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Theorem 16.8. We have\n\n(a)\n\n\[ \n{h}_{{K}_{m}} = \mathop{\prod }\limits_{{\chi \text{ odd }}}\left( {\mathop{\sum }\limits_{\substack{{a\text{ monic }} \\ {\deg a < {M}_{\chi }} }}\chi \left( a\right) }\right) \mathop{\prod }\limits_{\substack{{\chi \text{ even }} \\ {\chi \neq {\chi }_{o}} }}\left( {\mathop{\sum }\limits_{\substack{{a\text{ monic }} \\ {a\text{ monic }} }} - \deg {a\chi }\left( a\right) }\right) .\n\]\n\n(b)\n\n\[ \n{h}_{{K}_{m}^{ + }} = \mathop{\prod }\limits_{\substack{{\chi \text{ even }} \\ {\chi \neq {\chi }_{o}} }}\left( {\mathop{\sum }\limits_{\substack{{a\text{ monic }} \\ {a\text{ go } < {M}_{\chi }} }} - \deg {a\chi }\left( a\right) }\right) .\n\]
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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\[ \n{\zeta }_{{K}_{m}\left( w\right) } = \frac{{L}_{{K}_{m}}\left( {q}^{-w}\right) }{\left( {1 - {q}^{-w}}\right) \left( {1 - {q}^{1 - w}}\right) },\n\]\n\nwhere \( {L}_{{K}_{m}}\left( u\right) \) is a polynomial whose value at \( u = 1 \) is \( {h}_{{K}_{m}} \) .\n\nSince every prime in \( {S}_{m} \) has degree 1, we have \( 1 - N{\mathfrak{P}}^{-w} = 1 - {q}^{-w} \) for all \( \mathfrak{P} \in {S}_{m} \) . So, combining the last two equations and switching to the \
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No
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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 \( {\sigma }_{b}\lambda /\lambda \in {K}_{m}^{ + } \) .
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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\[ \n{\sigma }_{\alpha }\left( {{\sigma }_{b}\lambda /\lambda }\right) = {\sigma }_{b}{\sigma }_{\alpha }\lambda /{\sigma }_{\alpha }\lambda = {\sigma }_{b}\left( {\alpha \lambda }\right) /{\alpha \lambda } = {\sigma }_{b}\lambda /\lambda .\n\]\n\nIt follows that \( {\sigma }_{b}\lambda /\lambda \) is fixed for all elements in \( \left\{ {{\sigma }_{\alpha } \mid \alpha \in {\mathbb{F}}^{ * }}\right\} = J \), the Galois group of \( {K}_{m}/{K}_{m}^{ + } \) . The result follows.
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Yes
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Proposition 16.10. For \( a, b \in A \) monic and prime to \( m \) we have\n\n\[{\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }^{ + }}\left( {{\sigma }_{b}{\lambda }_{m}/{\lambda }_{m}}\right) = \frac{{f}_{m}\left( {ab}\right) - {f}_{m}\left( a\right) }{q - 1}.\]
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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\[= \left( {q - 1}\right) \left( {M - \deg \langle {ab}\rangle - 1}\right) - 1 = {f}_{m}\left( {ab}\right) .\]\n\nCombining Equations \( {3}^{\prime } \) and 4, we find\n\n\[{\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }}\left( {{\sigma }_{b}{\lambda }_{m}/{\lambda }_{m}}\right) = {\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }}\left( {{\sigma }_{b}\lambda }\right) - {\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }}\left( {\lambda }_{m}\right) = {f}_{m}\left( {ab}\right) - {f}_{m}\left( a\right) .\]\n\nFinally, since \( {\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty } \) is totally and tamely ramified over \( {\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }^{ + } \) and \( {\sigma }_{b}{\lambda }_{m}/{\lambda }_{m} \in {K}_{m}^{ + } \) by Lemma 16.9 we have\n\n\[{\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }^{ + }}\left( {{\sigma }_{b}{\lambda }_{m}/{\lambda }_{m}}\right) = \frac{1}{q - 1}{\operatorname{ord}}_{{\sigma }_{a}^{-1}{\mathfrak{P}}_{\infty }}\left( {{\sigma }_{b}{\lambda }_{m}/{\lambda }_{m}}\right) = \frac{{f}_{m}\left( {ab}\right) - {f}_{m}\left( a\right) }{q - 1}.\]
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Yes
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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\[ \n{T}_{m} = \left\{ {{\sigma }_{b}{\lambda }_{m}/{\lambda }_{m} \mid b\text{ monic,}0 < \deg b < \deg m,\left( {b, m}\right) = 1}\right\} .\n\]
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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 a cyclotomic unit, then\n\n\[ \nu = \mathop{\prod }\limits_{a}{\left( {\sigma }_{a}{\lambda }_{P}\right) }^{{n}_{a}} \]\n\nwhere the exponents \( {n}_{a} \) are in \( \mathbb{Z} \) . Rewrite this equation as\n\n\[ \nu = \mathop{\prod }\limits_{a}{\left( {\sigma }_{a}{\lambda }_{P}/{\lambda }_{P}\right) }^{{n}_{a}} \times {\lambda }_{P}^{\sum {n}_{a}}.\n\]\n\nConsider the fractional \( {\mathcal{O}}_{P} \) -ideal generated by both sides. We find, \( {\mathcal{O}}_{P} = \) \( {\left( {\lambda }_{P}\right) }^{\sum {n}_{a}} \) . Since \( \left( {\lambda }_{P}\right) \) is a prime ideal, this implies \( \sum {n}_{a} = 0 \) . It remains to show that we can restrict our attention to \( a \) monic.\n\nIf \( a \) is not monic, \( a = {\alpha b} \) with \( \alpha \in {\mathbb{F}}^{ * } \) and \( b \) monic. Then\n\n\[ {\sigma }_{a}{\lambda }_{P}/{\lambda }_{P} = {\sigma }_{b}{\sigma }_{\alpha }{\lambda }_{P}/{\lambda }_{P} = \alpha {\sigma }_{b}{\lambda }_{P}/{\lambda }_{P}.\n\]\n\nThe lemma now follows immediately.
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No
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Lemma 16.13. Assume \( q = \\left| \\mathbb{F}\\right| \) is odd and that \( P \) is a monic irreducible of even degree \( d \) . Then \( k\\left( \\sqrt{P}\\right) \\subseteq {K}_{P}^{ + } \) .
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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 \( {\\left( -1\\right) }^{{q}^{d} - 1}\\prod \\lambda = \\prod \\lambda = P \) , where the product is over all elements in \( {\\Lambda }_{P} - \\{ 0\\} \) .\n\nThe set of non-zero elements of \( {\\Lambda }_{P} \) coincides with \( \\left\\{ {{\\sigma }_{a}{\\lambda }_{P} \\mid a \\neq 0,\\deg a < }\\right. \) \( d\\} \) . Since every non-zero polynomial can be written uniquely as the product of a constant times a monic, we derive the following equation:\n\n\\[ \n{\\left( \\mathop{\\prod }\\limits_{{\\alpha \\in {\\mathbb{F}}^{ * }}}\\alpha \\right) }^{\\frac{{q}^{d} - 1}{q - 1}}\\mathop{\\prod }\\limits_{{a \\in \\mathcal{M}}}{\\left( {\\sigma }_{a}{\\lambda }_{P}\\right) }^{q - 1} = P.\n\\]\n\nThe product of all the non-zero elements in a finite field is -1 . Since \( q \) is assumed odd and \( d \) is even, \( \\frac{{q}^{d} - 1}{q - 1} \) is even. It follows that \( P \) is a \( q - 1 \) -power in \( {K}_{P} \), and so, a posteriori, a square. This shows that \( k\\left( \\sqrt{P}\\right) \\subseteq {K}_{P} \).\n\nTo show \( k\\left( \\sqrt{P}\\right) \\subseteq {K}_{P}^{ + } \), note that \( {G}_{P} \) is cyclic. This shows there is a unique quadratic extension of \( k \) inside \( {K}_{P} \) . We have just seen that \( \\frac{{q}^{d} - 1}{q - 1} = \) \( \\left| {G}_{P}^{ + }\\right| \) is even. It follows that \( k \) has a quadratic extension inside \( {K}_{P}^{ + } \) . By the uniqueness, it must be \( k\\left( \\sqrt{P}\\right) \) .
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Yes
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Lemma 16.14. The unit \( \eta \) is an element of \( k\left( \sqrt{P}\right) \) .
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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\n\n\[ \n{\sigma }_{b}\eta = \left( {\mathop{\prod }\limits_{{a \in {\left( A/PA\right) }^{ * }/{\mathbb{F}}^{ * }}}{\left( \frac{{\sigma }_{ba}{\lambda }_{P}}{\operatorname{sgn}\left( {ab}\right) {\lambda }_{P}}\right) }^{-\left( {a/P}\right) }}\right) \times {\left( \frac{{\sigma }_{b}{\lambda }_{P}}{\operatorname{sgn}\left( b\right) {\lambda }_{P}}\right) }^{\mathop{\sum }\limits_{a}\left( {a/P}\right) }.\n\]\n\nThe second factor is equal to 1, since \( \mathop{\sum }\limits_{a}\left( {a/P}\right) = 0 \) . If \( b \) is a square in \( {\left( A/PA\right) }^{ * }/{\mathbb{F}}^{ * } \), then \( \left( {b/P}\right) = 1 \) so \( \left( {a/P}\right) = \left( {{ba}/P}\right) \) . Under these conditions the above equation shows \( {\sigma }_{b}\eta = \eta \), which proves the lemma.
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Yes
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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 \( \epsilon > 0 \), we have \[ F\left( N\right) = \frac{1}{{\zeta }_{K}\left( 2\right) }\frac{{h}_{K}}{{q}^{g - 1}\left( {q - 1}\right) }{q}^{N} + {O}_{\epsilon }\left( {q}^{\left( {\frac{1}{4} + \epsilon }\right) N}\right) . \] Moreover, \( \operatorname{Ave}\left( f\right) = 1/{\zeta }_{K}\left( 2\right) \) .
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Proof. Recall that for divisors \( C \) and \( D \) we have \( N\left( {C + D}\right) = {NCND} \) . From this we calculate \[ {\zeta }_{f}\left( s\right) = \mathop{\sum }\limits_{D}\frac{f\left( D\right) }{N{D}^{s}} = \mathop{\sum }\limits_{{D\text{ square-free }}}\frac{1}{N{D}^{s}} = \mathop{\prod }\limits_{P}\left( {1 + \frac{1}{N{P}^{s}}}\right) = \frac{{\zeta }_{K}\left( s\right) }{{\zeta }_{K}\left( {2s}\right) }. \] By the function-field Riemann Hypothesis we know that all the zeros of \( {\zeta }_{K}\left( s\right) \) are on the line \( \Re \left( s\right) = \frac{1}{2} \) . Thus \( 1/{\zeta }_{K}\left( {2s}\right) \) has no poles in the region \( \Re \left( s\right) > \frac{1}{4} \) . On the other hand, we know that in this region \( {\zeta }_{K}\left( s\right) \) is holomorphic except for a simple pole at \( s = 1 \) . Choose an \( \epsilon > 0 \) and set \( {\delta }^{\prime } = \frac{1}{4} + \epsilon \) . Then all the hypotheses of Theorem 17.1 apply to \( {\zeta }_{f}\left( s\right) \) and we find \[ F\left( N\right) = \alpha \log \left( q\right) {q}^{N} + {O}_{\epsilon }\left( {q}^{\left( {\frac{1}{4} + \epsilon }\right) N}\right) , \] where \( \alpha \) is the residue of \( {\zeta }_{K}\left( s\right) /{\zeta }_{K}\left( {2s}\right) \) at \( s = 1 \) . We have seen in Chapter 5 that the residue of \( {\zeta }_{K}\left( s\right) \) at \( s = 1 \) is \[ {\rho }_{K} = \frac{{h}_{K}}{{q}^{g - 1}\left( {q - 1}\right) \log \left( q\right) }. \] It follows that \( \alpha = {\rho }_{K}/{\zeta }_{K}\left( 2\right) \) . Substituting this information into Equation 3 completes the proof of the first assertion of the proposition. To prove the second assertion recall that \( \operatorname{Ave}\left( f\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}F\left( N\right) /{b}_{N}\left( K\right) \) and that for all \( N > {2g} - 2,{b}_{N}\left( K\right) = {h}_{K}\left( {{q}^{N - g + 1} - 1}\right) /\left( {q - 1}\right) \) . By the first part of the proposition we find, for \( N \) in this range, \[ \frac{F\left( N\right) }{{b}_{N}\left( K\right) } = \frac{1}{{\zeta }_{K}\left( 2\right) }\frac{{q}^{N - g + 1}}{{q}^{N - g + 1} - 1} + {O}_{\epsilon }\left( {q}^{\left( {-\frac{3}{4} + \epsilon }\right) N}\right) . \] Now, simply pass to the limit as \( N \) tends to \( \infty \) .
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Yes
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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\[ \mathop{\sum }\limits_{{\deg D = N}}\sigma \left( D\right) = {\zeta }_{K}\left( 2\right) \frac{{h}_{K}}{{q}^{g - 1}\left( {q - 1}\right) }{q}^{2N} + {O}_{\epsilon }\left( {q}^{\left( {1 + \epsilon }\right) N}\right) . \]
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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 \( s \) . This yields \( {\zeta }_{\sigma }\left( {s + 1}\right) = {\zeta }_{K}\left( {s + 1}\right) {\zeta }_{K}\left( s\right) \) . This function has a simple pole at \( s = 1 \) and is otherwise holomorphic on the region \( \Re \left( s\right) > 0 \) . Choose an \( \epsilon > 0 \) and set \( {\delta }^{\prime } = \epsilon \) in Theorem 17.1. We have \( {\zeta }_{\sigma }\left( {s + 1}\right) \) is holomorphic on the region \( \Re \left( s\right) \geq \epsilon \) except for a simple pole at \( s = 1 \) with residue \( {\zeta }_{K}\left( 2\right) {\rho }_{K} \) .\n\nWe are all set to apply Theorem 17.1, except that we need the expansion of \( {\zeta }_{\sigma }\left( {s + 1}\right) \) as a power series in \( {q}^{-s} = u \) . This is easy,\n\n\[ {\zeta }_{\sigma }\left( {s + 1}\right) = \mathop{\sum }\limits_{D}\frac{\sigma \left( D\right) }{N{D}^{s + 1}} = \mathop{\sum }\limits_{{N = 0}}^{\infty }\left( {\mathop{\sum }\limits_{{\deg D = N}}\frac{\sigma \left( D\right) }{ND}}\right) {q}^{-{Ns}} \]\n\n\[ = \mathop{\sum }\limits_{{N = 0}}^{\infty }\left( {{q}^{-N}\mathop{\sum }\limits_{{\deg D = N}}\sigma \left( D\right) }\right) {u}^{N}. \]\n\nIt follows that\n\n\[ {q}^{-N}\mathop{\sum }\limits_{{\deg D = N}}\sigma \left( D\right) = {\zeta }_{K}\left( 2\right) {\rho }_{K}\log \left( q\right) {q}^{N} + {O}_{\epsilon }\left( {q}^{\epsilon N}\right) . \]\n\nMultiply both sides of this equation by \( {q}^{N} \) and use the explicit expression for \( {\rho }_{K} \) given by Equation 4. This finishes the proof.
