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Proposition 3.12. If \( E \) is a finite Galois extension of \( K, F \) is a field extension of \( K \), and the composite \( {EF} \) is defined, then \( {EF} \) is a finite Galois extension of \( F, E \) is a finite Galois extension of \( E \cap F \), and \( \operatorname{Gal}\left( {{EF} : F}\right) \cong \) \( \oper... | Proof. By 3.3,3.1, \( {EF} \) is a Galois extension of \( F \) and \( E \) is a Galois extension of \( E \cap F \subseteq E \) ; \( E \) is finite over \( E \cap F \) since \( E \) is finite over \( K \subseteq E \cap F \), and \( {EF} \) is finite over \( F \) by IV.3.8.\n\nSince \( E \) is normal over \( E \cap F \),... | Yes |
Proposition 4.1. Let \( E \) be Galois over \( K \) and let \( K \subseteq F \subseteq E \) . Then \( \left\lbrack {\operatorname{Gal}\left( {E : K}\right) : \operatorname{Gal}\left( {E : F}\right) }\right\rbrack = \left\lbrack {F : K}\right\rbrack \) . Moreover, \( \operatorname{Gal}\left( {E : F}\right) \) is normal ... | Proof. By 3.1, \( E \) is Galois over \( F \) . Every \( K \) -homomorphism of \( F \) into \( \bar{K} \supseteq E \) is the restriction to \( F \) of a \( K \) -automorphism of \( E \) . Now, \( \sigma \) and \( \tau \in \operatorname{Gal}\left( {E : K}\right) \) have the same restriction to \( F \) if and only if \( ... | No |
Proposition 4.2. Let \( E \) be a Galois extension of \( K \) and let \( \mathcal{F} \) be the set of all Galois groups \( \operatorname{Gal}\left( {E : F}\right) \subseteq \operatorname{Gal}\left( {E : K}\right) \) of finite extensions \( F \subseteq E \) of \( K \) . (1) Every \( H \in \mathcal{F} \) has finite index... | Proof. (1) follows from 4.1. (2). Let \( \sigma \in \mathop{\bigcap }\limits_{{H \in \mathcal{F}}}H \) . If \( \alpha \in E \), then \( K\left( \alpha \right) \subseteq E \) is finite over \( K \) , \( \operatorname{Gal}\left( {E : K\left( \alpha \right) }\right) \in \mathcal{F},\sigma \in \operatorname{Gal}\left( {E :... | Yes |
Proposition 4.3. Let \( E \) be a Galois extension of \( K \). Let \( \mathcal{N} \) be the set of all cosets of normal subgroups \( N \in \mathcal{F} \), let \( \mathcal{L} \) be the set of all left cosets of subgroups \( H \in \mathcal{F} \), and let \( \mathcal{R} \) be the set of all right cosets of subgroups \( H ... | Proof. Let \( H = \operatorname{Gal}\left( {E : F}\right) \in \mathcal{F} \), where \( F \subseteq E \) is finite over \( K \). Then \( F = K\left( S\right) \) for some finite subset \( S \) of \( E \). If \( \sigma ,\tau \in \operatorname{Gal}\left( {E : K}\right) \), then \( \tau \in V\left( {\sigma, S}\right) \) if ... | Yes |
Proposition 4.5. If \( E \) is a Galois extension of \( K \) and \( H \) is subgroup of \( \operatorname{Gal}\left( {E : K}\right) \), then \( E \) is a Galois extension of \( F = {\operatorname{Fix}}_{E}\left( H\right) \) and \( \operatorname{Gal}\left( {E : F}\right) \) is the closure of \( H \) in \( \operatorname{G... | Proof. By 3.1, \( E \) is Galois over \( F \) . Let \( \sigma \in \bar{H} \) and \( \alpha \in F \) . Then \( K\left( \alpha \right) \subseteq F \) is finite over \( K, U = \operatorname{Gal}\left( {E : K\left( \alpha \right) }\right) \in \mathcal{F},{\sigma U} \) is open, and there exists \( \tau \in H \cap {\sigma U}... | Yes |
Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be the distinct roots of \( f \in K\left\lbrack X\right\rbrack \) in \( \bar{K} \) . Every \( \tau \in \operatorname{Gal}\left( {f : K}\right) \) permutes the roots of \( f \) in \( \bar{K} \) ; hence \( \operatorname{Gal}\left( {f : K}\right) \) is isomorphic to a subgroup... | Let \( E \) be the splitting field of \( f \) . If \( \tau \) is a \( K \) -automorphism of \( E \) and \( f\left( \alpha \right) = 0 \), then \( f\left( {\tau \alpha }\right) = {}^{\tau }f\left( {\tau \alpha }\right) = {\tau f}\left( \alpha \right) = 0 \) ; hence \( \tau \) permutes the roots of \( f \) and induces a ... | Yes |
Proposition 5.2. If \( f \in K\left\lbrack X\right\rbrack \) and the field \( K \) does not have characteristic 2, then \( \operatorname{Gal}\left( {f : K}\right) \) induces an odd permutation if and only if \( \operatorname{Dis}\left( f\right) \) does not have a square root in \( K \). | Proof. The splitting field \( E \) of \( f \) contains all \( {\alpha }_{i} \) and contains Dis \( \left( f\right) \) . We see that Dis \( \left( f\right) = {d}^{2} \), where \( d = {a}_{n}^{n - 1}\mathop{\prod }\limits_{{1 \leqq i < j \leqq n}}\left( {{\alpha }_{i} - {\alpha }_{j}}\right) \) . If \( \tau \in \operator... | Yes |
Corollary 5.3. Let \( f \in K\left\lbrack X\right\rbrack \) be a separable irreducible polynomial of degree 3. If \( \operatorname{Dis}\left( f\right) \) has a square root in \( K \), then \( \operatorname{Gal}\left( {f : K}\right) \cong {A}_{3} \) ; otherwise, \( \operatorname{Gal}\left( {f : K}\right) \cong {S}_{3}. ... | Proof. We saw that \( \operatorname{Gal}\left( {f : K}\right) \) is isomorphic to either \( {A}_{3} \) or \( {S}_{3} \) . \( ▱ \) | No |
Proposition 5.6. Let \( f\left( X\right) = a{X}^{4} + b{X}^{3} + c{X}^{2} + {dX} + e \in K\left\lbrack X\right\rbrack \) be a separable irreducible polynomial of degree 4, where the field \( K \) does not have characteristic 2. Let \( F \subseteq \bar{K} \) be the splitting field of its resolvent. Then \( \left\lbrack ... | Proof. The resolvent \( s \) of \( f \) is separable, since \( {a}^{6}\operatorname{Dis}\left( s\right) = \operatorname{Dis}\left( f\right) \neq 0 \) . Hence its roots \( u, v, w \) are all distinct. Let \( E \subseteq \bar{K} \) and \( F = K\left( {u, v, w}\right) \subseteq \bar{K} \) be the splitting fields of \( f \... | Yes |
Proposition 6.1. For all integers \( n, q \geqq 2,{\Phi }_{n}\left( q\right) \in \mathbb{R} \) and \( {\Phi }_{n}\left( q\right) > q - 1 \) . | Proof. The number \( {\Phi }_{n}\left( q\right) \) is the product of \( \phi \left( n\right) \) complex numbers \( q - \varepsilon \) , where \( \left| \varepsilon \right| = 1,\varepsilon \neq 1 \) . Hence \( \left| {q - \varepsilon }\right| > q - 1 \geqq 1 \) and \( \left| {{\Phi }_{n}\left( q\right) }\right| > {\left... | Yes |
Proposition 6.3. \( {\Phi }_{n} \) is monic and has integer coefficients. | Proof. By induction. First, \( {\Phi }_{1}\left( X\right) = X - 1 \) is monic and has integer coefficients. If \( n > 1 \), then polynomial division in \( \mathbb{Z}\left\lbrack X\right\rbrack \) of \( {X}^{n} - 1 \in \mathbb{Z}\left\lbrack X\right\rbrack \) by the monic polynomial \( \mathop{\prod }\limits_{{d \mid n,... | Yes |
