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Problem 23 Let \( a, b, c \) be positive real numbers such that \( a + b + c = 3 \) .Prove that\n\n\[ k{\left( a + b + c\right) }^{4} \geq \left( {{a}^{3}b + {b}^{3}c + {c}^{3}a}\right) + \left( {{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2}}\right) + {abc}\left( {a + b + c}\right) . \]\n | Solution: Let \( a = 2, b = 1, c = 0 \Rightarrow k \geq \frac{4}{27} \) .\n\nWe ’ll prove \( k \geq \frac{4}{27} \) is better constand.\n\nThe inequality equivalent to\n\n\[ \frac{4}{27}{\left( a + b + c\right) }^{4} \geq \mathop{\sum }\limits_{{cyc}}{a}^{3}b + \mathop{\sum }\limits_{{sym}}{b}^{2}{c}^{2} + {abc}\mathop... | Yes |
Problem 24 Let \( a, b, c \) be nonnegative real numbers. Find the beter constand \( k \) to that inequality always true\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + k\frac{{ab} + {bc} + {ca}}{{a}^{2} + {b}^{2} + {c}^{2}} \geq 3 + k. \] | Solution: We have\n\n\[ 2\mathop{\sum }\limits_{{cyc}}\frac{a}{b} = \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{b} + \mathop{\sum }\limits_{{cyc}}\frac{b}{a}}\right) + \left( {\mathop{\sum }\limits_{{cyc}}\frac{a}{b} - \mathop{\sum }\limits_{{cyc}}\frac{b}{a}}\right) = \frac{\mathop{\sum }\limits_{{sym}}{a}^{2}\left(... | Yes |
Problem 25(Vo Thanh Van) Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \sqrt{\frac{{\left( a + b\right) }^{3}}{{8ab}\left( {{4a} + {4b} + c}\right) }} + \sqrt{\frac{{\left( b + c\right) }^{3}}{{8bc}\left( {{4b} + {4c} + a}\right) }} + \sqrt{\frac{{\left( c + a\right) }^{3}}{{8ca}\left( {{4c} + {4a} + ... | Solution: Let \( P = \sqrt{\frac{{\left( a + b\right) }^{3}}{{8ab}\left( {{4a} + {4b} + c}\right) }} + \sqrt{\frac{{\left( b + c\right) }^{3}}{{8bc}\left( {{4b} + {4c} + a}\right) }} + \sqrt{\frac{{\left( c + a\right) }^{3}}{{8ca}\left( {{4c} + {4a} + b}\right) }} \)\n\n\[ Q = {8ab}\left( {{4a} + {4b} + c}\right) + {8b... | Yes |
Problem 25(APMO 2004) Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \left( {{a}^{2} + 2}\right) \left( {{b}^{2} + 2}\right) \left( {{c}^{2} + 2}\right) \geq 9\left( {{ab} + {bc} + {ca}}\right) \] | Solution: The equivalent to\n\n\[ {a}^{2}{b}^{2}{c}^{2} + 2\left( {{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2}}\right) + 4\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) + 8 \geq 9\left( {{ab} + {bc} + {ca}}\right) .\n\nWe have \( {a}^{2} + {b}^{2} + {c}^{2} \geq {ab} + {bc} + {ca} \)\n\n\[ \left( {{a}^{2}{b}^{2} + 1}... | No |
Problem 26(Vo Thanh Van) Let \( a, b, c \) be nonnegative real numbers. Prove that\n\n\[ \frac{a}{{b}^{3} + {c}^{3}} + \frac{b}{{a}^{3} + {c}^{3}} + \frac{c}{{a}^{3} + {b}^{3}} \geq \frac{18}{5\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) - {ab} - {ac} - {bc}} \] | Solution: The equivalent to\n\n\[ \sum \frac{a\left( {a + b + c}\right) }{{b}^{3} + {c}^{3}} \geq \frac{{18}\left( {a + b + c}\right) }{5\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) - {ab} - {bc} - {ca}} \]\n\n\[ \Leftrightarrow \sum \frac{{a}^{2}}{{b}^{3} + {c}^{3}} + \frac{a}{{b}^{2} + {c}^{2} - {bc}} \geq \frac{{18}\... | Yes |
Problem 27( Moldova TST 2005) Let \( a, b, c \) be nonnegative real numbers such that \( {a}^{4} + {b}^{4} + {c}^{4} = 3 \) . Prove that\n\n\[ \frac{1}{4 - {ab}} + \frac{1}{4 - {bc}} + \frac{1}{4 - {ca}} \leq 1 \] | Solution: The equivalent to\n\n\[ {49} - 8\left( {{ab} + {bc} + {ca}}\right) + \left( {a + b + c}\right) {abc} \leq {64} - {16}\left( {{ab} + {bc} + {ca}}\right) + 4\left( {a + b + c}\right) {abc} - {a}^{2}{b}^{2}{c}^{2} \]\n\n\[ \Leftrightarrow {16} + 3\left( {a + b + c}\right) {abc} \geq {a}^{2}{b}^{2}{c}^{2} + 8\lef... | Yes |
Problem 28(Vasile Cirtoaje) Let \( a, b, c \) be nonnegative real numbers such that \( {ab} + {bc} + {ca} = 3 \) . Prove that\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {7abc} \geq {10} \n\] | Solution: Apply Schur inequality we have\n\n\[ \n{a}^{3} + {b}^{3} + {c}^{3} + {3abc} \geq {ab}\left( {a + b}\right) + {bc}\left( {b + c}\right) + {ca}\left( {c + a}\right) \n\]\n\n\[ \n\Leftrightarrow {a}^{3} + {b}^{3} + {c}^{3} + {6abc} \geq \left( {{ab} + {bc} + {ca}}\right) \left( {a + b + c}\right) = {pq} = {3p} \... | Yes |
Problem 29(Nguyen Phi Hung) Let \( a, b, c \) be nonnegative real numbers such that \( {a}^{2} + {b}^{2} + {c}^{2} = 8 \) . Prove that\n\n\[ 4\left( {a + b + c - 4}\right) \leq {abc} \] | Solution: From the condition we have \( {p}^{2} - {2q} = 8 \)\n\nApply Schur Inequality we have\n\n\[ r \geq \frac{\left( {{4q} - {p}^{2}}\right) \left( {{p}^{2} - q}\right) }{6p} = \frac{\left( {{p}^{2} - {16}}\right) \left( {{p}^{2} + 8}\right) }{12p} \]\n\nSo we need prove that\n\n\[ \frac{\left( {{p}^{2} - {16}}\ri... | Yes |
Problem 30 Let \( a, b, c > 0 \) and \( a + b + c = 1 \) .Prove that\n\n\[ \frac{\sqrt{{a}^{2} + {abc}}}{{ab} + c} + \frac{\sqrt{{b}^{2} + {abc}}}{{bc} + a} + \frac{\sqrt{{c}^{2} + {abc}}}{{ca} + b} \leq \frac{1}{2\sqrt{abc}} \]\n | Solution: Changes \( a, b, c \) to \( p, q, r \) we have \( r \leq \frac{{q}^{2}\left( {1 - q}\right) }{2\left( {2 - {3q}}\right) } \)\n\nApply Cauchy-Schwarz Inequality we have\n\n\[ {\left\lbrack \sum \frac{\sqrt{{a}^{2} + {abc}}}{\left( {b + c}\right) \left( {b + a}\right) }\right\rbrack }^{2} \leq \left\lbrack {\su... | Yes |
Problem 31 Let \( a, b, c \) be nonnegative real numbers, no two of which are zero. Show that\n\n\[ \left( {{ab} + {bc} + {ca}}\right) \left( {\frac{1}{{\left( a + b\right) }^{2}} + \frac{1}{{\left( b + c\right) }^{2}} + \frac{1}{{\left( c + a\right) }^{2}}}\right) \geq \frac{9}{4}. \] | Solution: We can rewrite inequality\n\n\[ 4{p}^{4}q - {17}{p}^{2}{q}^{2} + 4{q}^{3} + {34pqr} - 9{r}^{2} \geq 0 \]\n\n\[ \Leftrightarrow {pq}\left( {{p}^{3} - {4pqr} + {9r}}\right) + q\left( {{p}^{4} - 5{p}^{2}q + 4{q}^{2} + {6pr}}\right) + r\left( {{pq} - {9r}}\right) \geq 0 \]\n\nFrom Schur Inequality we have\n\n\[ {... | Yes |
Problem 34 Let \( a, b, c \) be nonnegative real numbers. Find the beter constand \( k \) to that inequality always true\n\n\[ \frac{{a}^{2} + {b}^{2} + {c}^{2}}{{\left( a + b + c\right) }^{2}} + k \cdot \frac{{a}^{2}b + {b}^{2}c + {c}^{2}a}{a{b}^{2} + b{c}^{2} + c{a}^{2}} \geq k + 1. \] | \( {k}_{\max } = {2.7775622}\ldots \ldots \) | Yes |
Problem 35 Let \( a, b, c \) be nonnegative real numbers. Find the beter constand \( k \) to that inequality always true\n\n\[ \frac{3\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) }{{\left( a + b + c\right) }^{2}} + k \cdot \frac{{a}^{4}b + {b}^{4}c + {c}^{4}a}{{a}^{3}{b}^{2} + {b}^{3}{c}^{2} + {c}^{3}{a}^{2}} \geq k + 1... | \( {k}_{\max } \approx 0,{89985223}\ldots \) | Yes |
Problem 36 Let \( a, b, c \) be nonnegative real numbers. Find the beter constand \( k \) to that inequality always true\n\n\[ \frac{{a}^{3}}{b} + \frac{{b}^{3}}{c} + \frac{{c}^{3}}{a} + k\left( {{ab} + {bc} + {ca}}\right) \geq k\left( {k + 1}\right) \left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) . \] | \( {k}_{\max } \approx {2.581412182}\ldots \) | No |
