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Theorem 16.8. We have\n\n(a)\n\n\[ \n{h}_{{K}_{m}} = \mathop{\prod }\limits_{{\chi \text{ odd }}}\left( {\mathop{\sum }\limits_{\substack{{a\text{ monic }} \\ {\deg a < {M}_{\chi }} }}\chi \left( a\right) }\right) \mathop{\prod }\limits_{\substack{{\chi \text{ even }} \\ {\chi \neq {\chi }_{o}} }}\left( {\mathop{\sum }... | Proof. Recall that\n\n\[ \n{\zeta }_{{\mathcal{O}}_{m}}\left( w\right) = \mathop{\prod }\limits_{{\mathfrak{P} \in {S}_{m}}}\left( {1 - N{\mathfrak{P}}^{-w}}\right) {\zeta }_{{K}_{m}}\left( w\right)\n\]\n\nand\n\n\[ \n{\zeta }_{{K}_{m}\left( w\right) } = \frac{{L}_{{K}_{m}}\left( {q}^{-w}\right) }{\left( {1 - {q}^{-w}}... | No |
Lemma 16.9. Let \( m \) be a monic polynomial and suppose that \( \lambda \) is a generator of \( {\Lambda }_{m} \) . If \( b \) is a polynomial prime to \( m \), then \( {\sigma }_{b}\lambda /\lambda \in {K}_{m}^{ + } \) . | Proof. By definition of the automorphism \( {\sigma }_{b} \) we have \( {\sigma }_{b}\lambda = {C}_{b}\left( \lambda \right) \) . In particular, when \( \alpha \in {\mathbb{F}}^{ * } \), we have \( {\sigma }_{\alpha }\lambda = {C}_{\alpha }\left( \lambda \right) = {\alpha \lambda } \) . Thus,\n\n\[ \n{\sigma }_{\alpha ... | Yes |
Proposition 16.10. For \( a, b \in A \) monic and prime to \( m \) we have\n\n\[{\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }^{ + }}\left( {{\sigma }_{b}{\lambda }_{m}/{\lambda }_{m}}\right) = \frac{{f}_{m}\left( {ab}\right) - {f}_{m}\left( a\right) }{q - 1}.\] | Proof. Using Equation \( {3}^{\prime } \) above, we find\n\n\[{\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }}\left( {{\sigma }_{b}{\lambda }_{m}}\right) = {\operatorname{ord}}_{{\sigma }_{b}^{-1}{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }}\left( {\lambda }_{m}\right)\]\n\n\[= \left( {q - 1}\right) \lef... | Yes |
Lemma 16.11. If \( m = {P}^{t} \) is a power of an irreducible \( P \), then the group of cyclotomic units, \( {\mathcal{E}}_{m} \), is generated by \( {\mathbb{F}}^{ * } \) and the set\n\n\[ \n{T}_{m} = \left\{ {{\sigma }_{b}{\lambda }_{m}/{\lambda }_{m} \mid b\text{ monic,}0 < \deg b < \deg m,\left( {b, m}\right) = 1... | Proof. We will give the proof when \( m = P \) is irreducible and leave the case \( m = {P}^{t}, t > 1 \), as an exercise.\n\nEvery non-zero element of \( {\Lambda }_{P} \) has the form \( {\sigma }_{a}{\lambda }_{P} \), where \( a \) varies over the non-zero polynomials of degree less that \( \deg P \) . If \( u \) is... | No |
Lemma 16.13. Assume \( q = \\left| \\mathbb{F}\\right| \) is odd and that \( P \) is a monic irreducible of even degree \( d \) . Then \( k\\left( \\sqrt{P}\\right) \\subseteq {K}_{P}^{ + } \) . | Proof. Recall the factorization of the Carlitz polynomial\n\n\\[ \n\\frac{{C}_{P}\\left( u\\right) }{u} = \\mathop{\\prod }\\limits_{\\substack{{\\lambda \\in {\\Lambda }_{P}} \\\\ {\\lambda \\neq 0} }}\\left( {u - \\lambda }\\right) \n\\]\n\nComparing constant terms on both sides shows \( {\\left( -1\\right) }^{{q}^{d... | Yes |
Lemma 16.14. The unit \( \eta \) is an element of \( k\left( \sqrt{P}\right) \) . | Proof. We have seen that \( k\left( \sqrt{P}\right) \subseteq {K}_{P}^{ + } \) . An element \( {\sigma }_{b} \in {G}_{P}^{ + } \) is in \( \operatorname{Gal}\left( {{K}_{P}^{ + }/k\left( \sqrt{P}\right) }\right) \) if and only if \( b \) is a square in \( {\left( A/PA\right) }^{ * }/{\mathbb{F}}^{ * } \) . Notice that\... | Yes |
Proposition 17.2. Let \( f : {\mathcal{D}}_{K}^{ + } \rightarrow \mathbb{C} \) be the characteristic function of the square-free effective divisors. Then \( F\left( N\right) = \mathop{\sum }\limits_{{{degD} = N}}f\left( D\right) \) is the number of square-free effective divisors of degree \( N \) . Given \( \epsilon > ... | Proof. Recall that for divisors \( C \) and \( D \) we have \( N\left( {C + D}\right) = {NCND} \) . From this we calculate \[ {\zeta }_{f}\left( s\right) = \mathop{\sum }\limits_{D}\frac{f\left( D\right) }{N{D}^{s}} = \mathop{\sum }\limits_{{D\text{ square-free }}}\frac{1}{N{D}^{s}} = \mathop{\prod }\limits_{P}\left( {... | Yes |
Proposition 17.3. Let \( \sigma : {\mathcal{D}}_{K}^{ + } \rightarrow \mathbb{C} \) be the sum of norms of divisors function defined above. Given an \( \epsilon > 0 \), we have\n\n\[ \mathop{\sum }\limits_{{\deg D = N}}\sigma \left( D\right) = {\zeta }_{K}\left( 2\right) \frac{{h}_{K}}{{q}^{g - 1}\left( {q - 1}\right) ... | Proof. Since \( {\zeta }_{\sigma }\left( s\right) = {\zeta }_{K}\left( s\right) {\zeta }_{K}\left( {s - 1}\right) \), it has a pole at \( s = 2 \), a double pole at \( s = 1 \), and a pole at \( s = 0 \) . The conditions of Theorem 17.1 do not hold! However, we can make progress by substituting \( s + 1 \) for \( s \) ... | Yes |
Theorem 17.4. Let \( f : {\mathcal{D}}_{K}^{ + } \rightarrow \mathbb{C} \) and let \( {\zeta }_{f}\left( s\right) \) be the corresponding Dirichlet series. Suppose this series converges absolutely in the region \( \Re \left( s\right) > 1 \) and is holomorphic in the region \( \{ s \in B \mid \Re \left( s\right) = 1\} \... | Proof. As in the proof of Theorem 17.1, we can find a \( \delta < 1 \) such that \( {Z}_{f}\left( u\right) \) is holomorphic on the disc \( \left\{ {u \in \mathbb{C}\left| \right| u \mid \leq {q}^{-\delta }}\right\} \) . We again let \( C \) be the boundary of this disc oriented counterclockwise and \( {C}_{\epsilon } ... | Yes |
Proposition 17.5. Let \( K/\mathbb{F} \) be a global function field and \( d\left( D\right) \) the divisor function on the effective divisors. Then, there exist constants \( {\mu }_{K} \) and \( {\lambda }_{K} \) such that for fixed \( \epsilon > 0 \) we have\n\n\[ \mathop{\sum }\limits_{{\deg D = N}}d\left( D\right) =... | Proof. We have already seen that \( {\zeta }_{d}\left( s\right) = {\zeta }_{K}{\left( s\right) }^{2} \), a function which has a double pole at \( s = 1 \) and is otherwise holomorphic for \( \Re \left( s\right) > 0 \) . Choose \( \epsilon > 0 \) . Notice that \( \mathop{\lim }\limits_{{s \rightarrow 1}}{\left( s - 1\ri... | Yes |
Proposition 17.6. With the notations introduced above, the integral closure of \( A \) in \( K \) is \( {\mathcal{O}}_{{m}_{0}} \). The ring \( {\mathcal{O}}_{m} \) is a subring of \( {\mathcal{O}}_{{m}_{0}} \) and the polynomial \( {m}_{1} \) is a generator of the annihilator of the A-module \( {\mathcal{O}}_{{m}_{0}}... | Proof. Since the characteristic of \( \mathbb{F} \) is odd, it is easy to see that \( K/k \) is a Galois extension. Let \( \sigma \) generate the Galois group.\n\nClearly, \( K = k\left( \sqrt{m}\right) = k\left( \sqrt{{m}_{0}}\right) \). Every element in \( K \) has the form \( r + \) \( s\sqrt{{m}_{0}} \) for suitabl... | Yes |