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Yes
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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 \( \{ s \in B \mid \Re \left( s\right) = 1\} \) except for a pole of order \( r \) at \( s = 1 \) . Let \( \alpha = \mathop{\lim }\limits_{{s \rightarrow 1}}{\left( s - 1\right) }^{r}{\zeta }_{f}\left( s\right) \) . Then, there is a \( \delta < 1 \) and constants \( {c}_{-i} \) with \( 1 \leq i \leq r \) such that\n\n\[ F\left( N\right) = \mathop{\sum }\limits_{{\deg D = N}}f\left( D\right) = {q}^{N}\left( {\mathop{\sum }\limits_{{i = 1}}^{r}{c}_{-i}\left( \begin{matrix} N + i - 1 \\ i - 1 \end{matrix}\right) {\left( -q\right) }^{i}}\right) + O\left( {q}^{\delta N}\right) . \]
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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 \( {C}_{\epsilon } \) a small circle about \( s = 0 \) oriented clockwise. By the Cauchy integral theorem, the integral\n\n\[ \frac{1}{2\pi i}{\oint }_{{C}_{\epsilon } + C}\frac{{Z}_{f}\left( u\right) }{{u}^{N + 1}}{du} \]\n\nis equal to the sum of the residues of the function \( {Z}_{f}\left( u\right) {u}^{-N - 1} \) in the region between the two circles. There is only one pole in this region. It is located at \( u = {q}^{-1} \) . To find the residue there, we expand both \( {Z}_{f}\left( u\right) \) and \( {u}^{-N - 1} \) in Laurent series about \( u = {q}^{-1} \), multiply the results together, and pick out the coefficient of \( {\left( u - {q}^{-1}\right) }^{-1} \) .\n\nBy using the Taylor series formula or the general binomial expansion theorem we find\n\n\[ {u}^{-N - 1} = {q}^{N + 1}\mathop{\sum }\limits_{{j = 0}}^{\infty }\left( \begin{matrix} - N - 1 \\ j \end{matrix}\right) {q}^{j}{\left( u - {q}^{-1}\right) }^{j}. \]\n\nThe Laurent series for \( {Z}_{f}\left( u\right) \) has the form\n\n\[ {Z}_{f}\left( u\right) = \mathop{\sum }\limits_{{i = - r}}^{\infty }{c}_{i}{\left( u - {q}^{-1}\right) }^{i} \]\n\nwith \( {c}_{-r} \neq 0 \) .\n\nMultiplying these two series together and isolating the coefficient of \( {\left( u - {q}^{-1}\right) }^{-1} \) in the result yields\n\n\[ {\operatorname{Res}}_{u = {q}^{-1}}{Z}_{f}\left( u\right) {u}^{-N - 1} = {q}^{N + 1}\mathop{\sum }\limits_{{i = - r}}^{{-1}}{c}_{i}\left( \begin{matrix} - N - 1 \\ - i - 1 \end{matrix}\right) {q}^{-i - 1} \]\n\n\[ = {q}^{N}\mathop{\sum }\limits_{{i = 1}}^{r}{c}_{-i}\left( \begin{matrix} - N - 1 \\ i - 1 \end{matrix}\right) {q}^{i}. \]\n\nTo get the last equality we simply transformed \( i \) to \( - i \) and redistributed one factor of \( q \) .\n\nIt is easy to see that \( \left( \begin{matrix} - N - 1 \\ k \end{matrix}\right) = {\left( -1\right) }^{k}\left( \begin{matrix} N + k \\ k \end{matrix}\right) \), so the residue can be rewrit-\n\n\[ - {q}^{N}\mathop{\sum }\limits_{{i = 1}}^{r}{c}_{-i}\left( \begin{matrix} N + i - 1 \\ i - 1 \end{matrix}\right) {\left( -q\right) }^{i}. \]\n\nAs in the proof of Theorem 17.1, it now follows that\n\n\[ F\left( N\right) = {q}^{N}\left( {\mathop{\sum }\limits_{{i = 1}}^{r}{c}_{-i}\left( \begin{matrix} N + i - 1 \\ i - 1 \end{matrix}\right) {\left( -q\right) }^{i}}\right) + O\left( {q}^{\delta N}\right) . \]
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Yes
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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\[ \mathop{\sum }\limits_{{\deg D = N}}d\left( D\right) = {q}^{N}\left( {{\lambda }_{K}N + {\mu }_{K}}\right) + {O}_{\epsilon }\left( {q}^{\epsilon N}\right) . \]\n\nMore explicitly, \( {\lambda }_{K} = {h}_{K}^{2}{q}^{2 - {2g}}{\left( q - 1\right) }^{-2} \) .
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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 \( \mathop{\lim }\limits_{{s \rightarrow 1}}{\left( s - 1\right) }^{2}{\zeta }_{K}{\left( s\right) }^{2} = {\rho }_{K}^{2} \) . Applying Theorem 17.4 and the remarks given after that theorem we find there are constants \( {\lambda }_{K} \) and \( {\mu }_{K} \) such that\n\n\[ \mathop{\sum }\limits_{{\deg D = N}}d\left( D\right) = {q}^{N}\left( {{\lambda }_{K}N + {\mu }_{K}}\right) + {O}_{\epsilon }\left( {q}^{\epsilon N}\right) . \]\n\nApplying the formula for the leading term of the polynomial in parenthesis given in the statement of Theorem 17.4, we find\n\n\[ {\lambda }_{K} = \frac{\log {\left( q\right) }^{r}}{\left( {r - 1}\right) !}\alpha = \frac{\log {\left( q\right) }^{2}}{1!}{\rho }_{K}^{2} = \frac{{h}_{K}^{2}}{{q}^{{2g} - 2}{\left( q - 1\right) }^{2}}. \]\n\nThis finishes the proof.
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Yes
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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 \( {\mathcal{O}}_{{m}_{0}}/{\mathcal{O}}_{m} \). Finally, if \( \mathcal{O} \) is any \( A \) -order in \( K \), then \( \mathcal{O} = {\mathcal{O}}_{m} \) for some \( m \in A \).
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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 suitable \( r, s \in k \). The automorphism \( \sigma \) takes \( \sqrt{{m}_{0}} \) to \( - \sqrt{{m}_{0}} \).\n\nSuppose \( r + s\sqrt{{m}_{0}} \) is integral over \( A \). Applying \( \sigma \) we see that \( r - s\sqrt{{m}_{0}} \) is also integral over \( A \). Thus, so is the sum and product of these two elements, i.e. \( {2r} \) and \( {r}^{2} - {m}_{0}{s}^{2} \) are integral over \( A \). Since \( A \) is integrally closed, we have \( {2r} \in A \) and \( {r}^{2} - {m}_{0}{s}^{2} \in A \). We may divide by 2, so \( r \in A \) and it follows that \( {m}_{0}{s}^{2} \in A \). Since \( {m}_{0} \) is square-free, we must have \( s \in A \). We have proved that if \( r + s\sqrt{{m}_{0}} \) is integral over \( A \), then \( r + s\sqrt{{m}_{0}} \in A + A\sqrt{{m}_{0}} \). The converse is clear, so our first assertion is established.\n\nFrom the definitions, \( {\mathcal{O}}_{m} = A + A\sqrt{m} = A + A{m}_{1}\sqrt{{m}_{0}} \subseteq A + A\sqrt{{m}_{0}} = \) \( {\mathcal{O}}_{{m}_{0}} \). It is then immediate that as \( A \) -modules\n\n\[{\mathcal{O}}_{{m}_{0}}/{\mathcal{O}}_{m} \cong A/{m}_{1}A\]\n\nwhich proves the second assertion.\n\nLet \( \mathcal{O} \) be any \( A \) -order in \( K \). One can easily show that every element of \( \mathcal{O} \) is integral over \( A \). Since \( 1 \in \mathcal{O} \) by definition, we have \( A \subset \mathcal{O} \). Since \( K \) is the quotient field of \( \mathcal{O} \) there is an element \( \alpha \in \mathcal{O} \) such that \( \alpha \notin A \). By the first part of the proof we can write \( \alpha = a + b\sqrt{{m}_{0}} \) with \( a, b \in A \) and \( b \neq 0 \). It follows that \( b\sqrt{{m}_{0}} \in \mathcal{O} \) with \( b \in A - \{ 0\} \). Choose \( {m}_{1} \in A - \{ 0\} \) to be a non-zero polynomial of least degree such that \( {m}_{1}\sqrt{{m}_{0}} \in \mathcal{O} \). Set \( m = {m}_{0}{m}_{1}^{2} \). We claim that \( \mathcal{O} = {\mathcal{O}}_{m} \). It is clear that \( {\mathcal{O}}_{m} \subseteq \mathcal{O} \), so we must show the reverse inclusion. Suppose \( a + b\sqrt{{m}_{0}} \in \mathcal{O} \) with \( a, b \in A \). By the division algorithm in \( A \) we can write \( b = c{m}_{1} + r \), where \( c, r \in A \) and either \( r = 0 \) or \( \deg r < \deg {m}_{1} \). Multiply this relation on the right by \( \sqrt{{m}_{0}} \) and we can deduce that \( r\sqrt{{m}_{0}} \in \mathcal{O} \). Since \( {m}_{1} \) is a non-zero polynomial of least degree with this property, we conclude that \( r = 0 \). Thus, \( \alpha = a + c{m}_{1}\sqrt{{m}_{0}} \in {\mathcal{O}}_{m} \) and we are done.
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Yes
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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 \( {\left( 1 + {q}^{-s}\right) }^{-1} \) if \( \infty \) is inert in \( K \) . Then\n\n\[ \n{L}_{\infty }\left( {s,{\chi }_{m}}\right) L\left( {s,{\chi }_{m}}\right)\n\]\n\nis the Artin L-function associated to the unique non-trivial character of \( \operatorname{Gal}\left( {K/k}\right) \) .
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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 \( L \) -function associated to the unique non-trivial character \( \chi \) of \( \operatorname{Gal}\left( {K/k}\right) \) . If \( P \) is a finite prime, \( \chi \left( P\right) = 0 \) if \( P \) is ramified and \( \chi \left( P\right) = \chi \left( \left( {P, K/k}\right) \right) \) if \( P \) is unramified. By the definition of the Artin symbol, \( \left( {P, K/k}\right) \) is \( e \) if \( P \) splits, and \( \sigma \) if \( P \) is inert \( (\sigma \) being the non-trivial element of the Galois group). Thus, \( \chi \left( P\right) = 1 \) if \( P \) splits and \( \chi \left( P\right) = - 1 \) if \( P \) is inert. By the decomposition law in quadratic extensions (take \( l = 2 \) in Proposition 10.5), \( P \) splits in \( K \) if and only if \( {\chi }_{m}\left( P\right) = 1 \) . Thus, for finite primes \( \chi \left( P\right) = {\chi }_{m}\left( P\right) \) . At \( \infty \) we know that \( \left| \infty \right| = q \) so \( {\left( 1 - \chi \left( \infty \right) {q}^{-s}\right) }^{-1} \) is 1 if \( \infty \) is ramified, \( {\left( 1 - {q}^{-s}\right) }^{-1} \) if \( \infty \) splits, and \( {\left( 1 + {q}^{-s}\right) }^{-1} \) if \( \infty \) is inert. Thus, \( {\left( 1 - \chi \left( \infty \right) {\left| \infty \right| }^{-s}\right) }^{-1} = {L}_{\infty }\left( {s,{\chi }_{m}}\right) \) . We have shown that \( L\left( {s,\chi }\right) \) and \( {L}_{\infty }\left( {s,{\chi }_{m}}\right) L\left( {s,{\chi }_{m}}\right) \) have the same Euler factors for all primes. Thus, they are equal.
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Yes
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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\[ \n{h}_{m} = {h}_{{m}_{0}}\frac{\left| {m}_{1}\right| }{\left\lbrack {\mathcal{O}}_{{m}_{0}}^{ * } : {\mathcal{O}}_{m}^{ * }\right\rbrack }\mathop{\prod }\limits_{{P \mid {m}_{1}}}\left( {1 - {\chi }_{{m}_{0}}\left( P\right) {\left| P\right| }^{-1}}\right) .\n\]
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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 \( \mathbb{Z} \) -rank 1 and one can show without much difficulty that the index is the same as the quotient of regulators \( {R}_{m}/{R}_{{m}_{0}} \) . We set \( {R}_{m} = {R}_{{m}_{0}} = 1 \) in the first two cases. Then the relationship given by Proposition 17.9 can be rewritten\n\n\[ \n\frac{{h}_{m}{R}_{m}}{\sqrt{\left| m\right| }} = \frac{{h}_{{m}_{0}}{R}_{{m}_{0}}}{\sqrt{\left| {m}_{0}\right| }}\mathop{\prod }\limits_{{P \mid {m}_{1}}}\left( {1 - {\chi }_{{m}_{0}}\left( P\right) {\left| P\right| }^{-1}}\right) .\n\]
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No
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Lemma 17.10. If \( m \notin {\mathbb{F}}^{ * } \) is not a square, \( {S}_{d}\left( {\chi }_{m}\right) = 0 \) for \( d \geq M = \) \( \deg \left( m\right) \) .
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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 \( {\chi }_{m}\left( n\right) = \) \( {c}_{d}\left( {n/m}\right) \) . Thus, if \( d \geq M \) ,\n\n\[ {S}_{d}\left( {\chi }_{m}\right) = {c}_{d}\mathop{\sum }\limits_{\substack{{n\text{ monic }} \\ {\deg \left( n\right) = d} }}\left( \frac{n}{m}\right) = 0 \]\n\nby the proof of Proposition 4.3.