Proposition 6.5. The field \( \mathbb{Q}\left( {\varepsilon }_{n}\right) \) is a Galois extension of \( \mathbb{Q};\left\lbrack {\mathbb{Q}\left( {\varepsilon }_{n}\right) : \mathbb{Q}}\right\rbrack = \) \( \phi \left( n\right) \) ; and \( \operatorname{Gal}\left( {\mathbb{Q}\left( {\varepsilon }_{n}\right) : \mathbb{Q... | Proof. First, \( \mathbb{Q}\left( {\varepsilon }_{n}\right) \), which contains all complex \( n \) th roots of unity, is a splitting field of \( {X}^{n} - 1 \) and is Galois over \( \mathbb{Q} \) . Next, \( {\Phi }_{n}\left( {\varepsilon }_{n}\right) = 0 \), whence\n\n\( {\Phi }_{n} = \operatorname{Irr}\left( {{\vareps... | Yes |
Theorem 6.6 (Wedderburn [1905]). Every finite division ring is a field. | Proof. Let \( D \) be a finite division ring. The center \( K = \{ x \in D \mid {xy} = {yx} \) for all \( y \in D\} \) of \( D \) is a subfield of \( D \) . Let \( n \) be the dimension of \( D \) as a vector space over \( K \) . We prove that \( n = 1 \) .\n\nLet \( \left| K\right| = q \), so that \( \left| D\right| =... | Yes |
Theorem 6.7 (Dirichlet). For every positive integer \( n \) there are infinitely many prime numbers \( p \equiv 1\left( {\;\operatorname{mod}\;n}\right) \) . | Proof. We start with a lemma. | No |
Lemma 6.8. Let \( p \) be prime and \( m, n > 0 \) . If \( p \) divides \( {\Phi }_{n}\left( m\right) \), then \( p \) does not divide \( m \), and either \( p \) divides \( n \) or \( p \equiv 1\left( {\;\operatorname{mod}\;n}\right) \) . | Proof. By 6.2, \( {\Phi }_{n}\left( m\right) \) divides \( {m}^{n} - 1 \) ; hence \( p \) divides \( {m}^{n} - 1 \), and does not divide \( m \) . Let \( k \) be the order of \( \bar{m} \) in the multiplicative group \( {\mathbb{Z}}_{p} \smallsetminus \{ 0\} ;k \) divides \( \left| {{\mathbb{Z}}_{p}\smallsetminus \{ 0\... | Yes |
Proposition 6.9. Every finite abelian group is the Galois group of a finite extension of \( \mathbb{Q} \) . | Proof. A finite abelian group \( G \) is a direct sum \( G = {C}_{{n}_{1}} \oplus {C}_{{n}_{2}} \oplus \cdots \oplus {C}_{{n}_{r}} \) of cyclic groups of orders \( {n}_{1},\ldots ,{n}_{r} \) . By 6.7 there exist distinct primes \( {p}_{1},\ldots ,{p}_{r} \) such that \( {p}_{i} \equiv 1\left( {{\;\operatorname{mod}\;{n... | Yes |
Lemma 7.1. If \( E \) is finite over \( K \) and \( \alpha \in E \), then \( \det \left( {{T}_{\alpha } - {XI}}\right) = \) \( {\left( -1\right) }^{n}q{\left( X\right) }^{\ell } \), where \( n = \left\lbrack {E : K}\right\rbrack, q = \operatorname{Irr}\left( {\alpha : K}\right) \), and \( \ell = \left\lbrack {E : K\lef... | Proof. We have \( {T}_{a\beta } = a{T}_{\beta },{T}_{\beta + \gamma } = {T}_{\beta } + {T}_{\gamma } \), and \( {T}_{\beta \gamma } = {T}_{\beta }{T}_{\gamma } \), for all \( a \in K \) and \( \beta ,\gamma \in E \) . Hence \( f\left( {T}_{\alpha }\right) = {T}_{f\left( \alpha \right) } \) for every \( f \in K\left\lbr... | Yes |
Proposition 7.2. Let \( E \) be a finite extension of \( K \) of degree \( n \) . Let \( {\alpha }_{1},\ldots \) , \( {\alpha }_{r} \in \bar{K} \) be the distinct conjugates of \( \alpha \in E \), and let \( {\varphi }_{1},\ldots ,{\varphi }_{t} \) be the distinct \( K \) -homomorphisms of \( E \) into \( \bar{K} \) . ... | Proof. The conjugates of \( \alpha \) are the roots of \( q = \operatorname{Irr}\left( {\alpha : K}\right) \), which by IV.5.1 all have the same multiplicity \( m \) . Hence\n\n\[ \nq\left( X\right) = {\left( X - {\alpha }_{1}\right) }^{m}\cdots {\left( X - {\alpha }_{r}\right) }^{m}\n\]\n\n\[ \n= {X}^{rm} - m\left( {{... | Yes |
Proposition 7.4. If \( E \) is finite over \( K \), then \( {\mathrm{N}}_{K}^{E}\left( {\alpha \beta }\right) = {\mathrm{N}}_{K}^{E}\left( \alpha \right) {\mathrm{N}}_{K}^{E}\left( \beta \right) \) and \( {\operatorname{Tr}}_{K}^{E}\left( {\alpha + \beta }\right) = {\operatorname{Tr}}_{K}^{E}\left( \alpha \right) + {\o... | Proof. In Proposition 7.2, \( {\varphi }_{1},\ldots ,{\varphi }_{t} \) are homomorphisms. | No |
Proposition 7.5 (Tower Property). If \( K \subseteq E \subseteq F \) are finite over \( K \), then \( {\mathrm{N}}_{K}^{F}\left( \alpha \right) = {\mathrm{N}}_{K}^{E}\left( {{\mathrm{\;N}}_{E}^{F}\left( \alpha \right) }\right) \) and \( {\operatorname{Tr}}_{K}^{F}\left( \alpha \right) = {\operatorname{Tr}}_{K}^{E}\left... | Proof. We may assume that \( F \subseteq \bar{K} \) and choose \( \bar{E} = \bar{K} \) . Let \( m = \left\lbrack {E : K}\right\rbrack \) and \( n = \left\lbrack {F : E}\right\rbrack \), let \( {\varphi }_{1},\ldots ,{\varphi }_{t} \) be the distinct \( K \) -homomorphisms of \( E \) into \( \bar{K} \), and let \( {\psi... | Yes |
Lemma 7.6. Let \( E \) and \( F \) be field extensions of \( K \) . Distinct \( K \) -homomorphisms of \( E \) into \( F \) are linearly independent over \( F \) . | Proof. Assume that there is an equality \( {\gamma }_{1}{\varphi }_{1} + \cdots + {\gamma }_{n}{\varphi }_{n} = 0 \), in which \( n > 0 \) , \( {\gamma }_{1},\ldots ,{\gamma }_{n} \in F \) are not all 0, and \( {\varphi }_{1},\ldots ,{\varphi }_{n} \) are distinct \( K \) -homomorphisms of \( E \) into \( F \) . Among ... | Yes |
Lemma 7.7 (Hilbert's Theorem 90 [1897]). Let \( E \) be a finite Galois extension of \( K \). If \( \operatorname{Gal}\left( {E : K}\right) \) is cyclic, \( \operatorname{Gal}\left( {E : K}\right) = \langle \tau \rangle \), then, for any \( \alpha \in E \): (1) \( {\mathrm{N}}_{K}^{E}\left( \alpha \right) = 1 \) if and... | Proof. If \( \gamma \in E,\gamma \neq 0 \), then \[ \mathrm{N}\left( {\tau \gamma }\right) = \mathop{\prod }\limits_{{\sigma \in G}}{\sigma \tau \gamma } = \mathop{\prod }\limits_{{\sigma \in G}}{\sigma \gamma } = \mathrm{N}\left( \gamma \right) \] by 7.3, where \( G = \operatorname{Gal}\left( {E : K}\right) \); hence ... | Yes |
Let \( n > 0 \) . Let \( K \) be a field whose characteristic is either 0 or not a divisor of \( n \), and that contains a primitive \( n \) th root of unity.\n\nIf \( E \) is a cyclic extension of \( K \) of degree \( n \), then \( E = K\left( \alpha \right) \), where \( {\alpha }^{n} \in K \) .\n\nIf \( E = K\left( \... | Proof. By the hypothesis, \( K \) contains a primitive \( n \) th root of unity \( \varepsilon \in \bar{K} \) .\n\nLet \( E \) be cyclic over \( K \) of degree \( n \) and \( \operatorname{Gal}\left( {E : K}\right) = \langle \tau \rangle \) . Since \( \mathrm{N}\left( \varepsilon \right) = \) \( {\varepsilon }^{n} = 1 ... | Yes |