Proposition 17.7. Suppose \( m \) is square-free. Consider the quadratic extension \( K = k\left( \sqrt{m}\right) \) of \( k \) . Let \( {L}_{\infty }\left( {s,{\chi }_{m}}\right) \) be 1 if \( \infty \) is ramified in \( K \) , \( {\left( 1 - {q}^{-s}\right) }^{-1} \) if \( \infty \) splits in \( K \), and \( {\left( ... | Proof. We have seen that \( A + A\sqrt{m} \) is the integral closure of \( A \) in \( K \) . The discriminant of this ring over \( A \) is \( {4m} \) . Since 4 is a non-zero constant, a prime \( P \) of \( A \) is ramified if and only if it divides \( m \) .\n\nLet \( L\left( {s,\chi }\right) \) be the Artin \( L \) -f... | Yes |
Proposition 17.9. Let \( m \in A \) be a non-square and write \( m = {m}_{0}{m}_{1}^{2} \) with \( {m}_{0} \) square-free. Then,\n\n\[ \n{h}_{m} = {h}_{{m}_{0}}\frac{\left| {m}_{1}\right| }{\left\lbrack {\mathcal{O}}_{{m}_{0}}^{ * } : {\mathcal{O}}_{m}^{ * }\right\rbrack }\mathop{\prod }\limits_{{P \mid {m}_{1}}}\left(... | Implicit in this result is that the index \( \left\lbrack {{\mathcal{O}}_{{m}_{0}}^{ * } : {\mathcal{O}}_{m}^{ * }}\right\rbrack \) is finite. If \( \infty \) either ramifies or is inert, both groups are equal to \( {\mathbb{F}}^{ * } \) and the index is 1 . If \( \infty \) splits, then both groups have \( \mathbb{Z} \... | No |
Lemma 17.10. If \( m \notin {\mathbb{F}}^{ * } \) is not a square, \( {S}_{d}\left( {\chi }_{m}\right) = 0 \) for \( d \geq M = \) \( \deg \left( m\right) \) . | Proof. By the reciprocity law, Theorem 3.5, we have\n\n\[ \left( \frac{m}{n}\right) \left( \frac{n}{m}\right) = {\left( -1\right) }^{\frac{q - 1}{2}{Md}}\operatorname{sgn}{\left( m\right) }^{d}. \]\n\nCall the quantity on the right of this equation \( {c}_{d} \) . Then, we have \( {\chi }_{m}\left( n\right) = \) \( {c}... | Yes |
Proposition 17.11. \( \Phi \left( {0, M}\right) = {q}^{M} \) and if \( M, N \geq 1 \), then\n\n\[ \Phi \left( {N, M}\right) = {q}^{M + N}\left( {1 - \frac{1}{q}}\right) . \] | Proof. From the definition, \( \Phi \left( {0, M}\right) \) is equal to the number of monic polynomials of degree \( M \) which we know is \( {q}^{M} \) . This proves the first assertion. To prove the second assertion, call two pairs \( \left( {n, m}\right) \) and \( \left( {{n}^{\prime },{m}^{\prime }}\right) \) equiv... | Yes |
Proposition 17.12. Suppose \( 1 \leq d \leq M - 1 \) . Then\n\n\[ \mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}{S}_{d}\left( {\chi }_{m}\right) = {\left( q - 1\right) }^{-1}\mathop{\sum }\limits_{{\deg \left( m\right) = M}}{S}_{d}\left( {\chi }_{m}\right) = \Phi \left( {d/2, M}\rig... | Proof. To begin with assume all sums are over monics. Then\n\n\[ \mathop{\sum }\limits_{{\deg \left( m\right) = M}}{S}_{d}\left( {\chi }_{m}\right) = \mathop{\sum }\limits_{{\deg \left( m\right) = M}}\mathop{\sum }\limits_{{\deg \left( n\right) = d}}\left( \frac{m}{n}\right) = \mathop{\sum }\limits_{{\deg \left( n\righ... | Yes |
Theorem 17.13. Let \( M \) be odd and positive. We have, for all \( s \in B \) with \( s \neq \frac{1}{2} \)\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {s,{\chi }_{m}}\right) = \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) } -... | Proof. By the corollary to Lemma 17.10, \( L\left( {s,{\chi }_{m}}\right) = \mathop{\sum }\limits_{{d = 0}}^{{M - 1}}{S}_{d}\left( {\chi }_{m}\right) {q}^{-{ds}} \) . From this, Proposition 17.11 and Proposition 17.12, we find\n\n\[ \n\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\... | Yes |
Corollary 1. If \( \Re \left( s\right) > \frac{1}{2} \), then\n\n\[ {q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {s,{\chi }_{m}}\right) \rightarrow \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) }\n\]\nas \( M \rightarrow \infty \) throu... | Proof. This follows immediately from the theorem together with the observation that if \( \Re \left( s\right) > \frac{1}{2} \) then \( \left| {q}^{1 - {2s}}\right| < 1 \) . | No |
Corollary 2. If \( M \) is odd and positive, then\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}{h}_{m} = \frac{{\zeta }_{A}\left( 2\right) }{{\zeta }_{A}\left( 3\right) }{q}^{\frac{M - 1}{2}} - {q}^{-1}.\n\] | Proof. We begin by substituting \( s = 1 \) into the identity given in the theorem. We find\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {1,{\chi }_{m}}\right) = \frac{{\zeta }_{A}\left( 2\right) }{{\zeta }_{A}\left( 3\right) } - \left( {1 - \frac{1}{q}}\rig... | Yes |
Corollary 1. If \( \operatorname{Re}\left( s\right) > \frac{1}{2} \), then as \( M \rightarrow \infty \) though even integers, | \[ {q}^{-M}\sum L\left( {s,{\chi }_{m}}\right) \rightarrow \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) }.\] | No |
Theorem 17.15. Let \( M \) be positive and even, and let \( \gamma \in {\mathbb{F}}^{ * } \) be a non-square constant. The following sum is over all non-square monic polynomials of degree \( M \) . For \( s \neq \frac{1}{2} \) we have\n\n\[ \n{q}^{-M}\sum L\left( {s,{\chi }_{\gamma m}}\right) = \n\] | \n\( \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) } - \left( {1 - \frac{1}{q}}\right) {\left( {q}^{1 - {2s}}\right) }^{\frac{M}{2}}{\zeta }_{A}\left( {2s}\right) - {q}^{-\frac{M}{2}}\left( {\frac{1 + {q}^{-s}}{1 + {q}^{1 - s}} - \left( {1 - \frac{1}{q}}\right) \frac{{\left( {q}^{1 - s}\ri... | Yes |
Corollary 1. If \( \Re \left( s\right) > \frac{1}{2} \), then as \( M \rightarrow \infty \) through even integers, | \[ {q}^{-M}\sum L\left( {s,{\chi }_{\gamma m}}\right) \rightarrow \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) }.\] | No |
Proposition 2.1. Let \( G \) be a group and let \( H, K \) be two subgroups such that \( H \cap K = e,{HK} = G \), and such that \( {xy} = {yx} \) for all \( x \in H \) and \( y \in K \) . Then the map\n\n\[ \nH \times K \rightarrow G \n\]\n\nsuch that \( \left( {x, y}\right) \mapsto {xy} \) is an isomorphism. | Proof. It is obviously a homomorphism, which is surjective since \( {HK} = G \) .\n\nIf \( \left( {x, y}\right) \) is in its kernel, then \( x = {y}^{-1} \), whence \( x \) lies in both \( H \) and \( K \), and \( x = e \) , so that \( y = e \) also, and our map is an isomorphism. | Yes |
Proposition 2.2. Let \( G \) be a group and \( H \) a subgroup. Then\n\n\[ \left( {G : H}\right) \left( {H : 1}\right) = \left( {G : 1}\right) ,\]\nin the sense that if two of these indices are finite, so is the third and equality holds as stated. If \( \left( {G : 1}\right) \) is finite, the order of \( H \) divides t... | Proof. Note that\n\n\[ H = \mathop{\bigcup }\limits_{i}{x}_{i}K\;\text{ (disjoint), \]\n\n\[ G = \mathop{\bigcup }\limits_{j}{y}_{j}H\;\text{ (disjoint). \]\n\nHence\n\n\[ G = \mathop{\bigcup }\limits_{{i, j}}{y}_{j}{x}_{i}K \]\n\nWe must show that this union is disjoint, i.e. that the \( {y}_{j}{x}_{i} \) represent di... | Yes |