Problem 37 Let \( a, b, c \) be nonnegative real numbers. Find the beter constand \( k \) to that inequality always true\n\n\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + k \geq \left( {k + 3}\right) \sqrt[3]{{\left( \frac{{a}^{2} + {b}^{2} + {c}^{2}}{{ab} + {bc} + {ca}}\right) }^{2}}. \] | \( {k}_{\max } \approx {0.3820494092}\ldots \) | Yes |
Lemma 1.3. If \( p \) is an odd prime and \( a \) is any integer such that \( p \) does not divide \( a \), then \( {a}^{\frac{1}{2}\left( {p - 1}\right) } \equiv \left( \frac{a}{p}\right) {\;\operatorname{mod}\;p} \) . | Proof. The multiplicative group \( {\mathbb{F}}_{p}^{ \times } \) being cyclic, let \( b \) be a generator. Write \( a \equiv {b}^{r}{\;\operatorname{mod}\;p} \) for some integer \( r \) . Since \( \left( \frac{a}{p}\right) = {\left( -1\right) }^{r} \) and \( {a}^{\frac{1}{2}\left( {p - 1}\right) } \equiv {\left( {b}^{... | Yes |
Lemma 1.4 (Gauss). Let \( p \) be an odd prime, and let \( a \) be any integer such that \( p \) does not divide \( a \) . Among the least positive residues modulo \( p \) of the integers \( a,{2a},{3a},\ldots ,\frac{1}{2}\left( {p - 1}\right) a \), let \( n \) denote the number of residues that exceed \( p/2 \) . Then... | Proof. Let \( {r}_{1},\ldots ,{r}_{n} \) be the least positive residues exceeding \( p/2 \), and let \( {s}_{1},\ldots ,{s}_{k} \) be those less than \( p/2 \), so that \( n + k = \frac{1}{2}\left( {p - 1}\right) \) . The residues \( {r}_{1},\ldots ,{r}_{n},{s}_{1},\ldots ,{s}_{k} \) are distinct, since no two of \( a,... | Yes |
Lemma 1.5. If \( p \) is an odd prime and \( a \) is a positive odd integer such that \( p \) does not divide \( a \), then \( \left( \frac{a}{p}\right) = {\left( -1\right) }^{t} \), where \( t = \mathop{\sum }\limits_{{u = 1}}^{{\frac{1}{2}\left( {p - 1}\right) }}\left\lbrack {{ua}/p}\right\rbrack \) . Here \( \left\l... | Proof. With notation as in Lemma 1.4 and its proof, we form each \( {ua} \) for \( 1 \leq \) \( u \leq \frac{1}{2}\left( {p - 1}\right) \) and reduce modulo \( p \), obtaining as least positive residue either some \( {r}_{i} \) for \( i \leq n \) or some \( {s}_{j} \) for \( j \leq k \) . Then \( {ua}/p = \left\lbrack ... | Yes |
Theorem 1.8. Fix a positive nonsquare discriminant \( D \) . (a) Each form of discriminant \( D \) is properly equivalent to some reduced form of discriminant \( D \) . | Proof of THEOREM 1.8a. If \( \left( {a, b, c}\right) \) is given and is not reduced, let \( m \) be the unique integer such that \[ \sqrt{D} - 2\left| c\right| < - b + {2cm} < \sqrt{D} \] \( \left( *\right) \) and define \( \left( {{a}^{\prime },{b}^{\prime },{c}^{\prime }}\right) = \left( {c, - b + {2cm}, a - {bm} + c... | Yes |
Proposition 1.9. Let \( \left( {{a}_{1}, b,{c}_{1}}\right) \) and \( \left( {{a}_{2}, b,{c}_{2}}\right) \) be two primitive forms with the same middle coefficient \( b \) and with the same nonsquare discriminant \( D \), hence with \( {a}_{1}{c}_{1} = {a}_{2}{c}_{2} \neq 0 \) . Suppose that \( j = {c}_{1}{a}_{2}^{-1} =... | Proof. The form \( \left( {{a}_{1}{a}_{2}, b, j}\right) \) is primitive because any prime that divides \( \operatorname{GCD}\left( {{a}_{1}{a}_{2}, b, j}\right) \) has to divide either \( \operatorname{GCD}\left( {{a}_{1}, b, j}\right) \) or \( \operatorname{GCD}\left( {{a}_{2}, b, j}\right) \) and then certainly has t... | Yes |
Lemma 1.10. If \( \left( {a, b, c}\right) \) is a primitive form of nonsquare discriminant and if \( m \neq 0 \) is an integer, then \( \left( {a, b, c}\right) \) primitively represents some integer relatively prime to \( m \) . | Proof. Let\n\n\( {w}_{0} = \) product of all primes dividing \( a, c \), and \( m \) ,\n\n\( {x}_{0} = \) product of all primes dividing \( a \) and \( m \) but not \( c \) ,\n\n\( {y}_{0} = \) product of all primes dividing \( m \) but not \( a \) .\n\nReferring to the definitions, we see that any prime dividing \( m ... | Yes |
Lemma 1.11. Suppose that \( \left( {{a}_{1}, b,{c}_{1}}\right) \) and \( \left( {{a}_{2}, b,{c}_{2}}\right) \) are properly equivalent forms of nonsquare discriminant. If \( l \) is an integer such that \( \operatorname{GCD}\left( {{a}_{1},{a}_{2}, l}\right) = 1 \) and such that \( l \) divides \( \operatorname{GCD}\le... | Proof. Since \( \left( {{a}_{1}, b,{c}_{1}}\right) \) and \( \left( {{a}_{2}, b,{c}_{2}}\right) \) are properly equivalent, there exists \( \left( \begin{array}{ll} \alpha & \beta \\ \gamma & \delta \end{array}\right) \) with\n\n\[ \left( \begin{matrix} \alpha & \gamma \\ \beta & \delta \end{matrix}\right) \left( \begi... | Yes |
Theorem 1.12. Let \( D \) be a nonsquare discriminant, and let \( {\mathcal{C}}_{1} \) and \( {\mathcal{C}}_{2} \) be proper equivalence classes of primitive forms of discriminant \( D \). (a) There exist aligned forms \( \left( {{a}_{1}, b,{c}_{1}}\right) \in {\mathcal{C}}_{1} \) and \( \left( {{a}_{2}, b,{c}_{2}}\rig... | Proof of THEOREM 1.12a. By two applications of Lemma 1.10, \( {\mathcal{C}}_{1} \) primitively represents some integer \( {a}_{1} \) prime to \( m \), and \( {\mathcal{C}}_{2} \) primitively represents some integer \( {a}_{2} \) prime to \( {a}_{1}m \) . Arguing as in the last part of the proof of Theorem 1.6b, we may ... | Yes |
Lemma 1.13. If \( \left( {a, b, c}\right) \) is a form with rational coefficients and with non-square discriminant \( D \) that takes on a nonzero value \( q \in \mathbb{Q} \) for some \( \left( {{x}_{0},{y}_{0}}\right) \) in \( \mathbb{Q} \times \mathbb{Q} \), then \( \left( {a, b, c}\right) \) is properly equivalent ... | Proof. Suppose that \( a{x}_{0}^{2} + b{x}_{0}{y}_{0} + c{y}_{0}^{2} = q \) . Put \( \left( \begin{array}{l} \alpha \\ \gamma \end{array}\right) = \left( \begin{matrix} {x}_{0} \\ {y}_{0} \end{matrix}\right) \) . Since \( {x}_{0} \) and \( {y}_{0} \) cannot both be 0, we can choose rationals \( \beta \) and \( \delta \... | Yes |
Theorem 1.14. For a fundamental discriminant \( D \), the principal genus \( P \) of primitive integer forms \( {}^{8} \) is a subgroup of the form class group \( H \), and the cosets of \( P \) are the various genera. Thus the set \( G \) of genera is exactly the set of cosets \( H/P \) and inherits a group structure ... | Proof. Let \( V\left( \mathcal{C}\right) \) denote the set of \( \mathbb{Q} \) values assumed by forms in the class \( \mathcal{C} \) at points \( \left( {x, y}\right) \) in \( \mathbb{Q} \times \mathbb{Q} \) . If \( S \) and \( {S}^{\prime } \) are two genera and if \( \mathcal{C} \) is a class in \( S \) and \( {\mat... | Yes |
Lemma 1.15. If \( \alpha \) is a real irrational number and if \( N > 0 \) is an integer, then there exist integers \( A \) and \( B \) with\n\n\[ \left| {{B\alpha } - A}\right| < \frac{1}{N}\;\text{ and }\;0 < B \leq N. \]\n\nFor this \( A \) and this \( B \) ,\n\n\[ \left| {\alpha - \frac{A}{B}}\right| < \frac{1}{{B}... | Proof. Put \( {\alpha }_{n} = {n\alpha } - \left\lbrack {n\alpha }\right\rbrack \), where \( \left\lbrack \cdot \right\rbrack \) denotes the greatest-integer function. Then \( 0 \leq {\alpha }_{n} < 1 \) . We partition the half-open interval \( \lbrack 0,1) \) into \( N \) subintervals \( \left\lbrack {\frac{t - 1}{N},... | Yes |