Proposition 7.9. A root of unity \( \varepsilon \in \bar{K} \) is a primitive nth root of unity for some \( n > 0 \) ; if \( K \) has characteristic \( p \neq 0 \), then \( p \) does not divide \( n;K\left( \varepsilon \right) \) is a Galois extension of \( K \) of degree at most \( n \) ; and \( \operatorname{Gal}\lef... | The proof is an enjoyable exercise. In 7.9, it may happen that \( \left\lbrack {K\left( \varepsilon \right) : K}\right\rbrack < n \) , and that \( \operatorname{Gal}\left( {K\left( \varepsilon \right) : K}\right) \) is not cyclic; this makes more fine exercises. | No |
Proposition 7.10 (Artin-Schreier). Let \( K \) be a field of characteristic \( p \neq 0 \). If \( E \) is a cyclic extension of \( K \) of degree \( p \), then \( E = K\left( \alpha \right) \), where \( {\alpha }^{p} - \alpha \in K \). If \( E = K\left( \alpha \right) \), where \( {\alpha }^{p} - \alpha \in K,\alpha \n... | Proof. Let \( E \) be cyclic over \( K \) of degree \( p \) and \( \operatorname{Gal}\left( {E : K}\right) = \langle \tau \rangle \). Since \( \operatorname{Tr}\left( 1\right) = {p1} = 0 \) we have \( {\tau \alpha } - \alpha = 1 \) for some \( \alpha \in E \), by 7.7. Then \( {\tau }^{i}\alpha = \alpha + i \) for all \... | Yes |
Proposition 8.4. In \( K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \left\lbrack X\right\rbrack \) , \n\n\[ \n\left( {X - {X}_{1}}\right) \left( {X - {X}_{2}}\right) \cdots \left( {X - {X}_{n}}\right) = \mathop{\sum }\limits_{{0 \leqq k \leqq n}}{\left( -1\right) }^{k}{\mathfrak{s}}_{k}\left( {{X}_{1},\ldots ,{X}_{n}}\righ... | Proof. Expanding \( \left( {X - {X}_{1}}\right) \left( {X - {X}_{2}}\right) \cdots \left( {X - {X}_{n}}\right) \) yields a sum whose terms are all products \( {t}_{1}{t}_{2}\cdots {t}_{n} \) in which, for every \( 1 \leqq i \leqq n \), either \( {t}_{i} = X \) or \( {t}_{i} = - {X}_{i} \) . A product \( {t}_{1}{t}_{2}\... | Yes |
Proposition 8.5. For any field \( K, K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is a Galois extension of \( K\left( {{\mathfrak{s}}_{1},\ldots ,{\mathfrak{s}}_{n}}\right) \), whose Galois group is isomorphic to the symmetric group \( {S}_{n} \) . | Proof. For every \( \sigma \in {S}_{n},\bar{\sigma } : f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \mapsto f\left( {{X}_{▟},{X}_{\blacksquare },\ldots ,{X}_{\sigma n}}\right) \) is an automorphism of \( K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) . Then \( G = \left\{ {\bar{\sigma } \mid \sigma \in {S}_{n}}\right\} \) is ... | Yes |
Corollary 8.6. Every symmetric rational fraction of \( {X}_{1},\ldots ,{X}_{n} \) is a rational function of the elementary symmetric polynomials in \( {X}_{1},\ldots ,{X}_{n} \) . | Proof. This follows from the equality \( S = L \) in the proof of 8.5. \( ▱ \) | No |
Corollary 8.7. The elementary symmetric polynomials \( {\mathfrak{s}}_{1},\ldots ,{\mathfrak{s}}_{n} \) are algebraically independent over \( K \) in \( K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) . | Proof. By IV.8.6, \( K\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) has transcendence degree \( n \) and a transcendence base \( B \subseteq \left\{ {{\mathfrak{s}}_{1},\ldots ,{\mathfrak{s}}_{n}}\right\} \) ; hence \( B = \left\{ {{\mathfrak{s}}_{1},\ldots ,{\mathfrak{s}}_{n}}\right\} \) . \( ▱ \) | Yes |
Corollary 8.8. Every finite group is isomorphic to a Galois group. | Proof. This follows from Proposition 8.5 since, by Cayley's theorem II.3.2, every finite group is isomorphic to a subgroup of some \( {S}_{n} \) . \( ▱ \) | Yes |
Proposition 8.9. The Galois group of the general polynomial of degree \( n \) is isomorphic to the symmetric group \( {S}_{n} \) . | Proof. We show that the general polynomial of degree \( n \) can also be defined by its roots. Let \( S = K\left( A\right) \left( {{\mathfrak{s}}_{1},\ldots ,{\mathfrak{s}}_{n}}\right) \subseteq K\left( A\right) \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) and\n\n\[ f\left( X\right) = A\left( {X - {X}_{1}}\right) \left( ... | Yes |
Proposition 9.1. Constructible complex numbers constitute a subfield of \( \mathbb{C} \) . | Proof. The numbers 0 and 1 are constructible. Let \( a, b \in \mathbb{C} \) be constructible and let \( A, B \) be the corresponding points. Then \( a + b \) corresponds to the fourth point \( C \) of the parallelogram \( {OABC} \) and is constructible; \( a - b \) corresponds to the fourth point \( D \) of the paralle... | Yes |
Theorem 9.3. A complex number is constructible (from 0 and 1) if and only if it is algebraic over \( \mathbb{Q} \) and its degree is a power of 2. | Proof. Call a complex number 2-constructible when it is algebraic over \( \mathbb{Q} \) and its degree over \( \mathbb{Q} \) is a power of 2. | No |
Lemma 9.4. A complex number \( z \) is 2-constructible if and only if it belongs to a finite extension of \( \mathbb{Q} \) whose degree is a power of 2, if and only if it belongs to a finite normal extension of \( \mathbb{Q} \) whose degree is a power of 2. | The proof is an exercise. In particular, 2-constructible complex numbers are those that can be reached from \( \mathbb{Q} \) by successive adjunctions of square roots. We want to show that a complex number is constructible if and only if it is 2-constructible. | No |
Lemma 9.5. The 2-constructible complex numbers constitute a subfield of \( \mathbb{C} \) . | Proof. Let \( a, b \in \mathbb{C} \) be 2-constructible. By 9.4, \( a \) and \( b \) belong to finite normal extensions \( E \) and \( F \) of \( \mathbb{Q} \) whose degrees are powers of 2 . By 3.2,3.11, the composite \( {EF} \) is a Galois extension of \( \mathbb{Q} \), whose degree is a power of 2 since the orders o... | Yes |
Corollary 9.7. There is no construction by straightedge and compass that can trisect angles (split any angle into three equal parts). | Proof. If a \( \pi /3 \) angle could be trisected, then the complex number \( \varepsilon = {e}^{{i\pi }/9} \) would be constructible. But \( \varepsilon \) is a primitive 18 th root of unity; by \( {6.5},\varepsilon \) has degree \( \phi \left( {18}\right) = 6 \) over \( \mathbb{Q} \), and is not constructible. \( ▱ \... | Yes |
Lemma 9.10. If \( {2}^{k} + 1 \) is prime, then \( k \) is a power of 2 . | Proof. If \( k \) is not a power of 2, then \( k = {2}^{i}j \), where \( j \) is odd. Then every \( {m}^{j} + 1 \) is divisible by \( m + 1 \), and \( {2}^{k} + 1 = {\left( {2}^{{2}^{i}}\right) }^{j} + 1 \) is divisible by \( {2}^{{2}^{i}} + 1 \) . \( ▱ \) | Yes |
Corollary 9.11 (Gauss [1801]). A regular polygon with \( n \) sides can be constructed from its radius by straightedge and compass if and only if \( n \) is the product of a power of 2 and distinct Fermat primes. | Proof. A regular polygon with \( n \) sides is constructible from its radius if and only if the primitive \( n \) th root of unity \( {\varepsilon }_{n} = {e}^{{2i\pi }/n} \) is constructible. By \( {6.5},{\varepsilon }_{n} \) has degree \( \phi \left( n\right) \) over \( \mathbb{Q} \) . Write \( n \) as the product \(... | Yes |