Let \( G \) be a finite group. An abelian tower of \( G \) admits a cyclic refinement. Let \( G \) be a finite solvable group. Then \( G \) admits a cyclic tower whose last element is \( \{ e\} \) . | The second assertion is an immediate consequence of the first, and it clearly suffices to prove that if \( G \) is finite, abelian, then \( G \) admits a cyclic tower ending with \( \{ e\} \) . We use induction on the order of \( G \) . Let \( x \) be an element of \( G \) . We may assume that \( x \neq e \) . Let \( X... | Yes |
Theorem 3.2. Let \( G \) be a group and \( H \) a normal subgroup. Then \( G \) is solvable if and only if \( H \) and \( G/H \) are solvable. | Proof. We prove that \( G \) solvable implies that \( H \) is solvable. Let \( G = {G}_{0} \supset {G}_{1} \supset \ldots \supset {G}_{r} = \{ e\} \) be a tower of groups with \( {G}_{i + 1} \) normal in \( {G}_{i} \) and such that \( {G}_{i}/{G}_{i + 1} \) is abelian. Let \( {H}_{i} = H \cap {G}_{i} \) . Then \( {H}_{... | No |
Lemma 3.3. (Butterfly Lemma.) (Zassenhaus) Let \( U, V \) be subgroups of a group. Let \( u, v \) be normal subgroups of \( U \) and \( V \), respectively. Then\n\n\[ u\\left( {U \\cap v}\\right) \\;\\text{ is normal in }\\;u\\left( {U \\cap V}\\right) ,\]\n\n\[ \\left( {u \\cap V}\\right) v\\;\\text{ is normal in }\\;... | Proof. The combination of groups and factor groups becomes clear if one visualizes the following diagram of subgroups (which gives its name to the lemma):\n\n\n\nIn this diagram, we are given \( U, u, V, v \) . All the... | Yes |
Theorem 3.4. (Schreier) Let \( G \) be a group. Two normal towers of subgroups ending with the trivial group have equivalent refinements. | Proof. Let the two towers be as above. For each \( i = 1,\ldots, r - 1 \) and \( j = 1,\ldots, s \) we define\n\n\[ \n{G}_{ij} = {G}_{i + 1}\left( {{H}_{j} \cap {G}_{i}}\right) \n\]\n\nThen \( {G}_{is} = {G}_{i + 1} \), and we have a refinement of the first tower:\n\n\[ \nG = {G}_{11} \supset {G}_{12} \supset \cdots \s... | Yes |
Theorem 3.5. (Jordan-Hölder) Let \( G \) be a group, and let\n\n\[ G = {G}_{1} \supset {G}_{2} \supset \cdots \supset {G}_{r} = \{ e\} \]\n\nbe a normal tower such that each group \( {G}_{i}/{G}_{i + 1} \) is simple, and \( {G}_{i} \neq {G}_{i + 1} \) for \( i = 1,\ldots, r - 1 \) . Then any other normal tower of \( G ... | Proof. Given any refinement \( \left\{ {G}_{ij}\right\} \) as before for our tower, we observe that for each \( i \), there exists precisely one index \( j \) such that \( {G}_{i}/{G}_{i + 1} = {G}_{ij}/{G}_{i, j + 1} \) . Thus the sequence of non-trivial factors for the original tower, or the refined tower, is the sam... | Yes |
Proposition 4.2. Let \( G \) be a cyclic group. Then every subgroup of \( G \) is cyclic. Iff is a homomorphism of \( G \), then the image of \( f \) is cyclic. | Proof. If \( G \) is infinite cyclic, it is isomorphic to \( \mathbf{Z} \), and we determined above all subgroups of \( \mathbf{Z} \), finding that they are all cyclic. If \( f : G \rightarrow {G}^{\prime } \) is a homomorphism, and \( a \) is a generator of \( G \), then \( f\left( a\right) \) is obviously a generator... | Yes |
Proposition 5.2. The number of conjugate subgroups to \( H \) is equal to the index of the normalizer of \( H \) . | Proof. Note that \( H \) is contained in its normalizer \( {N}_{H} \), so the index of \( {N}_{H} \) in \( G \) is 1 or 2. If it is 1, then we are done. Suppose it is 2. Let \( G \) operate by conjugation on the set of subgroups. The orbit of \( H \) has 2 elements, and \( G \) operates on this orbit. In this way we ge... | No |
Proposition 5.3. There exists a unique homomorphism \( \varepsilon : {S}_{n} \rightarrow \{ \pm 1\} \) such that for every transposition \( \tau \) we have \( \varepsilon \left( \tau \right) = - 1 \) . | Proof. Let \( \Delta \) be the function\n\n\[ \Delta \left( {{x}_{1},\ldots ,{x}_{n}}\right) = \mathop{\prod }\limits_{{i < j}}\left( {{x}_{j} - {x}_{i}}\right) \]\n\nthe product being taken for all pairs of integers \( i, j \) satisfying \( 1 \leqq i < j \leqq n \) . Let \( \tau \) be a transposition, interchanging th... | Yes |
Theorem 5.4. If \( n \geqq 5 \) then \( {S}_{n} \) is not solvable. | Proof. We shall first prove that if \( H, N \) are two subgroups of \( {S}_{n} \) such that \( N \subset H \) and \( N \) is normal in \( H \), if \( H \) contains every 3-cycle, and if \( H/N \) is abelian, then \( N \) contains every 3-cycle. To see this, let \( i, j, k, r, s \) be five distinct integers in \( {J}_{n... | Yes |
Theorem 5.5. If \( n \geqq 5 \) then the alternating group \( {A}_{n} \) is simple. | Proof. Let \( N \) be a non-trivial normal subgroup of \( {A}_{n} \) . We prove that \( N \) contains some 3-cycle, whence the theorem follows by (b). Let \( \sigma \in N,\sigma \neq {id} \) , be an element which has the maximal number of fixed points; that is, integers \( i \) such that \( \sigma \left( i\right) = i \... | Yes |
Lemma 6.1. Let \( G \) be a finite abelian group of order \( m \), let \( p \) be a prime number dividing \( m \) . Then \( G \) has a subgroup of order \( p \) . | Proof. We first prove by induction that if \( G \) has exponent \( n \) then the order of \( G \) divides some power of \( n \) . Let \( b \in G, b \neq 1 \), and let \( H \) be the cyclic subgroup generated by \( b \) . Then the order of \( H \) divides \( n \) since \( {b}^{n} = 1 \), and \( n \) is an exponent for \... | Yes |
Theorem 6.2. Let \( G \) be a finite group and \( {pa} \) prime number dividing the order of \( G \) . Then there exists a p-Sylow subgroup of \( G \) . | Proof. By induction on the order of \( G \) . If the order of \( G \) is prime, our assertion is obvious. We now assume given a finite group \( G \), and assume the theorem proved for all groups of order smaller than that of \( G \) . If there exists a proper subgroup \( H \) of \( G \) whose index is prime to \( p \),... | Yes |
Lemma 6.3. Let \( H \) be a p-group acting on a finite set \( S \). Then:\n\n(a) The number of fixed points of \( H \) is \( \equiv \# \left( S\right) {\;\operatorname{mod}\;p} \).\n\n(b) If \( H \) has exactly one fixed point, then \( \# \left( S\right) \equiv 1{\;\operatorname{mod}\;p} \).\n\n(c) If \( p \mid \# \lef... | Proof. We repeatedly use the orbit formula\n\n\[ \n\# \left( S\right) = \sum \left( {H : {H}_{{s}_{i}}}\right)\n\]\n\nFor each fixed point \( {s}_{i} \) we have \( {H}_{{s}_{i}} = H \). For \( {s}_{i} \) not fixed, the index \( \left( {H : {H}_{{s}_{i}}}\right) \) is divisible by \( p \), so (a) follows at once. Parts ... | Yes |
Theorem 6.4. Let \( G \) be a finite group.\n\n(i) If \( H \) is a p-subgroup of \( G \), then \( H \) is contained in some p-Sylow subgroup.\n\n(ii) All p-Sylow subgroups are conjugate.\n\n(iii) The number of \( p \) -Sylow subgroups of \( G \) is \( \equiv 1{\;\operatorname{mod}\;p} \) . | Proof. Let \( P \) be a \( p \) -Sylow subgroup of \( G \) . Suppose first that \( H \) is contained in the normalizer of \( P \) . We prove that \( H \subset P \) . Indeed, \( {HP} \) is then a subgroup of the normalizer, and \( P \) is normal in \( {HP} \) . But\n\n\[ \left( {{HP} : P}\right) = \left( {H : H \cap P}\... | Yes |
