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Proposition 9. The point \( {x}_{0} \) is the unique inertially finite \( {\mathbb{Z}}_{p} \) -valued point of \( X \) . | Proof: Let \( x \in S \) be a \( {\mathbb{Z}}_{p} \) -valued inertially finite point. Since the image of \( {I}_{p} \) under \( {\rho }_{x} \) is finite, it follows that its image is isomorphic to its image under \( \bar{\rho } \), the isomorphism being given by the natural projection \( {\mathrm{{GL}}}_{2}\left( {\mat... | Yes |
Proposition 11. The universal deformation ring \( {R}_{D} \) is a power series ring in two variables over \( {\mathbb{Z}}_{p} \) . The subscheme \( {X}_{D} \) is a smooth hypersurface in \( X \) . | Proof: Let \( s : {D}_{2h} \rightarrow \left( {\pm 1}\right) \) denote the natural surjective homomorphism given by the dihedral structure of \( {D}_{2h} \) . Let \( {\Gamma }_{ + },{\Gamma }_{ - } \) be two free pro- \( p \) - groups on one generator (i.e., \( \cong {\mathbb{Z}}_{p} \) ) endowed with continuous \( {D}... | No |
Proposition 13. The reduced Zariski tangent space of \( {X}^{ \circ } \) has dimension \( \leq 1 \) over \( k = {\mathbb{F}}_{p} \) . | Proof: Consider the diagram of subschemes (0.2) above. Since the reduced Zariski tangent space of \( X \) is of dimension 3, and the reduced Zariski tangent space of \( {X}_{D} \) is of dimension 2 (Prop. of \( §4 \) ) the transversality lemma above implies our proposition. | No |
Proposition 14. The natural mapping\n\n\[ \n{R}^{o} \rightarrow {\mathrm{T}}_{\mathrm{m}} \]\n\nis an isomorphism of rings, which are noncanonically isomorphic to power series rings in one variable over \( {\mathbb{Z}}_{p} \) . | Proof: Putting lemmas 8, 9, 10 together, we get, firstly, that the Krull dimension of \( {R}^{ \circ } \) is equal to 2 . Then, by lemma \( 8,{R}^{ \circ } \) is isomorphic to \( {\mathbb{Z}}_{p}\left\lbrack \left\lbrack t\right\rbrack \right\rbrack \) . But then, by lemma 9, there is a surjection\n\n\[ \n\left( {{R}^{... | Yes |
Proposition 15. The morphism\n\n\[ \mu : \Phi \times {\widehat{X}}^{o} \rightarrow {\widehat{Z}}^{o} \]\n\n is an isomorphism of formal affine schemes. The affine ring \( \mathcal{A}\left( {\widehat{Z}}^{o}\right) \) is a power series ring on two parameters over \( {\mathbb{Z}}_{p} \) . | Proof: The idea is simply that \( {\widehat{X}}^{o} \) is transversal to the orbits of \( \Phi \) in the sense that the natural morphism induced by \( \mu \) on \( k\left\lbrack \epsilon \right\rbrack \) -points yields an injection\n\n\[ \Phi \left( {k\left\lbrack \epsilon \right\rbrack }\right) \times {\widehat{X}}^{o... | Yes |
Proposition 16. We have an equality of formal subschemes of \( \widehat{X} \) :\n\n\[ \n{\widehat{Z}}^{o} \cap {\widehat{Z}}^{oo} = {\widehat{X}}_{i.a.} \n\] | Proof: Let \( A \) be an artinian object of the category \( \mathcal{C} \) and let\n\n\[ \n{\rho }_{A} : {G}_{\mathbb{Q}, S} \rightarrow {\mathrm{{GL}}}_{2}\left( A\right) \n\]\n\nbe a deformation of \( \bar{\rho } \) to \( A \) . Let \( M = A \times A \) be given a \( {G}_{\mathbf{Q}, S} \) -module structure via \( {\... | No |
Proposition 17. Let \( x \) be a \( {\mathbf{Z}}_{p} \) -valued point of \( X \) . Then \( x \) is inertially reducible if and only if \( x \) is a \( {\mathbb{Z}}_{p} \) -valued point of \( {\widehat{Z}}^{o} \cup {\widehat{Z}}^{oo} \) . | Proof: It is evident that any ordinary \( {\mathbb{Z}}_{p} \) -valued point is inertially reducible, and that inertially reducible points remain so after tensoring with one-dimensional characters. It follows that any \( {\mathbb{Z}}_{p} \) -valued point of \( {\widehat{Z}}^{o} \cup {\widehat{Z}}^{oo} \) is inertially r... | No |
Proposition 19. Any \( {\mathbb{Z}}_{p} \) -valued point of \( {X}_{i.m.} \) is either inertially reducible or inertially dihedral. | Since the group of upper triangular matrices in \( {\mathrm{{GL}}}_{2}\left( A\right) \) is metabelian, we have that any inertially reducible \( A \) -valued points of \( X \) is contained in \( {X}_{i.m.} \) | No |
Corollary 6. The closed subscheme \( {X}_{i.r.} \) is contained in \( {X}_{i.m.} \) . | Proof: It suffices to show that \( {Z}^{o} \) and \( {Z}^{oo} \) are separately contained in \( {X}_{i.m.} \) . But each are smooth hypersurfaces in which the \( {\mathbb{Z}}_{p} \) -valued points are Zariski-dense. Since the \( {\mathbb{Z}}_{p} \) -valued points of \( {Z}^{o} \) and of \( {Z}^{oo} \) are contained in ... | Yes |
Proposition 22 (Sen). The mapping \( \mathcal{S} : {X}^{an} \rightarrow {\mathbb{Q}}_{p} \times {\mathbb{Q}}_{p} \) is locally analytic. | Proof: This is a particular case of the main result in [Sen 2]. | No |
Proposition 23. Let \( \left( {b, c}\right) \in {\mathbb{Q}}_{p} \times {\mathbb{Q}}_{p} \) . The set \( X\left( {b, c}\right) \) inherits the structure of \( p \) -adic analytic variety from \( {X}^{an} \) . The analytic variety \( X\left( {b, c}\right) \) may be empty, but for each smooth point \( x \in X\left( {b, c... | Proof: By \ | No |
Proposition 24. The restriction of \( \mathcal{S} \) to \( {Z}^{o}\left( {\mathbf{Z}}_{p}\right) \) gives us a finite-to-one mapping\n\n\[ \mathcal{S} : {Z}^{o}\left( {\mathbf{Z}}_{p}\right) \rightarrow {\mathbf{Z}}_{p} \times {\mathbf{Z}}_{p} \subset {\mathbb{Q}}_{p} \times {\mathbb{Q}}_{p} \] | Proof: By Prop. 15 of \( §6 \), we have:\n\n\[ {Z}^{o}\left( {\mathbb{Z}}_{p}\right) \cong \operatorname{Hom}\left( {\Lambda \widehat{ \otimes }{R}^{o};{\mathbb{Z}}_{p}}\right) \]\n\nLet \( \Lambda \overset{\iota }{ \rightarrow }{R}^{o} \) be the structural homomorphism and consider the composition, \( {\mathcal{S}}^{\... | Yes |
Proposition 25. All points \( x \) in \( X\left( {0,0}\right) - \left\{ {x}_{0}\right\} \) have non-semi-simple \( p \) - adic Hodge type. None of these points are inertially dihedral. All but a finite number of points in \( X\left( {0,0}\right) \) are inertially ample. | Proof: Let \( x \in X\left( {0,0}\right) \) and let \( {V}_{x} \) denote the associated \( {G}_{{\mathbb{Q}}_{p}} \) -representation as in the discussion at the beginning of this section. By the theorem of Sen already quoted, \( {V}_{x} \) has semi-simple \( p \) -adic Hodge type if and only if the action of \( {I}_{p}... | Yes |