A field \( F \) can be ordered if and only if -1 is not a sum of squares of elements of \( F \), if and only if 0 is not a nonempty sum of nonzero squares of elements of \( F \) . | Proof. If \( - 1 = \mathop{\sum }\limits_{i}{x}_{i}^{2} \), then \( 0 = 1 + \mathop{\sum }\limits_{i}{x}_{i}^{2} \) is a sum of squares. Conversely, if \( 0 = \mathop{\sum }\limits_{i}{x}_{i}^{2} \) with, say, \( {x}_{k} \neq 0 \), then \( - {x}_{k}^{2} = \mathop{\sum }\limits_{{i\; \neq \;k}}{x}_{i}^{2} \) and \( - 1 ... | Yes |
Proposition 2.1. If \( F \) is a formally real field and \( {\alpha }^{2} \in F,{\alpha }^{2} > 0 \), then \( F\left( \alpha \right) \) is formally real. | Proof. We may assume that \( \alpha \notin F \) . Then every element of \( F\left( \alpha \right) \) can be written in the form \( x + {\alpha y} \) for some unique \( x, y \in F \) . If \( {\alpha }^{2} > 0 \) in \( F \), then \( F\left( \alpha \right) \) is formally real, since\n\n\[ - 1 = \mathop{\sum }\limits_{i}{\... | Yes |
Proposition 2.2. If \( F \) is a formally real field, then every finite extension of \( F \) of odd degree is formally real. | Proof. This is proved by induction on \( n \), simultaneously for all \( F \) and all \( E \supseteq F \) of odd degree \( n = \left\lbrack {E : F}\right\rbrack \) . There is nothing to prove if \( n = 1 \) . Let \( n > 1 \) . If \( \alpha \in E \smallsetminus F \) and \( F \subsetneqq F\left( \alpha \right) \subsetneq... | Yes |
Corollary 2.4. A real closed field \( R \) can be made into an ordered field in only one way. | Proof. Namely, \( x \leqq y \) if and only if \( y - x = {a}^{2} \) for some \( a \in R \), by 2.3. \( ▱ \) | Yes |
Corollary 2.5. If \( R \) is real closed, then \( f \in R\left\lbrack X\right\rbrack \) is irreducible if and only if either \( f \) has degree 1, or \( f \) has degree 2 and no root in \( R \) . | This is proved like the similar property of \( \mathbb{R} \), as readers will happily verify. | No |
Corollary 2.6. The field of all algebraic real numbers is real closed. | Proof. \ | No |
Proposition 2.7. Every ordered field has a real closure. | Proof. Let \( F \) be an ordered field. The subfield \( E \) of \( \bar{F} \) generated by all square roots of positive elements of \( F \) is formally real: if \( - 1 = \mathop{\sum }\limits_{i}{\beta }_{i}^{2} \) in \( E \) , then \( - 1 = \mathop{\sum }\limits_{i}{\beta }_{i}^{2} \) in \( F\left( {{\alpha }_{1},\ldo... | Yes |
Theorem 2.8 (Artin-Schreier [1926]). For a field \( K \neq \bar{K} \) the following conditions are equivalent: (1) \( K \) is real closed; (2) \( \left\lbrack {\bar{K} : K}\right\rbrack \) is finite; (3) there is an upper bound for the degrees of irreducible polynomials in \( K\left\lbrack X\right\rbrack \) . | Proof. We start with three lemmas. | No |
Lemma 2.9. If there is an upper bound for the degrees of irreducible polynomials in \( K\left\lbrack X\right\rbrack \), then \( K \) is perfect. | Proof. If \( K \) is not perfect, then \( K \) has characteristic \( p \neq 0 \), and some \( c \in K \) is not a \( p \) th power in \( K \) . We show that \( f\left( X\right) = {X}^{{p}^{r}} - c \in K\left\lbrack X\right\rbrack \) is irreducible for every \( r \geqq 0 \) . In \( K\left\lbrack X\right\rbrack, f \) is ... | Yes |
Lemma 2.11. If \( \left\lbrack {\bar{K} : K}\right\rbrack = n \) is finite, then every irreducible polynomial in \( K\left\lbrack X\right\rbrack \) has degree at most \( n, K \) is perfect, and \( \bar{K} = K\left( i\right) \), where \( {i}^{2} = - 1 \) . | Proof. Every irreducible polynomial \( q \in K\left\lbrack X\right\rbrack \) has a root \( \alpha \) in \( \bar{K} \) ; then \( q = \operatorname{Irr}\left( {\alpha : K}\right) \) has degree \( \left\lbrack {K\left( \alpha \right) : K}\right\rbrack \leqq n \) . Then \( K \) is perfect, by 2.9, and \( \bar{K} \) is Galo... | Yes |
Proposition 3.1. Let \( v \) and \( w \) be absolute values on a field \( F \) . The following conditons are equivalent:\n\n(1) \( v \) and \( w \) induce the same topology on \( F \) ;\n\n(2) \( {\left| x\right| }_{v} < 1 \) if and only if \( {\left| x\right| }_{w} < 1 \) ;\n\n(3) there exists \( c > 0 \) such that \(... | Proof. (1) implies (2). Assume \( {\left| x\right| }_{v} < 1 \) . Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left| {x}^{n}\right| }_{v} = 0 \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}^{n} = 0 \) in the topology induced by \( v \) and \( w \) . Hence the open set \( \left\{ {x \in F;{\left... | Yes |
Proposition 3.2. For an absolute value \( v \) the following are equivalent:\n\n(1) \( v \) is nonarchimedean;\n\n(2) \( {\left| n\right| }_{v} \leqq 1 \) for all \( n \in \mathbb{Z}, n > 1 \) ;\n\n(3) \( {\left| n\right| }_{v} \leqq 1 \) for all \( n \in \mathbb{Z} \) ;\n\n(d) \( {\left| x + y\right| }_{v} \leqq \max ... | Proof. (1) implies (2). If \( \left| m\right| > 1 \) for some \( m > 1 \), then \( \mathop{\lim }\limits_{{k \rightarrow \infty }}\left| {m}^{k}\right| = \infty \) and there is no \( x \in F \) such that \( \left| n\right| \leqq \left| x\right| \) for all \( n \in \mathbb{Z} \).\n\n(2) implies (3) since \( {\left| -1\r... | Yes |
Proposition 4.2. Let \( F \) and \( K \) be fields with absolute values. If \( K \) is complete, then every homomorphism of \( F \) into \( K \) that preserves absolute values extends uniquely to a homomorphism of any completion of \( F \) into \( K \) that preserves absolute values. | Proof. Let \( \widehat{F} \) be a completion of \( F \) and let \( \varphi : F \rightarrow K \) be a homomorphism that preserves absolute values. Since \( F \) is dense in \( \widehat{F} \), every element \( \alpha \) of the metric space \( \widehat{F} \) is the limit of a sequence \( a = {\left( {a}_{n}\right) }_{n > ... | Yes |
Proposition 4.4. Every Laurent series \( \mathop{\sum }\limits_{{n \geqq m}}{x}_{n}{p}^{n} \) with coefficients \( {x}_{n} \in \mathbb{Z} \) converges in \( {\widehat{\mathbb{Q}}}_{p} \) ; every \( p \) -adic integer \( x \) is the sum\n\n\[ x = {x}_{0} + {x}_{1}p + \cdots + {x}_{n}{p}^{n} + \cdots \]\n\nof a unique po... | Proof. First we prove a lemma.\n\nLemma 4.5. If \( x \in | No |