Proposition 1.17. Equivalence and strict equivalence are the same for ideals of \( R \) if and only if either\n\n(a) \( m > 0 \) and the fundamental unit \( {\varepsilon }_{1} \) has \( N\left( {\varepsilon }_{1}\right) = - 1 \) or\n\n(b) \( m < 0 \).\n\nIn the contrary case when \( m > 0 \) and the fundamental unit \(... | Proof. Suppose that \( m > 0 \) and \( N\left( {\varepsilon }_{1}\right) = - 1 \) . If \( \left( r\right) I = \left( s\right) J \) with \( N\left( {r{s}^{-1}}\right) < \) 0, then \( \left( {{\varepsilon }_{1}r}\right) I = \left( s\right) J \) with \( N\left( {{\varepsilon }_{1}r{s}^{-1}}\right) > 0 \) . Thus equivalenc... | Yes |
Lemma 1.18. Let \( \mathcal{H} \) be the set of strict equivalence classes of nonzero ideals in \( R \), with its inherited commutative associative multiplication. If \( \mathcal{H} \) is finite, then \( \mathcal{H} \) is a group under this multiplication. | Proof. The identity element of \( \mathcal{H} \) is the strict equivalence class of the ideal \( R = \left( 1\right) \), and we are to prove the existence of inverses. Thus let \( I \) be given. For the sequence of ideals \( I,{I}^{2},{I}^{3},\ldots \), the finiteness of \( \mathcal{H} \) shows that two of these ideals... | Yes |
Theorem 1.20. The set \( \mathcal{H} \) of strict equivalence classes of nonzero ideals relative to the field \( K = \mathbb{Q}\left( \sqrt{m}\right) \) is a finite abelian group. Moreover, the mapping \( \mathcal{F} \) that carries a positively oriented expression \( I = \left\langle {{r}_{1},{r}_{2}}\right\rangle \) ... | Proof. The proof proceeds in six steps.\n\nStep 1. We show that the proper equivalence class of the quadratic form \( \mathcal{F}\left( {I,{r}_{1},{r}_{2}}\right) \) depends only on the ideal \( I \), not the positively oriented expression \( I = \left\langle {{r}_{1},{r}_{2}}\right\rangle \) for it. Thus the class of ... | Yes |
Proposition 1.22. If \( \left| {a}_{n}\right| < 1 \) for all \( n \), then the following conditions are equivalent:\n\n(a) \( \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 + \left| {a}_{n}\right| }\right) \) converges,\n\n(b) \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {a}_{n}\right| \) converges,\n\n(c) \... | Proof. Condition (c) is equivalent to\n\n(c’) \( \mathop{\prod }\limits_{{n = 1}}^{\infty }{\left( 1 - \left| {a}_{n}\right| \right) }^{-1} \) converges.\n\nFor each of (a),(b), and \( \left( {\mathrm{c}}^{\prime }\right) \), convergence is equivalent to boundedness above. Since\n\n\[ 1 + \mathop{\sum }\limits_{{n = 1}... | Yes |
Proposition 1.23. Let \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{n}^{-s} \) be a Dirichlet series.\n\n(a) If the series is convergent for \( s = {s}_{0} \), then it is convergent uniformly on compact sets for \( \operatorname{Re}s > \operatorname{Re}{s}_{0} \), and the sum of the series is analytic in this reg... | Proof. For (a), we write \( {a}_{n}{n}^{-s} = {a}_{n}{n}^{-{s}_{0}} \cdot {n}^{-\left( {s - {s}_{0}}\right) } = {u}_{n}{v}_{n} \) and then apply the summation by parts formula \( \left( *\right) \) . The given convergence means that the sequence \( \left\{ {U}_{n}\right\} \) is convergent, and certainly \( {v}_{n} \) t... | Yes |
Proposition 1.24. The Riemann zeta function \( \zeta \left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{n}^{-s} \), initially defined and analytic for \( \operatorname{Re}s > 1 \), extends to be meromorphic for \( \operatorname{Re}s > 0 \) . Its only pole is at \( s = 1 \), and the pole is simple. | Proof. For \( \operatorname{Re}s > 1 \), we have\n\n\[ \frac{1}{s - 1} = {\int }_{1}^{\infty }{t}^{-s}{dt} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\int }_{n}^{n + 1}{t}^{-s}{dt}. \]\n\nThus \( \operatorname{Re}s > 1 \) implies\n\n\[ \zeta \left( s\right) = \frac{1}{s - 1} + \mathop{\sum }\limits_{{n = 1}}^{\infty }... | Yes |
Proposition 1.25. Let \( Z\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{n}^{-s} \) be a Dirichlet series with all \( {a}_{n} \geq 0 \) . Suppose that the series is convergent in some half-plane and that the sum extends to be analytic for \( \operatorname{Re}s > 0 \) . Then the series converges for ... | Proof. By assumption the series converges somewhere, and therefore \( {s}_{0} = \) \( \inf \left\{ {s \geq 0 \mid \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{n}^{-s}}\right. \) converges \( \} \) is a well-defined real number \( \geq 0 \) . Arguing by contradiction, suppose that \( {s}_{0} > 0 \) . Since \( \sum {... | Yes |
Proposition 1.27. A first-degree Euler product \( \prod {\left( 1 - {a}_{p}{p}^{-s}\right) }^{-1} \) with \( \left| {a}_{p}\right| \leq {p}^{c} \) for some real \( c \) and all primes \( p \) defines an absolutely convergent Dirichlet series for \( \operatorname{Re}s > c + 1 \) and hence a valid identity \( \mathop{\su... | Proof. The coefficients \( {a}_{n} \) are strictly multiplicative, and thus \( \left| {a}_{n}\right| \leq {n}^{c} \) for all \( n \) . The absolute convergence follows. | No |
Proposition 1.28. Fix \( m \), and let \( \chi \) be a Dirichlet character modulo \( m \) . (a) The Dirichlet series \( L\left( {s,\chi }\right) \) is absolutely convergent for \( \operatorname{Re}s > 1 \) and is given in that region by a first-degree Euler product | Proof. For (a), the boundedness of \( \chi \) implies that the series is absolutely convergent for \( \operatorname{Re}s > 1 \) . Since \( \chi \) is strictly multiplicative, \( L\left( {s,\chi }\right) \) has a first-degree Euler product by Proposition 1.26, and the product is convergent in the same region. | Yes |
Lemma 2.1. Let \( D \) be a division ring, let \( R = {M}_{n}\left( D\right) \), and let \( {D}^{n} \) be the simple left \( R \) module of column vectors. Each member of \( D \) acts on \( {D}^{n} \) by scalar multiplication on the right side, yielding a member of \( {\operatorname{End}}_{R}\left( {D}^{n}\right) \) . ... | Proof. Let \( \varphi : D \rightarrow {\operatorname{End}}_{R}\left( {D}^{n}\right) \) be the function given by \( \varphi \left( d\right) \left( v\right) = {vd} \) . Then \( \varphi \left( {d{d}^{\prime }}\right) \left( v\right) = v\left( {d{d}^{\prime }}\right) = \left( {vd}\right) {d}^{\prime } = \varphi \left( {d}^... | Yes |
For a ring \( R \), left semisimple coincides with right semisimple. | The theorem gives the form of any left semisimple ring, and each ring of this form is certainly right semisimple. | No |
Theorem 2.4 (Wedderburn). Let \( R \) be a finite-dimensional algebra with identity over a field \( F \) . If \( R \) is a simple ring, then \( R \) is semisimple and hence is isomorphic to \( {M}_{n}\left( D\right) \) for some integer \( n \geq 1 \) and some finite-dimensional division algebra \( D \) over \( F \) . T... | Proof. By finite dimensionality, \( R \) has a minimal left ideal \( V \) . For \( r \) in \( R \) , form the set \( {Vr} \) . This is a left ideal, and we claim that it is minimal or is 0 . In fact, the function \( v \mapsto {vr} \) is \( R \) linear from \( V \) onto \( {Vr} \) . Since \( V \) is simple as a left \( ... | Yes |