Proposition 1.11 Let \( S \) be a nonempty subset of a group \( G \) . If\n\nS1: \( a, b \in S \Rightarrow {ab} \in S \), and\n\nS2: \( a \in S \Rightarrow {a}^{-1} \in S \) ,\n\nthen the binary operation on \( G \) makes \( S \) into a group. | Proof. (S1) implies that the binary operation on \( G \) defines a binary operation \( S \times S \rightarrow S \) on \( S \), which is automatically associative. By assumption \( S \) contains at least one element \( a \) , its inverse \( {a}^{-1} \), and the product \( e = a{a}^{-1} \) . Finally (S2) shows that the i... | Yes |
Proposition 1.13 An intersection of subgroups of \( G \) is a subgroup of \( G \) . | Proof. It is nonempty because it contains \( e \), and (S1) and (S2) obviously hold. | No |
Proposition 1.15 For any subset \( X \) of a group \( G \), there is a smallest subgroup of \( G \) containing \( X \) . It consists of all finite products of elements of \( X \) and their inverses (repetitions allowed). | Proof. The intersection \( S \) of all subgroups of \( G \) containing \( X \) is again a subgroup containing \( X \), and it is evidently the smallest such group. Clearly \( S \) contains with \( X \), all finite products of elements of \( X \) and their inverses. But the set of such products satisfies (S1) and (S2) a... | Yes |
Proposition 1.25 Let \( H \) be a subgroup of a group \( G \) .\n\n(a) An element \( a \) of \( G \) lies in a left coset \( C \) of \( H \) if and only if \( C = {aH} \) .\n\n(b) Two left cosets are either disjoint or equal.\n\n(c) \( {aH} = {bH} \) if and only if \( {a}^{-1}b \in H \) .\n\n(d) Any two left cosets hav... | Proof. (a) Certainly \( a \in {aH} \) . Conversely, if \( a \) lies in the left coset \( {bH} \), then \( a = {bh} \) for some \( h \), and so\n\n\[ \n{aH} = {bhH} = {bH}.\n\]\n\n(b) If \( C \) and \( {C}^{\prime } \) are not disjoint, then they have a common element \( a \), and \( C = {aH} \) and \( {C}^{\prime } = {... | Yes |
Proposition 1.31 For any subgroups \( H \supset K \) of \( G \) , \[ \left( {G : K}\right) = \left( {G : H}\right) \left( {H : K}\right) \] (meaning either both are infinite or both are finite and equal). | Proof. Write \( G = \mathop{\bigsqcup }\limits_{{i \in I}}{g}_{i}H \) (disjoint union), and \( H = \mathop{\bigsqcup }\limits_{{j \in J}}{h}_{j}K \) (disjoint union). On multiplying the second equality by \( {g}_{i} \), we find that \( {g}_{i}H = \mathop{\bigsqcup }\limits_{{j \in J}}{g}_{i}{h}_{j}K \) (disjoint union)... | Yes |
Proposition 1.34 A subgroup \( N \) of \( G \) is normal if and only if every left coset of \( N \) in \( G \) is also a right coset, in which case, \( {gN} = {Ng} \) for all \( g \in G \) . | Proof. Clearly, \[ {gN}{g}^{-1} = N \Leftrightarrow {gN} = {Ng}. \] Thus, if \( N \) is normal, then every left coset is a right coset (in fact, \( {gN} = {Ng} \) ). Conversely, if the left coset \( {gN} \) is also a right coset, then it must be the right coset \( {Ng} \) by (1.25a). Hence \( {gN} = {Ng} \), and so \( ... | Yes |
Proposition 1.37 If \( H \) and \( N \) are subgroups of \( G \) and \( N \) is normal, then \( {HN} \) is a subgroup of \( G \) . If \( H \) is also normal, then \( {HN} \) is a normal subgroup of \( G \) . | Proof. The set \( {HN} \) is nonempty, and\n\n\[ \left( {{h}_{1}{n}_{1}}\right) \left( {{h}_{2}{n}_{2}}\right) \overset{1.35}{ = }{h}_{1}{h}_{2}{n}_{1}^{\prime }{n}_{2} \in {HN}, \]\n\nand so it is closed under multiplication. Since\n\n\[ {\left( hn\right) }^{-1} = {n}^{-1}{h}^{-1}\overset{1.35}{ = }{h}^{-1}{n}^{\prime... | Yes |
Proposition 1.41 The kernel of a homomorphism is a normal subgroup. | Proof. It is obviously a subgroup, and if \( a \in \operatorname{Ker}\left( \alpha \right) \), so that \( \alpha \left( a\right) = e \), and \( g \in G \), then\n\n\[ \alpha \left( {{ga}{g}^{-1}}\right) = \alpha \left( g\right) \alpha \left( a\right) \alpha {\left( g\right) }^{-1} = \alpha \left( g\right) \alpha {\left... | Yes |
Proposition 1.42 Every normal subgroup occurs as the kernel of a homomorphism. More precisely, if \( N \) is a normal subgroup of \( G \), then there is a unique group structure on the set \( G/N \) of cosets of \( N \) in \( G \) for which the natural map \( a \mapsto \left\lbrack a\right\rbrack : G \rightarrow G/N \)... | Proof. Write the cosets as left cosets, and define \( \left( {aN}\right) \left( {bN}\right) = \left( {ab}\right) N \) . We have to check (a) that this is well-defined, and (b) that it gives a group structure on the set of cosets. It will then be obvious that the map \( g \mapsto {gN} \) is a homomorphism with kernel \(... | Yes |
Proposition 1.43 The map \( a \mapsto {aN} : G \rightarrow G/N \) has the following universal property: for any homomorphism \( \alpha : G \rightarrow {G}^{\prime } \) of groups such that \( \alpha \left( N\right) = \{ e\} \), there exists a unique homomorphism \( G/N \rightarrow {G}^{\prime } \) making the diagram at ... | Proof. Note that for \( n \in N,\alpha \left( {gn}\right) = \alpha \left( g\right) \alpha \left( n\right) = \alpha \left( g\right) \), and so \( \alpha \) is constant on each left coset \( {gN} \) of \( N \) in \( G \) . It therefore defines a map\n\n\[ \bar{\alpha } : G/N \rightarrow {G}^{\prime },\;\bar{\alpha }\left... | Yes |
Proposition 1.50 A group \( G \) is a direct product of subgroups \( {H}_{1},{H}_{2} \) if and only if\n\n(a) \( G = {H}_{1}{H}_{2} \) ,\n\n(b) \( {H}_{1} \cap {H}_{2} = \{ e\} \), and\n\n(c) every element of \( {H}_{1} \) commutes with every element of \( {H}_{2} \) . | Proof. If \( G \) is the direct product of \( {H}_{1} \) and \( {H}_{2} \), then certainly (a) and (c) hold, and (b) holds because, for any \( g \in {H}_{1} \cap {H}_{2} \), the element \( \left( {g,{g}^{-1}}\right) \) maps to \( e \) under \( \left( {{h}_{1},{h}_{2}}\right) \mapsto {h}_{1}{h}_{2} \) and so equals \( \... | Yes |