Lemma 4.5. If \( x \in {\widehat{\mathbb{Q}}}_{p} \) and \( {\left| x\right| }_{p} \leqq {p}^{-m} \), then \( {\left| x - t{p}^{m}\right| }_{p} < {p}^{-m} \) for some unique integer \( 0 \leqq t < p \) ; if \( {\left| x\right| }_{p} = {p}^{-m} \), then \( t \neq 0 \) . | Proof. If \( {\left| x\right| }_{p} < {p}^{-m} \), then \( t = 0 \) serves. Now assume \( {\left| x\right| }_{p} = {p}^{-m} \) . Since \( \mathbb{Q} \) is dense in \( {\widehat{\mathbb{Q}}}_{p} \) we have \( {\left| x - y\right| }_{p} < {p}^{-m} \) for some \( y \in \mathbb{Q} \) . Then \( {\left| y\right| }_{p} = {p}^... | Yes |
Theorem 5.2. Let \( E \) be a finite extension of degree \( n \) of a field \( K \) that is complete with respect to an absolute value \( v \) . If there exists an absolute value \( w \) on \( E \) that extends \( v \), then \( w \) is unique and\n\n\[{\left| \alpha \right| }_{w} = {\left( {\left| {\mathrm{N}}_{K}^{E}\... | Proof. The definition of \( N\left( \alpha \right) \) shows that \( N\left( \alpha \right) \) is a polynomial function of the coordinates of \( \alpha \) in any basis of \( E \) over \( K \), hence continuous, by 5.1. Let \( \alpha \in E \) , \( \alpha \neq 0 \), and \( \beta = {\alpha }^{n}N{\left( \alpha \right) }^{-... | Yes |
Proposition 5.3. If \( K \) is a field that is complete with respect to an absolute value \( v \), and does not have characteristic 2, then \( v \) can be extended to every finite extension of \( K \) of degree 2. | Proof. Inspired by 5.2 we try \( {\left| \alpha \right| }_{w} = {\left( {\left| N\left( \alpha \right) \right| }_{v}\right) }^{1/2} \) . By V.6.1, a finite extension \( E \) of \( K \) of degree 2 is a Galois extension and has a nontrivial automorphism \( \alpha \mapsto \bar{\alpha } \) . Then \( N\left( \alpha \right)... | Yes |
Lemma 5.4. Let \( K \) be a field that is complete with respect to an absolute value \( v \) , and does not have characteristic 2. If \( {\left( {\left| b\right| }_{v}\right) }^{2} > 4{\left| c\right| }_{v} \), then \( {X}^{2} - {bX} + c \in K\left\lbrack X\right\rbrack \) has a root in \( K \) . | Proof. We may assume that \( c \neq 0 \) ; then \( b \neq 0 \) . We use successive approximations \( {x}_{n + 1} = f\left( {x}_{n}\right) \) to find a root, noting that \( {x}^{2} - {bx} + c = 0 \) if and only if \( x = b - \left( {c/x}\right) \) . Let \( {x}_{1} = \frac{1}{2}b \) and \( {x}_{n + 1} = b - \left( {c/{x}... | Yes |
Theorem 5.5 (Ostrowski [1918]). Up to isomorphisms that preserve absolute values, \( \mathbb{R} \) and \( \mathbb{C} \) are the only fields that are complete with respect to an archimedean absolute value. | Proof. Let \( F \) be complete with respect to an archimedean absolute value \( v \) . By 3.3, \( F \) has characteristic 0 . Hence \( \mathbb{Q} \subseteq F \), up to isomorphism. By 3.4, the valuation induced by \( F \) on \( \mathbb{Q} \) is equivalent to the usual absolute value. We may replace \( v \) by an equiva... | Yes |
Proposition 6.2. Let \( R \) be a subring of a field \( F \) and let \( \mathfrak{u} \) be the group of units of \( R \) . The following properties are equivalent:\n\n(1) \( R \) is the valuation ring of a valuation on \( F \) ;\n\n(2) \( F = Q\left( R\right) \) and the ideals of \( R \) form a chain;\n\n(3) when \( x ... | Proof. (1) implies (2) by 6.1.\n\n(2) implies (3). Let \( x = a/b \in F \), where \( a, b \in R, b \neq 0 \) . If \( {Ra} \subseteq {Rb} \), then \( a = {br} \) for some \( r \in R \) and \( x = r \in R \) . If \( {Rb} \subseteq {Ra} \), then \( b = {ar} \) for some \( r \in R \) and \( {x}^{-1} = r \in R \) .\n\n(3) i... | Yes |
Proposition 6.3. Every valuation is equivalent to the valuation induced by its valuation ring. In particular, two valuations on the same field are equivalent if and only if they have the same valuation ring. | Proof. Let \( v \) be a valuation on a field \( F \) and let \( \mathbf{o} \) be its valuation ring. The valuations \( v \) and \( {v}_{0} \) induce surjective homomorphisms of multiplicative groups:\n\n\n\nwhere \( ... | Yes |
Proposition 6.4. Let \( R \) be a domain and let \( F = Q\left( R\right) \) be its quotient field. Then \( R \) is the valuation ring of a discrete valuation on \( F \) if and only if \( R \) is a principal ideal domain with a unique nonzero prime ideal. | The proof is an exercise. | No |
Let \( F \) be a field with a discrete valuation \( v \) . Let \( \mathfrak{r} \) be a subset of \( \mathfrak{o} \) with \( 0 \in \mathfrak{r} \) and one element in every coset of \( \mathfrak{m} \) in \( \mathfrak{o} \) . Every element of \( \mathfrak{o} \) is the sum of a unique power series \( \mathop{\sum }\limits_... | Proof. By \( {6.4},\mathfrak{m} \) is the ideal of \( \mathfrak{o} \) generated by \( p \) . Let \( x \in F, x \neq 0 \) . Then \( x = u{p}^{m} \) for some unique \( u \in \mathfrak{u} \) and \( m \in \mathbb{Z} \) . By the choice of \( \mathfrak{r}, u \in {r}_{m} + \mathfrak{m} \) for some unique \( {r}_{m} \in \mathf... | No |
Proposition 6.6. Let \( F \) be a field with a discrete valuation \( v \) . Every Laurent series \( \mathop{\sum }\limits_{{k \geqq m}}{r}_{k}{p}^{k} \) with coefficients \( {r}_{k} \in \mathfrak{o} \) converges in \( \widehat{F} \) . Conversely, let \( \mathfrak{r} \) be a subset of \( \mathfrak{o} \) with \( 0 \in \m... | Proof. First we show that \( v \) and its extension \( \widehat{v} \) to \( \widehat{F} \) have the same value group \( {G}_{v} \) . We may assume that \( {G}_{v} \) is a subgroup of \( \mathbb{P} \) . Every nonzero \( x \in \widehat{F} \) is the limit of a Cauchy sequence \( {\left( {x}_{n}\right) }_{n > 0} \) of elem... | No |
Theorem 7.2. Let \( K \) be a subfield of \( E \) . Every valuation on \( K \) extends to a valuation on \( E \) . | Proof. Let \( v \) be a valuation on \( K \) ; let \( \mathfrak{o} \) be the valuation ring of \( v \), let \( \mathfrak{m} \) be the maximal ideal of \( \mathfrak{o} \), and let \( \mathfrak{u} \) be its group of units. Let \( L \) be the algebraic closure of the field \( \mathfrak{o}/\mathfrak{m} \) . By 7.1, the pro... | Yes |
Proposition 7.3. If \( E \) is a field extension of \( K, v \) is a valuation on \( K \), and \( w \) is a valuation on \( E \) that extends \( v \), then \( \mathrm{e}\left( {w : v}\right) \mathrm{f}\left( {w : v}\right) \leqq \left\lbrack {E : K}\right\rbrack \) . | Proof. Let \( {\left( {\alpha }_{i}\right) }_{i \in I} \) be elements of \( E \) . Let \( {\left( {\beta }_{j}\right) }_{j \in J} \) be elements of \( \mathfrak{O} \) whose residue classes are linearly independent over \( {F}_{v} \) .\n\nLet \( \gamma = \mathop{\sum }\limits_{{j \in J}}{x}_{j}{\beta }_{j} \in E \), whe... | Yes |