Proposition 2.5. Let \( R \) be a ring with identity, and let \( M \) be a finitely generated unital left \( R \) module. If \( R \) is left Noetherian, then \( M \) satisfies the ascending chain condition for its \( R \) submodules; if \( R \) is left Artinian, then \( M \) satisfies the descending chain condition for... | Proof. We prove the first conclusion by induction on the number of generators, and the proof of the second conclusion is completely similar. The result is trivial if \( M \) has 0 generators. If \( M = {Rx} \), then \( M \) is a quotient of the left \( R \) module \( R \) and satisfies the ascending chain condition for... | Yes |
Theorem 2.6 (E. Artin). If \( R \) is a simple ring, then the following conditions are equivalent:\n\n(a) \( R \) is left Artinian,\n\n(b) \( R \) is semisimple,\n\n(c) \( R \) has a minimal left ideal,\n\n(d) \( R \cong {M}_{n}\left( D\right) \) for some integer \( n \geq 1 \) and some division ring \( D \) . | Proof. It is evident from Wedderburn's Theorem (Theorem 2.2) that (b) and (d) are equivalent. For the rest we prove that (a) implies (c), that (c) implies (b), and that (b) implies (a).\n\nSuppose that (a) holds. Applying the minimum condition for left ideals in \( R \) , we obtain a minimal left ideal. Thus (c) holds.... | Yes |
Lemma 2.7. If \( {I}_{1} \) and \( {I}_{2} \) are nilpotent left ideals in a ring \( R \) with identity, then \( {I}_{1} + {I}_{2} \) is nilpotent. | Proof. Let \( {I}_{1}^{r} = 0 \) and \( {I}_{2}^{s} = 0 \) . Expand \( {\left( {I}_{1} + {I}_{2}\right) }^{k} \) as \( \sum {I}_{{i}_{1}}{I}_{{i}_{2}}\cdots {I}_{{i}_{k}} \) with each \( {i}_{j} \) equal to 1 or 2 . Take \( k = r + s \) . In any term of the sum, there are \( \geq r \) indices 1 or \( \geq s \) indices ... | Yes |
Lemma 2.8. If \( I \) is a nilpotent left ideal in a ring \( R \) with identity, then \( I \) is contained in a nilpotent two-sided ideal \( J \) . | Proof. Put \( J = \mathop{\sum }\limits_{{r \in R}}{Ir} \) . This is a two-sided ideal. For any integer \( k \geq 0 \) , \( {J}^{k} = {\left( \mathop{\sum }\limits_{{r \in R}}Ir\right) }^{k} \subseteq \mathop{\sum }\limits_{{{r}_{1},\ldots ,{r}_{k}}}I{r}_{1}I{r}_{2}\cdots I{r}_{k} \subseteq \mathop{\sum }\limits_{{r}_{... | Yes |
Lemma 2.9. If \( R \) is a ring with identity, then the sum of all nilpotent left ideals in a nil two-sided ideal. | Proof. Let \( K \) be the sum of all nilpotent left ideals in \( R \), and let \( a \) be a member of \( K \) . Write \( a = {a}_{1} + \cdots + {a}_{n} \) with \( {a}_{i} \in {I}_{i} \) for a nilpotent left ideal \( {I}_{i} \) . Lemma 2.7 shows that \( I = \mathop{\sum }\limits_{{i = 1}}^{n}{I}_{i} \) is a nilpotent le... | Yes |
Theorem 2.10. If \( R \) is a left Artinian ring, then any nil left ideal in \( R \) is nilpotent. | Proof. Let \( I \) be a nil left ideal of \( R \), and form the filtration\n\n\[ I \supseteq {I}^{2} \supseteq {I}^{3} \supseteq \cdots \]\n\nSince \( R \) is left Artinian, this filtration is constant from some point on, and we have \( {I}^{k} = {I}^{k + 1} = {I}^{k + 2} = \cdots \) for some \( k \geq 1 \) . Put \( J ... | Yes |
Corollary 2.11. If \( R \) is a left Artinian ring, then there exists a unique largest nilpotent two-sided ideal \( I \) in \( R \) . This ideal is the sum of all nilpotent left ideals and also is the sum of all nilpotent right ideals. | Proof. By Lemma 2.9 and Theorem 2.10 the sum of all nilpotent left ideals in \( R \) is a two-sided nilpotent ideal \( I \) . Lemma 2.8 shows that any nilpotent right ideal is contained in a nilpotent two-sided ideal \( J \) . Since \( J \) is in particular a nilpotent left ideal, the definition of \( I \) forces \( J ... | Yes |
Lemma 2.12 (Brauer’s Lemma). If \( R \) is any ring with identity and if \( V \) is a minimal left ideal in \( R \), then either \( {V}^{2} = 0 \) or \( V = {Re} \) for some element \( e \) of \( V \) with \( {e}^{2} = e \) . | Proof. Being a minimal left ideal, \( V \) is a simple left \( R \) module. Schur’s Lemma (Proposition 10.4b of Basic Algebra) shows that \( {\operatorname{End}}_{R}V \) is a division ring. If \( a \) is in \( V \), then the map \( v \mapsto {va} \) of \( V \) into itself lies in \( {\operatorname{End}}_{R}V \) and hen... | Yes |
Theorem 2.13. If \( R \) is a left Artinian ring and if the Wedderburn-Artin radical of \( R \) is 0, then \( R \) is a semisimple ring. | Proof. Let us see that any minimal left ideal \( I \) of \( R \) is a direct summand as a left \( R \) submodule. Since rad \( R = 0, I \) is not nilpotent. Thus \( {I}^{2} \neq 0 \), and Lemma 2.12 shows that \( I \) contains an idempotent \( e \) . This element satisfies \( I = {Re} \) . Put \( {I}^{\prime } = \{ r \... | Yes |
Lemma 2.16. If \( R \) is a semisimple ring, then every unital left \( R \) module \( M \) is semisimple. | Proof. For each \( m \in M \), let \( {R}_{m} \) be a copy of the left \( R \) module \( R \), and define \( \widetilde{M} = {\bigoplus }_{m \in M}{R}_{m} \) as a left \( R \) module. Since each \( {R}_{m} \) is semisimple, \( \widetilde{M} \) is semisimple. Define a function \( \varphi : \widetilde{M} \rightarrow M \)... | Yes |
Theorem 2.18. Let \( A \) be a left Artinian ring with Wedderburn-Artin radical rad \( A \), and suppose that \( A/\operatorname{rad}A \) is simple, i.e., is of the form \( A/\operatorname{rad}A \cong {M}_{n}\left( D\right) \) for some division ring \( D \) . Then \( A \) is isomorphic as a ring to \( {M}_{n}\left( R\r... | The idea behind the proof of Theorem 2.18 is to give an abstract characterization of a ring of matrices in terms of the elements \( {E}_{ij} \) that are 1 in the \( {\left( i, j\right) }^{\text{th }} \) place and are 0 elsewhere. In turn, these elements arise from the diagonal such elements \( {E}_{ii} \), which are id... | No |
Proposition 2.19. In a ring \( R \) with identity, let \( e \) be an element of \( R \) with \( {e}^{2} = e \) .\n\n(a) If \( I \) is a left ideal in \( {eRe} \), then \( {eRI} = I \) . Hence \( I \mapsto {RI} \) is a one-one inclusion-preserving map of the left ideals of \( {eRe} \) to those of \( R \) . | Proof. For (a), we have \( {eRI} = {eR}\left( {eI}\right) = \left( {eRe}\right) I = I \), the first equality holding because \( e \) is the identity in \( {eRe} \) and the third equality holding because \( {eRe} \) contains its identity \( e \) . The rest of (a) then follows. | Yes |
Corollary 2.20. In a left Artinian ring \( R \), let \( e \) be an element with \( {e}^{2} = e \) . Then the ring \( {eRe} \) is left Artinian, and\n\n\[ \operatorname{rad}\left( {eRe}\right) = {eRe} \cap \operatorname{rad}R = e\left( {\operatorname{rad}R}\right) e. \]\n\nIf \( \bar{R} \) denotes the quotient ring \( R... | Proof. The ring \( {eRe} \) is left Artinian as an immediate consequence of Proposition 2.19a. For the first display we may assume that \( R \) and \( {eRe} \) are both left Artinian. Then \( {eRe} \cap \operatorname{rad}R \) is a two-sided ideal of \( {eRe} \), and \( {\left( eRe \cap \operatorname{rad}R\right) }^{n} ... | Yes |