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Yes
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Proposition 17.11. \( \Phi \left( {0, M}\right) = {q}^{M} \) and if \( M, N \geq 1 \), then\n\n\[ \Phi \left( {N, M}\right) = {q}^{M + N}\left( {1 - \frac{1}{q}}\right) . \]
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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 \( \left( {{n}^{\prime },{m}^{\prime }}\right) \) equivalent if \( \gcd \left( {n, m}\right) = \gcd \left( {{n}^{\prime },{m}^{\prime }}\right) \) . Breaking the set \( \{ \left( {n, m}\right) \mid \deg \left( n\right) = N,\deg \left( m\right) = \) \( M\} \) into equivalence classes and counting leads to the identity\n\n\[ {q}^{N + M} = \mathop{\sum }\limits_{{d = 0}}^{{\min \left( {N, M}\right) }}{q}^{d}\Phi \left( {N - d, M - d}\right) . \]\n\nSuppose \( M, N \geq 1 \) . The proof now proceeds by induction on the number \( M + N \) . The smallest value this number can have is 2, in which case the formula yields \( {q}^{2} = \Phi \left( {1,1}\right) + {q\Phi }\left( {0,0}\right) \), or \( \Phi \left( {1,1}\right) = {q}^{2} - q = {q}^{2}\left( {1 - {q}^{-1}}\right) \) .\n\nNow suppose the formula is correct for all pairs \( {N}^{\prime },{M}^{\prime } \geq 1 \) with \( {N}^{\prime } + \) \( {M}^{\prime } < N + M \) . We may also suppose, by symmetry, that \( N \leq M \) . Then\n\n\[ {q}^{M + N} = \Phi \left( {N, M}\right) + \mathop{\sum }\limits_{{d = 1}}^{{N - 1}}{q}^{d}\Phi \left( {N - d, M - d}\right) + {q}^{N}\Phi \left( {0, M - N}\right) . \]\n\nFor \( 1 \leq d \leq N - 1 \) we have \( \Phi \left( {N - d, M - d}\right) = {q}^{M + N - {2d}}\left( {1 - {q}^{-1}}\right) \) whereas by the first part of the proof, \( \Phi \left( {0, M - N}\right) = {q}^{M - N} \) . Substituting into the above formula and simplifying slightly,\n\n\[ {q}^{M + N} = \Phi \left( {N, M}\right) + {q}^{M + N}\mathop{\sum }\limits_{{d = 1}}^{{N - 1}}{q}^{-d}\left( {1 - {q}^{-1}}\right) + {q}^{M} = \Phi \left( {N, M}\right) + {q}^{M + N - 1}. \]\n\nThe second assertion now follows immediately.
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Yes
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Proposition 17.12. Suppose \( 1 \leq d \leq M - 1 \) . Then\n\n\[ \mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}{S}_{d}\left( {\chi }_{m}\right) = {\left( q - 1\right) }^{-1}\mathop{\sum }\limits_{{\deg \left( m\right) = M}}{S}_{d}\left( {\chi }_{m}\right) = \Phi \left( {d/2, M}\right) . \]
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Proof. To begin with assume all sums are over monics. Then\n\n\[ \mathop{\sum }\limits_{{\deg \left( m\right) = M}}{S}_{d}\left( {\chi }_{m}\right) = \mathop{\sum }\limits_{{\deg \left( m\right) = M}}\mathop{\sum }\limits_{{\deg \left( n\right) = d}}\left( \frac{m}{n}\right) = \mathop{\sum }\limits_{{\deg \left( n\right) = d}}\mathop{\sum }\limits_{{\deg \left( m\right) = M}}\left( \frac{m}{n}\right) . \]\n\nIf \( n \) is not a square, \( \left( {*/n}\right) \) is a non-trivial character modulo \( n \) . Thus, in this case, since \( M > \deg n = d \),\n\n\[ \mathop{\sum }\limits_{{\deg \left( m\right) = M}}\left( \frac{m}{n}\right) = 0 \]\n\nby the proof of Proposition 4.3.\n\nNow, suppose that \( n = {n}_{1}^{2} \) is a square. Then \( \left( {m/n}\right) = {\left( m/{n}_{1}\right) }^{2} = 1 \) whenever \( \gcd \left( {m,{n}_{1}}\right) = 1 \) and \( {\left( m/{n}_{1}\right) }^{2} = 0 \) otherwise. It follows that\n\n\[ \mathop{\sum }\limits_{{\deg \left( m\right) = M}}\left( \frac{m}{n}\right) = \mathop{\sum }\limits_{{\deg \left( m\right) = M}}{\left( \frac{m}{{n}_{1}}\right) }^{2} = {\Phi }_{{n}_{1}}\left( M\right) . \]\n\nThus\n\n\[ \mathop{\sum }\limits_{{\deg \left( m\right) = M}}{S}_{d}\left( {\chi }_{m}\right) = \mathop{\sum }\limits_{{\deg \left( {n}_{1}\right) = d/2}}{\Phi }_{{n}_{1}}\left( M\right) = \Phi \left( {d/2, M}\right) . \]\n\nTo do the general case, let \( \alpha \in {\mathbb{F}}^{ * } \) and sum over all \( {\alpha m} \) as \( m \) runs through the monics of degree \( M \) . The above calculation shows the answer is again equal to \( \Phi \left( {d/2, M}\right) \) . It follows that if we sum over all polynomials of degree \( d \) the answer is \( \left( {q - 1}\right) \Phi \left( {d/2, M}\right) \) . This completes the proof.
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Yes
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Theorem 17.13. Let \( M \) be odd and positive. We have, for all \( s \in B \) with \( s \neq \frac{1}{2} \)\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {s,{\chi }_{m}}\right) = \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) } - \left( {1 - \frac{1}{q}}\right) {\left( {q}^{1 - {2s}}\right) }^{\frac{M + 1}{2}}{\zeta }_{A}\left( {2s}\right) .\n\]\n\nFor \( s = \frac{1}{2} \), we have\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {1/2,{\chi }_{m}}\right) = 1 + \left( {1 - \frac{1}{q}}\right) \left( \frac{M - 1}{2}\right) .\n\]
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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\[ \n\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {s,{\chi }_{m}}\right) = \mathop{\sum }\limits_{{d = 0}}^{{M - 1}}\left( {\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}{S}_{d}\left( {\chi }_{m}\right) }\right) {q}^{-{ds}}\n\]\n\n\[ \n= {q}^{M} + \Phi \left( {1, M}\right) {q}^{-{2s}} + \Phi \left( {2, M}\right) {q}^{-{4s}} + \cdots + \Phi \left( {\left( {M - 1}\right) /2, M}\right) {q}^{-\left( {M - 1}\right) s}\n\]\n\n\[ \n= {q}^{M}\left( {1 + \left( {1 - \frac{1}{q}}\right) \left\lbrack {{q}^{1 - {2s}} + {\left( {q}^{1 - {2s}}\right) }^{2} + \cdots + {\left( {q}^{1 - {2s}}\right) }^{\frac{M - 1}{2}}}\right\rbrack }\right) .\n\]\n\nThe result for \( s = \frac{1}{2} \) follows from this by substitution. For \( s \neq \frac{1}{2} \) we sum the geometric series to derive\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {s,{\chi }_{m}}\right) = 1 + \left( {1 - \frac{1}{q}}\right) {q}^{1 - {2s}}\frac{1 - {\left( {q}^{1 - {2s}}\right) }^{\frac{M - 1}{2}}}{1 - {q}^{1 - {2s}}}\n\]\n\n\[ \n= 1 + \left( {1 - \frac{1}{q}}\right) \frac{{q}^{1 - {2s}}}{1 - {q}^{1 - {2s}}} - \left( {1 - \frac{1}{q}}\right) {\left( {q}^{1 - {2s}}\right) }^{\frac{M + 1}{2}}{\zeta }_{A}\left( {2s}\right) .\n\]\n\nWe have used the fact that \( {\zeta }_{A}\left( s\right) = {\left( 1 - {q}^{1 - s}\right) }^{-1} \), a fact we will use again almost immediately.\n\nA close look at the last line shows that it only remains to identify the sum of the first two terms with a quotient of zeta values. This follows from the calculation\n\n\[ \n1 + \left( {1 - \frac{1}{q}}\right) \frac{{q}^{1 - {2s}}}{1 - {q}^{1 - {2s}}} = 1 + \frac{{q}^{1 - {2s}} - {q}^{-{2s}}}{1 - {q}^{1 - {2s}}} = \frac{1 - {q}^{-{2s}}}{1 - {q}^{1 - {2s}}} = \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) }.\n\]
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Yes
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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 \( M \rightarrow \infty \) through odd values.
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Proof. This follows immediately from the theorem together with the observation that if \( \Re \left( s\right) > \frac{1}{2} \) then \( \left| {q}^{1 - {2s}}\right| < 1 \) .
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No
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Corollary 2. If \( M \) is odd and positive, then\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}{h}_{m} = \frac{{\zeta }_{A}\left( 2\right) }{{\zeta }_{A}\left( 3\right) }{q}^{\frac{M - 1}{2}} - {q}^{-1}.\n\]
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Proof. We begin by substituting \( s = 1 \) into the identity given in the theorem. We find\n\n\[ \n{q}^{-M}\mathop{\sum }\limits_{\substack{{m\text{ monic }} \\ {\deg \left( m\right) = M} }}L\left( {1,{\chi }_{m}}\right) = \frac{{\zeta }_{A}\left( 2\right) }{{\zeta }_{A}\left( 3\right) } - \left( {1 - \frac{1}{q}}\right) {q}^{-\frac{M + 1}{2}}{\zeta }_{A}\left( 2\right) = \frac{{\zeta }_{A}\left( 2\right) }{{\zeta }_{A}\left( 3\right) } - {q}^{-\frac{M + 1}{2}}.\n\]\n\nThe last equality follows from \( {\zeta }_{A}\left( 2\right) = {\left( 1 - {q}^{-1}\right) }^{-1} \) .\n\nBy Theorem 17.8A and Theorem 17.8B, we see that \( L\left( {1,{\chi }_{m}}\right) = {h}_{m}\frac{\sqrt{q}}{\sqrt{\left| m\right| }} = \) \( {h}_{m}{q}^{-\frac{M - 1}{2}} \) . Substituting this information into the last equation yields the result.
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Yes
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Corollary 1. If \( \operatorname{Re}\left( s\right) > \frac{1}{2} \), then as \( M \rightarrow \infty \) though even integers,
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\[ {q}^{-M}\sum L\left( {s,{\chi }_{m}}\right) \rightarrow \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) }.\]
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No
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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\[ \n{q}^{-M}\sum L\left( {s,{\chi }_{\gamma m}}\right) = \n\]
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\n\( \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) } - \left( {1 - \frac{1}{q}}\right) {\left( {q}^{1 - {2s}}\right) }^{\frac{M}{2}}{\zeta }_{A}\left( {2s}\right) - {q}^{-\frac{M}{2}}\left( {\frac{1 + {q}^{-s}}{1 + {q}^{1 - s}} - \left( {1 - \frac{1}{q}}\right) \frac{{\left( {q}^{1 - s}\right) }^{M}}{1 + {q}^{1 - s}}}\right) . \)
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Yes
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Corollary 1. If \( \Re \left( s\right) > \frac{1}{2} \), then as \( M \rightarrow \infty \) through even integers,
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\[ {q}^{-M}\sum L\left( {s,{\chi }_{\gamma m}}\right) \rightarrow \frac{{\zeta }_{A}\left( {2s}\right) }{{\zeta }_{A}\left( {{2s} + 1}\right) }.\]
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No
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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.
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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.
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Yes
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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 the order of \( G \) .
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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 distinct cosets. Suppose\n\n\[ {y}_{j}{x}_{i}K = {y}_{{j}^{\prime }}{x}_{{i}^{\prime }}K \]\n\nfor a pair of indices \( \left( {j, i}\right) \) and \( \left( {{j}^{\prime },{i}^{\prime }}\right) \) . Multiplying by \( H \) on the right, and noting that \( {x}_{i},{x}_{{i}^{\prime }} \) are in \( H \), we get\n\n\[ {y}_{j}H = {y}_{{j}^{\prime }}H \]\n\nwhence \( {y}_{j} = {y}_{{j}^{\prime }} \) . From this it follows that \( {x}_{i}K = {x}_{{i}^{\prime }}K \) and therefore that \( {x}_{i} = {x}_{{i}^{\prime }} \), as was to be shown.
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Yes
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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 \( \{ e\} \) .
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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 \( X \) be the cyclic group generated by \( x \) . Let \( {G}^{\prime } = G/X \) . By induction, we can find a cyclic tower in \( {G}^{\prime } \), and its inverse image is a cyclic tower in \( G \) whose last element is \( X \) . If we refine this tower by inserting \( \{ e\} \) at the end, we obtain the desired cyclic tower.
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Yes
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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.
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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 \( {H}_{i + 1} \) is normal in \( {H}_{i} \) , and we have an embedding \( {H}_{i}/{H}_{i + 1} \rightarrow {G}_{i}/{G}_{i + 1} \), whence \( {H}_{i}/{H}_{i + 1} \) is abelian, whence proving that \( H \) is solvable. We leave the proofs of the other statements to the reader.
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No
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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\[ \\left( {u \\cap V}\\right) v\\;\\text{ is normal in }\\;\\left( {U \\cap V}\\right) v,\]\n\nand the factor groups are isomorphic, i.e.\n\n\[ u\\left( {U \\cap V}\\right) /u\\left( {U \\cap v}\\right) \\approx \\left( {U \\cap V}\\right) v/\\left( {u \\cap V}\\right) v.\]
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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 other points in the diagram correspond to certain groups which can be determined as follows. The intersection of two line segments going downwards represents the intersection of groups. Two lines going upwards meet in a point which represents the product of two subgroups (i.e. the smallest subgroup containing both of them).\n\nWe consider the two parallelograms representing the wings of the butterfly, and we shall give isomorphisms of the factor groups as follows:\n\n\[ \\frac{u\\left( {U \\cap V}\\right) }{u\\left( {U \\cap v}\\right) } \\approx \\frac{U \\cap V}{\\left( {u \\cap V}\\right) \\left( {U \\cap v}\\right) } \\approx \\frac{\\left( {U \\cap V}\\right) v}{\\left( {u \\cap V}\\right) v}.\]\n\nIn fact, the vertical side common to both parallelograms has \( U \\cap V \) as its top end point, and \( \\left( {u \\cap V}\\right) \\left( {U \\cap v}\\right) \) as its bottom end point. We have an isomorphism\n\n\[ \\left( {U \\cap V}\\right) /\\left( {u \\cap V}\\right) \\left( {U \\cap v}\\right) \\approx u\\left( {U \\cap V}\\right) /u\\left( {U \\cap v}\\right) .\]\n\nThis is obtained from the isomorphism theorem\n\n\[ H/\\left( {H \\cap N}\\right) \\approx {HN}/N\n\nby setting \( H = U \\cap V \) and \( N = u\\left( {U \\cap v}\\right) \) . This gives us the isomorphism on the left. By symmetry we obtain the corresponding isomorphism on the right, which proves the Butterfly lemma.
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Yes
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Theorem 3.4. (Schreier) Let \( G \) be a group. Two normal towers of subgroups ending with the trivial group have equivalent refinements.