Theorem 7.4. If \( E \) is a finite extension of \( K \), then every nonarchimedean absolute value on \( K \) extends to a nonarchimedean absolute value on \( E \) . | Proof. By 7.2, a nonarchimedean absolute value \( v \) on \( K \) extends to a valuation \( w \) on \( E \) . By 7.3, \( e = \mathrm{e}\left( {w : v}\right) \) is finite. Then \( {g}^{e} \in {G}_{v} \) for every \( g \in {G}_{w} \), and \( g \mapsto {g}^{e} \) is a homomorphism of \( {G}_{w} \) into \( {G}_{v} \subsete... | Yes |
Proposition 7.6. Let \( E \) be a finite extension of \( K \), let \( v \) be a discrete non-archimedean absolute value on \( K \), and let \( w \) be a discrete nonarchimedean absolute value on \( E \) that extends \( v \). If \( K \) is complete with respect to \( v \), then \( \mathrm{e}\left( {w : v}\right) \mathrm... | Proof. First, \( w \) exists, by 7.4 and 7.5. Let \( v\left( p\right) < 1 \) generate \( {G}_{v} \) and \( w\left( \rho \right) < 1 \) generate \( {G}_{w} \). Then \( v\left( p\right) = w{\left( \rho \right) }^{e} \) for some \( e > 0;{\rho }^{e} = {up} \) for some \( u \in \mathfrak{U} \); the cosets of \( {G}_{v} \) ... | Yes |
Proposition 8.1. Let \( \mathfrak{o} \) be a valuation ring of a field \( K \) and let \( \mathfrak{m} \) be its maximal ideal. Every nonzero polynomial \( f \in K\left\lbrack X\right\rbrack \) can be written in the form \( f\left( X\right) = t{f}^{ * }\left( X\right) \), where \( t \in K, t \neq 0 \), and \( {f}^{ * }... | Proof. The ring \( \mathfrak{o} \) is the valuation ring of a valuation \( v \) on \( K \) . For every \( f\left( X\right) = {a}_{0} + {a}_{1}X + \cdots + {a}_{n}{X}^{n} \in K\left\lbrack X\right\rbrack \), let \( V\left( f\right) = \max \left( {v\left( {a}_{0}\right) ,\ldots, v\left( {a}_{n}\right) }\right) \) . If \(... | Yes |
Lemma 8.2 (Gauss). Let \( \mathfrak{o} \) be a valuation ring of a field \( K \) . If \( f \) and \( g \in \mathfrak{o}\left\lbrack X\right\rbrack \) are primitive, then \( {fg} \) is primitive. | Proof. Let \( f\left( X\right) = {a}_{0} + {a}_{1}X + \cdots + {a}_{m}{X}^{m} \) and \( g\left( X\right) = {b}_{0} + {b}_{1}X + \cdots + \) \( {b}_{n}{X}^{n} \), so that \( \left( {fg}\right) \left( X\right) = {c}_{0} + {c}_{1}X + \cdots + {c}_{m + n}{X}^{m + n} \), where \( {c}_{k} = \mathop{\sum }\limits_{{i + j = k}... | Yes |
Corollary 8.3. In Proposition 8.1, \( f \) is irreducible in \( K\left\lbrack X\right\rbrack \) if and only if \( {f}^{ * } \) is irreducible in \( \mathfrak{o}\left\lbrack X\right\rbrack \) . | Proof. We may assume that \( \deg f \geqq 2 \) . If \( f \) is not irreducible, then \( f \) has a factorization \( f = {gh} \) in which \( \deg g,\deg h \geqq 1 \) . Let \( f\left( X\right) = a{f}^{ * }\left( X\right), g\left( X\right) \) \( = b{g}^{ * }\left( X\right), h\left( X\right) = c{h}^{ * }\left( X\right) \),... | Yes |
Corollary 8.6. Let \( K \) be complete for a valuation \( v \) and \( f \in {\mathfrak{o}}_{v}\left\lbrack X\right\rbrack \) . If \( f\left( a\right) \in {\mathfrak{m}}_{v},{f}^{\prime }\left( a\right) \notin {\mathfrak{m}}_{v} \) for some \( a \in {\mathfrak{o}}_{v} \), then \( f\left( b\right) = 0 \) for some \( b \i... | Proof. The polynomial \( f \) is primitive, since \( {f}^{\prime }\left( a\right) \notin \mathfrak{m} \) . We have \( \bar{f}\left( \bar{a}\right) = 0 \) , \( {\bar{f}}^{\prime }\left( \bar{a}\right) \neq 0 \) . Hence \( \bar{f}\left( X\right) = \left( {X - \bar{a}}\right) \bar{h}\left( X\right) \) for some \( h \in \m... | Yes |
Proposition 9.1. If \( R = S\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), where \( S \) is a commutative ring, and \( \mathfrak{a} \) is the ideal generated by \( {X}_{1},\ldots ,{X}_{n} \), then \( {\widehat{R}}_{\mathfrak{a}} \cong S\left\lbrack \left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \right\rbr... | Proof. We see that \( {\mathfrak{a}}^{i} \) consists of all polynomials of order at least \( i \) . For every \( f \in S\left\lbrack \left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \right\rbrack \), let \( {\varphi }_{i}\left( f\right) \in R/{\mathfrak{a}}^{i} \) be the coset of any polynomial with the same terms o... | Yes |
If \( \mathcal{A} : {\mathfrak{a}}_{1} \supseteq {\mathfrak{a}}_{2} \supseteq \cdots \) is a filtration on \( R \), then\n\n\[{\widehat{\mathfrak{a}}}_{j} = \left\{ {\left( {{x}_{1} + {\mathfrak{a}}_{1},\ldots ,{x}_{i} + {\mathfrak{a}}_{i}\ldots }\right) \in {\widehat{R}}_{\mathcal{A}}\;}\right| \;{x}_{j} \in {\mathfra... | Proof. First, \( {\widehat{\mathfrak{a}}}_{j} \) is an ideal of \( {\widehat{R}}_{\mathcal{A}} \), since it is the kernel of the homomorphism \( \left( {{x}_{1} + {\mathfrak{a}}_{1},{x}_{2} + {\mathfrak{a}}_{2},\ldots }\right) \mapsto {x}_{j} + {\mathfrak{a}}_{j} \) of \( {\widehat{R}}_{\mathcal{A}} \) into \( R/{\math... | Yes |
Proposition 9.3. If \( R \) is complete (relative to a filtration), then every Cauchy sequence of elements of \( R \) has a unique limit in \( R \) . | Proof. Let \( {\left( {x}_{n}\right) }_{n > 0} \) be a Cauchy sequence relative to \( \mathcal{A} : {\mathfrak{a}}_{1} \supseteq {\mathfrak{a}}_{2} \supseteq \cdots \) . Choose \( n\left( i\right) \) by induction so that \( n\left( {i + 1}\right) \geqq n\left( i\right) \) and \( {x}_{m} - {x}_{n} \in {\mathfrak{a}}_{i}... | No |
Corollary 9.4. (1) If \( R \) is complete relative to a filtration \( {\mathfrak{a}}_{1} \supseteq {\mathfrak{a}}_{2} \supseteq \cdots \) , then every family \( {\left( {x}_{t}\right) }_{t \in T} \) of elements of \( R \) such that\n\n\[ \n\text{for every}i > 0,{x}_{t} \in {\mathfrak{a}}_{i}\text{for almost all}t \in T... | Proof. (1). We see that \( {s}_{i} \) is a finite sum, and \( {\left( {s}_{i}\right) }_{i > 0} \) is a Cauchy sequence. | No |
Let \( R \) and \( S \) be commutative rings. If \( S \) is complete relative to an ideal \( \mathfrak{b} \), then for every ring homomorphism \( \varphi : R \rightarrow S \) and elements \( {b}_{1},\ldots ,{b}_{n} \) of \( \mathfrak{b} \) there exists a unique homomorphism \( \psi : R\left\lbrack \left\lbrack {{X}_{1}... | Proof. Let \( \mathfrak{a} \) be the ideal of \( R\left\lbrack \left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \right\rbrack \) generated by \( {X}_{1},\ldots ,{X}_{n} \) . For every \( f = \mathop{\sum }\limits_{m}{r}_{m}{X}_{1}^{{m}_{1}}\cdots {X}_{n}^{{m}_{n}} \in S\left\lbrack \left\lbrack {{X}_{1},\ldots ,{X}_... | Yes |