Proposition 1.51 A group \( G \) is a direct product of subgroups \( {H}_{1},{H}_{2} \) if and only if\n\n(a) \( G = {H}_{1}{H}_{2} \) ,\n\n(b) \( {H}_{1} \cap {H}_{2} = \{ e\} \), and\n\n(c) \( {\mathrm{H}}_{1} \) and \( {\mathrm{H}}_{2} \) are both normal in \( \mathrm{G} \) . | Proof. Certainly, these conditions are implied by those in the previous proposition, and so it remains to show that they imply that each element \( {h}_{1} \) of \( {H}_{1} \) commutes with each element \( {h}_{2} \) of \( {H}_{2} \) . Two elements \( {h}_{1},{h}_{2} \) of a group commute if or only if their commutator... | Yes |
Proposition 1.52 A group \( G \) is a direct product of subgroups \( {H}_{1},{H}_{2},\ldots ,{H}_{k} \) if and only if\n\n(a) \( G = {H}_{1}{H}_{2}\cdots {H}_{k} \),\n\n(b) for each \( j,{H}_{j} \cap \left( {{H}_{1}\cdots {H}_{j - 1}{H}_{j + 1}\cdots {H}_{k}}\right) = \{ e\} \), and\n\n(c) each of \( {H}_{1},{H}_{2},\l... | Proof. The necessity of the conditions being obvious, we shall prove only the sufficiency. For \( k = 2 \), we have just done this, and so we argue by induction on \( k \) . An induction argument using (1.37) shows that \( {H}_{1}\cdots {H}_{k - 1} \) is a normal subgroup of \( G \) . The conditions (a, b, c) hold for ... | Yes |
Proposition 2.2 Products of equivalent words are equivalent, i.e.,\n\n\[ w \sim {w}^{\prime },\;v \sim {v}^{\prime } \Rightarrow {wv} \sim {w}^{\prime }{v}^{\prime }. \] | Proof. Let \( {w}_{0} \) and \( {v}_{0} \) be the reduced forms of \( w \) and of \( v \) . To obtain the reduced form of \( {wv} \), we can first cancel as much as possible in \( w \) and \( v \) separately, to obtain \( {w}_{0}{v}_{0} \) and then continue cancelling. Thus the reduced form of \( {wv} \) is the reduced... | Yes |
Proposition 2.3 For any map of sets \( \alpha : X \rightarrow G \) from \( X \) to a group \( G \), there exists a unique homomorphism \( {FX} \rightarrow G \) making the following diagram commute: | Proof. Consider a map \( \alpha : X \rightarrow G \) . We extend it to a map of sets \( {X}^{\prime } \rightarrow G \) by setting \( \alpha \left( {a}^{-1}\right) = \alpha {\left( a\right) }^{-1} \) . Because \( G \) is, in particular, a monoid, \( \alpha \) extends to a homomorphism of monoids \( S{X}^{\prime } \right... | Yes |
Proposition 2.8 Let \( G \) be the group defined by the presentation \( \left( {X, R}\right) \) . For any group \( H \) and map of sets \( \alpha : X \rightarrow H \) sending each element of \( R \) to 1 (in the obvious sense \( {}^{4} \) ), there exists a unique homomorphism \( G \rightarrow H \) making the following ... | Proof. From the universal property of free groups (2.3), we know that \( \alpha \) extends to a homomorphism \( {FX} \rightarrow H \), which we again denote \( \alpha \) . Let \( {\iota R} \) be the image of \( R \) in \( {FX} \) . By assumption \( {\iota R} \subset \operatorname{Ker}\left( \alpha \right) \), and there... | Yes |
Proposition 3.10 The composition law above makes \( G \) into a group, in fact, the semidirect product of \( N \) and \( Q \) . | Proof. Write \( {}^{q}n \) for \( \theta \left( q\right) \left( n\right) \), so that the composition law becomes\n\n\[ \left( {n, q}\right) \left( {{n}^{\prime },{q}^{\prime }}\right) = \left( {n \cdot {}^{q}{n}^{\prime }, q{q}^{\prime }}\right) . \]\n\nThen\n\n\[ \left( {\left( {n, q}\right) ,\left( {{n}^{\prime },{q}... | Yes |
Proposition 3.22 An extension (16) splits if \( N \) is complete. In fact, \( G \) is then the direct product of \( N \) with the centralizer of \( N \) in \( G \), \[ {C}_{G}\left( N\right) \overset{\text{ def }}{ = }\{ g \in G \mid {gn} = {ng}\text{ all }n \in N\} . \] | Proof. Let \( H = {C}_{G}\left( N\right) \) . We shall check that \( N \) and \( H \) satisfy the conditions of Proposition 1.51.\n\nObserve first that, for any \( g \in G, n \mapsto {gn}{g}^{-1} : N \rightarrow N \) is an automorphism of \( N \), and (because \( N \) is complete), it must be the inner automorphism def... | Yes |
Proposition 4.7 If \( G \) acts transitively on \( X \), then for any \( {x}_{0} \in X \), the map\n\n\[ g\operatorname{Stab}\left( {x}_{0}\right) \mapsto g{x}_{0} : G/\operatorname{Stab}\left( {x}_{0}\right) \rightarrow X \]\n\nis an isomorphism of \( G \) -sets. | Proof. It is well-defined because, if \( h \in \operatorname{Stab}\left( {x}_{0}\right) \), then \( {gh}{x}_{0} = g{x}_{0} \) . It is injective because\n\n\( g{x}_{0} = {g}^{\prime }{x}_{0} \Rightarrow {g}^{-1}{g}^{\prime }{x}_{0} = {x}_{0} \Rightarrow g,{g}^{\prime } \) lie in the same left coset of \( \operatorname{S... | Yes |
Proposition 4.9 Let \( {x}_{0} \in X \) . If \( G \) acts transitively on \( X \), then\n\n\[ \n\operatorname{Ker}\left( {G \rightarrow \operatorname{Sym}\left( X\right) }\right)\n\]\n\nis the largest normal subgroup contained in \( \operatorname{Stab}\left( {x}_{0}\right) \) . | Proof. When\n\n\[ \n\operatorname{Ker}\left( {G \rightarrow \operatorname{Sym}\left( X\right) }\right) = \mathop{\bigcap }\limits_{{x \in X}}\operatorname{Stab}\left( x\right) = \mathop{\bigcap }\limits_{{g \in G}}\operatorname{Stab}\left( {g{x}_{0}}\right) \overset{\left( {4.4}\right) }{ = }\bigcap g \cdot \operatorna... | Yes |
Proposition 4.18 Every group of order \( {p}^{2} \) is commutative, and hence is isomorphic to \( {C}_{p} \times {C}_{p} \) or \( {C}_{{p}^{2}} \) . | Proof. We know that the centre \( Z \) is nontrivial, and that \( G/Z \) therefore has order 1 or \( p \) . In either case it is cyclic, and the next result implies that \( G \) is commutative. | No |
Every permutation can be written (in essentially one way) as a product of disjoint cycles. | Let \( \sigma \in {S}_{n} \), and let \( O \subset \{ 1,2,\ldots, n\} \) be an orbit for \( \langle \sigma \rangle \) . If \( \left| O\right| = r \), then for any \( i \in O \) ,\n\n\[ O = \left\{ {i,\sigma \left( i\right) ,\ldots ,{\sigma }^{r - 1}\left( i\right) }\right\} \]\n\nTherefore \( \sigma \) and the cycle \(... | Yes |
Proposition 4.30 Two elements \( \sigma \) and \( \tau \) of \( {S}_{n} \) are conjugate if and only if they define the same partitions of \( n \) . | Proof. \( \Rightarrow \) : We saw in (4.29) that conjugating an element preserves the type of its disjoint cycle decomposition.\n\n\( \leftarrow \) : Since \( \sigma \) and \( \tau \) define the same partitions of \( n \), their decompositions into products of disjoint cycles have the same type:\n\n\[ \sigma = \left( {... | Yes |
