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Theorem 3 (Hermite-Lindemann). If \( \alpha \neq 0 \) is an algebraic number, \( {e}^{\alpha } \) is transcendental. In particular, \( e \) is transcendental (and also \( \pi \), since \( {e}^{i\pi } = - 1 \) ). | Proof. If \( {e}^{\alpha } \) were algebraic, there would be a nontrivial relation\n\n\[ \n{a}_{0}{e}^{0} + {a}_{1}{e}^{\alpha } + {a}_{2}{e}^{2\alpha } + \cdots + {a}_{n}{e}^{n\alpha } = 0\;\text{ with }{a}_{i} \in \mathbb{Q}, \n\]\n\ncontradicting Theorem 2. | Yes |
Theorem 4 (Lindemann-Weierstrass). If \( {x}_{1},\ldots ,{x}_{m} \) are distinct algebraic numbers, \( {e}^{{x}_{1}},{e}^{{x}_{2}},\ldots ,{e}^{{x}_{m}} \) are linearly independent over \( {\mathbb{Q}}^{c} \) . | Proof. Suppose\n\n\[ \n{a}_{1}{e}^{{x}_{1}} + \cdots + {a}_{m}{e}^{{x}_{m}} = 0, \n\] \n\nwhere \( {a}_{1},\ldots ,{a}_{m} \) are nonvanishing algebraic numbers. Choose a finite Galois extension \( K/\mathbb{Q} \) containing all the \( {a}_{i} \) . Setting \( G = G\left( {K/\mathbb{Q}}\right) \) we form the product\n\n... | Yes |
Theorem 1. Every field extension \( E/K \) has a transcendence basis. More precisely: Given a subset \( M \) of \( E \) such that \( E/K\left( M\right) \) is algebraic, and given a subset \( C \) of \( M \) that is algebraically independent over \( K \), there exists a transcendence basis \( B \) of \( E/K \) such that... | Proof. We must enlarge \( C \) to make it a maximal algebraically independent subset \( B \) of \( M \) ; by F3, such a set is a transcendence basis of \( E/K \) . If \( M \) is finite, the existence of \( B \) is clear. If \( M \) is infinite, one resorts to Zorn’s Lemma, the argument being wholly similar to the one u... | No |
Theorem 2. Any two transcendence bases of a field extension \( E/K \) have the same cardinality. | Proof. Let \( B \) and \( {B}^{\prime } \) be transcendence bases of \( E/K \) . If \( B \) and \( {B}^{\prime } \) are both infinite sets, the desired assertion follows easily on set-theoretical grounds (see \( \$ {18.2} \) in the Appendix).\n\nWe now prove the assertion for the more interesting case, where \( E/K \) ... | Yes |
Theorem 3. Let \( F \) be an intermediate field of \( E/K \), and let \( B \) and \( {B}^{\prime } \) be transcendence bases for \( F/K \) and \( E/F \), respectively. Then \( B \cap {B}^{\prime } = \varnothing \) and \( B \cup {B}^{\prime } \) is a transcendence basis for \( E/K \). In particular,\n\n\[\n\operatorname... | Proof. Suppose \( \alpha \) lies in both \( B \) and \( {B}^{\prime } \) ; then \( \alpha \), being an element of \( F \), is algebraic over \( F\left( {{B}^{\prime }\smallsetminus \{ \alpha \} }\right) \) . But because \( \alpha \in {B}^{\prime } \), this contradicts the algebraic independence of \( {B}^{\prime } \) o... | Yes |
Theorem 5. Let \( E/K \) be a field extension, and suppose \( E \) is finitely generated as a \( K \) -algebra. Then \( E/K \) is algebraic. | Proof. By Theorem 4, \( E \) has as a subring some polynomial ring over \( K \) in finitely many indeterminates, say \( F = K\left\lbrack {{u}_{1},\ldots ,{u}_{m}}\right\rbrack \), with the further property that \( E/F \) is integral. Since \( E \) is a field, F11 in Chapter 16 says that \( F \) must also be a field. B... | Yes |
Theorem 2 (Hilbert’s Nullstellensatz, geometric form). Let \( K \) be a field and \( C \) an algebraically closed extension of \( K \) (for instance, an algebraic closure of \( K \) ). Let an ideal \( \mathfrak{a} \) of the polynomial ring \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be given. Provided th... | Proof. The assumption \( \mathfrak{a} \neq K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) implies (see F12 in Chapter 6) that \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) has a maximal ideal \( \mathfrak{m} \) that contains \( \mathfrak{a} \) . We look at the quotient homomorphism \[ K\left\lbrac... | Yes |
Lemma 1. Let the \( K \) -algebra \( A \) be an integral domain and assume that \( {\operatorname{TrDeg}}_{K}\left( A\right) \) is finite. For any nonzero prime ideal \( \mathfrak{p} \) of \( A \) we have\n\n\[ \n{\operatorname{TrDeg}}_{K}\left( {A/\mathfrak{p}}\right) < {\operatorname{TrDeg}}_{K}\left( A\right)\n\] | Proof. For notational simplicity set \( \bar{A} = A/\mathfrak{p} \) and denote by\n\n\[ \n\varphi : A \rightarrow \bar{A}\n\]\n\n\[ \nx \mapsto \bar{x}\n\]\n\nthe quotient map. If \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{s} \) are algebraically independent over \( K \), so are \( {x}_{1},\ldots ,{x}_{s} \) . Thus we already... | Yes |
Lemma 2. If \( R \) is a UFD, a prime ideal \( \mathfrak{p} \) has height 1 if and only if it is a principal ideal. | Proof. Suppose \( h\left( \mathfrak{p}\right) = 1 \) . Then \( \mathfrak{p} \neq 0 \), and we may take a nonzero \( f \in \mathfrak{p} \) . Since \( \mathfrak{p} \) is a prime ideal, at least one prime factor \( {f}_{1} \) of \( f \) lies in \( \mathfrak{p} \) . Now \( \left( {f}_{1}\right) \) is likewise a prime ideal... | Yes |
Lemma 3. Let \( R = K\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \) be a polynomial ring in the \( r \) indeterminates \( {y}_{1},\ldots ,{y}_{r} \) over the field \( K \) . For every prime ideal \( \mathfrak{p} \) of height \( h\left( \mathfrak{p}\right) = 1 \) in \( R \), the quotient ring \( \bar{R} = R/\mat... | Proof. By Lemma 2, \( \mathfrak{p} \) has the form \( \mathfrak{p} = \left( p\right) \), where \( p \in k\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \) is a nonconstant polynomial. If, say, the variable \( y \mathrel{\text{:=}} {y}_{r} \) really does appear in \( p \), we also have \( {\deg }_{y}p \geq 1 \) . I... | Yes |
Theorem 6. Let \( A = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an affine \( K \) -algebra, and assume that \( A \) is an integral domain. Let\n\n(26)\n\n\[ 0 = {\mathfrak{P}}_{0} \subset {\mathfrak{P}}_{1} \subset \cdots \subset {\mathfrak{P}}_{m} \]\n\nbe a nonrefinable strict chain of prime ideals i... | Proof. We proceed inductively. Let \( m = 0 \) . Then \( {\mathfrak{P}}_{m} = 0 \) is a maximal ideal of \( A \) , so \( A \) is a field. Using Chapter 18, Theorem 5 or equation (25), we conclude that \( \operatorname{TrDeg}\left( {A/K}\right) = 0 \) .\n\nNow suppose \( m \geq 1 \) . By the Noether Normalization Theore... | Yes |
Lemma 1. Suppose given a quadratic preorder \( T \) of a field \( K \) with \( - 1 \notin T \) . Let a be an element of \( K \) such that \( a \notin - T \) . Then\n\n\[ \n{T}^{\prime } \mathrel{\text{:=}} T + {aT} \n\]\n\nis a quadratic preorder of \( K \), also satisfying \( - 1 \notin {T}^{\prime } \) . (Note that \... | Proof. Obviously we have \( {T}^{\prime } + {T}^{\prime } \subseteq {T}^{\prime },{T}^{\prime }{T}^{\prime } \subseteq {T}^{\prime } \), and \( {K}^{2} \subseteq T \subseteq {T}^{\prime } \), so \( {T}^{\prime } \) is a quadratic preorder. Assume that \( - 1 \in {T}^{\prime } \), meaning that there are elements \( b, c... | Yes |