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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\[ \nG = {G}_{11} \supset {G}_{12} \supset \cdots \supset {G}_{1, s - 1} \supset {G}_{2} \n\]\n\n\[ \n= {G}_{21} \supset {G}_{22} \supset \cdots \supset {G}_{r - 1,1} \supset \cdots \supset {G}_{r - 1, s - 1} \supset \{ e\} . \n\]\n\nSimilarly, we define\n\n\[ \n{H}_{ji} = {H}_{j + 1}\left( {{G}_{i} \cap {H}_{j}}\right) \n\]\n\nfor \( j = 1,\ldots, s - 1 \) and \( i = 1,\ldots, r \) . This yields a refinement of the second tower. By the butterfly lemma, for \( i = 1,\ldots, r - 1 \) and \( j = 1,\ldots, s - 1 \) we have isomorphisms\n\n\[ \n{G}_{ij}/{G}_{i, j + 1} \approx {H}_{ji}/{H}_{j, i + 1} \n\]\n\nWe view each one of our refined towers as having \( \left( {r - 1}\right) \left( {s - 1}\right) + 1 \) elements, namely \( {G}_{ij}\left( {i = 1,\ldots, r - 1;j = 1,\ldots, s - 1}\right) \) and \( \{ e\} \) in the first case, \( {H}_{ji} \) and \( \{ e\} \) in the second case. The preceding isomorphism for each pair of indices \( \left( {i, j}\right) \) shows that our refined towers are equivalent, as was to be proved.
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Yes
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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 \( G \) having the same properties is equivalent to this one.
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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 same. This proves our theorem.
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Yes
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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.
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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 of \( f\left( G\right) \) , which is therefore cyclic, so the image of \( f \) is cyclic. Next let \( H \) be a subgroup of \( G \) . We want to show \( H \) cyclic. Let \( a \) be a generator of \( G \) . Then we have a surjective homomorphism \( f : \mathbf{Z} \rightarrow G \) such that \( f\left( n\right) = {a}^{n} \) . The inverse image \( {f}^{-1}\left( H\right) \) is a subgroup of \( \mathbf{Z} \), and therefore equal to \( m\mathbf{Z} \) for some positive integer \( m \) . Since \( f \) is surjective, we also have a surjective homomorphism \( m\mathbf{Z} \rightarrow H \) . Since \( m\mathbf{Z} \) is cyclic (generated additively by \( m \) ), it follows that \( H \) is cyclic, thus proving the proposition.
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Yes
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Proposition 5.2. The number of conjugate subgroups to \( H \) is equal to the index of the normalizer of \( H \) .
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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 get a homomorphism of \( G \) into the group of permutations of 2 elements. Since there is one conjugate of \( H \) unequal to \( H \) , then the kernel of our homomorphism is normal, of index 2, hence equal to \( H \) , which is normal, a contradiction which concludes the proof.
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No
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Proposition 5.3. There exists a unique homomorphism \( \varepsilon : {S}_{n} \rightarrow \{ \pm 1\} \) such that for every transposition \( \tau \) we have \( \varepsilon \left( \tau \right) = - 1 \) .
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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 the two integers \( r \) and \( s \) . Say \( r < s \) . We wish to determine\n\n\[ {\tau \Delta }\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \mathop{\prod }\limits_{{i < j}}\left( {{x}_{\tau \left( j\right) } - {x}_{\tau \left( i\right) }}\right) . \]\n\nFor one factor involving \( j = s, i = r \), we see that \( \tau \) changes the factor \( \left( {{x}_{s} - {x}_{r}}\right) \) to \( - \left( {{x}_{s} - {x}_{r}}\right) \) . All other factors can be considered in pairs as follows:\n\n\[ \left( {{x}_{k} - {x}_{s}}\right) \left( {{x}_{k} - {x}_{r}}\right) \;\text{if}k > s\text{,} \]\n\n\[ \left( {{x}_{s} - {x}_{k}}\right) \left( {{x}_{k} - {x}_{r}}\right) \;\text{if}r < k < s, \]\n\n\[ \left( {{x}_{s} - {x}_{k}}\right) \left( {{x}_{r} - {x}_{k}}\right) \;\text{if}k < r\text{.} \]\n\nEach one of these pairs remains unchanged when we apply \( \tau \) . Hence we see that \( {\tau \Delta } = - \Delta \) .\n\nLet \( \varepsilon \left( \sigma \right) \) be the sign 1 or -1 such that \( {\sigma \Delta } = \varepsilon \left( \sigma \right) \Delta \) for a permutation \( \sigma \) . Since \( \pi \left( {\sigma \tau }\right) = \pi \left( \sigma \right) \pi \left( \tau \right) \), it follows at once that \( \varepsilon \) is a homomorphism, and the proposition is proved.
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Yes
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Theorem 5.4. If \( n \geqq 5 \) then \( {S}_{n} \) is not solvable.
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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 \( {J}_{n} \), and let \( \sigma = \left\lbrack {ijk}\right\rbrack \) and \( \tau = \left\lbrack {krs}\right\rbrack \) . Then a direct computation gives their commutator\n\n\[ \n{\sigma \tau }{\sigma }^{-1}{\tau }^{-1} = \left\lbrack {rki}\right\rbrack \n\]\n\nSince the choice of \( i, j, k, r, s \) was arbitrary, we see that the cycles \( \left\lbrack {rki}\right\rbrack \) all lie in \( N \) for all choices of distinct \( r, k, i \), thereby proving what we wanted.\n\nNow suppose that we have a tower of subgroups\n\n\[ \n{S}_{n} = {H}_{0} \supset {H}_{1} \supset {H}_{2} \supset \cdots \supset {H}_{m} = \{ e\} \n\]\n\nsuch that \( {H}_{v} \) is normal in \( {H}_{v - 1} \) for \( v = 1,\ldots, m \), and \( {H}_{v}/{H}_{v - 1} \) is abelian. Since \( {S}_{n} \) contains every 3-cycle, we conclude that \( {H}_{1} \) contains every 3-cycle. By induction, we conclude that \( {H}_{m} = \{ e\} \) contains every 3-cycle, which is impossible, thus proving the theorem.
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Yes
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Theorem 5.5. If \( n \geqq 5 \) then the alternating group \( {A}_{n} \) is simple.
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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 \( \sigma \left( i\right) = i \) . It will suffice to prove that \( \sigma \) is a 3-cycle or the identity. Decompose \( {J}_{n} \) into disjoint orbits of \( \langle \sigma \rangle \) . Then some orbits have more than one element. Suppose all orbits have 2 elements (except for the fixed points). Since \( \sigma \) is even, there are at least two such orbits. On their union, \( \sigma \) is represented as a product of two transpositions \( \left\lbrack {ij}\right\rbrack \left\lbrack {rs}\right\rbrack \) . Let \( k \neq i, j, r, s \) . Let \( \tau = \left\lbrack {rsk}\right\rbrack \) . Let \( {\sigma }^{\prime } = {\tau \sigma }{\tau }^{-1}{\sigma }^{-1} \) . Then \( {\sigma }^{\prime } \) is a product of a conjugate of \( \sigma \) and \( {\sigma }^{-1} \), so \( {\sigma }^{\prime } \in N \) . But \( {\sigma }^{\prime } \) leaves \( i, j \) fixed, and any element \( t \in {J}_{n}, t \neq i, j, r, s, k \) left fixed by \( \sigma \) is also fixed by \( {\sigma }^{\prime } \), so \( {\sigma }^{\prime } \) has more fixed points than \( \sigma \), contradicting our hypothesis.\n\nSo we are reduced to the case when at least one orbit of \( \langle \sigma \rangle \) has \( \geqq 3 \) elements, say \( i, j, k,\ldots \) If \( \sigma \) is not the 3-cycle \( \left\lbrack {ijk}\right\rbrack \), then \( \sigma \) must move at least two other elements of \( {J}_{n} \), otherwise \( \sigma \) is an odd permutation [ijkr] for some \( r \in {J}_{n} \), which is impossible. Then let \( \sigma \) move \( r, s \) other than \( i, j, k \), and let \( \tau = \left\lbrack {krs}\right\rbrack \) . Let \( {\sigma }^{\prime } \) be the commutator as before. Then \( {\sigma }^{\prime } \in N \) and \( {\sigma }^{\prime }\left( i\right) = i \), and all fixed points of \( \sigma \) are also fixed points of \( {\sigma }^{\prime } \) whence \( {\sigma }^{\prime } \) has more fixed points than \( \sigma \), a contradiction which proves the theorem.
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Yes
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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 \( p \) .
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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 \( G/H \) . Hence the order of \( G/H \) divides a power of \( n \) by induction, and consequently so does the order of \( G \) because\n\n\[ \left( {G : 1}\right) = \left( {G : H}\right) \left( {H : 1}\right) . \]\n\nLet \( G \) have order divisible by \( p \) . By what we have just seen, there exists an element \( x \) in \( G \) whose period is divisible by \( p \) . Let this period be \( {ps} \) for some integer \( s \) . Then \( {x}^{s} \neq 1 \) and obviously \( {x}^{s} \) has period \( p \), and generates a subgroup of order \( p \), as was to be shown.
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Yes
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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 \( G \) .
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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 \( p \), then a \( p \) -Sylow subgroup of \( H \) will also be one of \( G \), and our assertion follows by induction. We may therefore assume that every proper subgroup has an index divisible by \( p \) . We now let \( G \) act on itself by conjugation. From the class formula we obtain\n\n\[ \left( {G : 1}\right) = \left( {Z : 1}\right) + \sum \left( {G : {G}_{x}}\right) . \]\n\nHere, \( Z \) is the center of \( G \), and the term \( \left( {Z : 1}\right) \) corresponds to the orbits having one element, namely the elements of \( Z \) . The sum on the right is taken over the other orbits, and each index \( \left( {G : {G}_{x}}\right) \) is then \( > 1 \), hence divisible by \( p \) . Since \( p \) divides the order of \( G \), it follows that \( p \) divides the order of \( Z \), hence in particular that \( G \) has a non-trivial center.\n\nLet \( a \) be an element of order \( p \) in \( Z \), and let \( H \) be the cyclic group generated by \( a \) . Since \( H \) is contained in \( Z \), it is normal. Let \( f : G \rightarrow G/H \) be the canonical map. Let \( {p}^{n} \) be the highest power of \( p \) dividing \( \left( {G : 1}\right) \) . Then \( {p}^{n - 1} \) divides the order of \( G/H \) . Let \( {K}^{\prime } \) be a \( p \) -Sylow subgroup of \( G/H \) (by induction) and let \( K = {f}^{-1}\left( {K}^{\prime }\right) \) . Then \( K \supset H \) and \( f \) maps \( K \) onto \( {K}^{\prime } \) . Hence we have an isomorphism \( K/H \approx {K}^{\prime } \) . Hence \( K \) has order \( {p}^{n - 1}p = {p}^{n} \), as desired.
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Yes
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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 \( p \mid \# \left( S\right) \), then the number of fixed points of \( H \) is \( \equiv 0{\;\operatorname{mod}\;p} \).
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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 (b) and (c) are special cases of (a), thus proving the lemma.
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Yes
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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 \( \equiv 1{\;\operatorname{mod}\;p} \) .
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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\[ \left( {{HP} : P}\right) = \left( {H : H \cap P}\right) ,\]\n\nso if \( {HP} \neq P \), then \( {HP} \) has order a power of \( p \), and the order is larger than \( \# \left( P\right) \) , contradicting the hypothesis that \( P \) is a Sylow group. Hence \( {HP} = P \) and \( H \subset P \) .\n\nNext, let \( S \) be the set of all conjugates of \( P \) in \( G \) . Then \( G \) operates on \( S \) by conjugation. Since the normalizer of \( P \) contains \( P \), and has therefore index prime to \( p \), it follows that \( \# \left( S\right) \) is not divisible by \( p \) . Now let \( H \) be any \( p \) -subgroup. Then \( H \) also acts on \( S \) by conjugation. By Lemma 6.3(a), we know that \( H \) cannot have 0 fixed points. Let \( Q \) be a fixed point. By definition this means that \( H \) is contained in the normalizer of \( Q \), and hence by the first part of the proof, that \( H \subset Q \), which proves the first part of the theorem. The second part follows immediately by taking \( H \) to be a \( p \) -Sylow group, so \( \# \left( H\right) = \# \left( Q\right) \), whence \( H = Q \) . In particular, when \( H \) is a \( p \) -Sylow group, we see that \( H \) has only one fixed point, so that (iii) follows from Lemma 6.3(b). This proves the theorem.
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Yes
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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.
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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\[ \left( {G : 1}\right) = \operatorname{card}\left( Z\right) + \sum \left( {G : {G}_{x}}\right) \]\n\nthe sum being taken over certain \( x \) for which \( \left( {G : {G}_{x}}\right) \neq 1 \) . Then \( p \) divides \( \left( {G : 1}\right) \) and also divides every term in the sum, so that \( p \) divides the order of the center, as was to be shown.
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Yes
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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 \( p \) .
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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 \( G/H \) . Taking the inverse image of this tower in \( G \) gives us the desired sequence in \( G \) .
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No
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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.
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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 \( G \) into the symmetric group on \( p \) elements, whose order is \( p \) !. Let \( K \) be the kernel. Then \( K \) is the intersection of the isotropy groups, and the isotropy group of \( H \) is \( H \) by assumption, so \( K \subset H \) . If \( K \neq H \), then from \[ \left( {G : K}\right) = \left( {G : H}\right) \left( {H : K}\right) = p\left( {H : K}\right) , \] and the fact that only the first power of \( p \) divides \( p \) !, we conclude that some prime dividing \( \left( {p - 1}\right) \) ! also divides \( \left( {H : K}\right) \), which contradicts the assumption that \( p \) is the smallest prime dividing the order of \( G \), and proves the lemma.
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Yes
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Proposition 6.8. Let \( p, q \) be distinct primes and let \( G \) be a group of order pq. Then \( G \) is solvable.
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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.
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No
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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 \( j \) .
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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 that our map \( f \) is a homomorphism. Furthermore, we clearly have \( f \\circ {\\lambda }_{j}\\left( x\\right) = {f}_{j}\\left( x\\right) \) for each \( j \) and each \( x \\in {A}_{j} \) . Thus \( f \) has the desired commutativity property. It is also clear that the map \( f \) is uniquely determined, as was to be shown.
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Yes
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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 \( C \) with \( {A}^{\prime } \), and such that \( A = B \oplus C \) .