Theorem 6.5. Let \( G \) be a finite p-group. Then \( G \) is solvable. If its order is \( > 1 \), then \( G \) has a non-trivial center. | Proof. The first assertion follows from the second, since if \( G \) has center \( Z \), and we have an abelian tower for \( G/Z \) by induction, we can lift this abelian tower to \( G \) to show that \( G \) is solvable. To prove the second assertion, we use the class equation\n\n\[ \left( {G : 1}\right) = \operatorna... | Yes |
Corollary 6.6. Let \( G \) be a p-group which is not of order 1 . Then there exists a sequence of subgroups\n\n\[ \n\\{ e\\} = {G}_{0} \\subset {G}_{1} \\subset {G}_{2} \\subset \\cdots \\subset {G}_{n} = G \n\]\n\nsuch that \( {G}_{i} \) is normal in \( G \) and \( {G}_{i + 1}/{G}_{i} \) is cyclic of order \( p \) . | Proof. Since \( G \) has a non-trivial center, there exists an element \( a \\neq e \) in the center of \( G \), and such that \( a \) has order \( p \) . Let \( H \) be the cyclic group generated by \( a \) . By induction, if \( G \\neq H \), we can find a sequence of subgroups as stated above in the factor group \( G... | No |
Lemma 6.7. Let \( G \) be a finite group and let \( p \) be the smallest prime dividing the order of \( G \) . Let \( H \) be a subgroup of index \( p \) . Then \( H \) is normal. | Proof. Let \( N\left( H\right) = N \) be the normalizer of \( H \) . Then \( N = G \) or \( N = H \) . If \( N = G \) we are done. Suppose \( N = H \) . Then the orbit of \( H \) under conjugation has \( p = \left( {G : H}\right) \) elements, and the representation of \( G \) on this orbit gives a homomorphism of \( G ... | Yes |
Proposition 6.8. Let \( p, q \) be distinct primes and let \( G \) be a group of order pq. Then \( G \) is solvable. | Proof. Say \( p < q \) . Let \( Q \) be a Sylow subgroup of order \( q \) . Then \( Q \) has index \( p \), so by the lemma, \( Q \) is normal and the factor group has order \( p \) . But a group of prime order is cyclic, whence the proposition follows. | No |
Proposition 7.1. Let \( \\left\\{ {{f}_{i} : {A}_{i} \rightarrow B}\\right\\} \) be a family of homomorphisms into an abelian group \( B \) . Let \( A = \\oplus {A}_{i} \) . There exists a unique homomorphism\n\n\[ f : A \\rightarrow B \]\n\n such that \( f \\circ {\\lambda }_{j} = {f}_{j} \) for all \( j \) . | Proof. We can define a map \( f : A \\rightarrow B \) by the rule\n\n\[ f\\left( {\\left( {x}_{i}\\right) }_{i \\in I}\\right) = \\mathop{\\sum }\\limits_{{i \\in I}}{f}_{i}\\left( {x}_{i}\\right) \]\n\nThe sum on the right is actually finite since all but a finite number of terms are 0 . It is immediately verified tha... | Yes |
Lemma 7.2. Let \( A\overset{f}{ \rightarrow }{A}^{\prime } \) be a surjective homomorphism of abelian groups, and assume that \( {A}^{\prime } \) is free. Let \( B \) be the kernel of \( f \) . Then there exists a subgroup \( C \) of \( A \) such that the restriction of \( f \) to \( C \) induces an isomorphism of \( C... | Proof. Let \( {\left\{ {x}_{i}^{\prime }\right\} }_{i \in I} \) be a basis of \( {A}^{\prime } \), and for each \( i \in I \), let \( {x}_{i} \) be an element of \( A \) such that \( f\left( {x}_{i}\right) = {x}_{i}^{\prime } \) . Let \( C \) be the subgroup of \( A \) generated by all elements \( {x}_{i}, i \in I \) .... | Yes |
Theorem 7.3. Let \( A \) be a free abelian group, and let \( B \) be a subgroup. Then \( B \) is also a free abelian group, and the cardinality of a basis of \( B \) is \( \leqq \) the cardinality of a basis for \( A \) . Any two bases of \( B \) have the same cardinality. | Proof. We shall give the proof only when \( A \) is finitely generated, say by a basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \left( {n \geqq 1}\right) \), and give the proof by induction on \( n \) . We have an expression of \( A \) as direct sum:\n\n\[ A = \mathbf{Z}{x}_{1} \oplus \cdots \oplus \mathbf{Z}{x}_{n... | Yes |
Theorem 8.1 Let \( A \) be a torsion abelian group. Then \( A \) is the direct sum of its subgroups \( A\left( p\right) \) for all primes \( p \) such that \( A\left( p\right) \neq 0 \) . | Proof. There is a homomorphism\n\n\[ \n{\bigoplus }_{p}A\left( p\right) \rightarrow A \n\] \n\nwhich to each element \( \left( {x}_{p}\right) \) in the direct sum associates the element \( \sum {x}_{p} \) in \( A \) . We prove that this homomorphism is both surjective and injective. Suppose \( x \) is in the kernel, so... | Yes |
Lemma 8.3. Let \( \bar{b} \) be an element of \( A/{A}_{1} \), of period \( {p}^{r} \) . Then there exists a representative a of \( \bar{b} \) in \( A \) which also has period \( {p}^{r} \) . | Proof. Let \( b \) be any representative of \( \bar{b} \) in \( A \) . Then \( {p}^{r}b \) lies in \( {A}_{1} \), say \( {p}^{r}b = n{a}_{1} \) with some integer \( n \geqq 0 \) . We note that the period of \( \bar{b} \) is \( \leqq \) the period of \( b \) . If \( n = 0 \) we are done. Otherwise write \( n = {p}^{k}\m... | Yes |
Theorem 8.4. Let \( A \) be a finitely generated torsion-free abelian group. Then \( A \) is free. | Proof. Assume \( A \neq 0 \) . Let \( S \) be a finite set of generators, and let \( {x}_{1},\ldots ,{x}_{n} \) be a maximal subset of \( S \) having the property that whenever \( {v}_{1},\ldots ,{v}_{n} \) are integers such that\n\n\[ \n{v}_{1}{x}_{1} + \cdots + {v}_{n}{x}_{n} = 0 \n\]\n\nthen \( {v}_{j} = 0 \) for al... | Yes |
Theorem 8.5. Let \( A \) be a finitely generated abelian group, and let \( {A}_{\text{tor }} \) be the subgroup consisting of all elements of \( A \) having finite period. Then \( {A}_{\text{tor }} \) is finite, and \( A/{A}_{\text{tor }} \) is free. There exists a free subgroup \( B \) of \( A \) such that \( A \) is ... | Proof. We recall that a finitely generated torsion abelian group is obviously finite. Let \( A \) be finitely generated by \( n \) elements, and let \( F \) be the free abelian group on \( n \) generators. By the universal property, there exists a surjective homomorphism\n\n\[ F\overset{\varphi }{ \rightarrow }A \]\n\n... | Yes |
If \( A \) is a finite abelian group, expressed as a product \( A = B \times C \), then \( {A}^{ \land } \) is isomorphic to \( {B}^{ \land } \times {C}^{ \land } \) (under the mapping described below). | Consider the two projections\n\n\n\nof \( B \times C \) on its two components. We get homomorphisms\n\n\n\nand we co... | Yes |