Theorem 1. Suppose a quadratic preorder \( T \) on a field \( K \) does not contain -1 . Then \( T \) is the intersection of all field orders \( P \) of \( K \) that contain \( T \) :\n\n\[ T = \mathop{\bigcap }\limits_{{T \subseteq P}}P \] | Proof. By Zorn’s Lemma, \( T \) lies in a maximal preorder \( P \) of \( K \) such that \( - 1 \notin P \) . We claim that \( P \) is actually an order of \( K \) . By F1, we just have to show that\n\n\[ P \cup - P = K\text{.} \]\n\nLet \( a \) be an element of \( K \) such that \( a \notin - P \) . We must show that \... | Yes |
Theorem 3. Let \( K \) be a field of characteristic different from 2. Given \( b \in K \), there is equivalence between:\n\n(i) \( b \) being a sum of squares in \( K \) ;\n\n(ii) \( b \) being a totally positive element, that is, positive for any field order on \( K \) . | Proof. We consider the quadratic preorder \( T = \mathrm{{SQ}}\left( K\right) \) and assume that \( b \notin T \) . Then \( T \neq K \), whence, by \( \mathrm{F}1, - 1 \notin T \) . By Theorem \( 1, K \) has some order \( P \) such that \( b \notin P \) . This proves the implication (ii) \( \Rightarrow \) (i). The conv... | No |
Theorem 4. Let \( \\left( {K, \\leq }\\right) \) be an ordered field and \( E/K \) a field extension. The following statements are equivalent:\n\n(i) The order \( \\leq \) of \( K \) can be extended to an order on \( E \) .\n\n(ii) -1 is not a sum of elements of the form \( a{\\alpha }^{2} \) with \( a \\geq 0 \) in \(... | Condition (iii) is a simple reformulation of (ii). The equivalence between (i) and (ii) follows immediately from a more general fact: | No |
Theorem 5. Let \( \\left( {K, \\leq }\\right) \) be an ordered field. If \( E/K \) is a field extension and \( \\beta \) is an element of \( E \), there is equivalence between\n\n(i) \( \\beta \) being of the form\n\n\[ \n\\beta = \\mathop{\\sum }\\limits_{i}{a}_{i}{\\alpha }_{i}^{2}\n\]\n\nwith each \( {a}_{i} \\in K ... | Proof. The implication (i) \( \\Rightarrow \) (ii) is trivial. Let \( T \) be the set of all sums of elements of the form \( a{\\alpha }^{2} \) with \( a \\geq 0 \) in \( K \) and \( \\alpha \) in \( E \) . Clearly \( T \) is a quadratic preorder of \( E \) . Assume that (i) does not hold, so that \( \\beta \\notin T \... | Yes |
Theorem 6. Let \( \left( {K, \leq }\right) \) be an ordered field.\n\n(I) There exists a real-closed extension \( R \) of \( K \) such that the (unique) order on \( R \) extends the given order \( \leq \) on \( K \) . Such an \( R \) is called a real closure of the ordered field \( \left( {K, \leq }\right) \).\n\n(II) ... | Proof. In an algebraic closure \( C \) of \( K \), let \( E \) be the subfield of \( C \) obtained from \( K \) by adjoining the square roots of all positive elements of \( \left( {K, \leq }\right) \) . By F4, \( E \) is a real field. By F6, then, \( E \) has a real closure \( R \) (which we can regard as a subfield of... | No |
Theorem 7 (Euler-Lagrange). Let \( \left( {R, \leq }\right) \) be an ordered field with the following properties:\n\n(a) Every polynomial of odd degree over \( R \) has a root in \( R \) .\n\n(b) Every positive element in \( R \) is a square in \( R \) .\n\nThen the field \( C = R\left( i\right) \) obtained from \( R \... | Proof of Theorem 7. Let \( E/C \) be a finite field extension. We must show that \( E = C \) . By passing to a normal closure if needed, we can assume that \( E/R \) is Galois. Let \( H \) be a Sylow 2-subgroup of the Galois group \( G\left( {E/R}\right) \) of \( E/R \) and let \( {R}^{\prime } \) be the corresponding ... | Yes |
Theorem 9 (Sylvester). Let \( \\left( {K, P}\\right) \) be an ordered field and \( R \) a real-closed extension of \( K \) whose order induces the given order \( P \) on \( K \) . Consider for a given polynomial \( f \\neq 0 \) in \( K\\left\\lbrack X\\right\\rbrack \) the \( K \) -algebra \( A = K\\left\\lbrack X\\rig... | Proof. For simplicity we give the proof only for the case where \( f \) has no multiple roots (but see \( §{20.11} \) ).\n\nBy Euler-Lagrange (Theorem 7) we have \( C = R\\left( i\\right) \) with \( {i}^{2} + 1 = 0 \) . Therefore if\n\n\[ \nf\\left( X\\right) = {f}_{1}\\left( X\\right) {f}_{2}\\left( X\\right) \\ldots ... | Yes |
Lemma 2. Let \( K \) be a real-closed field, and let \( {h}_{1}\left( x\right) ,\ldots ,{h}_{n}\left( x\right) \) be polynomials in one variable \( x \) over \( K \) . Let \( \leq \) be an order of the field \( K\left( x\right) \) and \( \operatorname{sgn} \) the sign map it determines. There are infinitely many elemen... | Proof. First let \( h\left( x\right) \in K\left\lbrack x\right\rbrack \) be any polynomial. Since \( K \) is real-closed, the prime factorization of \( h \) has the form\n\n\[ h\left( x\right) = u\left( {x - {c}_{1}}\right) \ldots \left( {x - {c}_{r}}\right) {q}_{1}\left( x\right) \ldots {q}_{s}\left( x\right) ,\]\n\nw... | Yes |
Theorem 1 (Artin). Let \( K \) be a real field admitting a unique order, and let \( R \) be a real closure of \( K \) . If \( f \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is a polynomial in \( n \) variables over \( K \) such that\n\n(1)\n\n\[ f\left( {{a}_{1},\ldots ,{a}_{n}}\right) \geq 0\;\text{ for... | Proof. Suppose that \( f \) is not a sum of squares in \( K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) . By Theorem 3 in Chapter 20, there is an order \( \leq \) of the field\n\n\[ F \mathrel{\text{:=}} K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \]\n\nsuch that\n\n(2)\n\n\[ f = f\left( {{X}_{1},\ldots ,{X}_{n}}\right) < 0... | Yes |
Theorem 2. Let \( K \) be a real field admitting a unique order, and let \( R \) be a real closure of \( K \) . Suppose given an affine \( K \) -variety \( V \) in \( {R}^{n} \), that is, an irreducible algebraic \( K \) -set \( V \) of \( {R}^{n} \) . Then we have, for all \( f \in K\left( V\right) \) ,\n\n\[ f\text{i... | Proof. In essence we can proceed as in the proof of Theorem 1. Let\n\n\[ f = g/h\;\text{ with }g, h \in K\left\lbrack V\right\rbrack, h \neq 0. \]\n\nAssume that \( f \notin \mathrm{{SQ}}\left( {K\left( V\right) }\right) \) . Then there is an order \( \leq \) of the field \( F \mathrel{\text{:=}} K\left( V\right) \) su... | Yes |
Theorem 3. Let \( K \) be a real field admitting a unique order, and let \( R \) be a real closure of \( K \) . If a prime ideal \( \mathfrak{p} \) of a polynomial ring \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) in \( n \) variables over \( K \) is real, then\n\n\[{\mathcal{N}}_{R}\left( \mathfrak{p}\ri... | Proof. Let \( {x}_{1},\ldots ,{x}_{n} \) be the images of \( {X}_{1},\ldots ,{X}_{n} \) under the quotient map defined by \( \mathfrak{p} \) . Then\n\n(9)\n\n\[K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p} = K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack\]\n\nSince \( \mathfrak{p} \) is assum... | Yes |
Theorem 4. Let the assumptions and the notation be as in Theorem 3, and let \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) denote, as in (9), the quotient algebra arising from \( \mathfrak{p} \), with fraction field \( F = K\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . Let \( \leq \) be an order of the field... | Proof. We merely have to choose \( {R}_{F} \) as a real closure of the ordered field \( \left( {F, \leq }\right) \) and apply Theorem 10 of Chapter 20 to the intermediate field\n\n\[ R\left( {{x}_{1},\ldots ,{x}_{n},\sqrt{-g}}\right) \]\n\nof \( {R}_{F}/R \) . We obtain a homomorphism of \( R \) -algebras\n\n\[ R\left\... | Yes |
