), then there exists a proper subgroup \( H \) of \( G \) and a simple \( H \) -space \( F \) such that \( E \) is induced by \( F \) . | Proof. Since a simple representation of an abelian group is 1-dimensional, our hypothesis implies that \( G \) is not abelian.\n\nWe shall first give the proof of our theorem under the additional hypothesis that \( E \) is faithful. (This means that \( {\sigma x} = x \) for all \( x \in E \) implies \( \sigma = 1 \) .)... | No |
Corollary 9.5. Let \( G \) be a product of a p-group and a cyclic group, and let \( k \) be algebraically closed. If \( E \) is a simple \( \left( {G, k}\right) \) -space and is not \( 1 \) -dimensional, then \( E \) is induced by a 1-dimensional representation of some subgroup. | Proof. We apply the theorem step by step using the transitivity of induced representations until we get a 1-dimensional representation of a subgroup. | No |
Lemma 10.1. \( {V}_{R}\left( G\right) \) is an ideal in \( {X}_{R}\left( G\right) \) . | Proof. This is immediate from Theorem 6.1. | No |
Proposition 10.2. Every subgroup and every factor group of a p-elementary group is p-elementary. If \( S \) is a subgroup of the p-elementary group \( P \times C \) , where \( P \) is a p-group, and \( C \) is cyclic, of order prime to \( p \), then | \[ S = \left( {S \cap P}\right) \times \left( {S \cap C}\right) . \]\n\nProof. Clear. | No |
Lemma 10.3. If \( d \in \mathbf{Z} \) and the constant function \( d.{1}_{G} \) belongs to \( {V}_{R} \) then \( d.{1}_{G} \) belongs to \( {V}_{\mathbf{z}} \) . | Proof. We contend that \( 1,\zeta ,\ldots ,{\zeta }^{N - 1} \) are linearly independent over \( {X}_{\mathbf{Z}}\left( G\right) \) . Indeed, a relation of linear dependence would yield\n\n\[ \mathop{\sum }\limits_{{v = 1}}^{s}\mathop{\sum }\limits_{{j = 0}}^{{N - 1}}{c}_{vj}{\chi }_{v}{\zeta }^{j} = 0 \]\n\nwith intege... | Yes |
Lemma 10.4. Let \( f \in {X}_{R}\left( G\right) \), and assume that \( f\left( \sigma \right) \in \mathbf{Z} \) for all \( \sigma \in G \) . Then \( f \) is constant \( {\;\operatorname{mod}\;p} \) on every \( p \) -class. | Proof. Let \( x = {\sigma \tau } \), where \( \sigma \) is \( p \) -singular, and \( \tau \) is \( p \) -regular, and \( \sigma ,\tau \) commute. It will suffice to prove that \[ f\left( x\right) \equiv f\left( \tau \right) \;\left( {\;\operatorname{mod}\;p}\right) \] Let \( H \) be the cyclic subgroup generated by \( ... | Yes |
Lemma 10.5. Let \( R = \mathbf{Z}\left\lbrack \zeta \right\rbrack \) be as before. If \( a \in \mathbf{Z} \) and \( a \in {pR} \) then \( a \in p\mathbf{Z} \) . | Proof. This is immediate from the fact that \( R \) has a basis over \( \mathbf{Z} \) such that 1 is a basis element. | No |
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