Theorem 9.2. Let \( A \times {A}^{\prime } \rightarrow C \) be a bilinear map of two abelian groups into a cyclic group \( C \) of order \( m \) . Let \( B,{B}^{\prime } \) be its respective kernels on the left and right. Assume that \( {A}^{\prime }/{B}^{\prime } \) is finite. Then \( A/B \) is finite, and \( {A}^{\pr... | Proof. The injection of \( A/B \) into \( \operatorname{Hom}\left( {{A}^{\prime }/{B}^{\prime }, C}\right) \) shows that \( A/B \) is finite. Furthermore, we get the inequalities\n\n\[ \operatorname{ord}A/B \leqq \operatorname{ord}{\left( {A}^{\prime }/{B}^{\prime }\right) }^{ \land } = \operatorname{ord}{A}^{\prime }/... | Yes |
Corollary 9.3. Let \( A \) be a finite abelian group, \( B \) a subgroup, \( {A}^{ \land } \) the dual group, and \( {B}^{ \bot } \) the set of \( \varphi \in {A}^{ \land } \) such that \( \varphi \left( B\right) = 0 \) . Then we have a natural isomorphism of \( {A}^{ \land }/{B}^{ \bot } \) with \( {B}^{ \land } \) . | Proof. This is a special case of Theorem 9.2. | Yes |
Proposition 12.1. Let \( S \) be a set. Then there exists a free group \( \left( {F, f}\right) \) determined by \( S \). Furthermore, \( f \) is injective, and \( F \) is generated by the image of \( f \) . | Proof. (I owe this proof to J. Tits.) We begin with a lemma.\n\nLemma 12.2. There exists a set \( I \) and a family of groups \( {\left\{ {G}_{i}\right\} }_{i \in I} \) such that, if \( g : S \rightarrow G \) is a map of \( S \) into a group \( G \), and \( g \) generates \( G \), then \( G \) is isomorphic to some \( ... | No |
Lemma 12.2. There exists a set \( I \) and a family of groups \( {\left\{ {G}_{i}\right\} }_{i \in I} \) such that, if \( g : S \rightarrow G \) is a map of \( S \) into a group \( G \), and \( g \) generates \( G \), then \( G \) is isomorphic to some \( {G}_{i} \) . | Proof. This is a simple exercise in cardinalities, which we carry out. If \( S \) is finite, then \( G \) is finite or denumerable. If \( S \) is infinite, then the cardinality of \( G \) is \( \leqq \) the cardinality of \( S \) because \( G \) consists of finite products of elements of \( g\left( S\right) \) . Let \(... | Yes |
Proposition 12.3. Coproducts exist in the category of groups. | Proof. Let \( {\left\{ {G}_{i}\right\} }_{i \in I} \) be a family of groups. We let \( \mathcal{C} \) be the category whose objects are families of group-homomorphisms\n\n\[ \n{\left\{ {g}_{i} : {G}_{i} \rightarrow G\right\} }_{i \in I} \n\] \nand whose morphisms are the obvious ones. We must find a universal element i... | Yes |
Proposition 12.4. Let \( G \) be a group and \( {\left\{ {G}_{i}\right\} }_{i \in I} \) a family of subgroups. Assume:\n\n(a) The family generates \( G \) .\n\n(b) If\n\n\[ x = {x}_{{i}_{1}}\cdots {x}_{{i}_{n}}\;\text{with}\;{x}_{{i}_{\nu }} \in {G}_{{i}_{\nu }},\;{x}_{{i}_{\nu }} \neq e\;\text{and}\;{i}_{\nu } \neq {i... | Proof. The homomorphism from the coproduct into \( G \) is surjective by the assumption that the family generates \( G \) . Suppose an element is in the kernel. Then such an element has a representation\n\n\[ {x}_{{i}_{1}}\cdots {x}_{{i}_{n}} \]\n\nas in (b), mapping to the identity in \( G \), so all \( {x}_{{i}_{\nu ... | Yes |
Corollary 12.6. Let \( F\\left( S\\right) \) be the free group on a set \( S \), and let \( {x}_{1},\\ldots ,{x}_{n} \) be distinct elements of \( S \) . Let \( {\\nu }_{1},\\ldots ,{\\nu }_{r} \) be integers \( \\neq 0 \) and let \( {i}_{1},\\ldots ,{i}_{r} \) be integers,\n\n\\[ \n1 \\leqq {i}_{1},\\ldots ,{i}_{r} \\... | Proof. Let \( {G}_{1},\\ldots ,{G}_{n} \) be the cyclic groups generated by \( {x}_{1},\\ldots ,{x}_{n} \) . Let \( G = {G}_{1} \\circ \\cdots \\circ {G}_{n} \) . Let\n\n\\[ \nF\\left( S\\right) \\rightarrow G \n\\]\n\nbe the homomorphism sending each \( {x}_{i} \) on \( {x}_{i} \), and all other elements of \( S \) on... | Yes |
Corollary 12.7. Let \( S \) be a set with \( n \) elements \( {x}_{1},\ldots ,{x}_{n}, n \geqq 1 \) . Let \( {G}_{1} \) , \( \ldots ,{G}_{n} \) be the infinite cyclic groups generated by these elements. Then the map\n\n\[ F\left( S\right) \rightarrow {G}_{1} \circ \cdots \circ {G}_{n} \]\n\nsending each \( {x}_{i} \) o... | Proof. It is obviously surjective and injective. | No |
Let \( {G}_{1},\ldots ,{G}_{n} \) be groups with \( {G}_{i} \cap {G}_{j} = \{ 1\} \) if \( i \neq j \) . The homomorphism\n\n\[ \n{G}_{1} \coprod \cdots \coprod {G}_{n} \rightarrow {G}_{1} \circ \cdots \circ {G}_{n} \n\]\n\nof their coproduct into \( {G}_{1} \circ \cdots \circ {G}_{n} \) induced by the natural inclusio... | Proof. Again, it is obviously injective and surjective. | No |
Proposition 1.1. Products exist in the category of rings. | In fact, let \( {\left\{ {A}_{i}\right\} }_{i \in I} \) be a family of rings, and let \( A = \prod {A}_{i} \) be their product as additive abelian groups. We define a multiplication in \( A \) in the obvious way: If \( {\left( {x}_{i}\right) }_{i \in I} \) and \( {\left( {y}_{i}\right) }_{i \in I} \) are two elements o... | Yes |
Theorem 2.1. (Chinese Remainder Theorem). Let \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) be ideals of \( A \) such that \( {\mathfrak{a}}_{i} + {\mathfrak{a}}_{j} = A \) for all \( i \neq j \) . Given elements \( {x}_{1},\ldots ,{x}_{n} \in A \), there exists \( x \in A \) such that \( x \equiv {x}_{i}\left( {... | Proof. If \( n = 2 \), we have an expression\n\n\[ 1 = {a}_{1} + {a}_{2} \]\n\nfor some elements \( {a}_{i} \in {\mathfrak{a}}_{i} \), and we let \( x = {x}_{2}{a}_{1} + {x}_{1}{a}_{2} \).\n\nFor each \( i \geqq 2 \) we can find elements \( {a}_{i} \in {\mathfrak{a}}_{1} \) and \( {b}_{i} \in {\mathfrak{a}}_{i} \) such... | Yes |
Corollary 2.2. Let \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) be ideals of \( A \) . Assume that \( {\mathfrak{a}}_{i} + {\mathfrak{a}}_{j} = A \) for \( i \neq j \) . Let\n\n\[ f : A \rightarrow \mathop{\prod }\limits_{{i = 1}}^{n}A/{\mathfrak{a}}_{i} = \left( {A/{\mathfrak{a}}_{1}}\right) \times \cdots \time... | Proof. That the kernel of \( f \) is what we said it is, is obvious. The surjectivity follows from the theorem. | No |
Theorem 2.3. Let \( A \) be a cyclic group of order \( n \) . For each \( k \in \mathbf{Z} \) let \( {f}_{k} : A \rightarrow A \) be the endomorphism \( x \mapsto {kx} \) (writing \( A \) additively). Then \( k \mapsto {f}_{k} \) induces a ring isomorphism \( \mathbf{Z}/n\mathbf{Z} \approx \operatorname{End}\left( A\ri... | Proof. Recall that the additive group structure on \( \operatorname{End}\left( A\right) \) is simply addition of mappings, and the multiplication is composition of mappings. The fact that \( k \mapsto {f}_{k} \) is a ring-homomorphism is then a restatement of the formulas\n\n\[ \n{1a} = a,\;\left( {k + {k}^{\prime }}\r... | Yes |