Lemma 1. Let \( T \) be a quadratic preorder of \( A \) such that \( - 1 \notin T \) . Let \( a, b \) be elements of \( A \) satisfying \( {ab} \in T \) . Then at least one of \( T + {aT} \) and \( T - {bT} \) is a quadratic preorder of \( A \) not containing -1 . | Proof. Assume, contrary to the conclusion, that there exist \( {t}_{1},{t}_{2},{t}_{3},{t}_{4} \in T \) such that\n\n\[ \n- 1 = {t}_{1} + a{t}_{2} = {t}_{3} - b{t}_{4} \n\]\n\nMultiplying \( - a{t}_{2} = 1 + {t}_{1} \) by \( b{t}_{4} = 1 + {t}_{3} \) we obtain \( - {ab}{t}_{2}{t}_{4} = 1 + {t}_{5} \), with \( {t}_{5} \... | Yes |
Lemma 2. Among all quadratic preorders of \( A \) that do not contain -1, let \( T \) be a maximal element. Then\n\n(a) \( T \cup - T = A \) and\n\n(b) \( T \cap - T \) is a prime ideal of \( A \) . | Proof. (a) Take \( a \in A \) . Apply Lemma 1 with \( a = b \) ; because \( T \) is maximal, we get \( a \in T \) or \( - a \in T \) . (b) By part (a), \( T \cap - T \) is an ideal of \( A \) . Lemma 1 and part (a) imply that this ideal is prime. | Yes |
Lemma 3. Let \( {T}_{0} \) be a quadratic preorder of \( A \) such that \( - 1 \notin {T}_{0} \) . There exists a quadratic preorder \( T \) of \( A \) with the following properties:\n\n\[ \n{T}_{0} \subseteq T;\;T \cup - T = A;\;T \cap - T\text{is a prime ideal of}A\text{.} \n\] | Proof. This follows immediately from Lemma 2 using Zorn's Lemma. | No |
Lemma 4 (Prestel). Let \( f \) be an element of \( A \) such that\n\n(12)\n\n\[ \n{tf} \neq 1 + s\;\text{ for all }s, t \in \mathrm{{SQ}}\left( A\right) .\n\]\n\nThere exist a prime ideal \( \mathfrak{p} \) of \( A \) and an order \( \leq \) of the fraction field of \( A/\mathfrak{p} \) satisfying\n\n(13)\n\n\[ \n\bar{... | Proof. Condition (12) means that the quadratic preorder \( {T}_{0} = \mathrm{{SQ}}\left( A\right) - f\mathrm{{SQ}}\left( A\right) \) does not contain -1 . The assertion then follows easily from Lemma 3: If \( T \) is as in the conclusion of Lemma 3, consider the image \( \bar{T} \) of \( T \) in \( \bar{A} = A/\mathfra... | Yes |
Theorem 5. Let the assumptions and the notation be as in Theorem 2. If a polynomial function \( f \in K\left\lbrack V\right\rbrack \) is strictly positive definite, that is, if\n\n\[ f\left( a\right) > 0\;\text{ for all }a \in V, \]\n\nthere are sums of squares \( s \) and \( t \) in the ring \( K\left\lbrack V\right\r... | Proof. Suppose not. Since \( K\left( V\right) \) is real, we can apply Lemma 4 to \( A = K\left\lbrack V\right\rbrack \) and \( f \in A \) . We obtain a prime ideal \( \mathfrak{P} \) of \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) such that \( \mathfrak{P} \supseteq \mathcal{I}\left( V\right) \), and an ... | Yes |
Theorem 7. Let \( K \) be a real field admitting a unique order, and let \( R \) be a real closure of \( K \) . Suppose given an affine \( K \) -variety \( W \) of \( {R}^{n} \) and finitely many distinct nonzero functions \( {d}_{1},\ldots ,{d}_{r} \) in \( K\left( W\right) \) . If a function \( f \in K\left( W\right)... | Proof. Consider the extension\n\n\[ F \mathrel{\text{:=}} K\left( W\right) \left( {\sqrt{{d}_{1}},\ldots ,\sqrt{{d}_{r}}}\right) \]\n\narising from \( K\left( W\right) \) by the adjunction of square roots of the \( {d}_{1},\ldots ,{d}_{r} \) . We claim that \( f \) is a sum of squares in \( F \) :\n\n(19)\n\n\[ f \in \... | Yes |
Theorem 8. Let \( K \) be a real field admitting a unique order, and let \( R \) be a real closure of \( K \) . Suppose given an affine \( K \) -variety \( W \) of \( {R}^{n} \) and finitely many distinct nonzero functions \( {d}_{1},\ldots ,{d}_{r} \) in \( K\left\lbrack W\right\rbrack \) . If a (polynomial) function ... | Proof. Introduce the abbreviations\n\n\[ A \mathrel{\text{:=}} K\left\lbrack W\right\rbrack \left\lbrack {1/{d}_{1},\ldots ,1/{d}_{r}}\right\rbrack \;\text{ and }\;T \mathrel{\text{:=}} {\mathrm{{SQ}}}_{M}\left( A\right) .\n\nIf -1 lies in \( T \), so does \( f - 1 = {\left( f/2\right) }^{2} - {\left( 1 - f/2\right) }^... | Yes |
Theorem 4. If \( K \) is not real, \( I\left( K\right) \) is the set of all nilpotent elements in \( W\left( K\right) \) and there exists \( n \in \mathbb{N} \) such that\n\n\[ {2}^{n} \times f = 0\;\text{ for all }f \in W\left( K\right) . \]\n\nThus for \( K \) not real the additive group of \( W\left( K\right) \) is ... | Proof. Suppose \( K \) is not real. By Theorem \( 2, I\left( K\right) \) is the only prime ideal of \( W\left( K\right) \) , and so it coincides with the nilradical of \( W\left( K\right) \) . Since \( 1 + 1 = \langle 1,1\rangle \in I\left( K\right) \), there is some \( n \in \mathbb{N} \) such that \( {\left( 1 + 1\ri... | Yes |
Lemma 1. Let \( L/K \) be a quadratic field extension, so \( L = K\left( \sqrt{d}\right), d \in {K}^{ \times } \smallsetminus {K}^{\times 2} \) . Then the kernel of \( {r}_{L/K} : W\left( K\right) \rightarrow W\left( L\right) \) is the principal ideal of \( W\left( K\right) \) generated by \( \langle 1, - d\rangle \) . | Proof. Since \( {r}_{L/K}\langle 1, - d\rangle = \langle 1, - d{\rangle }_{L} = \langle 1, - 1{\rangle }_{L} = 0 \), the ideal generated by \( \langle 1, - d\rangle \) is contained in the kernel of \( {r}_{L/K} \) . Now let \( q \) be an anisotropic quadratic space over \( K \) . We assume that \( {q}_{L} \) is isotrop... | Yes |
Theorem 5. If \( K \) is real, the following statements are equivalent for \( f \in W\left( K\right) \) :\n\n(i) \( {r}_{L/K}\left( f\right) = 0 \) for every real closure \( L \) of \( K \) .\n\n(ii) \( {\operatorname{sgn}}_{P}\left( f\right) = 0 \) for every order \( P \) of \( K \) .\n\n(iii) \( f \) is nilpotent.\n\... | Proof. If \( L \) is a real closed field, \( \operatorname{Sign}\left( L\right) \) consists of a single element, which we denote by \( {\operatorname{sgn}}^{L} \) ; obviously \( {\operatorname{sgn}}^{L} : W\left( L\right) \rightarrow \mathbb{Z} \) is an isomorphism.\n\nNow, the orders \( P \) of \( K \) are in correspo... | Yes |
Theorem 6. If \( f \in W\left( K\right) \) is a torsion element, its order is a power of 2. | Proof. This follows from Theorem 5 when \( K \) is real and from Theorem 4 when \( K \) is not. | No |
Lemma 2. Let \( R \) be a nonzero commutative ring with unity, and let \( N \) be the set of zero divisors of \( R \), including 0 . Then \( N \) is a union of prime ideals in \( R \):\n\n\[ N = \bigcup {\mathfrak{p}}_{i} \] | Proof. Let \( S = R \smallsetminus N \) be the multiplicative subset of elements of \( R \) that are not zero divisors. Each \( x \in N \) lies in a maximal ideal \( \mathfrak{P} \) of \( {S}^{-1}R \) . Then \( \mathfrak{p} = \mathfrak{P} \cap R \) is a prime ideal of \( R \), with \( x \in \mathfrak{p} \) and \( \math... | No |
Theorem 1. Every nontrivial absolute value \( \mid \mid \) of the field \( \mathbb{Q} \) is equivalent either to \( {\left. \left| \sim \text{ or to }\right| ⫫ \right| }_{p} \) for some prime \( p \) . | Proof. (a) First assume that the given absolute value || on \( \mathbb{Q} \) is nonarchimedean. Then \( \left| m\right| \leq 1 \) for every \( m \in \mathbb{N} \) . Since \( \left| \right| \) is assumed nontrivial, there must be some \( n \in \mathbb{N} \) such that \( \left| n\right| < 1 \) . Let \( p \) be the smalle... | Yes |
