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Theorem 57.20. Let \( K/F \) be a Galois extension of fields of degree \( n \) with the characteristic of \( F \) either zero or \( \operatorname{char}F//n \) . Suppose that \( K/F \) is cyclic with \( G\left( {K/F}\right) = \langle \sigma \rangle \) and \( {t}^{n} - 1 \) splits over \( F \) . Then there exists an elem... | Proof. Let \( \zeta \) be a primitive \( n \) th root of unity in \( F \) . By our previous work, we know that \( U = \langle \zeta \rangle \) is a cyclic group satisfying \( \left| U\right| = n \) , so \( \left| U\right| = \left\lbrack {K : F}\right\rbrack \) . As \( {\zeta }^{-1} \) lies in \( F \) and \( {\mathrm{N}... | Yes |
Let \( F \) be a field of characteristic zero and \( f \) a nonconstant polynomial in \( F\left\lbrack t\right\rbrack \) . Then \( f \) is solvable by radicals if and only if the Galois group of \( f \) is solvable. | Proof. \( \left( \Rightarrow \right) \) has already been done.\n\n\( \left( \Leftarrow \right) \) : Let \( K \) be a splitting field of \( f \) over \( F \) . By hypothesis, \( G\left( {K/F}\right) \) is solvable, so by the theorem there exists an extension \( L/K \) with \( L/F \) Galois and radical. Since \( f \) spl... | No |
Corollary 57.23. Let \( F \) be a field of characteristic zero and \( f \) a nonconstant polynomial in \( F\left\lbrack t\right\rbrack \) of degree at most four. Then \( f \) is solvable by radicals. | Proof. The Galois group of \( f \) is isomorphic to a subgroup of \( {S}_{4} \) , a solvable group. | Yes |
Lemma 59.1. Let \( G \) be a nontrivial finite solvable group. Then any minimal normal subgroup of \( G \) is an elementary p-group. | Proof. Let \( H \) be a minimal normal subgroup of \( G \) . As \( H \) is a subgroup of a solvable group, it is solvable. We also know that any characteristic subgroup of \( H \) is normal in \( G \) (cf. Exercise 11.9(18)). Since the commutator subgroup of any group is a characteristic subgroup, the series obtained b... | Yes |
Theorem 59.2. Let \( p \) be a prime and \( S \) a set with \( p \) elements. Suppose that \( G \) is a transitive subgroup of \( \sum \left( S\right) \cong {S}_{p} \) . Then the following are equivalent:\n\n(1) \( G \) is solvable.\n\n(2) Each non-identity element in \( G \) fixes at most one element in \( S \) .\n\n(... | Proof. We may assume that \( G \subset {S}_{p} \), so \( S = \{ 1,\ldots, p\} \) . As \( G \) is transitive, the orbit \( {Gs} \) for any \( s \in S \), has order \( p \), so \( p\left| \right| G \mid \) and all the Sylow \( p \) -groups of \( G \) are cyclic of order \( p \), i.e., is generated by a \( p \) -cycle.\n\... | Yes |
Theorem 59.3. (Galois) Suppose that \( F \) is a field of characteristic zero and \( f \) an irreducible polynomial in \( F\left\lbrack t\right\rbrack \) of prime degree \( p \) . Let \( K \) be a splitting field of \( f \) over \( F \) . Then \( f \) is solvable by radicals if and only if \( K = F\left( {\alpha ,\beta... | Proof. We know that \( G\left( {K/F}\right) \) can be viewed as a transitive subgroup of \( {S}_{p} \) and \( f \) is solvable by radicals if and only if \( G\left( {K/F}\right) \) is solvable. By the previous theorem, \( G\left( {K/F}\right) \) is solvable if and only if no non-identity element in \( G\left( {K/F}\rig... | Yes |