Proposition 3.1. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a homomorphism of monoids. Then there exists a unique homomorphism \( h : A\left\lbrack G\right\rbrack \rightarrow A\left\lbrack {G}^{\prime }\right\rbrack \) such that \( h\left( x\right) = \) \( \varphi \left( x\right) \) for all \( x \in G \) and \(... | Proof. In fact, let \( \alpha = \sum {a}_{x}x \in A\left\lbrack G\right\rbrack \) . Define\n\n\[ h\left( \alpha \right) = \sum {a}_{x}\varphi \left( x\right) \]\n\nThen \( h \) is immediately verified to be a homomorphism of abelian groups, and \( h\left( x\right) = \varphi \left( x\right) \) . Let \( \beta = \sum {b}_... | Yes |
Proposition 3.2. Let \( G \) be a monoid and let \( f : A \rightarrow B \) be a homomorphism of commutative rings. Then there is a unique homomorphism\n\n\[ h : A\left\lbrack G\right\rbrack \rightarrow B\left\lbrack G\right\rbrack \]\n\n such that\n\n\[ h\left( {\mathop{\sum }\limits_{{x \in G}}{a}_{x}x}\right) = \math... | Proof. Since every element of \( A\left\lbrack G\right\rbrack \) has a unique expression as a sum \( \sum {a}_{x}x \), the formula giving \( h \) gives a well-defined map from \( A\left\lbrack G\right\rbrack \) into \( B\left\lbrack G\right\rbrack \) . This map is obviously a homomorphism of abelian groups. As for mult... | Yes |
Proposition 5.1. Let \( A \) be a principal entire ring and \( a, b \in A, a, b \neq 0 \). Let \( \left( a\right) + \left( b\right) = \left( c\right) \). Then \( c \) is a greatest common divisor of \( a \) and \( b \). | Proof. Since \( b \) lies in the ideal \( \left( c\right) \), we can write \( b = {xc} \) for some \( x \in A \), so that \( c \mid b \). Similarly, \( c \mid a \). Let \( d \) divide both \( a \) and \( b \), and write \( a = {dy} \), \( b = {dz} \) with \( y, z \in A \). Since \( c \) lies in \( \left( {a, b}\right) ... | Yes |
Proposition 2.1. A sequence\n\n\\[ \n{X}^{\prime }\\overset{\\lambda }{ \\rightarrow }X \\rightarrow {X}^{\prime \\prime } \\rightarrow 0 \n\\]\n\nis exact if and only if the sequence\n\n\\[ \n{\\operatorname{Hom}}_{A}\\left( {{X}^{\prime }, Y}\\right) \\leftarrow {\\operatorname{Hom}}_{A}\\left( {X, Y}\\right) \\lefta... | Proof. This is an important fact, whose proof is easy. For instance, suppose the first sequence is exact. If \\( g : {X}^{\prime \\prime } \\rightarrow Y \\) is an \\( A \\) -homomorphism, its image in \\( {\\operatorname{Hom}}_{A}\\left( {X, Y}\\right) \\) is obtained by composing \\( g \\) with the surjective map of ... | No |
Proposition 2.2. A sequence\n\n\[ 0 \rightarrow {Y}^{\prime } \rightarrow Y \rightarrow {Y}^{\prime \prime } \]\n\nis exact if and only if\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{A}\left( {X,{Y}^{\prime }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {X, Y}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {X,{Y... | The verification will be left to the reader. It follows at once from the definitions. | No |
Let \( M \) be an \( A \) -module and \( n \) an integer \( \geqq 1 \) . For each \( i = 1,\ldots, n \) let \( {\varphi }_{i} : M \rightarrow M \) be an \( A \) -homomorphism such that\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{\varphi }_{i} = \mathrm{{id}}\;\text{ and }\;{\varphi }_{i} \circ {\varphi }_{j} = 0\;\text{ ... | For each \( j \), we have\n\n\[ {\varphi }_{j} = {\varphi }_{j} \circ \mathrm{{id}} = {\varphi }_{j} \circ \mathop{\sum }\limits_{{i = 1}}^{n}{\varphi }_{i} = {\varphi }_{j} \circ {\varphi }_{j} = {\varphi }_{j}^{2}, \]\n\nthereby proving the first assertion. It is clear that \( \varphi \) is an \( A \) -homomorphism. ... | Yes |
Proposition 3.2. Let \( 0 \rightarrow {M}^{\prime }\overset{f}{ \rightarrow }M\overset{g}{ \rightarrow }{M}^{\prime \prime } \rightarrow 0 \) be an exact sequence of modules. The following conditions are equivalent:\n\n1. There exists a homomorphism \( \varphi : {M}^{\prime \prime } \rightarrow M \) such that \( g \cir... | Proof. Let us write the homomorphisms on the right:\n\n\[ M\underset{\varphi }{\overset{g}{ \rightleftarrows }}{M}^{\prime \prime } \rightarrow 0 \]\n\nLet \( x \in M \). Then\n\n\[ x - \varphi \left( {g\left( x\right) }\right) \]\n\nis in the kernel of \( g \), and hence \( M = \operatorname{Ker}g + \operatorname{Im}\... | No |
Theorem 4.1. Let \( A \) be a ring and \( M \) a module over \( A \) . Let \( I \) be a non-empty set, and let \( {\left\{ {x}_{i}\right\} }_{i \in I} \) be a basis of \( M \) . Let \( N \) be an \( A \) -module, and let \( {\left\{ {y}_{i}\right\} }_{i \in I} \) be a family of elements of \( N \) . Then there exists a... | Proof. Let \( x \) be an element of \( M \) . There exists a unique family \( {\left\{ {a}_{i}\right\} }_{i \in I} \) of elements of \( A \) such that\n\n\[ x = \mathop{\sum }\limits_{{i \in I}}{a}_{i}{x}_{i} \]\n\nWe define\n\n\[ f\left( x\right) = \sum {a}_{i}{y}_{i} \]\n\nIt is then clear that \( f \) is a homomorph... | Yes |
Corollary 4.2. Let the notation be as in the theorem, and assume that \( {\left\{ {y}_{i}\right\} }_{i \in I} \) is a basis of \( N \) . Then the homomorphism \( f \) is an isomorphism, i.e. a module-isomorphism. | Proof. By symmetry, there exists a unique homomorphism\n\n\[ g : N \rightarrow M \]\n\nsuch that \( g\left( {y}_{i}\right) = {x}_{i} \) for all \( i \), and \( f \circ g \) and \( g \circ f \) are the respective identity mappings. | Yes |
Corollary 4.3. Two modules having bases whose cardinalities are equal are isomorphic. | Proof. Clear. | No |
Theorem 5.1. Let \( V \) be a vector space over a field \( K \), and assume that \( V \neq \{ 0\} \) . Let \( \Gamma \) be a set of generators of \( V \) over \( K \) and let \( S \) be a subset of \( \Gamma \) which is linearly independent. Then there exists a basis \( \mathfrak{G} \) of \( V \) such that \( S \subset... | Proof. Let \( \mathfrak{T} \) be the set whose elements are subsets \( T \) of \( \Gamma \) which contain \( S \) and are linearly independent. Then \( \mathfrak{T} \) is not empty (it contains \( S \) ), and we contend that \( \mathfrak{T} \) is inductively ordered. Indeed, if \( \left\{ {T}_{i}\right\} \) is a totall... | Yes |
Theorem 5.2. Let \( V \) be a vector space over a field \( K \). Then two bases of \( V \) over \( K \) have the same cardinality. | Proof. Let us first assume that there exists a basis of \( V \) with a finite number of elements, say \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\}, m \geqq 1 \) . We shall prove that any other basis must also have \( m \) elements. For this it will suffice to prove: If \( {w}_{1},\ldots ,{w}_{n} \) are elements of \( V... | No |
Theorem 5.3. Let \( V \) be a vector space over a field \( K \), and let \( W \) be a subspace. Then\n\n\[{\dim }_{K}V = {\dim }_{K}W + {\dim }_{K}V/W\]\n\nIff: \( V \rightarrow U \) is a homomorphism of vector spaces over \( K \), then\n\n\[ \dim V = \dim \operatorname{Ker}f + \dim \operatorname{Im}f. \] | Proof. The first statement is a special case of the second, taking for \( f \) the canonical map. Let \( {\left\{ {u}_{i}\right\} }_{i \in I} \) be a basis of \( \operatorname{Im}f \), and let \( {\left\{ {w}_{j}\right\} }_{j \in J} \) be a basis of Ker \( f \) . Let \( {\left\{ {v}_{i}\right\} }_{i \in I} \) be a fami... | Yes |