In a ring \( R \) with identity, let \( {e}_{1} \) and \( {e}_{2} \) be idempotents. Then the unital left \( R \) modules \( R{e}_{1} \) and \( R{e}_{2} \) are isomorphic as left \( R \) modules if and only if there exist elements \( {e}_{12} \) and \( {e}_{21} \) in \( R \) such that\n\n\[ \n{e}_{1}{e}_{12}{e}_{2} = {... | Proof. Let \( \varphi : R{e}_{1} \rightarrow R{e}_{2} \) be an \( R \) isomorphism. Define \( {e}_{12} = \varphi \left( {e}_{1}\right) \) and \( {e}_{21} = {\varphi }^{-1}\left( {e}_{2}\right) \) . Every element \( s \) of \( R{e}_{2} \) has the property that \( s{e}_{2} = s \) because \( {e}_{2}^{2} = {e}_{2} \) ; sin... | Yes |
Proposition 2.23. Let \( R \) be a left Artinian ring. For each \( r \) in \( R \), let \( \bar{r} \) be the element \( r + \operatorname{rad}R \) of \( R/\operatorname{rad}R \). If \( a \) is an element of \( R \) such that \( \bar{a} \) is idempotent in \( R/\operatorname{rad}R \), then there exists an idempotent \( ... | Proof. Set \( b = 1 - a \). The elements \( a \) and \( b \) commute, and \( {ab} = a\left( {1 - a}\right) \) maps to \( \bar{a} - {\bar{a}}^{2} = 0 \) in \( R/\operatorname{rad}R \), since \( \bar{a} \) is idempotent. Therefore \( {ab} \) lies in rad \( R \) and must satisfy \( {\left( ab\right) }^{n} = 0 \) for some ... | Yes |
Corollary 2.24. Let \( R \) be a left Artinian ring. For each \( r \) in \( R \), let \( \bar{r} \) be the coset \( r + \operatorname{rad}R \) in \( R/\operatorname{rad}R \) . If \( x \) and \( y \) are orthogonal idempotents in \( \bar{R} = R/\operatorname{rad}R \) and if \( e \) is an idempotent in \( R \) with \( \b... | Proof. By Proposition 2.23 choose an idempotent \( {f}_{0} \) in \( R \) with \( {\bar{f}}_{0} = y \) . Then \( {f}_{0}e \) has \( \overline{{f}_{0}e} = {yx} = 0 \) . Hence \( {f}_{0}e \) is in rad \( R \), and \( {\left( {f}_{0}e\right) }^{n + 1} = 0 \) for some \( n \) . Consequently \( 1 + {f}_{0}e + {\left( {f}_{0}... | Yes |
Corollary 2.25. Let \( R \) be a left Artinian ring. For each \( r \) in \( R \), let \( \bar{r} \) be the coset \( r + \operatorname{rad}R \) in \( R/\operatorname{rad}R \) . If \( \left\{ {{x}_{1},\ldots ,{x}_{N}}\right\} \) is a finite set of mutually orthogonal idempotents in \( \bar{R} = R/\operatorname{rad}R \), ... | Proof. For the existence of \( \left\{ {{x}_{1},\ldots ,{x}_{N}}\right\} \), we proceed by induction on \( N \), the case \( N = 1 \) being Proposition 2.23. Suppose we have found \( {e}_{1},\ldots ,{e}_{n} \) and we want to find \( {e}_{n + 1} \) . Let \( e \) be the idempotent \( {e}_{1} + \cdots + {e}_{n} \), and ap... | Yes |
Corollary 2.28. If \( A \) is a finite-dimensional associative algebra with identity over a field \( F \) and if \( A/\operatorname{rad}A \cong {M}_{{n}_{1}}\left( F\right) \times \cdots \times {M}_{{n}_{r}}\left( F\right) \), then there is a subalgebra \( S \) of \( A \) isomorphic as an algebra to \( A/\operatorname{... | Proof. For \( 1 \leq j \leq r \), let \( {x}_{j} \) be the identity matrix of \( {M}_{{n}_{j}}\left( F\right) \) when \( {M}_{{n}_{j}}\left( F\right) \) is regarded as a subalgebra of \( A/\operatorname{rad}A \) . The elements \( {x}_{j} \) are orthogonal idempotents in \( A/\operatorname{rad}A \) with sum 1, and Corol... | Yes |
Corollary 2.30. Let \( F \) be a field, let \( K \) be a finite separable algebraic extension of \( F \), and let \( L \) be another field extension of \( F \) . Then the algebra \( K{ \otimes }_{F}L \) is semisimple. | Proof. The Theorem of the Primitive Element (Theorem 9.34 of Basic Algebra) shows that \( K/F \) is a simple extension, say with \( K = F\left( \alpha \right) \) . Since this extension is assumed separable, the minimal polynomial over \( F \) of any element of \( K \) is a separable polynomial. The hypotheses of Propos... | Yes |
Lemma 2.32. The center of a finite-dimensional simple algebra \( A \) over a field \( F \) is a field that is a finite extension of \( F \) . | Proof. By Theorem 2.4, \( A \cong {M}_{n}\left( D\right) \) for some finite-dimensional division algebra \( D \) over \( F \) . Let \( Z \) be the center of \( A \) . By inspection this consists of the scalar matrices whose entries lie in the center of \( D \) . The center of \( D \) is a field. Hence \( Z \) is a fiel... | Yes |
Proposition 2.33. Let \( A \) be a finite-dimensional semisimple algebra over a field \( F \) of characteristic 0, and suppose that \( K \) is a field containing \( F \) . Then the algebra \( A{ \otimes }_{F}K \) over \( K \) is semisimple. | Proof. Since the tensor product of a finite direct sum is the direct sum of tensor products, we may assume without loss of generality that \( A \) is simple. Lemma 2.32 shows that the center \( Z \) of \( A \) is a finite extension field of \( F \) . By Corollary 2.30 and the assumption that \( F \) has characteristic ... | Yes |
Lemma 2.34. Suppose that \( A \) is a finite-dimensional algebra with identity over a field \( F \), and suppose that \( N \) is a nilpotent two-sided ideal of \( A \) such that the algebra \( A/N \) is semisimple. Then \( N = \operatorname{rad}A \) . | Proof. The algebra \( A \) is left Artinian, being finite-dimensional. Since \( N \) is nilpotent, we must have \( N \subseteq \operatorname{rad}A \) . The two-sided ideal \( \left( {\operatorname{rad}A}\right) /N \) of the semisimple algebra \( A/N \) is nilpotent and hence must be 0 . Therefore \( N = \operatorname{r... | Yes |
Lemma 2.35. Let \( A \) and \( B \) be algebras with identity over a field \( F \), and suppose that \( B \) is central. Then\n\n(a) the members of \( A{ \otimes }_{F}B \) commuting with \( 1 \otimes B \) are the members of \( A \otimes 1 \) ,\n\n(b) \( \operatorname{center}\left( {A{ \otimes }_{F}B}\right) = \left( {\... | Proof. For (a), suppose that \( z = \mathop{\sum }\limits_{i}{a}_{i} \otimes {b}_{i} \) commutes with \( 1 \otimes B \) and that the \( {a}_{i} \) are linearly independent over \( F \) . If \( b \) is in \( B \), then\n\n\[ 0 = \left( {1 \otimes b}\right) z - z\left( {1 \otimes b}\right) = \mathop{\sum }\limits_{i}{a}_... | Yes |
Proposition 2.36. Let \( A \) and \( B \) be algebras with identity over a field \( F \), and suppose that \( B \) is central simple. Then\n\n(a) \( A \) simple implies \( A{ \otimes }_{F}B \) simple,\n\n(b) \( A \) central simple implies \( A{ \otimes }_{F}B \) central simple. | Proof. For (a), Proposition 2.31 shows that any two-sided ideal of \( A{ \otimes }_{F}B \) is of the form \( I{ \otimes }_{F}B \) for some two-sided ideal \( I \) of \( A \) . Since \( A \) is assumed simple, the only \( I \)’s are 0 and \( A \) . Thus the only ideals in \( A{ \otimes }_{F}B \) are 0 and \( A{ \otimes ... | Yes |
Corollary 2.37. If \( A \) and \( B \) are finite-dimensional semisimple algebras over a field \( F \) and at least one of them is separable over \( F \), then \( A{ \otimes }_{F}B \) is semisimple. | Proof. Without loss of generality, we may assume that \( A \) and \( B \) are simple. For definiteness let us say that \( A \) is the given separable algebra over \( F \) . Let \( K = \operatorname{center}\left( B\right) \) . Lemma 2.32 shows that \( K \) is a field, and associativity of tensor products allows us to wr... | Yes |