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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 \( {x}_{i}, i \in I \) . If we have a relation\n\n\[ \mathop{\sum }\limits_{{i \in I}}{n}_{i}{x}_{i} = 0 \]\n\nwith integers \( {n}_{i} \), almost all of which are equal to 0, then applying \( f \) yields\n\n\[ 0 = \mathop{\sum }\limits_{{i \in I}}{n}_{i}f\left( {x}_{i}\right) = \mathop{\sum }\limits_{{i \in I}}{n}_{i}{x}_{i}^{\prime } \]\n\nwhence all \( {n}_{i} = 0 \) . Hence our family \( {\left\{ {x}_{i}\right\} }_{i \in I} \) is a basis of \( C \) . Similarly, one sees that if \( z \in C \) and \( f\left( z\right) = 0 \) then \( z = 0 \) . Hence \( B \cap C = 0 \) . Let \( x \in A \) . Since \( f\left( x\right) \in {A}^{\prime } \) there exist integers \( {n}_{i}, i \in I \), such that\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{i \in I}}{n}_{i}{x}_{i}^{\prime } \]\n\nApplying \( f \) to \( x - \mathop{\sum }\limits_{{i \in I}}{n}_{i}{x}_{i} \), we find that this element lies in the kernel of \( f \) , say\n\n\[ x - \mathop{\sum }\limits_{{i \in I}}{n}_{i}{x}_{i} = b \in B. \]\n\nFrom this we see that \( x \in B + C \), and hence finally that \( A = B \oplus C \) is a direct sum, as contended.
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Yes
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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.
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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\[ A = \mathbf{Z}{x}_{1} \oplus \cdots \oplus \mathbf{Z}{x}_{n} \]\n\nLet \( f : A \rightarrow \mathbf{Z}{x}_{1} \) be the projection, i.e. the homomorphism such that\n\n\[ f\left( {{m}_{1}{x}_{1} + \cdots + {m}_{n}{x}_{n}}\right) = {m}_{1}{x}_{1} \]\n\nwhenever \( {m}_{i} \in \mathbf{Z} \) . Let \( {B}_{1} \) be the kernel of \( f \mid B \) . Then \( {B}_{1} \) is contained in the free subgroup \( \left\langle {{x}_{2},\ldots ,{x}_{n}}\right\rangle \) . By induction, \( {B}_{1} \) is free and has a basis with \( \leqq n - 1 \) elements. By the lemma, there exists a subgroup \( {C}_{1} \) isomorphic to a subgroup of \( \mathbf{Z}{x}_{1} \) (namely the image of \( f \mid B \) ) such that\n\n\[ B = {B}_{1} \oplus {C}_{1} \]\n\nSince \( f\left( B\right) \) is either 0 or infinite cyclic, i.e. free on one generator, this proves that \( B \) is free.\n\n(When \( A \) is not finitely generated, one can use a similar transfinite argument. See Appendix 2, §2, the example after Zorn's Lemma.)\n\nWe also observe that our proof shows that there exists at least one basis of \( B \) whose cardinality is \( \leqq n \) . We shall therefore be finished when we prove the last statement, that any two bases of \( B \) have the same cardinality. Let \( S \) be one basis, with a finite number of elements \( m \) . Let \( T \) be another basis, and suppose that \( T \) has at least \( r \) elements. It will suffice to prove that \( r \leqq m \) (one can then use symmetry). Let \( p \) be a prime number. Then \( B/{pB} \) is a direct sum of cyclic groups of order \( p \), with \( m \) terms in the sum. Hence its order is \( {p}^{m} \) . Using the basis \( T \) instead of \( S \), we conclude that \( B/{pB} \) contains an \( r \) -fold product of cyclic groups of order \( p \), whence \( {p}^{r} \leqq {p}^{m} \), and \( r \leqq m \), as was to be shown. (Note that we did not assume a priori that \( T \) was finite.)
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Yes
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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 \( A\left( p\right) \neq 0 \) .
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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 \( \sum {x}_{p} = 0 \) . Let \( q \) be a prime. Then \n\n\[ \n{x}_{q} = \mathop{\sum }\limits_{{p \neq q}}\left( {-{x}_{p}}\right) \n\] \n\nLet \( m \) be the least common multiple of the periods of elements \( {x}_{p} \) on the righthand side, with \( {x}_{q} \neq 0 \) and \( p \neq q \) . Then \( m{x}_{q} = 0 \) . But also \( {q}^{r}{x}_{q} = 0 \) for some positive integer \( r \) . If \( d \) is the greatest common divisor of \( m,{q}^{r} \) then \( d{x}_{q} = 0 \) , but \( d = 1 \), so \( {x}_{q} = 0 \) . Hence the kernel is trivial, and the homomorphism is injective.\n\nAs for the surjectivity, for each positive integer \( m \), denote by \( {A}_{m} \) the kernel of multiplication by \( m \), i.e. the subgroup of \( x \in A \) such that \( {mx} = 0 \) . We prove:\n\nIf \( m = {rs} \) with \( r \), s positive relative prime integers, then \( {A}_{m} = {A}_{r} + {A}_{s} \) .\n\nIndeed, there exist integers \( u, v \) such that \( {ur} + {vs} = 1 \) . Then \( x = {urx} + {vsx} \) , and \( {urx} \in {A}_{s} \) while \( {vsx} \in {A}_{r} \), and our assertion is proved. Repeating this process inductively, we conclude:\n\n\[ \n\text{If}m = \mathop{\prod }\limits_{{p \mid m}}{p}^{e\left( p\right) }\text{then}{A}_{m} = \mathop{\sum }\limits_{{p \mid m}}{A}_{{p}^{e\left( p\right) }}\text{.} \n\] \n\nHence the map \( \bigoplus A\left( p\right) \rightarrow A \) is surjective, and the theorem is proved.
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Yes
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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 \( {p}^{r} \) .
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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 \( n = {p}^{k}\mu \) where \( \mu \) is prime to \( p \) . Then \( \mu {a}_{1} \) is also a generator of \( {A}_{1} \), and hence has period \( {p}^{{r}_{1}} \) . We may assume \( k \leqq {r}_{1} \) . Then \( {p}^{k}\mu {a}_{1} \) has period \( {p}^{{r}_{1} - k} \) . By our previous remarks, the element \( b \) has period\n\[ {p}^{r + {r}_{1} - k} \]\nwhence by hypothesis, \( r + {r}_{1} - k \leqq {r}_{1} \) and \( r \leqq k \) . This proves that there exists an element \( c \in {A}_{1} \) such that \( {p}^{r}b = {p}^{r}c \) . Let \( a = b - c \) . Then \( a \) is a representative for \( \bar{b} \) in \( A \) and \( {p}^{r}a = 0 \) . Since period \( \left( a\right) \leqq {p}^{r} \) we conclude that \( a \) has period equal to \( {p}^{r} \) .
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Yes
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Theorem 8.4. Let \( A \) be a finitely generated torsion-free abelian group. Then \( A \) is free.
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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 all \( j \) . (Note that \( n \geqq 1 \) since \( A \neq 0 \) ). Let \( B \) be the subgroup generated by \( {x}_{1},\ldots ,{x}_{n} \) . Then \( B \) is free. Given \( y \in S \) there exist integers \( {m}_{1},\ldots ,{m}_{n}, m \) not all zero such that\n\n\[ \n{my} + {m}_{1}{x}_{1} + \cdots + {m}_{n}{x}_{n} = 0, \n\]\n\nby the assumption of maximality on \( {x}_{1},\ldots ,{x}_{n} \) . Furthermore, \( m \neq 0 \) ; otherwise all \( {m}_{j} = 0 \) . Hence \( {my} \) lies in \( B \) . This is true for every one of a finite set of generators \( y \) of \( A \), whence there exists an integer \( m \neq 0 \) such that \( {mA} \subset B \) . The map\n\n\[ \nx \mapsto {mx} \n\]\n\nof \( A \) into itself is a homomorphism, having trivial kernel since \( A \) is torsion free. Hence it is an isomorphism of \( A \) onto a subgroup of \( B \) . By Theorem 7.3 of the preceding section, we conclude that \( {mA} \) is free, whence \( A \) is free.
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Yes
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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 the direct sum of \( {A}_{\text{tor }} \) and \( B \) .
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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\nof \( F \) onto \( A \) . The subgroup \( {\varphi }^{-1}\left( {A}_{\text{tor }}\right) \) of \( F \) is finitely generated by Theorem 7.3. Hence \( {A}_{\text{tor }} \) itself is finitely generated, hence finite.\n\nNext, we prove that \( A/{A}_{\text{tor }} \) has no torsion. Let \( \bar{x} \) be an element of \( A/{A}_{\text{tor }} \) such that \( m\bar{x} = 0 \) for some integer \( m \neq 0 \) . Then for any representative of \( x \) of \( \bar{x} \) in \( A \), we have \( {mx} \in {A}_{\text{tor }} \), whence \( {qmx} = 0 \) for some integer \( q \neq 0 \) . Then \( x \in {A}_{\text{tor }} \), so \( \bar{x} = 0 \), and \( A/{A}_{\text{tor }} \) is torsion free. By Theorem 8.4, \( A/{A}_{\text{tor }} \) is free. We now use the lemma of Theorem 7.3 to conclude the proof.\n\nThe rank of \( A/{A}_{\text{tor }} \) is also called the rank of \( A \) .
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Yes
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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).
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Consider the two projections\n\n\n\nof \( B \times C \) on its two components. We get homomorphisms\n\n\n\nand we contend that these homomorphisms induce an isomorphism of \( {B}^{ \land } \times {C}^{ \land } \) onto \( {\left( B \times C\right) }^{ \land } \) .\n\nIn fact, let \( {\psi }_{1},{\psi }_{2} \) be in \( \operatorname{Hom}\left( {B,{Z}_{m}}\right) \) and \( \operatorname{Hom}\left( {C,{Z}_{m}}\right) \) respectively. Then \( \left( {{\psi }_{1},{\psi }_{2}}\right) \in {B}^{ \land } \times {C}^{ \land } \), and we have a corresponding element of \( {\left( B \times C\right) }^{ \land } \) by defining\n\n\[ \left( {{\psi }_{1},{\psi }_{2}}\right) \left( {x, y}\right) = {\psi }_{1}\left( x\right) + {\psi }_{2}\left( y\right) \]\n\nfor \( \left( {x, y}\right) \in B \times C \) . In this way we get a homomorphism\n\n\[ {B}^{ \land } \times {C}^{ \land } \rightarrow {\left( B \times C\right) }^{ \land }.\]\n\nConversely, let \( \psi \in {\left( B \times C\right) }^{ \land } \) . Then\n\n\[ \psi \left( {x, y}\right) = \psi \left( {x,0}\right) + \psi \left( {0, y}\right) . \]\n\nThe function \( {\psi }_{1} \) on \( B \) such that \( {\psi }_{1}\left( x\right) = \psi \left( {x,0}\right) \) is in \( {B}^{ \land } \), and similarly the function \( {\psi }_{2} \) on \( C \) such that \( {\psi }_{2}\left( y\right) = \psi \left( {0, y}\right) \) is in \( {C}^{ \land } \) . Thus we get a homomorphism\n\n\[ {\left( B \times C\right) }^{ \land } \rightarrow {B}^{ \land } \times {C}^{ \land },\]\n\nwhich is obviously inverse to the one we defined previously. Hence we obtain an isomorphism, thereby proving the first assertion in our theorem.
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Yes
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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 \( {A}^{\prime }/{B}^{\prime } \) is isomorphic to the dual group of \( A/B \) (under our map \( \psi \) ).
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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\[ \operatorname{ord}A/B \leqq \operatorname{ord}{\left( {A}^{\prime }/{B}^{\prime }\right) }^{ \land } = \operatorname{ord}{A}^{\prime }/{B}^{\prime } \]\n\nand\n\n\[ \text{ord}{A}^{\prime }/{B}^{\prime } \leqq \operatorname{ord}{\left( A/B\right) }^{ \land } = \operatorname{ord}A/B\text{.} \]\n\nFrom this it follows that our map \( \psi \) is bijective, hence an isomorphism.
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Yes
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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 \( {B}^{ \land } \) .
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Proof. This is a special case of Theorem 9.2.
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Yes
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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 \( f \) .
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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 \( {G}_{i} \). \n\nProof. 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 \( T \) be a set which is infinite denumerable if \( S \) is finite, and has the same cardinality as \( S \) if \( S \) is infinite. For each non-empty subset \( H \) of \( T \), let \( {\Gamma }_{H} \) be the set of group structures on \( H \) . For each \( \gamma \in {\Gamma }_{H} \), let \( {H}_{\gamma } \) be the set \( H \), together with the group structure \( \gamma \) . Then the family \( \left\{ {H}_{\gamma }\right\} \) for \( \gamma \in {\Gamma }_{H} \) and \( H \) ranging over subsets of \( T \) is the desired family.\n\nWe return to the proof of the proposition. For each \( i \in I \) we let \( {M}_{i} \) be the set of mappings of \( S \) into \( {G}_{i} \) . For each map \( \varphi \in {M}_{i} \), we let \( {G}_{i,\varphi } \) be the set-theoretic product of \( {G}_{i} \) and the set with one element \( \{ \varphi \} \), so that \( {G}_{i,\varphi } \) is the \
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No
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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 \( {G}_{i} \) .
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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 \( T \) be a set which is infinite denumerable if \( S \) is finite, and has the same cardinality as \( S \) if \( S \) is infinite. For each non-empty subset \( H \) of \( T \), let \( {\Gamma }_{H} \) be the set of group structures on \( H \) . For each \( \gamma \in {\Gamma }_{H} \), let \( {H}_{\gamma } \) be the set \( H \), together with the group structure \( \gamma \) . Then the family \( \left\{ {H}_{\gamma }\right\} \) for \( \gamma \in {\Gamma }_{H} \) and \( H \) ranging over subsets of \( T \) is the desired family.
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Yes
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Proposition 12.3. Coproducts exist in the category of groups.
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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 in this category. For each index \( i \), we let \( {S}_{i} \) be the same set as \( {G}_{i} \) if \( {G}_{i} \) is infinite, and we let \( {S}_{i} \) be denumerable if \( {G}_{i} \) is finite. We let \( S \) be a set having the same cardinality as the set-theoretic disjoint union of the sets \( {S}_{i} \) (i.e. their coproduct in the category of sets). We let \( \Gamma \) be the set of group structures on \( S \), and for each \( \gamma \in \Gamma \), we let \( {\Phi }_{\gamma } \) be the set of all families of homomorphisms \n\n\[ \n\varphi = \left\{ {{\varphi }_{i} : {G}_{i} \rightarrow {S}_{\gamma }}\right\} \n\] \n\nEach pair \( \left( {{S}_{\gamma },\varphi }\right) \), where \( \varphi \in {\Phi }_{\gamma } \), is then a group, using \( \varphi \) merely as an index. We let \n\n\[ \n{F}_{0} = \mathop{\prod }\limits_{{\gamma \in \Gamma }}\mathop{\prod }\limits_{{\varphi \in {\Phi }_{\gamma }}}\left( {{S}_{\gamma },\varphi }\right) \n\] \n\nand for each \( i \), we define a homomorphism \( {f}_{i} : {G}_{i} \rightarrow {F}_{0} \) by prescribing the component of \( {f}_{i} \) on each factor \( \left( {{S}_{\gamma },\varphi }\right) \) to be the same as that of \( {\varphi }_{i} \) .\n\nLet now \( g = \left\{ {{g}_{i} : {G}_{i} \rightarrow G}\right\} \) be a family of homomorphisms. Replacing \( G \) if necessary by the subgroup generated by the images of the \( {g}_{i} \), we see that \( \operatorname{card}\left( G\right) \leqq \operatorname{card}\left( S\right) \), because each element of \( G \) is a finite product of elements in these images. Embedding \( G \) as a factor in a product \( G \times {S}_{\gamma } \) for some \( \gamma \), we may assume that \( \operatorname{card}\left( G\right) = \operatorname{card}\left( S\right) \) . There exists a homomorphism \( {g}_{ * } : {F}_{0} \rightarrow G \) such that \n\n\[ \n{g}_{ * } \circ {f}_{i} = {g}_{i} \n\] \n\nfor all \( i \) . Indeed, we may assume without loss of generality that \( G = {S}_{v} \) for some \( \gamma \) and that \( g = \psi \) for some \( \psi \in {\Phi }_{\gamma } \) . We let \( {g}_{ * } \) be the projection of \( {F}_{0} \) on the factor \( \left( {{S}_{\gamma },\psi }\right) \) .\n\nLet \( F \) be the subgroup of \( {F}_{0} \) generated by the union of the images of the maps \( {f}_{i} \) for all \( i \) . The restriction of \( {g}_{ * } \) to \( F \) is the unique homomorphism satisfying \( {f}_{i} \circ {g}_{ * } = {g}_{i} \) for all \( i \), and we have thus constructed our universal object.