Proposition 4.43 The group \( G \) acts imprimitively if and only if there is a proper subset \( A \) of \( X \) with at least 2 elements such that,\n\n\[ \n\\text{for each}g \\in G\\text{, either}{gA} = A\\text{or}{gA} \\cap A = \\varnothing \\text{.}\n\]\n\n(25) | Proof. \( \\Rightarrow \) : The partition \( \\pi \) stabilized by \( G \) contains such an \( A \) .\n\n\( \\Leftarrow \) : From such an \( A \), we can form a partition \( \\left\{ {A,{g}_{1}A,{g}_{2}A,\\ldots }\\right\} \) of \( X \), which is stabilized by \( G \) . | No |
Proposition 4.44 Let \( A \) be a block in \( X \) with \( \\left| A\\right| \\geq 2 \) and \( A \\neq X \) . For any \( x \\in A \) , | \[ \\operatorname{Stab}\\left( x\\right) \\subsetneqq \\operatorname{Stab}\\left( A\\right) \\subsetneqq G. \] Proof. We have \( \\operatorname{Stab}\\left( A\\right) \\supset \\operatorname{Stab}\\left( x\\right) \) because \[ {gx} = x \\Rightarrow {gA} \\cap A \\neq \\varnothing \\Rightarrow {gA} = A. \] Let \( y \\i... | Yes |
Proposition 6.6 (a) Every subgroup and every quotient group of a solvable group is solvable. | Proof. (a) Let \( G \vartriangleright {G}_{1} \vartriangleright \cdots \vartriangleright {G}_{n} \) be a solvable series for \( G \), and let \( H \) be a subgroup of \( G \) . The homomorphism\n\n\[ x \mapsto x{G}_{i + 1} : H \cap {G}_{i} \rightarrow {G}_{i}/{G}_{i + 1} \]\n\nhas kernel \( \left( {H \cap {G}_{i}}\righ... | Yes |
The commutator subgroup \( {G}^{\prime } \) is a characteristic subgroup of \( G \) ; it is the smallest normal subgroup of \( G \) such that \( G/{G}^{\prime } \) is commutative. | An automorphism \( \alpha \) of \( G \) maps the generating set for \( {G}^{\prime } \) into \( {G}^{\prime } \), and hence maps \( {G}^{\prime } \) into \( {G}^{\prime } \) . Since this is true for all automorphisms of \( G,{G}^{\prime } \) is characteristic.\n\nWrite \( g \mapsto \bar{g} \) for the homomorphism \( g ... | Yes |
Proposition 6.10 A group \( G \) is solvable if and only if its \( k \) th derived subgroup \( {G}^{\left( k\right) } = 1 \) for some \( k \) . | Proof. If \( {G}^{\left( k\right) } = 1 \), then the derived series is a solvable series for \( G \) . Conversely, let\n\n\[ G = {G}_{0} \vartriangleright {G}_{1} \vartriangleright {G}_{2} \vartriangleright \cdots \vartriangleright {G}_{s} = 1 \]\n\nbe a solvable series for \( G \) . Because \( G/{G}_{1} \) is commutat... | Yes |
Proposition 6.13 (a) A subgroup of a nilpotent group is nilpotent. | Proof. (a) Let \( H \) be a subgroup of a nilpotent group \( G \) . Clearly, \( Z\left( H\right) \supset Z\left( G\right) \cap H \) . Assume (inductively) that \( {Z}^{i}\left( H\right) \supset {Z}^{i}\left( G\right) \cap H \) ; then \( {Z}^{i + 1}\left( H\right) \supset {Z}^{i + 1}\left( G\right) \cap H \), because (f... | Yes |
Proposition 6.15 A group \( G \) is nilpotent of class \( \leq m \) if and only if\n\n\[ \left\lbrack {\ldots \left\lbrack {\left\lbrack {{g}_{1},{g}_{2}}\right\rbrack ,{g}_{3}}\right\rbrack ,\ldots ,,{g}_{m + 1}}\right\rbrack = 1 \]\n\nfor all \( {g}_{1},\ldots ,{g}_{m + 1} \in G \) . | Proof. Recall, \( g \in {Z}^{i}\left( G\right) \Leftrightarrow \left\lbrack {g, x}\right\rbrack \in {Z}^{i - 1}\left( G\right) \) for all \( x \in G \).\n\nAssume \( G \) is nilpotent of class \( \leq m \) ; then\n\n\[ G = {Z}^{m}\left( G\right) \Rightarrow \left\lbrack {{g}_{1},{g}_{2}}\right\rbrack \in {Z}^{m - 1}\le... | Yes |
Proposition 6.22 (Frattini’s Argument) Let \( H \) be a normal subgroup of a finite group \( G \), and let \( P \) be a Sylow \( p \) -subgroup of \( H \). Then \( G = H \cdot {N}_{G}\left( P\right) \). | Proof. Let \( g \in G \). Then \( {gP}{g}^{-1} \subset {gH}{g}^{-1} = H \), and both \( {gP}{g}^{-1} \) and \( P \) are Sylow \( p \) -subgroups of \( H \). According to Sylow II, there is an \( h \in H \) such that \( {gP}{g}^{-1} = {hP}{h}^{-1} \), and it follows that \( {h}^{-1}g \in {N}_{G}\left( P\right) \) and so... | Yes |
Proposition 7.9 If the characteristic of \( F \) does not divide \( \left| G\right| \), then every \( F\left\lbrack G\right\rbrack \) -module is a direct sum of simple submodules. | Proof. Let \( V \) be a \( F\left\lbrack G\right\rbrack \) -module. If \( V \) is simple, then there is nothing to prove. Otherwise, it contains a nonzero proper submodule \( W \) . According to Maschke’s theorem, \( V = W \oplus {W}^{\prime } \) with \( {W}^{\prime } \) an \( F\left\lbrack G\right\rbrack \) -submodule... | Yes |
Proposition 7.12 Let \( V \) be an \( A \) -module. If \( V \) is a sum of simple submodules, say \( V = \mathop{\sum }\limits_{{i \in I}}{S}_{i} \) (the sum need not be direct), then for any submodule \( W \) of \( V \), there is a subset \( J \) of \( I \) such that | Proof. Let \( J \) be maximal among the subsets of \( I \) such the sum \( {S}_{J}\overset{\text{ def }}{ = }\mathop{\sum }\limits_{{j \in J}}{S}_{j} \) is direct and \( W \cap {S}_{J} = 0 \) . I claim that \( W + {S}_{J} = V \) (hence \( V \) is the direct sum of \( W \) and the \( {S}_{j} \) with \( j \in J \) ). For... | Yes |
Proposition 7.15 Let \( V \) be a semisimple \( A \) -module. A submodule of \( V \) is stable under all endomorphisms of \( V \) if and only if it is a sum of isotypic components of \( V \) . | Proof. The sufficiency follows from the above statement. For the necessity, let \( W \) be a submodule of \( V \) stable under all endomorphisms of \( V \), and let \( S \) be a simple submodule of \( W \) . If \( {S}^{\prime } \) is a submodule of \( V \) isomorphic to \( S \), then the endomorphism\n\n\[ \nV\xrightar... | Yes |
Proposition 7.17 Let \( A \) be a semisimple \( F \) -algebra. The isotypic components of the \( A \) -module \( {}_{A}A \) are the minimal two-sided ideals of \( A \) . Every two-sided ideal of \( A \) is a direct sum of minimal two-sided ideals. | Proof. The two-sided ideals of \( A \) are the submodules of \( {AA} \) stable under right multiplication by the elements of \( A \), i.e., by the endomorphisms of \( {}_{A}A\left( {7.16}\right) \), and so they are the sums of isotypic components of \( {}_{A}A \) (7.15). In particular, the minimal two-sided ideals are ... | Yes |