Corollary 5.4. Let \( V \) be a vector space and \( W \) a subspace. Then\n\n\[ \n\dim W \leqq \dim V \n\]\n\nIf \( V \) is finite dimensional and \( \dim W = \dim V \) then \( W = V \) . | Proof. Clear. | No |
Theorem 6.1. Let \( E \) be a finite free module over the commutative ring \( A \) , of finite dimension \( n \) . Then \( {E}^{ \vee } \) is also free, and \( \dim {E}^{ \vee } = n \) . If \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is a basis for \( E \), and \( {f}_{i} \) is the functional such that \( {f}_{i}\l... | Proof. Let \( f \in {E}^{ \vee } \) and let \( {a}_{i} = f\left( {x}_{i}\right) \left( {i = 1,\ldots, n}\right) \) . We have\n\n\[ f\left( {{c}_{1}{x}_{1} + \cdots + {c}_{n}{x}_{n}}\right) = {c}_{1}f\left( {x}_{1}\right) + \cdots + {c}_{n}f\left( {x}_{n}\right) . \]\n\nHence \( f = {a}_{1}{f}_{1} + \cdots + {a}_{n}{f}_... | Yes |
Corollary 6.2. When \( E \) is free finite dimensional, then the map \( E \rightarrow {E}^{\vee \vee } \) which to each \( x \in V \) associates the functional \( f \mapsto \langle x, f\rangle \) on \( {E}^{ \vee } \) is an isomorphism of \( E \) onto \( {E}^{\vee \vee } \). | Proof. Note that since \( \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) is a basis for \( {E}^{ \vee } \), it follows from the definitions that \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is the dual basis in \( E \), so \( E = {E}^{\vee \vee } \). | No |
Theorem 6.3. Let \( U, V, W \) be finite free modules over the commutative ring \( A \), and let\n\n\[ 0 \rightarrow W\overset{\lambda }{ \rightarrow }V\overset{\varphi }{ \rightarrow }U \rightarrow 0 \]\n\nbe an exact sequence of \( A \) -homomorphisms. Then the induced sequence\n\n\[ 0 \rightarrow {\operatorname{Hom}... | Proof. This is a consequence of \( \mathbf{{P2}} \), because a free module is projective. | No |
Theorem 6.4. Let \( V \times {V}^{\prime } \rightarrow K \) be a bilinear map, let \( W,{W}^{\prime } \) be its kernels on the left and right respectively, and assume that \( {V}^{\prime }/{W}^{\prime } \) is finite dimensional. Then the induced homomorphism \( {V}^{\prime }/{W}^{\prime } \rightarrow {\left( V/W\right)... | Proof. By symmetry, we have an induced homomorphism\n\n\[ V/W \rightarrow {\left( {V}^{\prime }/{W}^{\prime }\right) }^{ \vee } \]\n\nwhich is injective. Since\n\n\[ \dim {\left( {V}^{\prime }/{W}^{\prime }\right) }^{ \vee } = \dim {V}^{\prime }/{W}^{\prime } \]\n\nit follows that \( V/W \) is finite dimensional. From ... | Yes |
Corollary 7.2. Let \( E \) be a finitely generated module and \( {E}^{\prime } \) a submodule.\n\nThen \( {E}^{\prime } \) is finitely generated. | Proof. We can represent \( E \) as a factor module of a free module \( F \) with a finite number of generators: If \( {v}_{1},\ldots ,{v}_{n} \) are generators of \( E \), we take a free module \( F \) with basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) and map \( {x}_{i} \) on \( {v}_{i} \) . The inverse image ... | Yes |
Theorem 7.3. Let \( E \) be finitely generated. Then \( E/{E}_{\text{tor }} \) is free. There exists a free submodule \( F \) of \( E \) such that \( E \) is a direct sum\n\n\[ E = {E}_{\text{tor }} \oplus F. \] | Proof. We first prove that \( E/{E}_{\text{tor }} \) is torsion free. If \( x \in E \), let \( \bar{x} \) denote its residue class \( {\;\operatorname{mod}\;{E}_{\text{tor }}} \) . Let \( b \in R, b \neq 0 \) be such that \( b\bar{x} = 0 \) . Then \( {bx} \in {E}_{\text{tor }} \) , and hence there exists \( c \in R, c ... | Yes |
Lemma 7.4. Let \( E,{E}^{\prime } \) be modules, and assume that \( {E}^{\prime } \) is free. Let \( f : E \rightarrow {E}^{\prime } \) be a surjective homomorphism. Then there exists a free submodule \( F \) of \( E \) such that the restriction off to \( F \) induces an isomorphism of \( F \) with \( {E}^{\prime } \),... | Proof. Let \( {\left\{ {x}_{i}^{\prime }\right\} }_{i \in I} \) be a basis of \( {E}^{\prime } \) . For each \( i \), let \( {x}_{i} \) be an element of \( E \) such that \( f\left( {x}_{i}\right) = {x}_{i}^{\prime } \) . Let \( F \) be the submodule of \( E \) generated by all the elements \( {x}_{i} \) , \( i \in I \... | Yes |
Theorem 7.5. Let \( E \) be a finitely generated torsion module \( \neq 0 \) . Then \( E \) is the direct sum\n\n\[ E = {\bigoplus }_{p}E\left( p\right) \]\n\ntaken over all primes \( p \) such that \( E\left( p\right) \neq 0 \) . Each \( E\left( p\right) \) can be written as a direct sum\n\n\[ E\left( p\right) = R/\le... | Proof. Let \( a \) be an exponent for \( E \), and suppose that \( a = {bc} \) with \( \left( {b, c}\right) = \left( 1\right) \) . Let \( x, y \in R \) be such that\n\n\[ 1 = {xb} + {yc}. \]\n\nWe contend that \( E = {E}_{b} \oplus {E}_{c} \) . Our first assertion then follows by induction, expressing \( a \) as a prod... | Yes |
Lemma 7.6. Let \( E \) be a torsion module of exponent \( {p}^{r}\left( {r \geqq 1}\right) \) for some prime element \( p \) . Let \( {x}_{1} \in E \) be an element of period \( {p}^{r} \) . Let \( \bar{E} = E/\left( {x}_{1}\right) \) . Let \( {\bar{y}}_{1},\ldots ,{\bar{y}}_{m} \) be independent elements of \( \bar{E}... | Proof. Let \( \bar{y} \in \bar{E} \) have period \( {p}^{n} \) for some \( n \geqq 1 \) . Let \( y \) be a representative of \( \bar{y} \) in \( E \) . Then \( {p}^{n}y \in \left( {x}_{1}\right) \), and hence\n\n\[ {p}^{n}y = {p}^{s}c{x}_{1},\;c \in R, p \nmid c, \]\n\nfor some \( s \leqq r \) . If \( s = r \), we see ... | Yes |
Theorem 7.7. Let \( E \) be a finitely generated torsion module, \( E \neq 0 \) . Then \( E \) is isomorphic to a direct sum of non-zero factors\n\n\[ R/\left( {q}_{1}\right) \oplus \cdots \oplus R/\left( {q}_{r}\right) \]\n\nwhere \( {q}_{1},\ldots ,{q}_{r} \) are non-zero non-units of \( R \), and \( {q}_{1}\left| {q... | Proof. Using Theorem 7.5, decompose \( E \) into a direct sum of \( p \) -submodules, say \( E\left( {p}_{1}\right) \oplus \cdots \oplus E\left( {p}_{l}\right) \), and then decompose each \( E\left( {p}_{i}\right) \) into a direct sum of cyclic submodules of periods \( {p}_{i}^{{r}_{ij}} \) . We visualize these symboli... | Yes |
Theorem 7.9. Assume that the elementary matrices in \( R \) generate \( G{L}_{n}\left( R\right) \) . Let \( \left( {x}_{ij}\right) \) be a non-zero matrix with components in \( R \) . Then with a finite number of row and column operations, it is possible to bring the matrix to the form\n\n\[ \left( \begin{matrix} {a}_{... | We leave the proof for the reader. Either Theorem 7.9 can be viewed as equivalent to Theorem 7.8, or a direct proof may be given. | No |