Lemma 9.6. If \( f = {r}_{1}X + \cdots + {r}_{i}{X}^{i} + \cdots \in R\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \), then \( \eta : g\left( X\right) \mapsto g\left( f\right) \) is an automorphism of \( R\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) if and only if \( {r}_{1} \) is a unit of \( R ... | Proof. If \( \eta \) is an automorphism, then \( X = \eta \left( g\right) \) for some \( g;g = {hX} \) for some \( h \), since \( g \) and \( g\left( f\right) \) have the same constant term; \( X \) is a multiple of \( \eta \left( X\right) = f \) ; and \( {r}_{1} \) is a unit. Moreover, \( k\left( X\right) = {\eta }^{-... | Yes |
Theorem 9.7 (Hensel’s Lemma). Let \( f \in R\left\lbrack X\right\rbrack \), where \( R \) is complete relative to an ideal \( \mathfrak{a} \). If \( f\left( a\right) \in {f}^{\prime }{\left( a\right) }^{2}\mathfrak{a} \), then \( f\left( b\right) = 0 \) for some \( b \in a + {f}^{\prime }\left( a\right) \mathfrak{a} \)... | Proof. Let \( r = {f}^{\prime }\left( a\right) \). We have\n\n\[ f\left( {a + {rX}}\right) = f\left( a\right) + \left( {rX}\right) {f}^{\prime }\left( a\right) + {\left( rX\right) }^{2}h\left( X\right) = f\left( a\right) + {r}^{2}\left( {X + {X}^{2}h\left( X\right) }\right) \]\n\nfor some \( h \in R\left\lbrack X\right... | Yes |
Proposition 1.2. In a commutative ring \( R \), the product of ideals is commutative and associative, and distributes sums and unions of chains. Moreover, \( R\mathfrak{a} = \mathfrak{a} \) and \( \mathfrak{a}\mathfrak{b} \subseteq \mathfrak{a} \cap \mathfrak{b} \), for all ideals \( \mathfrak{a} \) and \( \mathfrak{b}... | By 1.2, all products of ideals \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) (in any order) are equal. The resulting product \( {\mathfrak{a}}_{1}\cdots {\mathfrak{a}}_{n} \) is written without parentheses; readers will verify that it is the ideal generated by all products \( {a}_{1}\cdots {a}_{n} \) in which \( ... | No |
Proposition 1.4. \( \operatorname{Rad}\mathfrak{a} = \left\{ {x \in R \mid {x}^{n} \in \mathfrak{a}}\right. \) for some \( \left. {n > 0}\right\} \) . | Proof. Let \( x \in R \) and let \( \mathfrak{r} \) be the intersection of all prime ideals that contain \( \mathfrak{a} \) . If \( x \in R \smallsetminus \mathfrak{r} \), then \( x \notin \mathfrak{p} \) for some prime ideal \( \mathfrak{p} \subseteq \mathfrak{a},{x}^{n} \notin \mathfrak{p} \) for all \( n > 0 \) sinc... | Yes |
Lemma 1.8. Every ideal of a Noetherian ring \( R \) is the intersection of finitely many irreducible ideals of \( R \) . | Proof. \ | No |
Theorem 1.9. In a Noetherian ring, every ideal is the intersection of finitely many primary ideals. | Proof. By 1.8 we need only show that every irreducible ideal \( \mathbf{i} \) of a Noetherian ring \( R \) is primary. Assume that \( {ab} \in \mathfrak{i} \) and \( b \notin \operatorname{Rad}\mathfrak{i} \) . Let \( {\mathfrak{a}}_{n} = \mathfrak{i} : {b}^{n} \) . Then \( a \in {\mathfrak{a}}_{1},\mathfrak{i} \subset... | Yes |
Theorem 1.10 (Noether-Lasker). In a Noetherian ring, every ideal is a reduced intersection of finitely many primary ideals, whose radicals are unique. | Proof. The associated prime ideals of an ideal \( \\mathfrak{a} \) are the prime ideals of the form \( \\mathfrak{a} : c \), where \( c \\notin \\mathfrak{a} \) . We show that in every reduced primary decomposition \( \\mathfrak{a} = {\\mathfrak{q}}_{1} \\cap \\cdots \\cap {\\mathfrak{q}}_{r} \) of an ideal \( \\mathfr... | Yes |
Proposition 2.1. Let \( E \) be a ring extension of \( R \) and let \( S \) be a subset of \( E \) . The subring \( R\left\lbrack S\right\rbrack \) of \( E \) generated by \( R \cup S \) is the set of all linear combinations with coefficients in \( R \) of products of powers of elements of \( S \) . | Proof. This is proved like IV.1.9. Let \( {\left( {X}_{s}\right) }_{s \in S} \) be a family of indeterminates, one for each \( s \in S \) . Let \( \psi : R\left\lbrack {\left( {X}_{s}\right) }_{s \in S}\right\rbrack \rightarrow E \) be the evaluation homomorphism that sends \( {X}_{s} \) to \( s \) for all \( s \in S \... | Yes |
Proposition 2.4. For an element \( \alpha \) of a ring extension \( E \) of a commutative ring \( R \) the following conditions are equivalent:\n\n(1) \( f\left( \alpha \right) = 0 \) for some monic polynomial \( f \in R\left\lbrack X\right\rbrack \) ;\n\n(2) \( R\left\lbrack \alpha \right\rbrack \) is a finitely gener... | Proof. (1) implies (2). Let \( f\left( \alpha \right) = 0 \), where \( f \) is monic; let \( n = \deg f \) . We show that \( 1,\alpha ,\ldots ,{\alpha }^{n - 1} \) generate \( R\left\lbrack \alpha \right\rbrack \) as a submodule of \( E \) . Indeed, let \( \beta \in R\left\lbrack \alpha \right\rbrack \) . By 2.2, \( \b... | Yes |
Lemma 2.5. Let \( M \) be an \( R \) -module and let \( {m}_{1},\ldots ,{m}_{n} \in M \) . If \( {x}_{ij} \in R \) for all \( i, j = 1,\ldots, n \) and \( \mathop{\sum }\limits_{{1 \leqq j \leqq n}}{x}_{ij}{m}_{j} = 0 \) for all \( i \), then \( D = \det \left( {x}_{ij}\right) \) satisfies \( D{m}_{i} = 0 \) for all \(... | Proof. If \( R \) is a field this is standard linear algebra. In general, we expand \( D \) by columns, which yields cofactors \( {c}_{ik} \) such that \( \mathop{\sum }\limits_{k}{c}_{ik}{x}_{kj} = D \) if \( i = j \) , \( \mathop{\sum }\limits_{k}{c}_{ik}{x}_{kj} = 0 \) if \( i \neq j \) ; hence\n\n\[ D{m}_{i} = \mat... | Yes |
Proposition 3.1. Let \( E \) be a ring extension of a commutative ring \( R \). (1) If \( E \) is a finitely generated \( R \)-module, then \( E \) is integral over \( R \). | Proof. (1). Every \( \alpha \in E \) satisfies condition (3) in Proposition 2.4. | No |
Proposition 3.4. If \( E \) is a ring extension of \( R \) and \( \mathfrak{A} \subseteq E \) lies over \( \mathfrak{a} \subseteq R \) , then \( R/\mathfrak{a} \) may be identified with a subring of \( E/\mathfrak{A} \) ; if \( E \) is integral over \( R \), then \( E/\mathfrak{A} \) is integral over \( R/\mathfrak{a} ... | Proof. The inclusion homomorphism \( R \rightarrow E \) induces a homomorphism \( R \rightarrow E/\mathfrak{A} \) whose kernel is \( \mathfrak{A} \cap R = \mathfrak{a} \), and an injective homomorphism \( R/\mathfrak{a} \rightarrow E/\mathfrak{A}, r + \mathfrak{a} \mapsto r + \mathfrak{A} \) . Hence \( R/\mathfrak{a} \... | Yes |
Proposition 3.5 (Lying Over). Let \( E \) be an integral extension of \( R \). For every prime ideal \( \mathfrak{p} \) of \( R \) there exists a prime ideal \( \mathfrak{P} \) of \( E \) that lies over \( \mathfrak{p} \). In fact, for every ideal \( \mathfrak{A} \) of \( E \) such that \( \mathfrak{p} \) contains \( \... | Proof. Let \( \mathfrak{A} \) be an ideal of \( E \) such that \( \mathfrak{A} \cap R \subseteq \mathfrak{p} \) (for instance,0 ). In the set of all ideals \( \mathfrak{B} \) of \( E \) such that \( \mathfrak{A} \subseteq \mathfrak{B} \) and \( \mathfrak{B} \cap R \subseteq \mathfrak{p} \), there is a maximal element \... | Yes |