Proposition 7.26 Every simple \( F \) -algebra \( A \) is semisimple. | Proof. It suffices to show that the \( A \) -module \( {}_{A}A \) is semisimple. After Theorem 7.25, we may assume that \( A = {M}_{n}\left( D\right) \) for some division algebra \( D \) . We saw in 7.19 that the sets \( L\left( i\right) \) are minimal left ideals in \( {M}_{n}\left( D\right) \), and that \( {M}_{n}\le... | Yes |
Proposition 7.31 The only division algebra over an algebraically closed field \( F \) is \( F \) itself. | Proof. Let \( D \) be division algebra over \( F \) . For any element \( \alpha \) of \( D \), the \( F \) -subalgebra \( F\left\lbrack \alpha \right\rbrack \) of \( D \) generated by \( \alpha \) is a field because it is an integral domain of finite degree over \( F \) . As \( F \) is algebraically closed, \( \alpha \... | Yes |
The dimension of the centre of \( F\left\lbrack G\right\rbrack \) as an \( F \) -vector space is the number of conjugacy classes in \( G \) . | Let \( {C}_{1},\ldots ,{C}_{t} \) be the conjugacy classes in \( G \), and, for each \( i \), let \( {c}_{i} \) be the element \( \mathop{\sum }\limits_{{a \in {C}_{i}}}a \) in \( F\left\lbrack G\right\rbrack \) . We shall prove the stronger statement,\n\n\[ \text{centre of}F\left\lbrack G\right\rbrack = F{c}_{1} \oplu... | Yes |
Proposition 7.40 The group algebra \( F\left\lbrack G\right\rbrack \) is isomorphic to a product of matrix algebras over \( F \) . | Proof. Recall that, when \( F \) has characteristic zero, Maschke’s theorem (7.9) implies that \( F\left\lbrack G\right\rbrack \) is semisimple, and so is a product of simple algebras (7.36). Each of these is a matrix algebra over a division algebra (7.25), but the only division algebra over an algebraically closed fie... | Yes |
Proposition 7.43 The functions \( {\chi }_{1},\ldots ,{\chi }_{t} \) are linearly independent over \( F \), i.e., if \( {c}_{1},\ldots ,{c}_{t} \in F \) are such that \( \mathop{\sum }\limits_{i}{c}_{i}{\chi }_{i}\left( g\right) = 0 \) for all \( g \in G \), then the \( {c}_{i} \) are all zero. | Proof. Write \( F\left\lbrack G\right\rbrack \approx {M}_{{f}_{1}}\left( F\right) \times \cdots \times {M}_{{f}_{t}}\left( F\right) \), and let \( {e}_{j} = \left( {0,\ldots ,0,1,0,\ldots ,0}\right) \) . Then \( {e}_{j} \) acts as 1 on \( {S}_{j} \) and as 0 on \( {S}_{i} \) for \( i \neq j \), and so\n\n\[ \n{\chi }_{... | Yes |
Proposition 7.44 Two \( F\left\lbrack G\right\rbrack \) -modules are isomorphic if and only if their characters are equal. | Proof. We have already observed that the character of a representation depends only on its isomorphism class. Conversely, if \( V = {\bigoplus }_{1 \leq i \leq t}{c}_{i}{S}_{i},{c}_{i} \in \mathbb{N} \), then its character is \( {\chi }_{V} = \mathop{\sum }\limits_{{1 \leq i \leq t}}{c}_{i}{\chi }_{i} \), and (34) show... | Yes |
Proposition 7.46 The simple characters of \( G \) form a \( \mathbb{Z} \) -basis for the virtual characters of \( G \) . | Proof. Let \( {\chi }_{1},\ldots ,{\chi }_{t} \) be the simple characters of \( G \) . Then the characters of \( G \) are exactly the class functions that can be expressed in the form \( \sum {m}_{i}{\chi }_{i},{m}_{i} \in \mathbb{N} \), and so the virtual characters are exactly the class functions that can be expresse... | Yes |
Proposition 7.47 The simple characters of \( G \) form an \( F \) -basis for the class functions on \( G \) . | Proof. The class functions are the functions from the set of conjugacy classes in \( G \) to \( F \) . As this set has \( t \) elements, they form an \( F \) -vector space of dimension \( t \) . As the simple characters are a set of \( t \) linearly independent elements of this vector space, they must form a basis. | Yes |
Proposition 7.50 For any \( F\left\lbrack G\right\rbrack \) -module \( V \) ,\n\n\[ \n{\dim }_{F}{V}^{G} = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{a \in G}}{\chi }_{V}\left( a\right) \n\] | Proof. Let \( \pi \) be as in Lemma 7.49. Because \( {\pi }_{V} \) is a projector, \( V \) is the direct sum of its 0-eigenspace and its 1-eigenspace, and we showed that the latter is \( {V}^{G} \) . Therefore, \( {\operatorname{Tr}}_{V}\left( {\pi }_{V}\right) = {\dim }_{F}{V}^{G} \) . On the other hand, because the t... | Yes |
Corollary 1.4. If \( \mathfrak{a} \neq \left( 1\right) \) is an ideal of \( A \), there exists a maximal ideal of \( A \) containing a. | proof of (1.3). | No |
Corollary 1.5. Every non-unit of \( A \) is contained in a maximal ideal. | ∎ lemma: the set of all ideals \( \neq \left( 1\right) \) has a maximal element. | No |
Let \( A \) be a ring and \( \mathfrak{m} \) a maximal ideal of \( A \), such that every element of \( 1 + \mathfrak{m} \) (i.e., every \( 1 + x \), where \( x \in \mathfrak{m} \) ) is a unit in \( A \) . Then \( A \) is a local ring. Hence \( \mathfrak{m} \) is the only maximal ideal of \( A \) . | Let \( x \in A - m \) . Since \( m \) is maximal, the ideal generated by \( x \) and \( m \) is ), hence there exist \( y \in A \) and \( t \in \mathfrak{m} \) such that \( {xy} + t = 1 \) ; hence \( {xy} = 1 - t \) belongs to \( 1 + m \) and therefore is a unit. Now use i). - | No |
Proposition 1.7. The set \( \mathfrak{R} \) of all nilpotent elements in a ring \( A \) is an ideal. | Proof. If \( x \in \mathfrak{N} \), clearly \( {ax} \in \mathfrak{N} \) for all \( a \in A \) . Let \( x, y \in \mathfrak{N} \) : say \( {x}^{m} = 0,{y}^{n} = 0 \) . By the binomial theorem (which is valid in any commutative ring), \( {\left( x + y\right) }^{m + n - 1} \) is a sum of integer multiples of products \( {x... | Yes |
Proposition 1.9. \( x \in \Re \Leftrightarrow 1 - {xy} \) is a unit in \( A \) for all \( y \in A \). | \( \Leftarrow \) : Suppose \( x \notin \mathfrak{m} \) for some maximal ideal \( \mathfrak{m} \) . Then \( \mathfrak{m} \) and \( x \) generate the Hence \( 1 - {xy} \in \mathfrak{m} \) and is therefore not a unit. | No |