Theorem 8.1. Let \( \varphi \) be a rule which to each simple module associates an element of a commutative group \( \Gamma \), and such that if \( M \approx {M}^{\prime } \) then\n\n\[ \varphi \left( M\right) = \varphi \left( {M}^{\prime }\right) \]\n\nThen \( \varphi \) has a unique extension to an Euler-Poincaré map... | Proof. Given a simple filtration\n\n\[ M = {M}_{1} \supset {M}_{2} \supset \cdots \supset {M}_{r} = 0 \]\n\nwe define\n\n\[ \varphi \left( M\right) = \mathop{\sum }\limits_{{i = 1}}^{{r - 1}}\varphi \left( {{M}_{i}/{M}_{i + 1}}\right) \]\n\nThe Jordan-Hölder theorem shows immediately that this is well-defined, and that... | Yes |
Lemma 9.1. (Snake Lemma). Given a snake diagram as above, the map\n\n\\[ \n\\delta : \\operatorname{Ker}{d}^{\\prime \\prime } \\rightarrow \\text{Coker}{d}^{\\prime }\n\\]\n\ninduced by \\( \\delta {z}^{\\prime \\prime } = {f}^{-1} \\circ d \\circ {g}^{-1}{z}^{\\prime \\prime } \\) is well defined, and we have an exac... | Proof. It is a routine verification that the class of \\( {z}^{\\prime }{\\;\\operatorname{mod}\\;\\operatorname{Im}}{d}^{\\prime } \\) is independent of the choices made when taking inverse images, whence defining the map \\( \\delta \\) . The proof of the exactness of the sequence is then routine, and consists in cha... | No |
Theorem 10.1. Direct limits exist in the category of abelian groups, or more generally in the category of modules over a ring. | Proof. Let \( \left\{ {M}_{i}\right\} \) be a directed system of modules over a ring. Let \( M \) be their direct sum. Let \( N \) be the submodule generated by all elements\n\n\[ \n{x}_{ij} = \left( {\ldots ,0, x,0,\ldots , - {f}_{j}^{i}\left( x\right) ,0,\ldots }\right) \n\]\n\nwhere, for a given pair of indices \( \... | No |
Theorem 10.2. Inverse limits exist in the category of groups, in the category of modules over a ring, and also in the category of rings. | Proof. Let \( \left\{ {G}_{i}\right\} \) be a directed family of groups, for instance, and let \( \Gamma \) be their inverse limit as defined in Chapter I,§10. Let \( {p}_{i} : \Gamma \rightarrow {G}_{i} \) be the projection (defined as the restriction from the projection of the direct product, since \( \Gamma \) is a ... | No |
Proposition 10.3. Assume that \( \left( {A}_{n}\right) \) satisfies ML. Given an exact sequence\n\n\[ 0 \rightarrow \left( {A}_{n}\right) \rightarrow \left( {B}_{n}\right) \overset{g}{ \rightarrow }\left( {C}_{n}\right) \rightarrow 0 \]\n\nof inverse systems, then\n\n\[ 0 \rightarrow \underline{\lim }{A}_{n} \rightarro... | Proof. The only point is to prove the surjectivity on the right. Let \( \left( {c}_{n}\right) \) be an element of the inverse limit. Then each inverse image \( {g}^{-1}\left( {c}_{n}\right) \) is a coset of \( {A}_{n} \), so in bijection with \( {A}_{n} \) . These inverse images form an inverse system, and the ML condi... | Yes |
Proposition 10.4. Let \( \\left( {C}_{n}\\right) \) be an inverse system of abelian groups satisfying ML, and let \( \\left( {u}_{m, n}\\right) \) be the system of connecting maps. Then we have an exact sequence\n\n\[ 0 \\rightarrow \\underline{\\lim }{C}_{n} \\rightarrow \\prod {C}_{n}\\overset{1 - u}{ \\rightarrow }\... | Proof. For each positive integer \( N \) we have an exact sequence with a finite product\n\n\[ 0 \\rightarrow \\mathop{\\lim }\\limits_{{1 \\leqq n \\leqq N}}{C}_{n} \\rightarrow \\mathop{\\prod }\\limits_{{n = 1}}^{N}{C}_{n}\\overset{1 - u}{ \\rightarrow }\\mathop{\\prod }\\limits_{{n = 1}}^{N}{C}_{n} \\rightarrow 0. ... | Yes |
Theorem 1.1. Let \( A \) be a commutative ring, let \( f, g \in A\left\lbrack X\right\rbrack \) be polynomials in one variable, of degrees \( \geqq 0 \), and assume that the leading coefficient of \( g \) is a unit in \( A \) . Then there exist unique polynomials \( q, r \in A\left\lbrack X\right\rbrack \) such that \[... | Proof. Write \[ f\left( X\right) = {a}_{n}{X}^{n} + \cdots + {a}_{0} \] \[ g\left( X\right) = {b}_{d}{X}^{d} + \cdots + {b}_{0} \] where \( n = \deg f, d = \deg g \) so that \( {a}_{n},{b}_{d} \neq 0 \) and \( {b}_{d} \) is a unit in \( A \) . We use induction on \( n \) . If \( n = 0 \), and \( \deg g > \deg f \), we ... | Yes |
Theorem 1.2. Let \( k \) be a field. Then the polynomial ring in one variable \( k\\left\\lbrack X\\right\\rbrack \) is principal. | Proof. Let \( \\mathfrak{a} \) be an ideal of \( k\\left\\lbrack X\\right\\rbrack \), and assume \( \\mathfrak{a} \\neq 0 \) . Let \( g \) be an element of \( \\mathfrak{a} \) of smallest degree \( \\geqq 0 \) . Let \( f \) be any element of \( \\mathfrak{a} \) such that \( f \\neq 0 \) . By the Euclidean algorithm we ... | Yes |
Theorem 1.4. Let \( k \) be a field and \( f \) a polynomial in one variable \( X \) in \( k\left\lbrack X\right\rbrack \), of degree \( n \geqq 0 \) . Then \( f \) has at most \( n \) roots in \( k \), and if \( a \) is a root of \( f \) in \( k \), then \( X - a \) divides \( f\left( X\right) \) . | Proof. Suppose \( f\left( a\right) = 0 \) . Find \( q, r \) such that\n\n\[ f\left( X\right) = q\left( X\right) \left( {X - a}\right) + r\left( X\right) \]\n\nand \( \deg r < 1 \) . Then\n\n\[ 0 = f\left( a\right) = r\left( a\right) \]\n\nSince \( r = 0 \) or \( r \) is a non-zero constant, we must have \( r = 0 \), wh... | Yes |
Corollary 1.6. Let \( k \) be a field, and let \( {S}_{1},\ldots ,{S}_{n} \) be infinite subsets of \( k \) . Let \( f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) be a polynomial in \( n \) variables over \( k \) . If \( f\left( {{a}_{1},\ldots ,{a}_{n}}\right) = 0 \) for all \( {a}_{i} \in {S}_{i}\left( {i = 1,\ldots, n... | Proof. By induction. We have just seen the result is true for one variable. Let \( n \geqq 2 \), and write\n\n\[ f\left( {{X}_{1},\ldots ,{X}_{n}}\right) = \mathop{\sum }\limits_{j}{f}_{j}\left( {{X}_{1},\ldots ,{X}_{n - 1}}\right) {X}_{n}^{j} \]\n\nas a polynomial in \( {X}_{n} \) with coefficients in \( k\left\lbrack... | Yes |
Corollary 1.8. Let \( k \) be a finite field with \( q \) elements. Let \( f \) be a polynomial in \( n \) variables over \( k \) such that the degree of \( f \) in each variable is \( < q \) . If \( f \) induces the zero function on \( {k}^{\left( n\right) } \), then \( f = 0 \) . | Proof. By induction. If \( n = 1 \), then the degree of \( f \) is \( < q \), and hence \( f \) cannot have \( q \) roots unless it is 0 . The inductive step is carried out just as we did for the proof of Corollary 1.6 above. | No |
Theorem 1.9. Let \( k \) be a field and let \( U \) be a finite multiplicative subgroup of \( k \) . Then \( U \) is cyclic. | Proof. Write \( U \) as a product of subgroups \( U\left( p\right) \) for each prime \( p \), where \( U\left( p\right) \) is a \( p \) -group. By Proposition 4.3(v) of Chapter I, it will suffice to prove that \( U\left( p\right) \) is cyclic for each \( p \) . Let \( a \) be an element of \( U\left( p\right) \) of max... | Yes |
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