Theorem 4. Let \( K \) be complete with respect to an absolute value ||. If \( E/K \) is a finite extension of degree \( n \), we can extend \( \mid \mid \) to a unique absolute value on \( E \) , again denoted by \( \left| \right| \) . For every \( \alpha \) in \( E \), we have\n\n\[ \left| \alpha \right| = {\left| {N... | Proof. (1) The uniqueness and the completeness of the extension follow from F11.\n\n(2) If \( \left| \right| \) is archimedean, there is nothing more to show in view of Ostrowski’s Theorem (page 55).\n\n(3) Let \( \left| \right| \) be nonarchimedean. Equation (56) defines a function \( \left| \right| : E \rightarrow {\... | Yes |
Theorem 5. Let \( E/K \) be a finite extension of degree \( n \) and let \( \mid \mid \) be an absolute value on \( K \). (a) || can be extended to an absolute value on \( E \). (b) There are at most \( n \) extensions of \( \mid \mid \) to distinct absolute values on \( E \). | Proof. (1) Let \( C \) be an algebraic closure of \( \widehat{K} \). By Theorems 4 and \( {4}^{\prime },\left| \right| \) has a unique extension to an absolute value on \( C \), which we again denote by \( \left| \right| \). Let \( {\sigma }_{1},\ldots ,{\sigma }_{m} \) be all the distinct \( K \)-homomorphisms from \(... | Yes |
Theorem 1. Let \( E/K \) be a field extension and \( \\left| \\right| \) a discrete absolute value on \( E \) . Assume that \( E/K \) has finite ramification index \( e \) and finite residue class degree \( f \) with respect to \( \\left| \\right| \) . If \( K \) is \( \\left| \\right| \)-complete, the extension \( E/K... | Proof. Let \( \\pi \) and \( \\Pi \) be \( \\left| \\;\\right| \)-prime elements of \( K \) and \( E \), respectively. The ideal \( {\\pi A} \) of \( A \) has the form \( {\\pi A} = {\\Pi }^{m}A \), where \( m \\in \\mathbb{N} \) ; see Remark (b) after Definition 2. Because \( \\left| {E}^{ \\times }\\right| : \\left| ... | Yes |
Theorem 2. Let \( E/K \) be a separable, finite field extension. Let \( {\left. \left| {}_{1},\ldots ,\right| \right| }_{r} \) be all the distinct extensions of a discrete absolute value \( \mid \mid \) on \( K \) to absolute values on \( E \) . If we denote by \( {e}_{i} \) and \( {f}_{i} \) the residue class degree a... | Proof. By Equation (60) in Chapter 23 we have \( n = \mathop{\sum }\limits_{{i = 1}}^{r}{n}_{i} \), and Theorem 1 gives \( {n}_{i} = {e}_{i}{f}_{i} \) . Taking into account Remark (b) after Definition 1, we obtain the desired equality (see also the remark preceding Definition 2). | Yes |
Theorem 3. Let \( K \) be \( \left| \cdot \right| \) -complete, where \( \left| \cdot \right| \) is nonarchimedean. On the algebraic closure \( C \) of \( K \), consider the unique extension of \( \left| \right| \) (also denoted by \( \left| \right| \) ).\n\n(i) The residue field \( \bar{C} \) of \( C \) is an algebrai... | Proof. (a) Let \( \bar{\alpha } \), with \( \alpha \in C \) and \( \left| \alpha \right| \leq 1 \), be an arbitrary element of \( \bar{C} \) . The minimal polynomial \( f \) of \( \alpha \) over \( K \) has all its coefficients in the valuation ring \( R \) of \( K \) (Chapter 23, Theorem \( {4}^{\prime } \) ). Hence w... | Yes |
Theorem 4. Suppose that \( K \) is complete with respect to a nonarchimedean absolute value | | and that the corresponding residue field \( \bar{K} \) is a finite field with q elements. Let the notation be as in Theorem 3.\n\n(i) For every \( n \in \mathbb{N} \) there exists exactly one unramified extension of degree \... | Proof. (a) \( \bar{K} \) has a unique extension \( F \) of degree \( n \) in \( \bar{C} \), namely the field \( F = {\mathbb{F}}_{{q}^{n}} \) with \( {q}^{n} \) elements (Chapter 9, Theorem 1 in vol. I). Since every extension of a finite field is separable, Theorem 3(iii) yields the existence of a unique unramified ext... | Yes |
Theorem 1. Let \( K \) be a field with an absolute value || having properties (i),(ii) and (iii) from F1.\n\n(I) If \( K \) has characteristic zero, it is a finite extension of \( {\mathbb{Q}}_{p} \), where \( p = \operatorname{char}\bar{K} \) , and \( {\left. \left| \right| \text{is equivalent to the unique extension ... | Proof. (I) When char \( K = 0 \) we can regard \( \mathbb{Q} \) as a subfield of \( K \) . Set \( p = \operatorname{char}\bar{K} \) . Then \( \left| p\right| < 1 \) ; by substituting an equivalent absolute value we can assume that \( \left| p\right| = 1/p \) . But now the restriction of \( \left| \right| \) to \( \math... | Yes |
Theorem 2. Local fields are precisely the completions of global fields with respect to nontrivial absolute values. | Proof. We go down the list of local fields: For \( \mathbb{R} \) and \( \mathbb{C} \) the assertion is justified by considering the global fields \( \mathbb{Q} \) and \( \mathbb{Q}\left( i\right) \), respectively, with respect to \( {\left. \mid \right| }_{\infty } \) . Also \( {\mathbb{F}}_{q}\left( \left( X\right) \r... | No |
Theorem 3. Let \( K \) be complete with respect to a nonarchimedean absolute value \( \left| \;\right| \) and let \( f \in K\left\lbrack X\right\rbrack \) be a normalized, irreducible, separable polynomial of degree \( n \) . There exists a constant \( \delta > 0 \) with the following property: If \( g \in K\left\lbrac... | Proof. (a) First let a root \( \beta \) of \( g \) be given; we claim that \( \left| \beta \right| \leq \left| g\right| \) . For, letting \( {b}_{i} \) denote the coefficients of \( g \), we’d get if \( \left| \beta \right| > \left| g\right| \) the inequality \( \left| {\beta }^{n}\right| > \left| {b}_{i}\right| \left|... | Yes |
Theorem 4. The completion \( {\mathbb{C}}_{p} \) of the algebraic closure \( C \) of \( {\mathbb{Q}}_{p} \) with respect to \( {\left| \right| }_{p} \) is algebraically closed. | Proof. Let \( f \in {\mathbb{C}}_{p}\left\lbrack X\right\rbrack \) be a normalized and irreducible polynomial over \( {\mathbb{C}}_{p} \), and let \( \alpha \) be a root of \( f \) (in the algebraic closure of \( {\mathbb{C}}_{p} \) ). The field \( C \) is dense in \( {\mathbb{C}}_{p} \) . By Theorem 3, applied to the ... | Yes |
Theorem 5. Every Galois extension \( E/K \) of local fields has a solvable Galois group. | Proof. Since \( G\left( {\mathbb{C}/\mathbb{R}}\right) = \mathbb{Z}/2 \), we only have to worry about the nonarchimedean case. Consider the maximal intermediate field \( L \) of \( E/K \) that is unramified over \( K \) (Chapter 24, Theorem 3). The extension \( E/L \) is purely ramified and \( G\left( {L/K}\right) \) i... | Yes |
Theorem 6. A local field \( K \) of characteristic 0 admits (inside a fixed algebraic closure) only finitely many extensions of a given degree. | Proof. We can assume that \( K \) is nonarchimedean and fix a prime element \( \pi \) of \( K \) . In view of Theorems 3 and 4 of Chapter 24, we need only concern ourselves with purely ramified extensions \( E/K \) of prescribed degree \( n = E : K \) . By Chapter 24, F3, any such extension arises by the adjunction of ... | Yes |