Proposition 1.10. i) If \( {\mathfrak{a}}_{\mathfrak{i}},\mathfrak{a} \), are coprime whenever \( i \neq j \), then \( \Pi {\mathfrak{a}}_{\mathfrak{i}} = \bigcap {\mathfrak{a}}_{\mathfrak{i}} \) . | Proof. i) by induction on \( n \) . The case \( n = 2 \) is dealt with above. Suppose \( n > 2 \) and the result true for \( {a}_{1},\ldots ,{a}_{n - 1} \), and let \( \mathfrak{b} = \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}{a}_{i} = \mathop{\bigcap }\limits_{{i = 1}}^{{n - 1}}{a}_{i} \) . Since \( {a}_{i} + {a}_{n} =... | Yes |
Let \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{n} \) be prime ideals and let \( \mathfrak{a} \) be an ideal | \[ a \neq {\mathfrak{p}}_{1}\left( {1 \leq i \leq n}\right) \Rightarrow a \neq \mathop{\bigcup }\limits_{{i = 1}}^{n}{\mathfrak{p}}_{i}. \] each \( i \) there exists \( {x}_{i} \in a \) such that \( {x}_{i} \notin {\mathfrak{p}}_{j} \) whenever \( j \neq i \) . If for some \( i \) we have \( {x}_{i} \notin {\mathfrak{p... | Yes |
Proposition 1.14. The radical of an ideal \( \mathfrak{a} \) is the intersection of the prime | Proof. Apply (1.8) to \( A/\mathfrak{a} \) . ∎ | No |
Proposition 1.15. \( D = \) set of zero-divisors of \( A = \mathop{\bigcup }\limits_{{x \neq 0}}r\left( {\operatorname{Ann}\left( x\right) }\right) \) . | Proof. \( D = r\left( D\right) = r\left( {\mathop{\bigcup }\limits_{{x \neq 0}}\operatorname{Ann}\left( x\right) }\right) = \mathop{\bigcup }\limits_{{x \neq 0}}r\left( {\operatorname{Ann}\left( x\right) }\right) \) . ∎ | Yes |
Proposition 1.17. i) \( \mathfrak{a} \subseteq {\mathfrak{a}}^{ec},\mathfrak{b} \supseteq {\mathfrak{b}}^{ce} \) ;\n\nii) \( {\mathfrak{b}}^{c} = {\mathfrak{b}}^{cec},{\mathfrak{a}}^{e} = {\mathfrak{a}}^{ece} \) ;\n\niii) If \( C \) is the set of contracted ideals in \( A \) and if \( E \) is the set of extended ideals... | Proof. i) is trivial, and ii) follows from i).\n\niii) If \( a \in C \), then \( a = {b}^{c} \cdot = {b}^{cec} = {a}^{ec} \) ; conversely if \( a = {a}^{ec} \) then \( a \) is the contraction of \( {\alpha }^{e} \) . Similarly for \( E \) . ∎ | No |
Proposition 2.3. \( M \) is a finitely generated \( A \) -module \( \Leftrightarrow M \) is isomorphic to a quotient of \( {A}^{n} \) for some integer \( n > 0 \) . | Proof. \( \Rightarrow : \) Let \( {x}_{1},\ldots ,{x}_{n} \) generate \( M \) . Define \( \phi : {A}^{n} \rightarrow M \) by \( \phi \left( {{a}_{1},\ldots ,{a}_{n}}\right) = \) \( {a}_{1}{x}_{1} + \cdots + {a}_{n}{x}_{n} \) . Then \( \phi \) is an \( A \) -module homomorphism onto \( M \), and there-\n\n\( \Leftarrow ... | Yes |
Proposition 2.4. Let \( M \) be a finitely generated \( A \) -module, let \( \mathfrak{a} \) be an ideal of \( A \), and let \( \phi \) be an \( A \) -module endomorphism of \( M \) such that \( \phi \left( M\right) \subseteq \mathfrak{a}M \) . Then\n\n\[{\phi }^{n} + {a}_{1}{\phi }^{n - 1} + \cdots + {a}_{n} = 0\] | Proof. Let \( {x}_{1},\ldots ,{x}_{n} \) be a set of generators of \( M \) . Then each \( \phi \left( {x}_{i}\right) \in {aM} \), so that we have say \( \phi \left( {x}_{i}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j}\left( {1 \leq i \leq n;{a}_{ij} \in a}\right) \), i.e.,\n\n\[ \mathop{\sum }\limits_{{j... | Yes |
Proposition 2.6. (Nakayama’s lemma). Let \( M \) be a finitely generated \( A \) -module and a an ideal of \( A \) contained in the Jacobson radical \( \Re \) of \( A \) . Then | First Proof. By (2.5) we have \( {xM} = 0 \) for some \( x \equiv 1\left( {\;\operatorname{mod}\;\Re }\right) \) . By (1.9) \( x \) is a unit in \( A \), hence \( M = {x}^{-1}{xM} = 0 \) . ∎ | No |
Corollary 2.7. Let \( M \) be a finitely generated \( A \) -module, \( N \) a submodule of \( M \). | Proof. Apply (2.6) to \( M/N \), observing that \( \mathfrak{a}\left( {M/N}\right) = \left( {\mathfrak{a}M + N}\right) /N \) . ∎ | No |
Proposition 2.8. Let \( {x}_{i}\left( {1 \leq i \leq n}\right) \) be elements of \( M \) whose images in | Proof. Let \( N \) be the submodule of \( M \) generated by the \( {x}_{i} \) . Then the composite map \( N \rightarrow M \rightarrow M/\mathfrak{m}M \) maps \( N \) onto \( M/\mathfrak{m}M \), hence \( N + \mathfrak{m}M = M \), hence | No |
Proposition 2.10. Let\n\n\\[ \n0 \rightarrow {M}^{\prime }\xrightarrow[]{u}M\xrightarrow[]{v}{M}^{\prime \prime } \rightarrow 0 \n\\]\n\n\\[ \n\begin{matrix} r \downarrow \;s \downarrow \\ 0 \rightarrow {N}^{\prime } \Rightarrow N \Rightarrow {N}^{\prime \prime } \rightarrow 0 \end{matrix} \n\\]\n\nbe a commutative dia... | The boundary homomorphism \\( d \\) is defined as follows: if \\( {x}^{n} \in \operatorname{Ker}\left( {f}^{n}\right) \\), we have \\( {x}^{\prime \prime } = v\\left( x\\right) \\) for some \\( x \in M \\), and \\( {v}^{\prime }\\left( {f\\left( x\\right) }\\right) = {f}^{\prime \prime }\\left( {v\\left( x\\right) }\\r... | No |
Proposition 2.11. Let \( 0 \rightarrow {M}_{0} \rightarrow {M}_{1} \rightarrow \cdots \rightarrow {M}_{n} \rightarrow 0 \) be an exact se-homomorphisms belong to \( C \) . Then for any additive function \( \lambda \) on \( C \) we have\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{n}{\left( -1\right) }^{i}\lambda \left( {M}_{... | Proof. Split up the sequence into short exact sequences\n\n\[ 0 \rightarrow {N}_{i} \rightarrow {M}_{i} \rightarrow {N}_{i + 1} \rightarrow 0 \]\n\n\( \left( {{N}_{0} = {N}_{n + 1} = 0}\right) \) . Then we have \( \lambda \left( {M}_{i}\right) = \lambda \left( {N}_{i}\right) + \lambda \left( {N}_{i + 1}\right) \) . Now... | Yes |
Proposition 2.12. Let \( M, N \) be \( A \) -modules. Then there exists a pair \( \left( {T, g}\right) \) consisting of an A-module \( T \) and an A-bilinear mapping \( g : M \times N \rightarrow T \), with the following property: Given any A-module \( P \) and any A-bilinear mapping \( f : M \times N \rightarrow P \),... | Proof. i) Uniqueness. Replacing \( \left( {P, f}\right) \) by \( \left( {{T}^{\prime },{g}^{\prime }}\right) \) we get a unique \( j : T \rightarrow {T}^{\prime } \) such that \( {g}^{\prime } = j \circ g \) . Interchanging the roles of \( T \) and \( {T}^{\prime } \), we get \( {j}^{\prime } : {T}^{\prime } \rightarro... | Yes |