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Yes
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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\[ x = {x}_{{i}_{1}}\cdots {x}_{{i}_{n}}\;\text{with}\;{x}_{{i}_{\nu }} \in {G}_{{i}_{\nu }},\;{x}_{{i}_{\nu }} \neq e\;\text{and}\;{i}_{\nu } \neq {i}_{\nu + 1}\;\text{for all}\;\nu ,\]\n\nthen \( x \neq e \) .\n\nThen the natural homomorphism of the coproduct of the family into \( G \) sending\n\n\( {G}_{i} \) on itself by the identity mapping is an isomorphism. In other words, simply put, \( G \) is the coproduct of the family of subgroups.
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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 \( {x}_{{i}_{\nu }} = e \) and the element itself is equal to \( e \), whence the homomorphism from the coproduct into \( G \) is injective, thereby proving the proposition.
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Yes
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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} \\leqq n \n\\]\n\nsuch that \( {i}_{j} \\neq {i}_{j + 1} \) for \( j = 1,\\ldots, r - 1 \) . Then\n\n\\[ \n{x}_{{i}_{1}}^{{v}_{1}}\\cdots {x}_{{i}_{r}}^{{v}_{r}} \\neq 1 \n\\]
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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 the unit element of \( G \) . Our assertion follows at once.
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Yes
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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 \( {x}_{i} \) on itself is an isomorphism.
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Proof. It is obviously surjective and injective.
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No
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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 inclusion \( {G}_{i} \rightarrow {G}_{1} \circ \cdots \circ {G}_{n} \) is an isomorphism.
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Proof. Again, it is obviously injective and surjective.
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No
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Proposition 1.1. Products exist in the category of rings.
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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 of \( A \), we define their product to be \( {\left( {x}_{i}{y}_{i}\right) }_{i \in I} \), i.e. we define multiplication componentwise, just as we did for addition. The multiplicative unit is simply the element of the product whose \( i \) -th component is the unit element of \( {A}_{i} \) . It is then clear that we obtain a ring structure on \( A \), and that the projection on the \( i \) -th factor is a ring-homomorphism. Furthermore, \( A \) together with these projections clearly satisfies the required universal property.
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Yes
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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 \( x \equiv {x}_{i}\left( {\;\operatorname{mod}\;{\mathfrak{a}}_{i}}\right) \) for all \( i \) .
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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 that\n\n\[ {a}_{i} + {b}_{i} = 1,\;i \geqq 2. \]\n\nThe product \( \mathop{\prod }\limits_{{i = 2}}^{n}\left( {{a}_{i} + {b}_{i}}\right) \) is equal to 1, and lies in\n\n\[ {\mathfrak{a}}_{1} + \mathop{\prod }\limits_{{i = 2}}^{n}{\mathfrak{a}}_{i} \]\n\n i.e. in \( {\mathfrak{a}}_{1} + {\mathfrak{a}}_{2}\cdots {\mathfrak{a}}_{n} \) . Hence\n\n\[ {\mathfrak{a}}_{1} + \mathop{\prod }\limits_{{i = 2}}^{n}{\mathfrak{a}}_{i} = A \]\n\nBy the theorem for \( n = 2 \), we can find an element \( {y}_{1} \in A \) such that\n\n\[ {y}_{1} \equiv 1\;\left( {\;\operatorname{mod}\;{\mathfrak{a}}_{1}}\right) \]\n\n\[ {y}_{1} = 0\;\left( {\;\operatorname{mod}\;\mathop{\prod }\limits_{{i = 2}}^{n}{\mathfrak{a}}_{i}}\right) \]\n\nWe find similarly elements \( {y}_{2},\ldots ,{y}_{n} \) such that\n\n\[ {y}_{j} \equiv 1\;\left( {\;\operatorname{mod}\;{\mathfrak{a}}_{j}}\right) \;\text{ and }\;{y}_{j} \equiv 0\;\left( {\;\operatorname{mod}\;{\mathfrak{a}}_{i}}\right) \;\text{ for }i \neq j. \]\n\nThen \( x = {x}_{1}{y}_{1} + \cdots + {x}_{n}{y}_{n} \) satisfies our requirements.
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Yes
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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\[ f : A \rightarrow \mathop{\prod }\limits_{{i = 1}}^{n}A/{\mathfrak{a}}_{i} = \left( {A/{\mathfrak{a}}_{1}}\right) \times \cdots \times \left( {A/{\mathfrak{a}}_{n}}\right) \]\n\nbe the map of \( A \) into the product induced by the canonical map of \( A \) onto \( A/{\mathfrak{a}}_{i} \) for each factor. Then the kernel of \( f \) is \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{a}}_{i} \), and \( f \) is surjective, thus giving an isomorphism\n\n\[ A/\bigcap {\mathfrak{a}}_{i} \cong \prod A/{\mathfrak{a}}_{i} \]
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Proof. That the kernel of \( f \) is what we said it is, is obvious. The surjectivity follows from the theorem.
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No
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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 \( \mathbf{Z}/n\mathbf{Z} \approx \operatorname{End}\left( A\right) \), and a group isomorphism \( {\left( \mathbf{Z}/n\mathbf{Z}\right) }^{ * } \approx \operatorname{Aut}\left( A\right) \)
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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\[ \n{1a} = a,\;\left( {k + {k}^{\prime }}\right) a = {ka} + {k}^{\prime }a,\;\text{ and }\;\left( {k{k}^{\prime }}\right) a = k\left( {{k}^{\prime }a}\right) \n\]\n\nfor \( k,{k}^{\prime } \in \mathbf{Z} \) and \( a \in A \) . If \( a \) is a generator of \( A \), then \( {ka} = 0 \) if and only if \( k \equiv 0{\;\operatorname{mod}\;n} \), so \( \mathbf{Z}/n\mathbf{Z} \) is embedded in \( \operatorname{End}\left( A\right) \) . On the other hand, let \( f : A \rightarrow A \) be an endomorphism. Again for a generator \( a \), we have \( f\left( a\right) = {ka} \) \n\nfor some \( k \), whence \( f = {f}_{k} \) since every \( x \in A \) is of the form ma for some \( m \in Z \), and\n\n\[ \nf\left( x\right) = f\left( {ma}\right) = {mf}\left( a\right) = {mka} = {kma} = {kx}. \n\]\n\nThis proves the isomorphism \( \mathbf{Z}/n\mathbf{Z} \approx \operatorname{End}\left( A\right) \) . Furthermore, if \( k \in {\left( \mathbf{Z}/n\mathbf{Z}\right) }^{ * } \) then there exists \( {k}^{\prime } \) such that \( k{k}^{\prime } \equiv 1{\;\operatorname{mod}\;n} \), so \( {f}_{k} \) has the inverse \( {f}_{{k}^{\prime }} \) and \( {f}_{k} \) is an automorphism. Conversely, given any automorphism \( f \) with inverse \( g \), we know from the first part of the proof that \( f = {f}_{k}, g = {g}_{{k}^{\prime }} \) for some \( k,{k}^{\prime } \), and \( f \circ g = \) id means that \( k{k}^{\prime } \equiv 1{\;\operatorname{mod}\;n} \), so \( k,{k}^{\prime } \in {\left( \mathbf{Z}/n\mathbf{Z}\right) }^{ * } \) . This proves the isomorphism \( {\left( \mathbf{Z}/n\mathbf{Z}\right) }^{ * } \approx \operatorname{Aut}\left( A\right) \) .
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Yes
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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 \( h\left( a\right) = a \) for all \( a \in A \) .
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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 \( \beta = \sum {b}_{y}y \) . Then\n\n\[ h\left( {\alpha \beta }\right) = \mathop{\sum }\limits_{z}\left( {\mathop{\sum }\limits_{{{xy} = z}}{a}_{x}{b}_{y}}\right) \varphi \left( z\right) \]\n\nWe get \( h\left( {\alpha \beta }\right) = h\left( \alpha \right) h\left( \beta \right) \) immediately from the hypothesis that \( \varphi \left( {xy}\right) = \) \( \varphi \left( x\right) \varphi \left( y\right) \) . If \( e \) is the unit element of \( G \), then by definition \( \varphi \left( e\right) = {e}^{\prime } \), so Proposition 3.1 follows.
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Yes
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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\[ h\left( {\mathop{\sum }\limits_{{x \in G}}{a}_{x}x}\right) = \mathop{\sum }\limits_{{x \in G}}f\left( {a}_{x}\right) x. \]
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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 multiplication, let\n\n\[ \alpha = \sum {a}_{x}x\;\text{ and }\;\beta = \sum {b}_{y}y. \]\n\nThen\n\n\[ h\left( {\alpha \beta }\right) = \mathop{\sum }\limits_{{z \in G}}f\left( {\mathop{\sum }\limits_{{{xy} = z}}{a}_{x}{b}_{y}}\right) z \]\n\n\[ = \mathop{\sum }\limits_{{z \in G}}\mathop{\sum }\limits_{{{xy} = z}}f\left( {a}_{x}\right) f\left( {b}_{y}\right) z \]\n\n\[ = h\left( \alpha \right) h\left( \beta \right) \text{.} \]\n\nWe have trivially \( h\left( 1\right) = 1 \), so \( h \) is a ring-homomorphism, as was to be shown.
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Yes
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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 \( b \).
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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 \( \left( {a, b}\right) \) we can write \[ c = {wa} + {tb} \] with some \( w, t \in A \). Then \( c = {wdy} + {tdz} = d\left( {{wy} + {tz}}\right) \), whence \( d \mid c \), and our proposition is proved.
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Yes
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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) \\leftarrow {\\operatorname{Hom}}_{A}\\left( {{X}^{\prime \\prime }, Y}\\right) \\leftarrow 0 \n\\]\n\nis exact for all \\( Y \\) .
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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 \\( X \\) on \\( {X}^{\prime \\prime } \\) . If this composition is 0, it follows that \\( g = 0 \\) because \\( X \\rightarrow {X}^{\prime \\prime } \\) is surjective. As another example, consider a homomorphism \\( g : X \\rightarrow Y \\) such that the composition\n\n\\[ \n{X}^{\prime }\\overset{\\lambda }{ \\rightarrow }X\\overset{g}{ \\rightarrow }Y \n\\]\n\nis 0 . Then \\( g \\) vanishes on the image of \\( \\lambda \\) . Hence we can factor \\( g \\) through the factor module,\n\n\n\nSince \\( X \\rightarrow {X}^{\prime \\prime } \\) is surjective, we have an isomorphism\n\n\\[ \nX/\\operatorname{Im}\\lambda \\leftrightarrow {X}^{\prime \\prime }\\text{.} \n\\]\n\nHence we can factor \\( g \\) through \\( {X}^{\prime \\prime } \\), thereby showing that the kernel of\n\n\\[ \n{\\operatorname{Hom}}_{A}\\left( {{X}^{\prime }, Y}\\right) \\leftarrow {\\operatorname{Hom}}_{A}\\left( {X, Y}\\right) \n\\]\n\nis contained in the image of\n\n\\[ \n{\\operatorname{Hom}}_{A}\\left( {X, Y}\\right) \\leftarrow {\\operatorname{Hom}}_{A}\\left( {{X}^{\prime \\prime }, Y}\\right) . \n\\]\n\nThe other conditions needed to verify exactness are left to the reader. So is the converse.
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No
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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\[ 0 \rightarrow {\operatorname{Hom}}_{A}\left( {X,{Y}^{\prime }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {X, Y}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {X,{Y}^{\prime \prime }}\right) \]\n\nis exact for all \( X \) .
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The verification will be left to the reader. It follows at once from the definitions.
|
No
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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\[ \mathop{\sum }\limits_{{i = 1}}^{n}{\varphi }_{i} = \mathrm{{id}}\;\text{ and }\;{\varphi }_{i} \circ {\varphi }_{j} = 0\;\text{ if }i \neq j. \]\n\nThen \( {\varphi }_{i}^{2} = {\varphi }_{i} \) for all \( i \) . Let \( {M}_{i} = {\varphi }_{i}\left( M\right) \), and let \( \varphi : M \rightarrow \prod {M}_{i} \) be such that\n\n\[ \varphi \left( x\right) = \left( {{\varphi }_{1}\left( x\right) ,\ldots ,{\varphi }_{n}\left( x\right) }\right) . \]\n\nThen \( \varphi \) is an \( A \) -isomorphism of \( M \) onto the direct product \( \prod {M}_{i} \) .
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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. Let \( x \) be in its kernel. Since\n\n\[ x = \operatorname{id}\left( x\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{\varphi }_{i}\left( x\right) \]\nwe conclude that \( x = 0 \), so \( \varphi \) is injective. Given elements \( {y}_{i} \in {M}_{i} \) for each \( i = 1,\ldots, n \), let \( x = {y}_{1} + \cdots + {y}_{n} \) . We obviously have \( {\varphi }_{j}\left( {y}_{i}\right) = 0 \) if \( i \neq j \) . Hence\n\n\[ {\varphi }_{j}\left( x\right) = {y}_{j} \]\n\nfor each \( j = 1,\ldots, n \) . This proves that \( \varphi \) is surjective, and concludes the proof of our proposition.