Theorem 7. Let \( K \) be any local field. For every natural number \( m \) ,\n\n\[ \frac{{K}^{ \times } : {K}^{\times m}}{\operatorname{Card}{W}_{m}\left( K\right) } = \frac{m}{\parallel m{\parallel }_{K}}. \] | Proof. For \( K = \mathbb{R},\mathbb{C} \) the assertion is clear; therefore let \( K \) be nonarchimedean. First suppose that \( m \) is prime to \( p = \operatorname{char}\bar{K} \) ; then \( {U}^{\left( 1\right) } = {U}^{\left( 1\right) m} \), as can be seen by applying F13 in Chapter 23 (see also §23.13). Thus the ... | Yes |
For char \( K = 0 \), the group \( {U}_{K}^{\left( 1\right) } \) of 1 -units has, as a \( {\mathbb{Z}}_{p} \) -module (and a fortiori as an abelian group), the structure\n\n\[ {U}_{K}^{\left( 1\right) } \simeq {W}_{p} \propto \left( K\right) \times {\mathbb{Z}}_{p}^{n},\;\text{ where }n = K : {\mathbb{Q}}_{p}. \] | Proof. Suppose \( r > e/p - 1 \) . The \( {\mathbb{Z}}_{p} \) -submodule \( {U}^{\left( r\right) } \) of \( {U}^{\left( 1\right) } \) is isomorphic to \( {\mathfrak{p}}^{r} = {\pi }^{r}{R}_{K} \simeq {R}_{K} \) . At the same time, the \( {\mathbb{Z}}_{p} \) -module \( {R}_{K} \) has a basis with \( n = K : {\mathbb{Q}}... | Yes |
Theorem 1. Let \( K \) be complete with respect to a discrete valuation \( w \) . Assume that the associated residue field \( \bar{K} \) is perfect of characteristic \( p > 0 \) . Then there exists a unique set \( S \) of representatives of \( \bar{K} \) satisfying\n\n(1)\n\n\[{S}^{p} = S\text{.}\]\n\nThis set is multi... | Proof. Let \( \pi \) be a prime element of \( w \) . For \( \alpha ,\beta \in R \) and \( i \geq 1 \) we have\n\n(2)\n\n\[ \alpha \equiv \beta {\;\operatorname{mod}\;{\pi }^{i}} \Rightarrow {\alpha }^{p} \equiv {\beta }^{p}{\;\operatorname{mod}\;{\pi }^{i + 1}}, \]\n\nas can be seen by writing \( {\alpha }^{p} - {\beta... | Yes |
Lemma 2. Let integers \( r \geq 1 \) and \( n \geq 0 \) be given. If \( A = \left( {{A}_{0},{A}_{1},\ldots }\right) \) and \( B = \left( {{B}_{0},{B}_{1},\ldots }\right) \) are vectors with components in \( \mathbb{Z}\left\lbrack {{X}_{0},{Y}_{0},{X}_{1},{Y}_{1},\ldots }\right\rbrack \), the congruences\n\n(22)\n\n\[ \... | Proof. We apply induction on \( n \) . For \( n = 0 \) the statement is correct. Take \( n > 0 \) . By the induction hypothesis we can assume that congruences (22) and (23) both hold for all \( i \leq n - 1 \) . By raising to the \( p \) -th power one then clearly gets\n\n\[ \n{A}_{i}^{p} \equiv {B}_{i}^{p}{\;\operator... | Yes |
Theorem 5. Let \( K \) be a field if characteristic \( p \) .\n\n(i) Take \( a = \left( {{a}_{0},\ldots ,{a}_{n - 1}}\right) \in {\mathcal{W}}_{n}\left( K\right) \) and let \( C \) be as in \( \mathrm{F}2 \) . If \( \alpha \in {\mathcal{W}}_{n}\left( C\right) \) is a solution of the equation \( \wp \left( \alpha \right... | Proof. We must show (ii) and (iii). Let \( L/K \) be as in (ii) and \( G \mathrel{\text{:=}} G\left( {L/K}\right) = \langle \sigma \rangle \) . Then \( L : K = {p}^{m} \), with \( m \leq n \) . Applying the trace \( {\operatorname{Tr}}_{L/K} \) to \( c \mathrel{\text{:=}} {p}^{n - m}e \in {\mathcal{W}}_{n}\left( {\math... | Yes |
Theorem 6 (Kummer theory of abelian extensions of exponent \( {p}^{n} \) in characteristic \( p \) ). Let \( K \) be a field of characteristic \( p \) and \( n \) a natural number. Then, with the notations above, the map\n\n(67)\n\n\[ \nW \mapsto K\left( {{\wp }^{-1}W}\right) \n\] \n\nis a bijection between the set of ... | Proof. We need only show that the correspondence (67) is bijective; everything else is covered by F3 and the subsequent remarks. Let \( E/K \) be an abelian extension of exponent \( {p}^{n} \) . Setting \( W \mathrel{\text{:=}} \wp {\mathcal{W}}_{n}\left( E\right) \cap {\mathcal{W}}_{n}\left( K\right) \) we get \( K\le... | Yes |
Theorem 3 (Tsen 1933). Every rational function field \( K\\left( X\\right) \) in one variable over an algebraically closed ground field \( K \) has Tsen rank 1. | Tsen's methods actually yield a more general statement (Theorem 4), from which Theorem 3 follows immediately since algebraically closed fields have Tsen rank zero (Theorem 1). | No |
Theorem 4 (Tsen 1936). If \( K \) is a \( {T}_{i} \) -field, the rational function field \( K\left( X\right) \) in one variable over \( K \) is a \( {T}_{i + 1} \) -field. In other words, if \( K \) has Tsen rank \( i \), then \( K\left( X\right) \) has Tsen rank at most \( i + 1 \) . | Proof. Consider a system of \( m \) polynomial equations in \( n \) unknowns over \( \mathbf{K}\left( \mathbf{X}\right) \) :\n\n(7)\n\n\[ \n{f}_{\mu }\left( {{X}_{1},\ldots ,{X}_{n}}\right) = 0,\;\text{ where }1 \leq \mu \leq m.\n\]\n\nAssume that all the \( {f}_{\mu } \) vanish at the point \( 0 \in {K}^{n} \), and th... | Yes |
Theorem 5. If a field \( K \) has Tsen rank \( i \), any algebraic extension \( E \) of \( K \) has Tsen rank at most \( i \) . | Proof. Since the coefficients of a given system over \( E \) all lie in some finite extension over \( K \), we may as well take \( E/K \) to be finite. So let a system\n\n\[ \n{f}_{\mu }\left( {{X}_{1},\ldots ,{X}_{n}}\right) = 0,\;\text{ with }1 \leq \mu \leq m,\n\]\n\nbe given over \( E \), and suppose that\n\n(14)\n... | Yes |
Theorem 7. If \( K \) is a \( {C}_{i} \) -field and \( E/K \) is an extension with transcendence degree \( j \), then \( E \) is a \( {C}_{i + j} \) -field. | Proof. We cannot derive Theorem 7 as a simple logical consequence of Theorem 6, because one of the assumptions has been weakened. But in any case it is enough to consider two cases:\n\n(a) \( E/K \) is purely transcendental of transcendence degree 1 .\n\n(b) \( E/K \) is algebraic.\n\nFor case (a) we proceed exactly as... | Yes |
Lemma 1. If a reduced polynomial \( g \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) vanishes at all points of \( {K}^{n} \) , then \( g = 0 \) . | Proof. We work by induction on \( n \) . For \( n = 0 \) there is nothing to show. Suppose \( n \geq 1 \) . A reduced polynomial \( g \) has a representation\n\n\[ g\left( {{X}_{1},\ldots ,{X}_{n}}\right) = \mathop{\sum }\limits_{{i = 0}}^{{q - 1}}{g}_{i}\left( {{X}_{1},\ldots ,{X}_{n - 1}}\right) {X}_{n}^{i},\]\n\nwhe... | Yes |
Lemma 2. Given \( f \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), there exists a unique reduced polynomial \( {f}^{ * } \) such that\n\n(28)\n\n\[ f \equiv {f}^{ * }{\;\operatorname{mod}\;\mathfrak{a}}. \]\n\nMoreover \( {f}^{ * } \) satisfies\n\n(29)\n\n\[ \deg {f}^{ * } \leq \deg f \]\n\nIf \( f \) van... | Proof. We already know that there exists a reduced polynomial \( {f}_{ * } \) such that \( f \equiv {f}_{ * } \) \( {\;\operatorname{mod}\;\mathfrak{a}} \) and \( \deg {f}_{ * } \leq \deg f \) . Suppose \( {f}^{ * } \) is another reduced polynomial satisfying (28). Then \( {f}^{ * } - {f}_{ * } \) is also a reduced pol... | Yes |
Lemma 3. If \( u \in K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is reduced, vanishes at every point of \( {K}^{n} \smallsetminus \{ 0\} \) , and takes the value 1 at the point 0 , it must equal\n\n(30)\n\n\[ u = \left( {1 - {X}_{1}^{q - 1}}\right) \left( {1 - {X}_{2}^{q - 1}}\right) \ldots \left( {1 - {X}_... | Proof. The polynomial on the right-hand side of (30) - call it \( v \) - satisfies the same assumptions as \( u \) . Now apply Lemma 1 to \( g = u - v \) to obtain \( u = v \) . | Yes |