Corollary 2.38. Let \( A \) be a central simple algebra of finite dimension \( n \) over a field \( F \), and let \( {A}^{o} \) be the opposite algebra. Then \( A{ \otimes }_{F}{A}^{o} \cong {M}_{n}\left( F\right) \) . | Proof. Let \( V \) be \( A \) considered as a vector space. For each \( {a}_{0} \in A \), we associate the members \( l\left( {a}_{0}\right) \) and \( r\left( {a}_{0}\right) \) of \( {\operatorname{End}}_{F}\left( V\right) \) given by \( l\left( {a}_{0}\right) a = {a}_{0}a \) and \( r\left( {a}_{0}\right) a = \) \( a{a... | Yes |
Corollary 2.39. Let \( A \) be a central simple algebra of finite dimension \( d \) over a field \( F \) . Then \( d \) is the square of an integer. | Proof. Let \( \bar{F} \) be an algebraic closure of \( F \) . Proposition 2.36a shows that the algebra \( \bar{F}{ \otimes }_{F}A \) is simple, and its dimension over \( \bar{F} \) is \( d \) . A simple finite-dimensional algebra over an algebraically closed field is a full matrix algebra over that field, and thus \( \... | Yes |
Corollary 2.40. If \( D \) is a division algebra finite-dimensional over its center \( F \), then \( {\dim }_{F}D \) is the square of an integer. | Proof. The algebra \( D \) is central simple over its center \( F \), and the result is immediate from Corollary 2.39. | No |
Theorem 2.41 (Skolem-Noether Theorem). Let \( A \) be a finite-dimensional central simple algebra over the field \( F \), and let \( B \) be any simple algebra over \( F \) . Suppose that \( f \) and \( g \) are \( F \) algebra homomorphisms of \( B \) into \( A \) carrying the identity to the identity. Then there exis... | Proof. Let us observe that the homomorphisms \( f \) and \( g \) are one-one because \( B \) is simple, and the finite dimensionality of \( A \) therefore forces \( B \) to be finite-dimensional.\n\nWe consider first the special case that \( A = {M}_{n}\left( F\right) \) for some \( n \) . The homomorphism \( f \) make... | Yes |
Corollary 2.42. If \( A \) is a finite-dimensional central simple algebra over the field \( F \), then every \( F \) automorphism of \( A \) is inner in the sense of being given by conjugation by an invertible element of \( A \) . | Proof. This is the special case of Theorem 2.41 in which \( B = A \) and \( g \) is the identity map on \( B \) . | Yes |
Lemma 2.44. Let \( A \) and \( {A}^{\prime } \) be algebras with identity over a field \( F \), let \( B \) and \( {B}^{\prime } \) be subalgebras of them, and let \( C \) and \( {C}^{\prime } \) be the centralizers of \( B \) and \( {B}^{\prime } \) in \( A \) and \( {A}^{\prime } \), respectively. Then the centralize... | Proof. Expand an element of \( A{ \otimes }_{F}{A}^{\prime } \) for the moment as \( x = \mathop{\sum }\limits_{i}{a}_{i} \otimes {a}_{i}^{\prime } \) with the elements \( {a}_{i}^{\prime } \) linearly independent over \( F \) . If \( x \) satisfies \( x\left( {b \otimes 1}\right) = \left( {b \otimes 1}\right) x \) for... | Yes |
Lemma 2.45. Let \( B \) be a finite-dimensional simple algebra over a field \( F \), and write \( V \) for the algebra \( B \) considered as a vector space. For \( b \) in \( B \) and \( v \) in \( V \) , define members \( l\left( b\right) \) and \( r\left( b\right) \) of \( {\operatorname{End}}_{F}\left( V\right) \) b... | Proof. Let \( K \) be the center of \( B \) . This is an extension field of \( F \) by Lemma 2.32, and \( B \) is central simple over \( K \) . Let us see that any member \( a \) of \( {\operatorname{End}}_{F}\left( V\right) \) that centralizes \( l\left( B\right) \) is actually in \( {\operatorname{End}}_{K}\left( V\r... | Yes |
Corollary 2.46. Let \( D \) be a central finite-dimensional division algebra over the field \( F \) . If \( K \) is any maximal subfield of \( D \), then \( {\dim }_{F}D = {\left( {\dim }_{F}K\right) }^{2} \) . | Proof. Apply the Double Centralizer Theorem (Theorem 2.43) with \( A = \) \( D \) . Let \( Z\left( K\right) \) be the centralizer of the simple subalgebra \( K \) in \( D \) . Since \( K \) is commutative, \( K \subseteq Z\left( K\right) \) . If \( a \) is in \( Z\left( K\right) \) but not \( K \), then \( K\left( a\ri... | Yes |
Corollary 2.47. Let \( A \) be a finite-dimensional central simple algebra over a field \( F \), and let \( K \) be a subfield of \( A \) . Then the following are equivalent:\n\n(a) \( K \) is its own centralizer,\n\n(b) \( {\dim }_{F}A = {\left( {\dim }_{F}K\right) }^{2} \),\n\n(c) \( K \) is a maximal commutative sub... | Proof. Let \( Z\left( K\right) \) be the centralizer of \( K \) in \( A \) . The Double Centralizer Theorem (Theorem 2.43) gives the equality\n\n\[ \n{\dim }_{F}A = \left( {{\dim }_{F}K}\right) \left( {{\dim }_{F}Z\left( K\right) }\right) .\n\]\n\n\( \left( *\right) \)\n\nIf (a) holds, then \( Z\left( K\right) = K \), ... | Yes |
Lemma 2.49. If \( G \) is a finite group and \( H \) is a proper subgroup, then \( \mathop{\bigcup }\limits_{{g \in G}}{gH}{g}^{-1} \) does not exhaust \( G \) . | Proof. In the union \( \mathop{\bigcup }\limits_{{g \in G}}{gH}{g}^{-1} \), the terms corresponding to \( g \) and to \( {gh} \), for \( h \) in \( H \), are the same because \( \left( {gh}\right) H{\left( gh\right) }^{-1} = g\left( {{hH}{h}^{-1}}\right) {g}^{-1} = {gH}{g}^{-1} \) . Thus the union can be rewritten as \... | Yes |
Lemma 3.1. If \( F \) is a field, then\n\n(a) \( {M}_{n}\left( R\right) \cong R{ \otimes }_{F}{M}_{n}\left( F\right) \) for any algebra \( R \) with identity over \( F \) , | Proof. For (a), the \( F \) bilinear map \( \left( {r,\left\lbrack {a}_{ij}\right\rbrack }\right) \mapsto \left\lbrack {r{a}_{ij}}\right\rbrack \) of \( R \times {M}_{n}\left( F\right) \) into \( {M}_{n}\left( R\right) \) has a unique linear extension \( \varphi \) to an \( F \) linear map of \( R{ \otimes }_{F}{M}_{n}... | Yes |
Proposition 3.2. For the field \( F \), the operation of tensor product on finite-dimensional central simple algebras over \( F \) descends to an operation on the set of Brauer equivalence classes of such algebras and makes this set into an abelian group. | Proof. The tensor product of two finite-dimensional algebras over \( F \) is again a finite-dimensional algebra, and Proposition 2.36 shows that the tensor product of two central simple algebras is again central simple. Hence tensor product is well defined as an operation on finite-dimensional central simple algebras o... | Yes |
Theorem 3.3. Let \( K/F \) be a finite extension of fields. Then \( K \) is a splitting field for a given member \( X \) of \( \mathcal{B}\left( {K/F}\right) \) if and only if there exists an algebra \( A \) over \( F \) in the Brauer equivalence class \( X \) containing a subfield \( {K}^{\prime } \) isomorphic to \( ... | Proof. Suppose that \( A \) is a central simple algebra in the Brauer equivalence class \( X \) containing a subfield \( {K}^{\prime } \) isomorphic to \( K \) such that \( {\dim }_{F}A = {\left( {\dim }_{F}{K}^{\prime }\right) }^{2} \). We are to prove that \( {K}^{\prime } \) splits \( A \). Write \( n \) for \( {\di... | Yes |