Proposition 3.6. Let \( E \) be an integral extension of \( R \) and let \( \mathfrak{P},\mathfrak{Q} \subseteq E \) be prime ideals of \( E \) that lie over \( \mathfrak{p} \subseteq R \) . If \( \mathfrak{P} \subseteq \mathfrak{Q} \), then \( \mathfrak{P} = \mathfrak{Q} \) . | Proof. Let \( \alpha \in \mathfrak{Q} \) . We have \( f\left( \alpha \right) = 0 \in \mathfrak{P} \) for some monic polynomial \( f \in R\left\lbrack X\right\rbrack \) . Let \( f\left( X\right) = {X}^{n} + {r}_{n - 1}{X}^{n - 1} + \cdots + {r}_{0} \in R\left\lbrack X\right\rbrack \) be a monic polynomial of minimal deg... | Yes |
Proposition 3.7. If \( E \) is an integral extension of \( R \) and the prime ideal \( \mathfrak{P} \subseteq E \) lies over \( \mathfrak{p} \subseteq R \), then \( \mathfrak{P} \) is a maximal ideal of \( E \) if and only if \( \mathfrak{p} \) is a maximal ideal of \( R \) . | Proof. By 3.4 we may identify \( R/\mathfrak{p} \) with a subring of \( E/\mathfrak{P} \), and then \( E/\mathfrak{P} \) is integral over \( R/\mathfrak{p} \) . If \( \mathfrak{p} \) is maximal, then \( R/\mathfrak{p} \) is a field, \( E/\mathfrak{P} \) is a field by 3.3, and \( \mathfrak{P} \) is maximal. But if \( \m... | Yes |
Proposition 3.9. Every unique factorization domain is integrally closed. | Proof. Let \( R \) be a UFD and let \( a/b \in Q\left( R\right) \) . We may assume that \( a \) and \( b \) are relatively prime (no irreducible element of \( R \) divides both \( a \) and \( b) \) . If \( a/b \) is integral over \( R \), then \( f\left( {a/b}\right) = 0 \) for some monic polynomial \( f\left( X\right)... | Yes |
Proposition 3.10. Let \( R \) be a domain and let \( E \) be an algebraic extension of its quotient field. The integral closure \( \bar{R} \) of \( R \) in \( E \) is an integrally closed domain whose quotient field is \( E \) . | Proof. Every \( \alpha \in E \) is algebraic over \( Q\left( R\right) \) ; by 2.6, \( {r\alpha } \) is integral over \( R \) for some \( r \in R \) ; hence \( E = Q\left( \bar{R}\right) \) . If \( \alpha \in E \) is integral over \( \bar{R} \), then \( \alpha \) is integral over \( R \) by 3.1 and \( \alpha \in \bar{R}... | Yes |
Proposition 3.11. If \( m \in \mathbb{Z} \) is square free and not congruent to \( 1\left( {\;\operatorname{mod}\;4}\right) \) , then \( \mathbb{Z}\left\lbrack \sqrt{m}\right\rbrack \) is integrally closed; and then, for all \( x, y \in \mathbb{Q}, x + y\sqrt{m} \) is an algebraic integer in \( \mathbb{Q}\left( \sqrt{m... | Proof. We show that \( \mathbb{Z}\left\lbrack \sqrt{m}\right\rbrack \) is the integral closure of \( \mathbb{Z} \) in \( \mathbb{Q}\left\lbrack \sqrt{m}\right\rbrack \) ; hence \( \mathbb{Z}\left\lbrack \sqrt{m}\right\rbrack \) is integrally closed, by 3.10. First, \( \mathbb{Z}\left\lbrack \sqrt{m}\right\rbrack = \{ x... | Yes |
Lemma 4.1. Let \( S \) be a proper multiplicative subset of a commutative ring \( R \) . The relation\n\n\[ \left( {a, s}\right) \equiv \left( {b, t}\right) \text{if and only if at}u = {bsu}\text{for some}u \in S \]\n\nis an equivalence relation on \( R \times S \), and \( {S}^{-1}R = \left( {R \times S}\right) / \equi... | By definition, \( a/s = b/t \) if and only if \( {atu} = {bsu} \) for some \( u \in S \) . In particular, \( a/s = {at}/{st} \) for all \( t \in S;s/s = 1\left( { = 1/1}\right) \) for all \( s \in S \) ; and \( a/s = 0 \) \( \left( { = 0/1}\right) \) if and only if \( {at} = 0 \) for some \( t \in S \) . | Yes |
Proposition 4.2. Let \( S \) be a proper multiplicative subset of a commutative ring \( R \). Every homomorphism \( \varphi \) of \( R \) into a ring \( {R}^{\prime } \) in which \( \varphi \left( s\right) \) is a unit for every \( s \in S \) factors uniquely through \( \iota : R \rightarrow {S}^{-1}R \): there is a ho... | Proof. If \( a/s = b/t \), then \( {atu} = {bsu} \) for some \( u \in S,\varphi \left( a\right) \varphi \left( t\right) \varphi \left( u\right) = \)\n\( \varphi \left( b\right) \varphi \left( b\right) \varphi \left( u\right) ,\varphi \left( a\right) \varphi \left( t\right) = \varphi \left( b\right) \varphi \left( s\rig... | Yes |
Corollary 4.3. If \( R \) is a domain, then \( {S}^{-1}R \) is isomorphic to the subring \( \left\{ {a{s}^{-1} \mid a \in R, s \in S}\right\} \) of \( Q\left( R\right) \) . | Proof. Up to isomorphism, \( R \) is a subring of \( Q\left( R\right) \), and 4.2 provides a homomorphism \( \psi : {S}^{-1}R \rightarrow Q\left( R\right) \) that sends \( a/s \in {S}^{-1}R \) to \( a{s}^{-1}( = a/s \) as calculated in \( Q\left( R\right) \) ). We see that \( \psi \) is injective. | No |
Proposition 4.4. For all ideals \( \mathfrak{a},\mathfrak{b} \) of \( R \) and \( \mathfrak{A} \) of \( {S}^{-1}R \) :\n\n(1) \( {\mathfrak{a}}^{E} = {S}^{-1}R \) if and only if \( \mathfrak{a} \cap S \neq \varnothing \) ;\n\n(2) if \( \mathfrak{a} = {\mathfrak{A}}^{C} \), then \( \mathfrak{A} = {\mathfrak{a}}^{E} \) ;... | The proofs make good exercises. | No |
Proposition 4.5. Let \( S \) be a proper multiplicative subset of \( R \) . Contraction and expansion induce a one-to-one correspondence between prime ideals of \( {S}^{-1}R \) and prime ideals of \( R \) disjoint from \( S \) . | Proof. If \( \mathfrak{p} \subseteq R \smallsetminus S \) is a prime ideal of \( R \), then \( a/s \in {\mathfrak{p}}^{E} \) implies \( a/s = b/t \) for some \( b \in \mathfrak{p}, t \in S \), at \( u = {bsu} \in \mathfrak{p} \) for some \( u \in S \), and \( a \in \mathfrak{p} \), since \( \mathfrak{p} \) is prime and... | Yes |
Proposition 4.6. Let \( S \) be a proper multiplicative subset of \( R \) . Contraction and expansion induce a one-to-one correspondence, which preserves radicals, between primary ideals of \( {S}^{-1}R \) and primary ideals of \( R \) disjoint from \( S \) . | This is proved like 4.5. | No |
Proposition 4.8. If \( \mathfrak{p} \) is a prime ideal of \( R \), then \( {R}_{\mathfrak{p}} \) has only one maximal ideal, \( \mathfrak{M} = {\mathfrak{p}}^{E} = \left\{ {a/s \in {R}_{\mathfrak{p}} \mid a \in \mathfrak{p}}\right\} \) ; moreover, \( x \in {R}_{\mathfrak{p}} \) is a unit if and only if \( x \notin \ma... | Proof. If \( a/s \in \mathfrak{M} \), then \( a/s = b/t \) for some \( b \in \mathfrak{p}, t \notin \mathfrak{p} \), at \( u = {bsu} \in \mathfrak{p} \) for some \( u \notin \mathfrak{p} \), and \( a \in \mathfrak{p} \) since \( \mathfrak{p} \) is a prime ideal and \( {tu} \notin \mathfrak{p} \) . Thus \( a/s \in \math... | Yes |
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