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Yes
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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 \( g \circ \varphi = \mathrm{{id}} \).\n\n2. There exists a homomorphism \( \psi : M \rightarrow {M}^{\prime } \) such that \( \psi \circ f = \mathrm{{id}} \).\n\nIf these conditions are satisfied, then we have isomorphisms:\n\n\[ M = \operatorname{Im}f \oplus \operatorname{Ker}\psi ,\;M = \operatorname{Ker}g \oplus \operatorname{Im}\varphi ,\]\n\n\[ M \approx {M}^{\prime } \oplus {M}^{\prime \prime } \]
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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 \( M = \operatorname{Ker}g + \operatorname{Im}\varphi \).\n\nThis sum is direct, for if\n\n\[ x = y + z \]\n\nwith \( y \in \operatorname{Ker}g \) and \( z \in \operatorname{Im}\varphi, z = \varphi \left( w\right) \) with \( w \in {M}^{\prime \prime } \), and applying \( g \) yields \( g\left( x\right) = w \). Thus \( w \) is uniquely determined by \( x \), and therefore \( z \) is uniquely determined by \( x \). Hence so is \( y \), thereby proving the sum is direct.\n\nThe arguments concerning the other side of the sequence are similar and will be left as exercises, as well as the equivalence between our conditions. When these conditions are satisfied, the exact sequence of Proposition 3.2 is said to split. One also says that \( \psi \) splits \( f \) and \( \varphi \) splits \( g \).
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No
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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 unique homomorphism \( f : M \rightarrow N \) such that \( f\left( {x}_{i}\right) = {y}_{i} \) for all \( i \) .
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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 homomorphism satisfying our requirements, and that it is the unique such, because we must have\n\n\[ f\left( x\right) = \sum {a}_{i}f\left( {x}_{i}\right) \]
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Yes
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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.
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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.
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Yes
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Corollary 4.3. Two modules having bases whose cardinalities are equal are isomorphic.
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Proof. Clear.
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No
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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 \( S \subset \mathfrak{G} \subset \Gamma \) .
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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 totally ordered subset of \( \mathfrak{T} \) (by ascending inclusion), then \( \bigcup {T}_{i} \) is again linearly independent and contains \( S \) . By Zorn’s lemma, let \( \mathfrak{B} \) be a maximal element of \( \mathfrak{T} \) . Then \( \mathfrak{B} \) is linearly independent. Let \( W \) be the subspace of \( V \) generated by \( \mathfrak{B} \) . If \( W \neq V \), there exists some element \( x \in \Gamma \) such that \( x \notin W \) . Then \( \mathcal{B} \cup \{ x\} \) is linearly independent, for given a linear combination\n\n\[ \mathop{\sum }\limits_{{y \in \mathbb{G}}}{a}_{y}y + {bx} = 0,\;{a}_{y}, b \in K, \]\n\nwe must have \( b = 0 \), otherwise we get\n\n\[ x = - \mathop{\sum }\limits_{{y \in \mathcal{B}}}{b}^{-1}{a}_{y}y \in W. \]\n\nBy construction, we now see that \( {a}_{y} = 0 \) for all \( y \in \mathfrak{B} \), thereby proving that \( \mathcal{B} \cup \{ x\} \) is linearly independent, and contradicting the maximality of \( \mathcal{B} \) . It follows that \( W = V \), and furthermore that \( \mathcal{B} \) is not empty since \( V \neq \{ 0\} \) . This proves our theorem.
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Yes
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Theorem 5.2. Let \( V \) be a vector space over a field \( K \). Then two bases of \( V \) over \( K \) have the same cardinality.
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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 \( V \) which are linearly independent over \( K \), then \( n \leqq m \) (for we can then use symmetry). We proceed by induction. There exist elements \( {c}_{1},\ldots ,{c}_{m} \) of \( K \) such that\n\n(1)\n\n\[ \n{w}_{1} = {c}_{1}{v}_{1} + \cdots + {c}_{m}{v}_{m} \n\] \n\nand some \( {c}_{i} \), say \( {c}_{1} \), is not equal to 0 . Then \( {v}_{1} \) lies in the space generated by \( {w}_{1},{v}_{2},\ldots ,{v}_{m} \) over \( K \), and this space must therefore be equal to \( V \) itself. Furthermore, \( {w}_{1},{v}_{2},\ldots ,{v}_{m} \) are linearly independent, for suppose \( {b}_{1},\ldots ,{b}_{m} \) are elements of \( K \) such that \n\n\[ \n{b}_{1}{w}_{1} + {b}_{2}{v}_{2} + \cdots + {b}_{m}{v}_{m} = 0. \n\] \n\nIf \( {b}_{1} \neq 0 \), divide by \( {b}_{1} \) and express \( {w}_{1} \) as a linear combination of \( {v}_{2},\ldots ,{v}_{m} \) . Subtracting from (1) would yield a relation of linear dependence among the \( {v}_{i} \), which is impossible. Hence \( {b}_{1} = 0 \), and again we must have all \( {b}_{i} = 0 \) because the \( {v}_{i} \) are linearly independent.\n\nSuppose inductively that after a suitable renumbering of the \( {v}_{i} \), we have found \( {w}_{1},\ldots ,{w}_{r}\left( {r < n}\right) \) such that \n\n\[ \n\left\{ {{w}_{1},\ldots ,{w}_{r},{v}_{r + 1},\ldots ,{v}_{m}}\right\} \n\] \n\nis a basis of \( V \) . We express \( {w}_{r + 1} \) as a linear combination \n\n(2) \n\n\[ \n{w}_{r + 1} = {c}_{1}{w}_{1} + \cdots + {c}_{r}{w}_{r} + {c}_{r + 1}{v}_{r + 1} + \cdots + {c}_{m}{v}_{m} \n\] \n\nwith \( {c}_{i} \in K \) . The coefficients of the \( {v}_{i} \) in this relation cannot all be 0 ; otherwise there would be a linear dependence among the \( {w}_{j} \) . Say \( {c}_{r + 1} \neq 0 \) . Using an argument similar to that used above, we can replace \( {v}_{r + 1} \) by \( {w}_{r + 1} \) and still have a basis of \( V \) . This means that we can repeat the procedure until \( r = n \), and therefore that \( n \leqq m \), thereby proving our theorem.\n\nWe shall leave the general case of an infinite basis as an exercise to the reader. [Hint: Use the fact that a finite number of elements in one basis is contained in the space generated by a finite number of elements in another basis.]
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No
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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\[ \dim V = \dim \operatorname{Ker}f + \dim \operatorname{Im}f. \]
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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 family of elements of \( V \) such that \( f\left( {v}_{i}\right) = {u}_{i} \) for each \( i \in I \) . We contend that\n\n\[{\left\{ {v}_{i},{w}_{j}\right\} }_{i \in I, j \in J}\]\n\n is a basis for \( V \) . This will obviously prove our assertion.\n\nLet \( x \) be an element of \( V \) . Then there exist elements \( {\left\{ {a}_{i}\right\} }_{i \in I} \) of \( K \) almost all of which are 0 such that\n\n\[f\left( x\right) = \mathop{\sum }\limits_{{i \in I}}{a}_{i}{u}_{i}\]\n\nHence \( f\left( {x-\sum {a}_{i}{v}_{i}}\right) = f\left( x\right) - \sum {a}_{i}f\left( {v}_{i}\right) = 0 \) . Thus\n\n\[x - \sum {a}_{i}{v}_{i}\]\n\nis in the kernel of \( f \), and there exist elements \( {\left\{ {b}_{j}\right\} }_{j \in J} \) of \( K \) almost all of which are 0 such that\n\n\[x - \sum {a}_{i}{v}_{i} = \sum {b}_{j}{w}_{j}\]\n\nFrom this we see that \( x = \sum {a}_{i}{v}_{i} + \sum {b}_{j}{w}_{j} \), and that \( \left\{ {{v}_{i},{w}_{j}}\right\} \) generates \( V \) . It remains to be shown that the family \( \left\{ {{v}_{i},{w}_{j}}\right\} \) is linearly independent. Suppose that there exist elements \( {c}_{i},{d}_{j} \) such that\n\n\[0 = \sum {c}_{i}{v}_{i} + \sum {d}_{j}{w}_{j}\]\n\nApplying \( f \) yields\n\n\[0 = \sum {c}_{i}f\left( {v}_{i}\right) = \sum {c}_{i}{u}_{i}\]\n\nwhence all \( {c}_{i} = 0 \) . From this we conclude at once that all \( {d}_{j} = 0 \), and hence that our family \( \left\{ {{v}_{i},{w}_{j}}\right\} \) is a basis for \( V \) over \( K \), as was to be shown.
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Yes
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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 \( W = V \) .
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Proof. Clear.
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No
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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 \( {f}_{i}\left( {x}_{j}\right) = {\delta }_{ij} \), then \( \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) is a basis for \( {E}^{ \vee } \) .
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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 \( f = {a}_{1}{f}_{1} + \cdots + {a}_{n}{f}_{n} \), and we see that the \( {f}_{i} \) generate \( {E}^{ \vee } \) . Furthermore, they are linearly independent, for if\n\n\[ {b}_{1}{f}_{1} + \cdots + {b}_{n}{f}_{n} = 0 \]\n\nwith \( {b}_{i} \in K \), then evaluating the left-hand side on \( {x}_{i} \) yields\n\n\[ {b}_{i}{f}_{i}\left( {x}_{i}\right) = 0 \]\n\nwhence \( {b}_{i} = 0 \) for all \( i \) . This proves our theorem.
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Yes
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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 \( {E}^{\vee \vee } \).
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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 \( E = {E}^{\vee \vee } \).
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No
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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\[ 0 \rightarrow {\operatorname{Hom}}_{A}\left( {U, A}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {V, A}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {W, A}\right) \rightarrow 0 \]\n\ni.e.\n\n\[ 0 \rightarrow {U}^{ \vee } \rightarrow {V}^{ \vee } \rightarrow {W}^{ \vee } \rightarrow 0 \]\n\nis also exact.
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Proof. This is a consequence of \( \mathbf{{P2}} \), because a free module is projective.
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No
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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 \( {V}^{\prime }/{W}^{\prime } \rightarrow {\left( V/W\right) }^{ \vee } \) is an isomorphism.
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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 the above injective homomorphism and the other, namely\n\n\[ 0 \rightarrow {V}^{\prime }/{W}^{\prime } \rightarrow {\left( V/W\right) }^{ \vee } \]\n\nwe get the inequalities\n\n\[ \dim V/W \leqq \dim {V}^{\prime }/{W}^{\prime } \]\n\nand\n\n\[ \dim {V}^{\prime }/{W}^{\prime } \leqq \dim V/W \]\n\nwhence an equality of dimensions. Hence our homomorphisms are surjective and inverse to each other, thereby proving the theorem.
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Yes
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Corollary 7.2. Let \( E \) be a finitely generated module and \( {E}^{\prime } \) a submodule.\n\nThen \( {E}^{\prime } \) is finitely generated.
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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 of \( {E}^{\prime } \) in \( F \) is a submodule, which is free, and finitely generated, by the theorem. Hence \( {E}^{\prime } \) is finitely generated. The assertion also follows using simple properties of Noetherian rings and modules.
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Yes
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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\[ E = {E}_{\text{tor }} \oplus F. \]
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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 \( c \in R, c \neq 0 \), such that \( {cbx} = 0 \) . Hence \( x \in {E}_{\text{tor }} \) and \( \bar{x} = 0 \), thereby proving that \( E/{E}_{\text{tor }} \) is torsion free. It is also finitely generated.\n\nAssume now that \( M \) is a torsion free module which is finitely generated. Let \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a maximal set of elements of \( M \) among a given finite set of generators \( \left\{ {{y}_{1},\ldots ,{y}_{m}}\right\} \) such that \( \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) is linearly independent. If \( y \) is one of the generators, there exist elements \( a,{b}_{1},\ldots ,{b}_{n} \in R \) not all 0, such that\n\n\[ {ay} + {b}_{1}{v}_{1} + \cdots + {b}_{n}{v}_{n} = 0. \]\n\nThen \( a \neq 0 \) (otherwise we contradict the linear independence of \( {v}_{1},\ldots ,{v}_{n} \) ). Hence \( {ay} \) lies in \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) . Thus for each \( j = 1,\ldots, m \) we can find \( {a}_{j} \in R \) , \( {a}_{j} \neq 0 \), such that \( {a}_{j}{y}_{j} \) lies in \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) . Let \( a = {a}_{1}\cdots {a}_{m} \) be the product. Then \( {aM} \) is contained in \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \), and \( a \neq 0 \) . The map\n\n\[ x \mapsto {ax} \]\n\n is an injective homomorphism, whose image is contained in a free module. This image is isomorphic to \( M \), and we conclude from Theorem 7.1 that \( M \) is free, as desired.\n\nTo get the submodule \( F \) we need a lemma.
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Yes
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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 \( {E}^{\prime } \), and such that \( E = F \oplus \operatorname{Ker}f \) .
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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 \( {x}_{i} \) , \( i \in I \) . Then one sees at once that the family of elements \( {\left\{ {x}_{i}\right\} }_{i \in I} \) is linearly independent, and therefore that \( F \) is free. Given \( x \in E \), there exist elements \( {a}_{i} \in R \) such that\n\n\[ f\left( x\right) = \sum {a}_{i}{x}_{i}^{\prime } \]\n\nThen \( x - \sum {a}_{i}{x}_{i} \) lies in the kernel of \( f \), and therefore \( E = \operatorname{Ker}f + F \) . It is clear that \( \operatorname{Ker}f \cap F = 0 \), and hence that the sum is direct, thereby proving the lemma.
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Yes
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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\[ E\left( p\right) = R/\left( {p}^{{v}_{1}}\right) \oplus \cdots \oplus R/\left( {p}^{{v}_{s}}\right) \]\n\nwith \( 1 \leqq {v}_{1} \leqq \cdots \leqq {v}_{s} \) . The sequence \( {v}_{1},\ldots ,{v}_{s} \) is uniquely determined.
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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 product of prime powers. Let \( v \in E \) . Then\n\n\[ v = {xbv} + {ycv}. \]\n\nThen \( {xbv} \in {E}_{c} \) because \( {cxbv} = {xav} = 0 \) . Similarly, \( {ycv} \in {E}_{b} \) . Finally \( {E}_{b} \cap {E}_{c} = 0 \) , as one sees immediately. Hence \( E \) is the direct sum of \( {E}_{b} \) and \( {E}_{c} \) .\n\nWe must now prove that \( E\left( p\right) \) is a direct sum as stated. If \( {y}_{1},\ldots ,{y}_{m} \) are elements of a module, we shall say that they are independent if whenever we have a relation\n\n\[ {a}_{1}{y}_{1} + \cdots + {a}_{m}{y}_{m} = 0 \]\n\nwith \( {a}_{i} \in R \), then we must have \( {a}_{i}{y}_{i} = 0 \) for all \( i \) . (Observe that independent does not mean linearly independent.) We see at once that \( {y}_{1},\ldots ,{y}_{m} \) are independent if and only if the module \( \left( {{y}_{1},\ldots ,{y}_{m}}\right) \) has the direct sum decomposition\n\n\[ \left( {{y}_{1},\ldots ,{y}_{m}}\right) = \left( {y}_{1}\right) \oplus \cdots \oplus \left( {y}_{m}\right) \]\n\nin terms of the cyclic modules \( \left( {y}_{i}\right), i = 1,\ldots, m \) .