Theorem 2 (Chevalley). Let \( K \) be a finite field, and let \( {f}_{1},\ldots ,{f}_{m} \) be (nonzero) polynomials in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), all vanishing at the point \( 0 \in {K}^{n} \). Suppose the sum of their degrees \( {d}_{\mu } \) is less than the number \( n \) of variabl... | Proof. As before, let \( q \) be the cardinality of \( K \). The polynomial\n\n(31)\n\n\[ f \mathrel{\text{:=}} \left( {1 - {f}_{1}^{q - 1}}\right) \left( {1 - {f}_{2}^{q - 1}}\right) \ldots \left( {1 - {f}_{m}^{q - 1}}\right) \]\n\nhas degree\n\n(32)\n\n\[ \deg f = \left( {q - 1}\right) \mathop{\sum }\limits_{{\mu = 1... | Yes |
Lemma 4. Let \( K \) be a finite field with \( q \) elements and let \( a \geq 0 \) be an integer. If \( a \) is zero or is not divisible by \( q - 1 \), then\n\n\[ \mathop{\sum }\limits_{{x \in K}}{x}^{a} = 0 \]\n\nIf, contrarywise, \( a \) is nonzero and divisible by \( q - 1 \), the sum on the left has value -1. | Proof. For \( a = 0 \), all terms in the sum equal 1, so the sum is worth \( q \cdot 1 \), which vanishes since \( q \) is a power of the characteristic. Suppose \( a \geq 1 \) . Then \( {0}^{a} = 0 \), and we are left with a sum over the elements of the multiplicative group \( {K}^{ \times } \) :\n\n\[ S\left( a\right... | Yes |
Theorem 1 (Jordan-Hölder Theorem for modules). If\n\n\\[ \n0 = {M}_{0} \subseteq {M}_{1} \subseteq \cdots \subseteq {M}_{n} = M\\;\\text{ and }\\;0 = {L}_{0} \subseteq {L}_{1} \subseteq \cdots \subseteq {L}_{m} = M \n\\]\n\nare composition series of the same module \\( M \\), then \\( m = n \\), and there is a permutat... | Proof. Consider the sequences\n\n(37)\n\n\\[ \n0 = {L}_{0} \\cap {M}_{n - 1} \\subseteq \\cdots \\subseteq {L}_{m} \\cap {M}_{n - 1} = {M}_{n - 1}, \n\\]\n\n(38)\n\n\\[ \n{M}_{n - 1} = {L}_{0} + {M}_{n - 1} \\subseteq \\cdots \\subseteq {L}_{m} + {M}_{n - 1} = {M}_{n}. \n\\]\n\nFor each \\( j \\) we have an exact seque... | Yes |
Theorem 2 (Krull-Remak-Schmidt). Let \( M \) be an artinian and noetherian \( A \) - module. Then \( M \) is a direct sum of finitely many indecomposable submodules (compare \( {F28} \) ). If\n\n\[ M = {N}_{1} \oplus \cdots \oplus {N}_{m} = {N}_{1}^{\prime } \oplus \cdots \oplus {N}_{n}^{\prime }, \]\n\nwhere the \( {N... | Proof. Let \( {\pi }_{i} : M \rightarrow {N}_{i} \) be the projections and \( {\iota }_{i} : {N}_{i} \rightarrow M \) the injections associated with the direct decomposition \( M = {N}_{1} \oplus \cdots \oplus {N}_{m} \) . The endomorphisms \( {p}_{i} \mathrel{\text{:=}} {\iota }_{i}{\pi }_{i} \) of \( M \) are called ... | Yes |
Theorem 3. The radical of an algebra \( A \) consists precisely of the elements \( x \) in \( A \) such that\n\n\[ 1 + {ax} \in {A}^{ \times }\;\text{ for all }a \in A. \] | Proof. Apply F34 to the \( A \) -module \( M = {A}_{l} \), with generator \( {x}_{1} = 1 \) . If (44) is fulfilled for \( x \in A \), every element of the form \( 1 + {ax} \) is a unit in \( A \), hence a generator of \( {A}_{l} \) . By F34, then, \( x \) belongs to \( \mathfrak{R}\left( A\right) \) .\n\nConversely, ta... | Yes |
Theorem 4. If \( A \) is artinian, \( \Re \left( A\right) \) is nilpotent, that is, there exists \( k \in \mathcal{N} \) such that all \( k \) -fold products in \( \Re \left( A\right) \) vanish - in short, \( \Re {\left( A\right) }^{k} = 0 \) . | Proof of Theorem 4. Set \( I = \Re \left( A\right) \) . Then \( {I}^{0} \mathrel{\text{:=}} A \supseteq {I}^{1} \supseteq {I}^{2} \supseteq \cdots \) is a descending chain of ideals in \( A \) . If \( A \) is artinian, the chain is stationary after some \( k \), so\n\n(45)\n\n\[ I{I}^{k} = {I}^{k} \]\n\nIf we knew that... | Yes |
Theorem 1. A nonzero semisimple algebra \( A \) has only finitely many distinct minimal ideals \( {A}_{1},\ldots ,{A}_{n} \) . Each \( {A}_{i} \) is itself an algebra under the addition and multiplication induced from A. Moreover\n\n\[ A = {A}_{1} \times {A}_{2} \times \cdots \times {A}_{n} \]\n\nis the direct product ... | Proof. We use F18 from the last chapter, which says that, as a left \( A \) -module, \( A \) is the direct sum of all the (finitely many) distinct minimal ideals \( {A}_{1},\ldots ,{A}_{n} \) of \( A \) :\n\n\[ A = {A}_{1} \oplus {A}_{2} \oplus \cdots \oplus {A}_{n} \]\n\nSince the \( {A}_{i} \) are two-sided ideals of... | Yes |
Theorem 2. A commutative artinian algebra having no nonzero nilpotent elements is semisimple (so F7 applies). | Theorem 2 can be proved directly as well. One starts by showing that if \( A \) is an algebra as in the statement and \( P \) is a prime ideal of it, \( A/P \) is a field. An application of the Chinese Remainder Theorem (Chapter 4, F16 in vol. 1) and the result in \( §{4.14} \) conclude the proof. | No |
Theorem 3. Let \( A \) be a simple artinian algebra and let \( N \) be a simple \( A \) -module with endomorphism algebra \( D \) . (I) The natural map \( A \rightarrow {\operatorname{End}}_{D}\left( N\right) \) is an isomorphism. The \( D \) -vector space \( N \) is finite-dimensional, and its dimension coincides with... | Proof. Step 1. A is semisimple because it is a simple artinian algebra (F1). Thus every \( A \) -module is a direct sum of simple \( A \) -modules (F13 in Chapter 28). But all simple \( A \) -modules are isomorphic (F2), so every \( A \) -module \( M \) is isomorphic to \( {N}^{\left( I\right) } \) for some set \( I \)... | Yes |
Theorem 5. A finite-dimensional K-algebra \( A \) is simple if and only if it is isomorphic to a matrix algebra \( {M}_{n}\left( D\right) \) over a \( K \) -division algebra \( D \) . The number \( n \) and the isomorphism class of \( D \) are uniquely determined by the condition\n\n\[ A \simeq {M}_{n}\left( D\right) \... | Proof. Since any finite-dimensional \( K \) -algebra is artinian, Wedderburn’s Theorem applies. The \( K \) -isomorphism (21) implies \( A : K = {n}^{2}\left( {D : K}\right) \), so \( D : K < \infty \) . The centers of \( A \) and \( D \) are \( K \) -isomorphic, by Theorem 3 . | Yes |
Theorem 6. If \( K \) is algebraically closed, every finite-dimensional simple \( K \) -algebra \( A \) is isomorphic to a matrix algebra \( {M}_{n}\left( K\right) \) over the field \( K \) ; in particular, the center of \( A \) is \( K \) and the dimension \( A : K = {n}^{2} \) is a square. | Proof. This follows from Theorem 5 and the next lemma. | No |
Lemma 1. Let \( D \) be a finite-dimensional division algebra over a field \( K \) . Every commutative subalgebra \( E \) of \( D \) is a field. If \( K \) is algebraically closed, \( D \) coincides with \( K \) . | Proof. \( E \) is clearly an integral domain and has finite dimension over \( K \) . Therefore \( E \) is a field, by F2 in Chapter 2 (vol. 1). If \( K \) is algebraically closed, \( E \) coincides with \( K \) . Applying this to the subalgebra \( E = K\left\lbrack a\right\rbrack \) of \( D \) generated by an arbitrary... | Yes |