Corollary 2.13. Let \( {x}_{i} \in M,{y}_{i} \in N \) be such that \( \sum {x}_{i} \otimes {y}_{i} = 0 \) in \( M \otimes N \) . \( \sum {x}_{i} \otimes {y}_{i} = 0 \) in \( {M}_{0} \otimes {N}_{0} \) . | Proof. If \( \sum {x}_{i} \otimes {y}_{i} = 0 \) in \( M \otimes N \), then in the notation of the proof of (2.11) Let \( {M}_{0} \) be the submodule of \( M \) generated by the \( {x}_{i} \) and all the elements of \( M \) which occur as first coordinates in these generators of \( D \), and define \( {N}_{0} \) simila... | Yes |
Proposition 2.14. Let \( M, N, P \) be \( A \) -modules. Then there exist unique\ni) \( M \otimes N \rightarrow N \otimes M \)\nii) \( \left( {M \otimes N}\right) \otimes P \rightarrow M \otimes \left( {N \otimes P}\right) \rightarrow M \otimes N \otimes P \)\niii) \( \left( {M \oplus N}\right) \otimes P \rightarrow \l... | Proof. In each case the point is to show that the mappings so described are well defined. The technique is to construct suitable bilinear or multilinear mappings, and use the defining property (2.12) or \( \left( {2.12}^{ * }\right) \) to infer the existence of homomorphisms of tensor products. We shall prove half of i... | No |
Proposition 2.16. Suppose \( N \) is finitely generated as a B-module and that \( B \) is Proof. Let \( {y}_{1},\ldots ,{y}_{n} \) generate \( N \) over \( B \), and let \( {x}_{1},\ldots ,{x}_{m} \) generate \( B \) as an \( A \) -module. Then the \( {mn} \) products \( {x}_{i}{y}_{j} \) generate \( N \) over \( A \) ... | ∎ | No |
Proposition 2.17. If \( M \) is finitely generated as an \( A \) -module, then \( {M}_{B} \) is finitely generated as a B-module. | Proof. If \( {x}_{1},\ldots ,{x}_{m} \) generate \( M \) over \( A \), then the \( 1 \otimes {x}_{i} \) generate \( {M}_{B} \) over \( B \) . ∎ | Yes |
Proposition 2.18. Let\n\n\\[ \n{M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \n\\]\n\nbe an exact sequence of \\( A \\) -modules and homomorphisms, and let \\( N \\) be any\n\n\\[ \n{M}^{\prime } \otimes N\\xrightarrow[]{f \otimes 1}M \otimes N\\xrightarrow[]{g \otimes 1}{M}^{\prime \prime ... | Let \\( P \\) be any \\( A \\) -module. Since (2) is exact, the sequence Hom \\( \\left( {E,\\operatorname{Hom}\\left( {N, P}\\right) }\\right) \\) is exact by (2.9); hence by (1) the sequence Hom \\( \\left( {E \otimes N, P}\\right) \\) is exact. By (2.9) | No |
Proposition 3.1. Let \( g : A \rightarrow B \) be a ring homomorphism such that \( g\left( s\right) \) is a unit in \( B \) for all \( s \in S \) . Then there exists a unique ring homomorphism \( h : {S}^{-1}A \rightarrow B \) such that \( g = h \circ f \) . | Proof. i) Uniqueness. If \( h \) satisfies the conditions, then \( h\left( {a/1}\right) = {hf}\left( a\right) = g\left( a\right) \) for all \( a \in A \) ; hence, if \( s \in S \) , \[ h\left( {1/s}\right) = h\left( {\left( s/1\right) }^{-1}\right) = h{\left( s/1\right) }^{-1} = g{\left( s\right) }^{-1} \] and therefor... | Yes |
Proposition 3.3. The operation \( {S}^{-1} \) is exact, i.e., if \( {M}^{\prime }\xrightarrow[]{f}M\xrightarrow[]{g}{M}^{\prime \prime } \) is exact at \( M \), then \( {S}^{-1}{M}^{\prime }\xrightarrow[]{s - {1f}}{S}^{-1}M\xrightarrow[]{s - {1g}}{S}^{-1}{M}^{\prime \prime } \) is exact at \( {S}^{-1}M \) . | Proof. We have \( g \circ f = 0 \), hence \( {S}^{-1}g \circ {S}^{-1}f = {S}^{-1}\left( 0\right) = 0 \), hence \( \operatorname{Im}\left( {{S}^{-1}f}\right) \) \( \subseteq \operatorname{Ker}\left( {{S}^{-1}g}\right) \) . To prove the reverse inclusion, let \( m/s \in \operatorname{Ker}\left( {{S}^{-1}g}\right) \), the... | No |
Corollary 3.4. Formation of fractions commutes with formation of finite of an \( A \) -module \( M \), then\ni) \( {S}^{-1}\left( {N + P}\right) = {S}^{-1}\left( N\right) + {S}^{-1}\left( P\right) \)\niii) the \( {S}^{-1}A \) -modules \( {S}^{-1}\left( {M/N}\right) \) and \( \left( {{S}^{-1}M}\right) /\left( {{S}^{-1}N... | Proof. i) follows readily from the definitions and ii) is easy to verify: \( w = {uty} = {usz} \in N \cap P \) and therefore \( y/s = w/{stu} \in {S}^{-1}\left( {N \cap P}\right) \) . Consequently \( {S}^{-1}N \cap {S}^{-1}P \subseteq {S}^{-1}\left( {N \cap P}\right) \), and the reverse inclusion is obvious.\n\niii) Ap... | No |
Proposition 3.5. Let \( M \) be an \( A \) -module. Then the \( {S}^{-1}A \) modules \( {S}^{-1}M \) and \( {S}^{-1}A{ \otimes }_{A}M \) are isomorphic; more precisely, there exists a unique isomorphism \( f : {S}^{-1}A{ \otimes }_{A}M \rightarrow {S}^{-1}M \) for which \( f\left( {\left( {a/s}\right) \otimes m}\right)... | Proof. The mapping \( {S}^{-1}A \times M \rightarrow {S}^{-1}M \) defined by is \( A \) -bilinear, and therefore by the universal property (2.12) of the tensor product induces an \( A \) -homomorphism \( f : {S}^{-1}A{ \otimes }_{A}M \rightarrow {S}^{-1}M \) satisfying (1). Clearly \( f \) is surjective, and is uniquel... | Yes |
Proposition 3.7. If \( M, N \) are \( A \) -modules, there is a unique isomorphism of \( {S}^{-1}A \) -modules \( f : {S}^{-1}M{ \otimes }_{{S}^{-1}A}{S}^{-1}N \rightarrow {S}^{-1}\left( {M{ \otimes }_{A}N}\right) \) such that \[ f\left( {\left( {m/s}\right) \otimes \left( {n/t}\right) }\right) = \left( {m \otimes n}\r... | Proof. Use (3.5) and the canonical isomorphisms of Chapter 2. ∎ | No |
Proposition 3.10. For any A-module \( M \), the following statements are\ni) \( M \) is a flat \( A \) -module:\nii) \( {M}_{\mathfrak{p}} \) is a flat \( {A}_{\mathfrak{p}} \) -module for each prime ideal \( \mathfrak{p} \) ; | Proof. i) \( \Rightarrow \) ii) by (3.5) and (2.20).\n\nii) \( \Rightarrow \) iii) O.K. maximal ideal of \( A \), then\n\n\( N \rightarrow P \) injective \( \Rightarrow {N}_{\mathrm{m}} \rightarrow {P}_{\mathrm{m}} \) injective, by (3.9)\n\n\( \Rightarrow {N}_{\mathfrak{m}}{ \otimes }_{{A}_{\mathfrak{m}}}{M}_{\mathfrak... | No |
Proposition 3.11. i) Every ideal in \( {S}^{-1}A \) is an extended ideal. | i) Let \( \mathfrak{b} \) be an ideal in \( {S}^{-1}A \), and let \( x/s \in \mathfrak{b} \) . Then \( x/1 \in \mathfrak{b} \), hence \( x \in {\mathfrak{b}}^{c} \) and therefore \( x/s \in {\mathfrak{b}}^{ce} \) . Since \( \mathfrak{b} \supseteq {\mathfrak{b}}^{ce} \) in any case (1.17), it follows that | No |
Corollary 3.12. If \( \mathfrak{R} \) is the nilradical of \( A \), the nilradical of \( {S}^{-1}A \) is \( {S}^{-1}\mathfrak{N} \) . | ∎ | No |