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Yes
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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 \( \bar{E} \) . Then for each \( i \) there exists a representative \( {y}_{i} \in E \) of \( {\bar{y}}_{i} \), such that the period of \( {y}_{i} \) is the same as the period of \( {\bar{y}}_{i} \) . The elements \( {x}_{1},{y}_{1},\ldots ,{y}_{m} \) are independent.
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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 that \( y \) has the same period as \( \bar{y} \) . If \( s < r \), then \( {p}^{s}c{x}_{1} \) has period \( {p}^{r - s} \), and hence \( y \) has period \( {p}^{n + r - s} \) . We must have\n\n\[ n + r - s \leqq r \]\n\nbecause \( {p}^{r} \) is an exponent for \( E \) . Thus we obtain \( n \leqq s \), and we see that\n\n\[ y - {p}^{s - n}c{x}_{1} \]\n\nis a representative for \( \bar{y} \), whose period is \( {p}^{n} \) .\n\nLet \( {y}_{i} \) be a representative for \( {\bar{y}}_{i} \) having the same period. We prove that \( {x}_{1},{y}_{1},\ldots ,{y}_{m} \) are independent. Suppose that \( a,{a}_{1},\ldots ,{a}_{m} \in R \) are elements such that\n\n\[ a{x}_{1} + {a}_{1}{y}_{1} + \cdots + {a}_{m}{y}_{m} = 0. \]\n\nThen\n\n\[ {a}_{1}{\bar{y}}_{1} + \cdots + {a}_{m}{\bar{y}}_{m} = 0. \]\n\nBy hypothesis, we must have \( {a}_{i}{\bar{y}}_{i} = 0 \) for each \( i \) . If \( {p}^{{r}_{i}} \) is the period of \( {\bar{y}}_{i} \), then \( {p}^{{r}_{i}} \) divides \( {a}_{i} \) . We then conclude that \( {a}_{i}{y}_{i} = 0 \) for each \( i \), and hence finally that \( a{x}_{1} = 0 \), thereby proving the desired independence.
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Yes
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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 \( {q}_{1}\left| {q}_{2}\right| \cdots \mid {q}_{r} \) . The sequence of ideals \( \left( {q}_{1}\right) ,\ldots ,\left( {q}_{r}\right) \) is uniquely determined by the above conditions.
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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 symbolically as described by the following diagram:\n\n\[ E\left( {p}_{1}\right) : {r}_{11} \leqq {r}_{12} \leqq \cdots \]\n\n\[ E\left( {p}_{2}\right) : {r}_{21} \leqq {r}_{22} \leqq \cdots \]\n\n\[ E\left( {p}_{l}\right) : \;{r}_{l1} \leqq {r}_{l2} \leqq \cdots \]\n\nA horizontal row describes the type of the module with respect to the prime at the left. The exponents \( {r}_{ij} \) are arranged in increasing order for each fixed \( i = 1,\ldots, l \) . We let \( {q}_{1},\ldots ,{q}_{r} \) correspond to the columns of the matrix of exponents, in other words\n\n\[ {q}_{1} = {p}_{1}^{{r}_{11}}{p}_{2}^{{r}_{21}}\cdots {p}_{l}^{{r}_{l1}} \]\n\n\[ {q}_{2} = {p}_{1}^{{r}_{12}}{p}_{2}^{{r}_{22}}\cdots {p}_{l}^{{r}_{l2}} \]\n\nThe direct sum of the cyclic modules represented by the first column is then isomorphic to \( R/\left( {q}_{1}\right) \), because, as with abelian groups, the direct sum of cyclic modules whose periods are relatively prime is also cyclic. We have a similar remark for each column, and we observe that our proof actually orders the \( {q}_{j} \) by increasing divisibility, as was to be shown.
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Yes
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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\[ \left( \begin{matrix} {a}_{1} & 0 & \cdots & \cdot & \cdot & \cdots & 0 \\ 0 & {a}_{2} & \cdots & \cdot & \cdot & \cdots & 0 \\ \vdots & & \ddots & & & & \vdots \\ 0 & \cdot & \cdots & {a}_{m} & \cdot & \cdots & 0 \\ 0 & \cdot & \cdots & \cdot & 0 & \cdots & 0 \\ \vdots & & & & & & \vdots \\ 0 & \cdot & \cdots & \cdot & \cdot & \cdots & 0 \end{matrix}\right) . \]\n\nwith \( {a}_{1}\cdots {a}_{m} \neq 0 \) and \( {a}_{1}\left| {a}_{2}\right| \cdots \mid {a}_{m} \) .
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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.
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No
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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 mapping defined on all modules of finite length.
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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 this extension of \( \varphi \) is an Euler-Poincaré map.
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Yes
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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 exact sequence\n\n\\[ \n\\operatorname{Ker}{d}^{\\prime } \\rightarrow \\operatorname{Ker}d \\rightarrow \\operatorname{Ker}{d}^{\\prime \\prime }\\overset{\\delta }{ \\rightarrow }\\operatorname{Coker}{d}^{\\prime } \\rightarrow \\operatorname{Coker}d \\rightarrow \\operatorname{Coker}{d}^{\\prime \\prime }\n\\]\n\nwhere the maps besides \\( \\delta \\) are the natural ones.
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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 chasing around diagrams. It should be carried out in full detail by the reader who wishes to acquire a feeling for this type of triviality. As an example, we shall prove that\n\n\\[ \n\\operatorname{Ker}\\delta \\subset \\operatorname{Im}{g}_{ * }\n\\]\n\nwhere \\( {g}_{ * } \\) is the induced map on kernels. Suppose the image of \\( {z}^{\\prime \\prime } \\) is 0 in Coker \\( {d}^{\\prime } \\) . By definition, there exists \\( {u}^{\\prime } \\in {M}^{\\prime } \\) such that \\( {z}^{\\prime } = {d}^{\\prime }{u}^{\\prime } \\) . Then\n\n\\[ \n{dz} = f{z}^{\\prime } = f{d}^{\\prime }{u}^{\\prime } = {df}{u}^{\\prime }\n\\]\n\nby commutativity. Hence\n\n\\[ \nd\\left( {z - f{u}^{\\prime }}\\right) = 0\n\\]\n\nand \\( z - f{u}^{\\prime } \\) is in the kernel of \\( d \\) . But \\( g\\left( {z - f{u}^{\\prime }}\\right) = {gz} = {z}^{\\prime \\prime } \\) . This means that \\( {z}^{\\prime \\prime } \\) is in the image of \\( {g}_{ * } \\), as desired. All the remaining cases of exactness will be left to the reader.
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No
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Theorem 10.1. Direct limits exist in the category of abelian groups, or more generally in the category of modules over a ring.
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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 \( \left( {i, j}\right) \) with \( j \geqq i,{x}_{ij} \) has component \( x \) in \( {M}_{i} \) , \( {f}_{j}^{i}\left( x\right) \) in \( {M}_{j} \), and component 0 elsewhere. Then we leave to the reader the verification that the factor module \( M/N \) is a direct limit, where the maps of \( {M}_{i} \) into \( M/N \) are the natural ones arising from the composite homomorphism\n\n\[ \n{M}_{i} \rightarrow M \rightarrow M/N \n\]
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No
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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.
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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 subgroup of \( \prod {G}_{i} \) ). It is routine to verify that these data give an inverse limit in the category of groups. The same construction also applies to the category of rings and modules.
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No
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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\[ 0 \rightarrow \underline{\lim }{A}_{n} \rightarrow \underline{\lim }{B}_{n} \rightarrow \underline{\lim }{C}_{n} \rightarrow 0 \]\nis exact.
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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 condition on \( \left( {A}_{n}\right) \) implies ML on \( \left( {{g}^{-1}\left( {c}_{n}\right) }\right) \) . Let \( {S}_{n} \) be the stable subset\n\n\[ {S}_{n} = \mathop{\bigcap }\limits_{{m \geqq n}}{u}_{m, n}^{B}\left( {{g}^{-1}\left( {c}_{m}\right) }\right) \]\n\nThen the connecting maps in the inverse system \( \left( {S}_{n}\right) \) are surjective, and so there is an element \( \left( {b}_{n}\right) \) in the inverse limit. It is immediate that \( g \) maps this element on the given \( \left( {c}_{n}\right) \), thereby concluding the proof of the Proposition.
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Yes
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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\[ 0 \\rightarrow \\underline{\\lim }{C}_{n} \\rightarrow \\prod {C}_{n}\\overset{1 - u}{ \\rightarrow }\\prod {C}_{n} \\rightarrow 0. \]
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Proof. For each positive integer \( N \) we have an exact sequence with a finite product\n\n\[ 0 \\rightarrow \\mathop{\\lim }\\limits_{{1 \\leqq n \\leqq N}}{C}_{n} \\rightarrow \\mathop{\\prod }\\limits_{{n = 1}}^{N}{C}_{n}\\overset{1 - u}{ \\rightarrow }\\mathop{\\prod }\\limits_{{n = 1}}^{N}{C}_{n} \\rightarrow 0. \]\n\nThe map \( u \) is the natural one, whose effect on a vector is\n\n\[ \\left( {0,\\ldots ,0,{c}_{m},0,\\ldots ,0}\\right) \\mapsto \\left( {0,\\ldots ,0,{u}_{m, m - 1}{c}_{m},0,\\ldots ,0}\\right) . \]\n\nOne sees immediately that the sequence is exact. The infinite products are inverse limits taken over \( N \) . The hypothesis implies at once that ML is satisfied for the inverse limit on the left, and we can therefore apply Proposition 10.3 to conclude the proof.
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Yes
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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 \[ f = {gq} + r \] and \( \deg r < \deg g \) .
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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 let \( q = 0, r = f \) . If \( \deg g = \deg f = 0 \), then we let \( r = 0 \) and \( q = {a}_{n}{b}_{d}^{-1} \) . Assume the theorem proved for polynomials of degree \( < n \) (with \( n > 0 \) ). We may assume \( \deg g \leqq \deg f \) (otherwise, take \( q = 0 \) and \( r = f \) ). Then \[ f\left( X\right) = {a}_{n}{b}_{d}^{-1}{X}^{n - d}g\left( X\right) + {f}_{1}\left( X\right) \] where \( {f}_{1}\left( X\right) \) has degree \( < n \) . By induction, we can find \( {q}_{1}, r \) such that \[ f\left( X\right) = {a}_{n}{b}_{d}^{-1}{X}^{n - d}g\left( X\right) + {q}_{1}\left( X\right) g\left( X\right) + r\left( X\right) \] and \( \deg r < \deg g \) . Then we let \[ q\left( X\right) = {a}_{n}{b}_{d}^{-1}{X}^{n - d} + {q}_{1}\left( X\right) \] to conclude the proof of existence for \( q, r \) . As for uniqueness, suppose that \[ f = {q}_{1}g + {r}_{1} = {q}_{2}g + {r}_{2} \] with \( \deg {r}_{1} < \deg g \) and \( \deg {r}_{2} < \deg g \) . Subtracting yields \[ \left( {{q}_{1} - {q}_{2}}\right) g = {r}_{2} - {r}_{1} \] Since the leading coefficient of \( g \) is assumed to be a unit, we have \[ \deg \left( {{q}_{1} - {q}_{2}}\right) g = \deg \left( {{q}_{1} - {q}_{2}}\right) + \deg g. \] Since \( \deg \left( {{r}_{2} - {r}_{1}}\right) < \deg g \), this relation can hold only if \( {q}_{1} - {q}_{2} = 0 \), i.e. \( {q}_{1} = {q}_{2} \), and hence finally \( {r}_{1} = {r}_{2} \) as was to be shown.
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Yes
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Theorem 1.2. Let \( k \) be a field. Then the polynomial ring in one variable \( k\\left\\lbrack X\\right\\rbrack \) is principal.
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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 can find \( q, r \\in k\\left\\lbrack X\\right\\rbrack \) such that\n\n\[ f = {qg} + r \]\n\nand \( \\deg r < \\deg g \) . But \( r = f - {qg} \), whence \( r \) is in a. Since \( g \) had minimal degree \( \\geqq 0 \) it follows that \( r = 0 \), hence that a consists of all polynomials \( {qg} \) (with \( q \\in k\\left\\lbrack X\\right\\rbrack \) ). This proves our theorem.
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Yes
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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 \( f\left( X\right) \) .
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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 \( r = 0 \), whence \( X - a \) divides \( f\left( X\right) \) . If \( {a}_{1},\ldots ,{a}_{m} \) are distinct roots of \( f \) in \( k \), then inductively we see that the product\n\n\[ \left( {X - {a}_{1}}\right) \cdots \left( {X - {a}_{m}}\right) \]\ndivides \( f\left( X\right) \), whence \( m \leqq n \), thereby proving the theorem.
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Yes
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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 \( {a}_{i} \in {S}_{i}\left( {i = 1,\ldots, n}\right) \), then \( f = 0 \) .
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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 \( k\left\lbrack {{X}_{1},\ldots ,{X}_{n - 1}}\right\rbrack \) . If there exists\n\n\[ \left( {{b}_{1},\ldots ,{b}_{n - 1}}\right) \in {S}_{1} \times \cdots \times {S}_{n - 1} \]\n\nsuch that for some \( j \) we have \( {f}_{j}\left( {{b}_{1},\ldots ,{b}_{n - 1}}\right) \neq 0 \), then\n\n\[ f\left( {{b}_{1},\ldots ,{b}_{n - 1},{X}_{n}}\right) \]\n\nis a non-zero polynomial in \( k\left\lbrack {X}_{n}\right\rbrack \) which takes on the value 0 for the infinite set of elements \( {S}_{n} \) . This is impossible. Hence \( {f}_{j} \) induces the zero function on \( {S}_{1} \times \cdots \times {S}_{n - 1} \) for all \( j \), and by induction we have \( {f}_{j} = 0 \) for all \( j \) . Hence \( f = 0 \), as was to be shown.
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Yes
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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 \( f = 0 \) .
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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.
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No
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Theorem 1.9. Let \( k \) be a field and let \( U \) be a finite multiplicative subgroup of \( k \) . Then \( U \) is cyclic.
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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 maximal period \( {p}^{r} \) for some integer \( r \) . Then \( {x}^{{p}^{r}} = 1 \) for every element \( x \in U\left( p\right) \), and hence all elements of \( U\left( p\right) \) are roots of the polynomial\n\n\[ \n{X}^{{p}^{r}} - 1\text{.} \n\]\n\nThe cyclic group generated by \( a \) has \( {p}^{r} \) elements. If this cyclic group is not equal to \( U\left( p\right) \), then our polynomial has more than \( {p}^{r} \) roots, which is impossible. Hence \( a \) generates \( U\left( p\right) \), and our theorem is proved.
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Yes
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