Theorem 7. Let \( A, B \) be simple \( K \) -algebras, and suppose \( A \) or \( B \) is central over \( K \) . Then \( A{ \otimes }_{K}B \) is simple. | Proof. Let \( A \) be a simple and central \( K \) -algebra. We will show, more generally, that for any \( K \) -algebra \( B \) every ideal \( T \) of \( A \otimes B \) is of the form \( A \otimes I \), where we have set \( I = T \cap B \) . Indeed, any \( t \in T \) can be written as\n\n\[ t = \mathop{\sum }\limits_{... | Yes |
Theorem 8. If \( A \) and \( B \) are central-simple \( K \) -algebras, so is \( A{ \otimes }_{K}B \) . | Proof. \( A{ \otimes }_{K}B \) is simple, by Theorem 7. By (32) we have\n\n\[ Z\left( {A{ \otimes }_{K}B}\right) = Z\left( A\right) { \otimes }_{K}Z\left( B\right) = K{ \otimes }_{K}K = K, \]\n\nso \( A{ \otimes }_{K}B \) is also central. Finally, \( \left( {A{ \otimes }_{K}B}\right) : K = \left( {A : K}\right) \left( ... | Yes |
Theorem 9. The set \( \operatorname{Br}K \) of similarity classes of central-simple \( K \) -algebras enjoys an abelian group structure provided by the tensor product operation. | Proof. One must first check that the tensor product is compatible with the similarity equivalence relation:\n\n(37)\n\n\[ A \sim {A}^{\prime }, B \sim {B}^{\prime }\; \Rightarrow \;A \otimes B \sim {A}^{\prime } \otimes {B}^{\prime }.\]\n\nOnly then is the multiplication in \( \operatorname{Br}K \) given by\n\n(38)\n\n... | Yes |
Theorem 11. Let \( A \) be a simple and central \( K \) -algebra. For any \( K \) -algebra \( B \), the ideals \( T \) of \( A \otimes B \) are in one-to-one correspondence with the ideals \( I \) of \( B \), through the assignments \( T = A \otimes I \) and \( I = T \cap B \) . | Proof. The second part follows from the first and from the following considerations: If a \( K \) -algebra \( C = {C}_{1} \times \cdots \times {C}_{n} \) is the direct product of simple algebras, each ideal of \( C \) is a direct product of some of the \( {C}_{i} \) . The \( {C}_{i} \) are precisely the minimal ideals ... | No |
Theorem 12. Let \( A \) and \( B \) be simple \( K \) -algebras. The ideals of \( A{ \otimes }_{K}B \) are in one-to-one correspondence with those of \( Z\left( A\right) { \otimes }_{K}Z\left( B\right) \) . Further, \( A{ \otimes }_{K}B \) is a finite direct product of simple algebras if and only if the same is true of... | Proof. Theorem \( {11}^{\prime } \) reduces everything to the study of the \( K \) -algebra \( Z\left( A\right) { \otimes }_{K}B \) . Since \( B \) is simple, we can apply Theorem \( {11}^{\prime } \) to the algebra \( B{ \otimes }_{K}Z\left( A\right) = \) \( Z\left( A\right) { \otimes }_{K}B \), thus reducing the prob... | Yes |
Theorem 13. Let \( A \) and \( B \) be semisimple \( K \) -algebras, of which at least one is finite-dimensional and at least one is separable. Then the \( K \) -algebra \( A{ \otimes }_{K}B \) is semisimple and its simple components are in one-to-one correspondence with the simple components of the commutative semisim... | Proof. Without loss of generality, we can assume in addition that \( A \) and \( B \) are simple (compare the proof of Theorem 10). Then \( E \mathrel{\text{:=}} Z\left( A\right) \) and \( F \mathrel{\text{:=}} Z\left( B\right) \) are fields. By statement (c) in the remark on page \( {162}, A{ \otimes }_{K}B \) is cert... | Yes |
Theorem 14 (Centralizer Theorem). Let \( A \) be a simple, artinian and central \( K \) - algebra, let \( B \) be a simple subalgebra of \( A \) such that \( B : K < \infty \), and let \( C = {Z}_{A}\left( B\right) \) be its centralizer in \( A \) .\n\n(a) \( C \) is a simple artinian subalgebra of \( A \) .\n\n(b) \( ... | Proof. We start from the relation\n\n(52)\n\n\[ \n{\operatorname{End}}_{B \otimes {A}^{ \circ }}\left( A\right) = {Z}_{A}\left( B\right) = C;\n\]\n\nsee (30). We next apply our Theorem 3, with the role of \( A \) (a simple artinian algebra) in that statement being played by \( B \otimes {A}^{ \circ } \), and the role o... | Yes |
Theorem 15. Let \( A \) be an artinian, simple and central \( K \) -algebra, and let \( L \) be a finite-dimensional subalgebra \( L \) of \( A \) that is a field. There is equivalence between:\n\n(i) \( L \) is a maximal commutative subalgebra of \( A \), that is, \( {Z}_{A}\left( L\right) = L \) .\n\n(ii) The dimensi... | Proof. We apply Theorem 14 to \( B = L \) . If \( {Z}_{A}\left( L\right) = L \) we obtain condition (ii), using (50). Conversely, if \( A : K = {\left( L : K\right) }^{2} < \infty \), by virtue of (50) the inclusion \( L \subseteq {Z}_{A}\left( L\right) \) must be an equality. If (i) is satisfied, we obtain from (b) th... | Yes |
Theorem 16. Let \( D \) be a division algebra having finite dimension over its center \( K \) . Then \( D \) has maximal subfields, and any of them is a splitting field of \( D \) . The degree \( L : K \) of every maximal subfield \( L \) of \( D \) equals \( \sqrt{D : K} \), the index of \( D \). | Proof. A maximal subfield \( L \) of \( D \) is commutative and obviously contains \( K \) . By Lemma 1, every commutative \( K \) -subalgebra of \( D \) is a subfield of \( D \) . Thus being a maximal commutative \( K \) -subalgebra and a maximal subfield of \( D \) mean the same. Now the theorem follows immediately f... | No |
Theorem 17. Every central-simple K-algebra A admits a splitting field \( L \) such that \( L : K = s\left( A\right) \) . | Proof. We have \( A \sim D \), with \( D \) as Theorem 16. Every maximal subfield \( L \) of \( D \) is a splitting field of \( D \) and hence also of \( A \) . Such an \( L \) satisfies \( L : K = \sqrt{D : K} = \) \( s\left( D\right) = s\left( A\right) . \) | Yes |
Theorem 18. If \( L \) is a splitting field of a central-simple \( K \) -algebra \( A \) and \( L : K \) is \( {fi} \) - nite, there is some algebra \( {A}^{\prime } \) similar to \( A \) that contains \( L \) as a maximal commutative subalgebra. | Proof. Since \( A \) splits over \( L \), there is a finite-dimensional \( L \) -vector space \( V \) such that \( {A}_{L} = A{ \otimes }_{K}L \) is isomorphic to the \( L \) -algebra \( {\operatorname{End}}_{L}\left( V\right) \) . We view \( A \) as embedded in \( {\operatorname{End}}_{L}\left( V\right) \) by means of... | Yes |
Theorem 19. Let \( A \) be a central-simple K-algebra and \( s \) the Schur index of \( A \). (a) If \( L \) is a splitting field of \( A \), then \( s \) divides \( L : K \). | Proof. Let \( D \) be the division algebra such that \( A \sim D \) (it is well defined up to isomorphism). Let \( L \) be a splitting field of \( A \). Assume \( L : K \) is finite. By Theorem 18, there exists \( r \) such that \( L \) is isomorphic to a maximal commutative subalgebra of \( {M}_{r}\left( D\right) \). ... | Yes |
Lemma 2. Let \( D \) be a division algebra having finite dimension over its center \( K \) . If no subfield \( E \supsetneqq K \) of \( D \) is separable over \( K \), then \( D = K \) . | Proof. Let \( L \) be a maximal subfield of \( D \) . Then, by Theorem 16,\n\n\[ D : K = {n}^{2},\;\text{ with }n = L : K.\]\n\nIn contradiction to the conclusion of the lemma, assume that \( D \neq K \), so \( n > 1 \) . By assumption, \( L/K \) is purely inseparable, and therefore \( n \) is a power of the characteri... | Yes |