Corollary 3.4. If \( D \) is a finite-dimensional central division algebra over the field \( F \), then any maximal subfield \( K \) of \( D \) splits \( D \) . | Proof. This is the special case of Theorem 3.3 in which \( A = D \) . The formula for the dimensions holds by Corollary 2.47. | Yes |
Corollary 3.5. If \( F \) is a field, then the Brauer group \( \mathcal{B}\left( F\right) \) is the union of all relative Brauer groups \( \mathcal{B}\left( {K/F}\right) \) as \( K \) ranges over all finite extensions of \( F \) . | Proof. Any member of \( \mathcal{B}\left( F\right) \) has some central division algebra \( D \) as a representative, and Corollary 3.4 identifies an extension field \( K \) of \( F \) that splits \( D \), namely any maximal subfield of \( D \) . | Yes |
Corollary 3.6. Let \( D \) be a finite-dimensional central division algebra over a field \( F \), and let \( {\dim }_{F}D = {n}^{2} \) . If \( K \) is a splitting field for \( D \), then \( {\dim }_{F}K \) is a multiple of \( n \) . | Proof. If \( K \) is a splitting field for \( D \), then Theorem 3.3 says that there exists an integer \( r \) such that \( {M}_{r}\left( D\right) \) contains a subfield \( {K}^{\prime } \) isomorphic to \( K \) with \( {\dim }_{F}{M}_{r}\left( D\right) = {\left( {\dim }_{F}{K}^{\prime }\right) }^{2} \) . Thus \( {r}^{... | Yes |
Corollary 3.8. If \( D \) is a noncommutative finite-dimensional central division algebra over the field \( F \) and if \( K \) is a subfield of \( D \) that is separable over \( F \), then there exists a maximal subfield \( L \) of \( D \) containing \( K \) such that \( L \) is separable over \( F \) . | Proof. Because of the finite dimensionality, we may assume without loss of generality that \( K \) is not properly contained in any larger subfield of \( D \) that is separable over \( F \) . Arguing by contradiction, we may assume that \( K \) is not a maximal subfield of \( D \) . Let \( E \) be the centralizer of \(... | Yes |
Corollary 3.9. If \( F \) is a field, then the Brauer group \( \mathcal{B}\left( F\right) \) is the union of all relative Brauer groups \( \mathcal{B}\left( {K/F}\right) \) as \( K \) ranges over all finite Galois extensions of \( F \) . | Proof. If \( D \) is a central division algebra over \( F \), then Corollaries 3.4 and 3.8 together show that some finite separable extension \( {K}^{\prime } \) of \( F \) splits \( D \) . That is, the Brauer equivalence class of \( D \) lies in \( \mathcal{B}\left( {{K}^{\prime }/F}\right) \) . Let us write \( {K}^{\... | Yes |
Lemma 3.11. With hypotheses as above, the set \( \left\{ {{x}_{\sigma } \mid \sigma \in \operatorname{Gal}\left( {K/F}\right) }\right\} \) is a vector-space basis of \( A \) over \( {K}^{\prime } \) . | Proof. Let \( G = \operatorname{Gal}\left( {K/F}\right) \) . Since \( \left| G\right| = {\dim }_{F}K = {\dim }_{F}{K}^{\prime } = {\dim }_{{K}^{\prime }}A \), it is enough to prove linear independence. Arguing by contradiction, assume that the set \( \left\{ {{x}_{\sigma } \mid \sigma \in G}\right\} \) is linearly depe... | Yes |
Proposition 3.12. Let \( K/F \) be a finite Galois extension, and let \( a = a\left( {\sigma ,\tau }\right) \) be in \( {Z}^{2}\left( {\operatorname{Gal}\left( {K/F}\right) ,{K}^{ \times }}\right) \). Then there exist a central simple algebra \( A \) over \( F \) with \( {\dim }_{F}A = {\left( {\dim }_{F}K\right) }^{2}... | Proof. Let \( G = \operatorname{Gal}\left( {K/F}\right) \), form a set \( \left\{ {{x}_{\sigma } \mid \sigma \in G}\right\} \), and let \( A \) be the \( K \) vector space (free \( K \) module) with basis \( \left\{ {x}_{\sigma }\right\} \). Then \( A = {\bigoplus }_{\sigma \in G}K{x}_{\sigma } \). Define a multiplicat... | Yes |
Theorem 3.14. If \( K \) is a finite Galois extension of the field \( F \), then the map \( \mathcal{B}\left( {K/F}\right) \rightarrow {H}^{2}\left( {\operatorname{Gal}\left( {K/F}\right) ,{K}^{ \times }}\right) \) defined via factor sets is a group isomorphism. | Proof (Chase). Let \( G, a, b, A, B \), and \( C \) be as in the remarks. We can regard \( A \) and \( B \) as vector spaces over \( K \) with \( K \) acting on the left in each case. We define an \( F \) vector space \( M \) to be the quotient of \( A{ \otimes }_{F}B \) by the \( F \) vector subspace \( I \) generated... | Yes |
Corollary 3.16. If \( F \) is any field, then every element of \( \mathcal{B}\left( F\right) \) has finite order. | Proof. If \( A \) is any central simple algebra over \( F \), then Theorem 2.4 shows that \( A \cong {M}_{l}\left( D\right) \) for some integer \( l \geq 1 \) and some central division algebra \( D \) over \( F \) . Corollary 3.15 shows that the Brauer equivalence class of \( D \) has finite order in \( \mathcal{B}\lef... | Yes |
Theorem 3.17. If \( K/F \) is any finite Galois extension of fields, then \( {H}^{1}\left( {\operatorname{Gal}\left( {K/F}\right) ,{K}^{ \times }}\right) = 0. \) | Proof. Let \( G = \operatorname{Gal}\left( {K/F}\right) \), put \( n = {\dim }_{F}K \), and enumerate \( G \) as \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) . By the Theorem of the Primitive Element, we can write \( K = F\left( \alpha \right) \) for some \( \alpha \) in \( K \), and then \( \left\{ {1,\alpha ,{\alpha }^{... | Yes |
Corollary 3.18. If \( K/F \) is a finite Galois extension with cyclic Galois group and if \( \sigma \) is a generator of the Galois group, then every member \( x \) of \( K \) with \( {N}_{K/F}\left( x\right) = 1 \) is of the form \( x = \sigma \left( y\right) {y}^{-1} \) for some \( y \in {K}^{ \times } \) . | Proof. Let \( G = \left\{ {1,\sigma ,{\sigma }^{2},\ldots ,{\sigma }^{n - 1}}\right\} \) be the Galois group, and define a function \( F : \mathbb{Z} \rightarrow {K}^{ \times } \) by \( F\left( 0\right) = 1 \) and\n\n\[ F\left( k\right) = {x\sigma }\left( x\right) {\sigma }^{2}\left( x\right) \cdots {\sigma }^{k - 1}\l... | Yes |
For \( n \geq 0 \), the left \( \mathbb{Z}G \) module \( {F}_{n} \) is a free \( \mathbb{Z}G \) module with \( \mathbb{Z}G \) basis consisting of all \( \left( {n + 1}\right) \) -tuples \( \left( {1,{g}_{1},\ldots ,{g}_{n}}\right) \), i.e., all \( \mathbb{Z} \) basis elements with \( {g}_{0} = 1 \) . | Proof. The formula \( {g}_{0}\left( {1,{g}_{0}^{-1}{g}_{1},\ldots ,{g}_{0}^{-1}{g}_{n}}\right) = \left( {{g}_{0},{g}_{1},\ldots ,{g}_{n}}\right) \) shows that all members of the \( \mathbb{Z} \) basis defining \( {F}_{n} \) are \( \mathbb{Z}G \) images of the asserted \( \mathbb{Z}G \) basis; hence the asserted \( \mat... | Yes |
Lemma 3.21. The sequence\n\n\[ \cdots \xrightarrow[]{{\partial }_{n + 1}}{F}_{n + 1}\xrightarrow[]{{\partial }_{n}}{F}_{n}\xrightarrow[]{{\partial }_{n - 1}}\cdots \xrightarrow[]{{\partial }_{0}}{F}_{0}\xrightarrow[]{\varepsilon }\mathbb{Z} \rightarrow 0 \]\n\nin \( {\mathcal{C}}_{\mathbb{Z}G} \) is a complex, i.e., \(... | Proof. With the understanding that the symbol ^ indicates an expression to be omitted, we have\n\n\[ {\partial }_{n - 1}{\partial }_{n}\left( {{g}_{0},\ldots ,{g}_{n}}\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{\left( -1\right) }^{i}{\partial }_{n - 1}\left( {{g}_{0},\ldots ,{\widehat{g}}_{i},\ldots ,{g}_{n}}\right) ... | Yes |