Corollary 3.13. If \( \mathfrak{p} \) is a prime ideal of \( A \), the prime ideals of the local ring \( {A}_{\mathfrak{p}} \) are in one-to-one correspondence with the prime ideals of \( A \) contained in \( \mathfrak{p} \) . | Proof. Take \( S = A - \mathfrak{p} \) in (3.11) (iv). ∎ | No |
Proposition 3.14. Let \( M \) be a finitely generated \( A \) -module, \( S \) a multiplicatively closed subset of \( A \) . Then \( {S}^{-1}\left( {\operatorname{Ann}\left( M\right) }\right) = \operatorname{Ann}\left( {{S}^{-1}M}\right) \) . | Proof. If this is true for two \( A \) -modules, \( M, N \), it is true for \( M + N \) :\n\n\[ \n{S}^{-1}\left( {\operatorname{Ann}\left( {M + N}\right) }\right) = {S}^{-1}\left( {\operatorname{Ann}\left( M\right) \cap \operatorname{Ann}\left( N\right) }\right) \text{by (2.2)} \n\]\n\n\[ \n= {S}^{-1}\left( {\operatorn... | Yes |
Corollary 3.15. If \( N, P \) are submodules of an \( A \) -module \( M \) and if \( P \) is finitely generated, then \( {S}^{-1}\left( {N : P}\right) = \left( {{S}^{-1}N : {S}^{-1}P}\right) \) . | Proof. \( \left( {N : P}\right) = \operatorname{Ann}\left( {\left( {N + P}\right) /N}\right) \) by (2.2); now apply (3.14). ∎ | No |
Proposition 3.16. Let \( A \rightarrow B \) be a ring homomorphism and let \( \mathfrak{p} \) be a prime ideal of \( A \) . Then \( \mathfrak{p} \) is the contraction of a prime ideal of \( B \) if and only if \( {\mathfrak{p}}^{ec} = \mathfrak{p} \) . | Proof. If \( \mathfrak{p} = {\mathfrak{q}}^{c} \) then \( {\mathfrak{p}}^{ec} = \mathfrak{p} \) by (1.17). Conversely, if \( {\mathfrak{p}}^{ec} = \mathfrak{p} \), let \( S \) be the image of \( A - \mathfrak{p} \) in \( B \) . Then \( {\mathfrak{p}}^{e} \) does not meet \( S \), therefore by (3.11) its extension in \(... | Yes |
Proposition 4.1. Let \( \\mathfrak{q} \) be a primary ideal in a ring \( A \) . Then \( r\\left( \\mathfrak{q}\\right) \) is the smallest prime ideal containing \( \\mathfrak{q} \) . | Proof. By (1.8) it is enough to show that \( \\mathfrak{p} = r\\left( \\mathfrak{q}\\right) \) is prime. Let \( {xy} \\in r\\left( \\mathfrak{q}\\right) \), then \( {\\left( xy\\right) }^{m} \\in \\mathfrak{q} \) for some \( m > 0 \), and therefore either \( {x}^{m} \\in \\mathfrak{q} \) or \( {y}^{mn} \\in \\mathfrak{... | Yes |
Lemma 4.3. If \( {\mathfrak{q}}_{i}\left( {1 \leq i \leq n}\right) \) are \( \mathfrak{p} \) -primary, then \( \mathfrak{q} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) is \( \mathfrak{p} \) -primary. | Proof. \( r\left( \mathfrak{q}\right) = r\left( {\mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i}}\right) = \bigcap r\left( {\mathfrak{q}}_{i}\right) = \mathfrak{p} \) . Let \( {xy} \in \mathfrak{q}, y \notin \mathfrak{q} \) . Then for some \( i \) | No |
Lemma 4.4. Let \( \mathfrak{q} \) be a \( \mathfrak{p} \) -primary ideal, \( x \) an element of \( A \) . Then\ni) if \( x \in \mathfrak{q} \) then \( \left( {\mathfrak{q} : x}\right) = \left( 1\right) \) ;\niii) if \( x \notin \mathfrak{p} \) then \( \left( {\mathfrak{q} : x}\right) = \mathfrak{q} \) . | Proof. i) and iii) follow immediately from the definitions. \( \left( {\mathfrak{q} : x}\right) \subseteq \mathfrak{p} \) ; taking radicals, we get \( r\left( {\mathfrak{q} : x}\right) = \mathfrak{p} \) . Let \( {yz} \in \left( {\mathfrak{q} : x}\right) \) with \( y \notin \mathfrak{p} \) ; then \( {xyz} \in \mathfrak{... | No |
Theorem 4.5. (1st uniqueness theorem). Let \( \mathfrak{a} \) be a decomposable ideal and let \( \mathfrak{a} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) be a minimal primary decomposition of \( \mathfrak{a} \) . Let \( {\mathfrak{p}}_{i} = r\left( {\mathfrak{q}}_{i}\right) \) \( \left( {1 \leq i \leq... | Proof. \( \mathop{\bigcap }\limits_{{i = 1}}^{n}r\left( {{\mathfrak{q}}_{i} : x}\right) = \mathop{\bigcap }\limits_{{x \notin {\mathfrak{q}}_{j}}}{\mathfrak{p}}_{j} \) by (4.4). Suppose \( r\left( {a : x}\right) \) is prime; then by (1.11) we have \( r\left( {a : x}\right) = {\mathfrak{p}}_{j} \) for some \( j \) . Hen... | Yes |
Proposition 4.8. Let \( S \) be a multiplicatively closed subset of \( A \), and let \( \mathfrak{q} \). i) If \( S \cap \mathfrak{p} \neq \varnothing \), then \( {S}^{-1}\mathfrak{q} = {S}^{-1}A \). ii) If \( S \cap \mathfrak{p} = \varnothing \), then \( {S}^{-1}\mathfrak{q} \) is \( {S}^{-1}\mathfrak{p} \)-primary an... | Proof. i) If \( s \in S \cap \mathfrak{p} \), then \( {s}^{n} \in S \cap \mathfrak{q} \) for some \( n > 0 \) ; hence \( {S}^{-1}\mathfrak{q} \) contains \( {s}^{n}/1 \), which is a unit in \( {S}^{-1}A \). ii) If \( S \cap \mathfrak{p} = \varnothing \), then \( s \in S \) and \( {as} \in \mathfrak{q} \) imply \( a \in... | Yes |
Proposition 4.9. Let \( S \) be a multiplicatively closed subset of \( A \) and let \( \mathfrak{a} \) be a decomposable ideal. Let \( \mathfrak{a} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) be a minimal primary decomposition of a. Let \( {\mathfrak{p}}_{i} = r\left( {\mathfrak{q}}_{i}\right) \) and ... | Proof. \( {S}^{-1}a = \mathop{\bigcap }\limits_{{i = 1}}^{n}{S}^{-1}{q}_{i} \) by (3.11) \( = \mathop{\bigcap }\limits_{{i = 1}}^{m}{S}^{-1}{q}_{i} \) by (4.8), and \( {S}^{-1}{q}_{i} \) is \( {S}^{-1}{\mathfrak{p}}_{i} \) -primary for \( i = 1,\ldots, m \) . Since the \( {\mathfrak{p}}_{i} \) are distinct, so are the ... | Yes |
Corollary 4.11. The isolated primary components (i.e., the primary components \( {q}_{i} \) corresponding to minimal prime ideals \( {p}_{i} \) ) are uniquely determined by \( a \) . | Proof of (4.10). We have \( {\mathfrak{q}}_{{i}_{1}} \cap \cdots \cap {\mathfrak{q}}_{{i}_{m}} = S\left( \mathfrak{a}\right) \) where \( S = A - {\mathfrak{p}}_{{i}_{1}} \cup \cdots \cup {\mathfrak{p}}_{{i}_{m}} \) , hence depends only on \( a \) (since the \( {\mathfrak{p}}_{i} \) depend only on \( a \) ). ∎ | No |
Example 5.0. \( A = \mathbf{Z}, B = \mathbf{Q} \) . If a rational number \( x = r/s \) is integral over \( \mathbf{Z} \) , | \[ {r}^{n} + {a}_{1}{r}^{n - 1}s + \cdots + {a}_{n}{s}^{n} = 0 \] the \( {a}_{i} \) being rational integers. Hence \( s \) divides \( {r}^{n} \), hence \( s = \pm 1 \), hence \( x \in \mathbf{Z} \). | Yes |
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