Theorem 20 (Skolem-Noether). Let \( A \) and \( B \) be simple K-algebras, and suppose the center of \( A \) coincides with \( K \) . Assume further that \( B \) has finite dimension over \( K \) and \( A \) is artinian. Then, for any two \( K \) -algebra homomorphisms \( f, g : B \rightarrow A \) , there exists a unit... | Proof. Consider the simple artinian \( K \) -algebra\n\n\[ A{ \otimes }_{K}{B}^{ \circ } \]\n\n(see Theorem 7 and statement (c) in the remark thereto). The maps \( f \) and \( g \) endow \( A \) with two \( A{ \otimes }_{K}{B}^{ \circ } \) -module structures:\n\n\[ \left( {a \otimes b}\right) x = {axf}\left( b\right) \... | Yes |
Theorem 21 (Wedderburn). Every finite division algebra is commutative. In other words, if a field \( K \) is finite, its Brauer group is trivial. | Proof. Let \( D \) be a finite division algebra, with center \( K \) . Every element of \( D \) lies in a maximal subfield of \( D \) . By Theorem 16, all maximal subfields of \( D \) have the same degree. But for a finite field \( K \), all extensions of the same degree are \( K \) -isomorphic (Theorem \( {1}^{\prime ... | Yes |
Lemma 3. In the situation just described, there exist homogeneous polynomials \( F\\left( {{t}_{1},\\ldots ,{t}_{m}}\\right), G\\left( {{t}_{1},\\ldots ,{t}_{m}}\\right), H\\left( {{t}_{1},\\ldots ,{t}_{m}, X}\\right) \) of respective degrees \( 1, n, n \), with coefficients in \( \\mathbf{K} \), such that, for every \... | Proof. Take an \( L \) -representation \( h : A \\rightarrow {M}_{n}\\left( L\\right) \) of \( A \) with \( L/K \) Galois. Consider the field of rational functions \( L\\left( t\\right) \\mathrel{\\text{:=}} L\\left( {{t}_{1},\\ldots ,{t}_{m}}\\right) \) in \( m \) variables \( {t}_{1},\\ldots ,{t}_{m} \) over \( K \) ... | Yes |
Theorem 22. Let a central-simple K-algebra \( A \) contain a field \( L \) as a subalgebra. Any element \( c \) in the centralizer \( C = {Z}_{A}\left( L\right) \) satisfies\n\n\[ \n{N}_{A/K}^{0}\left( c\right) = {N}_{L/K}{N}_{C/L}^{0}\left( c\right) ,\;{\operatorname{Tr}}_{A/K}^{0}\left( c\right) = {\operatorname{Tr}}... | Proof. By the Centralizer Theorem (page 168) we have\n\n\[ \nA : K = \left( {L : K}\right) \left( {C : K}\right) .\n\]\n\nAlso, \( C \) is a central-simple \( L \) -algebra. Consider the \( C \) -module \( A \) . Since \( C \) is simple, we seem from (87) by comparing dimensions that there is an isomorphism of \( C \) ... | Yes |
Theorem 23. If \( K \) is a \( {C}_{1} \) -field (see Definition 3 in Chapter 27), then \( \operatorname{Br}K = 1 \) . | Proof. Take \( \left\lbrack D\right\rbrack \in \operatorname{Br}K \), where \( D \) is a division algebra with \( D : K = m = {n}^{2} > 1 \) . The reduced norm \( {N}_{D/K}^{0} \) is then given by a homogeneous polynomial of degree \( n \) in \( K\left\lbrack {{X}_{1},\ldots ,{X}_{m}}\right\rbrack \), by Lemma 3 above.... | Yes |
Theorem 1. Let \( L/K \) be a finite Galois extension with Galois group \( G \), and let \( \Gamma \) be a \( K \)-algebra containing \( L \) as a subalgebra. Suppose that \( \Gamma \) is generated by the elements \( \lambda \in L \) and by certain elements \( {u}_{\sigma }\left( {\sigma \in G}\right) \), with \( {u}_{... | Proof. (a) For \( \tau = {\sigma }^{-1} \) we obtain from (16)\n\n\[ {u}_{\sigma }{u}_{{\sigma }^{-1}} = {u}_{1}{c}_{\sigma ,{\sigma }^{-1}} \]\n\nSince \( {u}_{1} \) is assumed to be in \( {L}^{ \times } \), therefore, each \( {u}_{\sigma } \) is invertible in \( \Gamma \) ; in particular, \( {u}_{\sigma } \neq 0 \).\... | Yes |
For any field \( K \), the Brauer group \( \operatorname{Br}K \) is a torsion group. More precisely, every \( \left\lbrack A\right\rbrack \in \operatorname{Br}K \) having Schur index \( s = s\left( A\right) \) satisfies\n\n\[{\left\lbrack A\right\rbrack }^{s} = 1\]\n\nthat is, the \( s \) -fold tensor product \( A \oti... | In view of Theorem 2 (and (59) in the previous chapter) what we must show is this: If \( \Gamma = \left( {L, G, c}\right) \) is any crossed product with Schur index \( s = s\left( \Gamma \right) \), the class \( \gamma \) of the cocycle \( c = {c}_{\sigma ,\tau } \) in \( {H}^{2}\left( {G,{L}^{ \times }}\right) \) sati... | Yes |
Theorem 5. Let \( L/K \) be cyclic with Galois group \( G = \langle \tau \rangle \), and let a be any element of \( {K}^{ \times } \). (a) If \( E \) is an intermediate field of \( L/K \), then \( E/K \) is also cyclic with Galois group \( G\left( {E/K}\right) = \left\langle {\tau }_{E}\right\rangle \), and \[ \left( {... | Proof. (a) Since \( G = G\left( {L/K}\right) \) is abelian, every intermediate field \( E \) is Galois over \( K \) . Hence \( \bar{G} = G\left( {E/K}\right) \), a homomorphic image of \( G \), is cyclic, and generated by the automorphism \( \bar{\tau } = {\tau }_{E} \) arising from \( E \) . Let \( c \) denote a unifo... | Yes |
Theorem 1. Take \( \left\lbrack D\right\rbrack \in \operatorname{Br}K \), where \( D \) is a division algebra. Among the maximal subfields of \( D \) there exists \( L \) such that \( L/K \) is unramified. Thus every central-simple \( K \) -algebra \( A \) possesses unramified splitting fields over \( K \) of degree \(... | Proof. We start just as in the proof of F21 in Chapter 29, with \ | No |
Lemma 1. Take \( \left\lbrack D\right\rbrack \in \operatorname{Br}K \), where \( D \) is a division algebra. If every subfield \( F \supsetneqq K \) of \( D \) is ramified over \( K \), then \( D = K \) . | Proof. Let \( x \) be any element of \( D \) such that \( v\left( x\right) \geq 0 \), and consider the subfield \( E = K\left\lbrack x\right\rbrack \) of \( D \) . If \( \bar{E} \neq \bar{K} \), there exists by Theorem 3(iv) in Chapter 24 an intermediate field \( F \) of \( E/K \) such that \( F/K \) is unramified of d... | Yes |
Lemma 2. In the situation just described, \( A \) is finitely generated as an \( R \) -module. | Proof. There exists a \( K \) -basis \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) of \( D \) consisting only of elements of \( A \) . Any \( x \in A \) has a unique representation\n\n\[ x = {x}_{1}{\alpha }_{1} + \cdots + {x}_{m}{\alpha }_{m},\;\text{ with }{x}_{i} \in K. \]\n\nMultiply from the right by \( {\alpha }_{j} ... | Yes |
Lemma 3. In the preceding situation, assume further that \( \bar{K} \) is perfect. Then\n\n(17)\n\n\[ n = {f}_{0}c. \] | Proof. We consider maximal subfields \( \Lambda \) of the division algebra \( \bar{D} \) . Every such \( \Lambda \) must contain the center \( Z\left( \bar{D}\right) \) of \( \bar{D} \), and hence (by Chapter 29, Theorem 16) have dimension\n\n(18)\n\n\[ \Lambda : \bar{K} = {f}_{0}c \]\n\nin the notation of (12). Since ... | Yes |
Theorem 3. Let the situation be as in Theorem \( {2}^{\prime } \), and let \( {\varphi }_{L/K} \) be the Frobenius automorphism of \( L/K \) (see Theorem 4(iii) in Chapter 24). Then every \( \alpha \in \operatorname{Br}\left( {L/K}\right) \) is represented by a cyclic algebra of the form (34) \[ \left( {L,{\varphi }_{L... | Proof. By (83) in Chapter 29, it suffices to consider the case of a division algebra \( A = D \). By Theorem 1, such an algebra admits a maximal subfield \( L \) unramified over \( K \). Since \( {N}^{0} \) agrees with \( {N}_{L/K} \) on \( L \), by (98) in Chapter 29, we conclude from Theorem 2 that all units in \( K ... | Yes |
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