Lemma 3.22. Fix \( s \) in \( G \) . For \( n \geq 0 \), define a homomorphism \( {h}_{n} : {F}_{n} \rightarrow {F}_{n + 1} \) of abelian groups to be the additive extension of the function with\n\n\[ \n{h}_{n}\left( {{g}_{0},\ldots ,{g}_{n}}\right) = \left( {s,{g}_{0},\ldots ,{g}_{n}}\right) ,\n\]\n\nand define \( {h}... | Proof. On the \( \mathbb{Z} \) basis of \( \left( {n + 1}\right) \) -tuples in \( {F}_{n} \), we have\n\n\[ \n{\partial }_{n}{h}_{n}\left( {{g}_{0},\ldots ,{g}_{n}}\right) = {\partial }_{n}\left( {s,{g}_{0},\ldots ,{g}_{n}}\right) \n\]\n\n\[ \n= \left( {{g}_{0},\ldots ,{g}_{n}}\right) + \mathop{\sum }\limits_{{i = 0}}^... | Yes |
Proposition 3.23. A cochain map \( f : X \rightarrow Y \) as in Figure 3.1 induces an \( R \) homomorphism on cohomology \( {H}^{n}\left( X\right) \rightarrow {H}^{n}\left( Y\right) \) in each degree. | Proof. Suppose that \( {x}_{n} \) is in \( \ker {d}_{n} \), i.e., that \( {d}_{n}\left( {x}_{n}\right) = 0 \) . The commutativity of the right square gives \( {d}_{n}^{\prime }\left( {{f}_{n}\left( {x}_{n}\right) }\right) = {f}_{n + 1}\left( {{d}_{n}\left( {x}_{n}\right) }\right) = 0 \), and hence \( f\left( {x}_{n}\ri... | Yes |
Proposition 3.24. In the situation of Figure 3.1 if \( f = \left\{ {f}_{n}\right\} \) and \( g = \left\{ {g}_{n}\right\} \) are two cochain maps of \( X \) into \( Y \) and if \( f \) and \( g \) are homotopic, then \( f \) and \( g \) induce identical maps \( {H}^{n}\left( X\right) \rightarrow {H}^{n}\left( Y\right) \... | Proof. Suppose that \( {d}_{n}\left( {x}_{n}\right) = 0 \) . Then \( {f}_{n}\left( {x}_{n}\right) - {g}_{n}\left( {x}_{n}\right) = {d}_{n - 1}^{\prime }\left( {{h}_{n}\left( {x}_{n}\right) }\right) + \) \( {h}_{n + 1}\left( {{d}_{n}\left( {x}_{n}\right) }\right) = {d}_{n - 1}^{\prime }\left( {{h}_{n}\left( {x}_{n}\righ... | No |
Proposition 3.25. For the diagram\nin\n\nin \( {\mathcal{C}}_{R} \), suppose that the top and bottom rows are exact at \( M \) and \( {M}^{\prime } \), suppose that the square on the right commutes, and suppose that ... | Proof. If \( x \) is a free generator of \( F \), then \( 0 = {f}_{1}{\partial }_{1}\partial \left( x\right) = {\partial }_{1}^{\prime }\left( {f\partial x}\right) \) . By exactness at \( {M}^{\prime }, f\partial x \) lies in image \( \left( {\partial }^{\prime }\right) \) . Choose any \( y \in {F}^{\prime } \) with \(... | Yes |
Corollary 3.26. In the category \( {\mathcal{C}}_{\mathbb{Z}G} \), if the rows of the diagram\n\n\n\nare free resolutions and the vertical identity map \( 1 : \mathbb{Z} \rightarrow \mathbb{Z} \) is given, then the r... | Proof. There is no harm in including a vertical 0 map at the right between the two 0 modules. Certainly the square whose verticals are the identity map \( 1 : \mathbb{Z} \rightarrow \mathbb{Z} \) and the 0 map commutes. Proposition 3.25 is to be applied first to this square and the second square from the right (with ve... | Yes |
Corollary 3.28. In the category \( {\mathcal{C}}_{\mathbb{Z}G} \), if a free resolution \( X = \left\{ {X}_{n}\right\} \) of \( \mathbb{Z} \) and a chain map \( f = \left\{ {f}_{n}\right\} \) of \( X \) with itself are given such that the map from \( \mathbb{Z} \) to itself is 0, then the chain map \( f \) is homotopic... | PROOF. We are given the diagram\n\n\n\nin the category \( {\mathcal{C}}_{\mathbb{Z}G} \) with the two rows as free resolutions and all squares commuting. We are to construct maps \( {h}_{n} : {X}_{n} \rightarrow {X}_... | Yes |
Corollary 3.29. In the category \( {\mathcal{C}}_{\mathbb{Z}G} \), if a free resolution \( X = \left\{ {X}_{n}\right\} \) of \( \mathbb{Z} \) and a chain map \( f = \left\{ {f}_{n}\right\} \) of \( X \) with itself are given such that the map from \( \mathbb{Z} \) to itself is the identity 1, then the chain map \( f \)... | Proof. Apply Corollary 3.28 to \( f - 1 \) . | No |
Corollary 3.30. In the category \( {\mathcal{C}}_{\mathbb{Z}G} \), if two free resolutions \( X = \left\{ {X}_{n}\right\} \) of \( \mathbb{Z} \) and \( Y = \left\{ {Y}_{n}\right\} \) of \( \mathbb{Z} \) are given and if two chain maps \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) are given such that the map f... | Proof. Apply Corollary 3.29 to \( {fg} \) and then to \( {gf} \) . | No |
Theorem 3.31. If\n\n\\[ \n\\cdots \\xrightarrow[]{{\\partial }_{n + 1}^{\\prime }}{F}_{n + 1}^{\\prime }\\xrightarrow[]{{\\partial }_{n}^{\\prime }}{F}_{n}^{\\prime }\\xrightarrow[]{{\\partial }_{n - 1}^{\\prime }}\\cdots \\xrightarrow[]{{\\partial }_{0}^{\\prime }}{F}_{0}^{\\prime }\\xrightarrow[]{\\varepsilon }\\math... | Proof. Let the resolution in the statement of the theorem be \\( Y \\), and let \\( X \\) be the standard free resolution of \\( \\mathbb{Z} \\) in the category \\( {\\mathcal{C}}_{\\mathbb{Z}G} \\) . Two applications of Corollary 3.26 produce chain maps \\( f : X \\rightarrow Y \\) and \\( g : Y \\rightarrow X \\) ove... | Yes |
Proposition 3.32. If \( G \) is a finite cyclic group, then the sequence\n\n\[ \cdots \overset{T}{ \rightarrow }\mathbb{Z}G\overset{N}{ \rightarrow }\mathbb{Z}G\overset{T}{ \rightarrow }\cdots \overset{T}{ \rightarrow }\mathbb{Z}G\overset{N}{ \rightarrow }\mathbb{Z}G\overset{T}{ \rightarrow }\mathbb{Z}G\overset{\vareps... | Proof. We still need to check exactness at the first \( \mathbb{Z}G \) from the right. The map \( \varepsilon \) is the \( \mathbb{Z}G \) homomorphism with \( \varepsilon \left( \left( 1\right) \right) = 1 \) . Hence \( \varepsilon \left( \left( {s}^{j}\right) \right) = 1 \) for all \( j \) , and \( \varepsilon \left( ... | Yes |
Corollary 3.33. If \( G \) is a finite cyclic group and \( M \) is an abelian group on which \( G \) acts by automorphisms, then\n\n\[ \n{H}^{2}\left( {G, M}\right) \cong {M}^{G}/\left( {\left( 1\right) + \left( s\right) + \cdots + \left( {s}^{n - 1}\right) }\right) M,\n\]\n\nwhere \( {M}^{G} \) is the subgroup of all ... | Proof. Let us number the terms \( \mathbb{Z}G \) in the resolution of Proposition 3.32 starting with index 0 from the right. Combining Proposition 3.32 with Theorem 3.31, we see that we may compute \( {H}^{2}\left( {G, M}\right) \) as the cohomology of the complex obtained by applying the functor \( {\operatorname{Hom}... | Yes |
Corollary 3.34. If \( K/F \) is a finite Galois extension of fields such that \( \operatorname{Gal}\left( {K/F}\right) \) is cyclic, then\n\n\[ \n{H}^{2}\left( {\operatorname{Gal}\left( {K/F}\right) ,{K}^{ \times }}\right) \cong {F}^{ \times }/{N}_{K/F}\left( {K}^{ \times }\right) .\n\] | Corollary 3.34 considerably simplifies the proofs of Frobenius's Theorem about division algebras over the reals (Theorem 2.50) and Wedderburn's Theorem about finite division rings (Theorem 2.48), and thus the theory in Chapter III has added something to the theory of Chapter II even in these very special situations. In... | Yes |
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