In the computation and the notation there, we let \( n = 2 \) . Then\n\n\[ f = \left( {t - {\alpha }_{1}}\right) \left( {t - {\alpha }_{2}}\right) = {t}^{2} - \left( {{\alpha }_{1} + {\alpha }_{2}}\right) t + {\alpha }_{1}{\alpha }_{2} = {t}^{2} - {s}_{1}t + {s}_{2}. \] | So \( {s}_{1} = {\alpha }_{1} + {\alpha }_{2} \) and \( {s}_{2} = {\alpha }_{1}{\alpha }_{2} \), hence \( {e}_{1} = {s}_{1} = {\alpha }_{1} + {\alpha }_{2} \) and \( {e}_{2} = \) \( {\alpha }_{1}^{2} + {\alpha }_{2}^{2} = {\left( {\alpha }_{1} + {\alpha }_{2}\right) }^{2} - 2{\alpha }_{1}{\alpha }_{2} = {s}_{1}^{2} - 2... | Yes |
Proposition 60.6. Let \( f \) be an irreducible and separable polynomial in \( F\left\lbrack t\right\rbrack \) of degree \( n, L/F \) a splitting field of \( f,\alpha = {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \) the (distinct) roots of \( f \) in \( L, K = F\left( \alpha \right) \), and \( {T}_{\alpha } : L \... | Proof. (1) and (2) are left as new exercises and (3) is Exercise \( {57.24}\left( 3\right) \) . | No |
In Proposition 60.6, set \( F = \mathbb{Q} \) and \( L = K = \mathbb{Q}\left( \zeta \right) \), where \( \zeta \) a primitive \( p \) th root of unity in \( \mathbb{C} \) with \( p \) an odd prime. The \( p \) th cyclotomic polynomial \( {\Phi }_{p} = {t}^{p - 1} + \cdots + 1 = {m}_{\mathbb{Q}}\left( \zeta \right) \). ... | We compute the right hand side of this equation. Since \( {t}^{p} - 1 = \left( {t - 1}\right) {\Phi }_{p} \), taking the derivative yields \( p{t}^{p - 1} = {\Phi }_{p} + \left( {t - 1}\right) {\Phi }_{p}^{\prime } \). Hence \[ {\Phi }_{p}^{\prime }\left( \zeta \right) = \frac{p{\zeta }^{p - 1}}{\zeta - 1} \] By Propos... | Yes |
Lemma 60.8. Let \( \zeta \) be a primitive nth root of unity in \( \mathbb{C} \) and \( K = \) \( \mathbb{Q}\left( \zeta \right) \) . Then \( \Delta \left( \zeta \right) \mid {n}^{\varphi \left( n\right) } \) . | Proof. Let \( {\Phi }_{n} = {m}_{K/F}\left( \zeta \right) \), then \( {t}^{n} - 1 = {\Phi }_{n}g \) for some \( g \in \mathbb{Z}\left\lbrack t\right\rbrack \) . Taking derivatives, we have \( n{t}^{n - 1} = {\Phi }_{n}{g}^{\prime } + {\Phi }_{n}g \) . Evaluating at \( t = \zeta \) and taking norms give\n\n\[ \n{n}^{\va... | Yes |
Lemma 61.1. Let \( F \) be a field and \( x \) an element transcendental over \( F \) . Suppose that \( u \) is an element in \( F\left( x\right) \) not lying in \( F \), say \( u = \) \( f/g \) with \( f \) and \( g \) nonzero non-constant relatively prime polynomials in \( F\left\lbrack t\right\rbrack \), and \( r = ... | Proof. Let \( h = f - {ug} \) in \( F\left\lbrack u\right\rbrack \left\lbrack t\right\rbrack \subset F\left( u\right) \left\lbrack t\right\rbrack \) with \( f \) and \( g \) as in the lemma. Suppose that \( f = \sum {a}_{i}{t}^{i} \) and \( g = \sum {b}_{i}{t}^{i} \) . If \( {b}_{j} \) is nonzero then so is \( {a}_{j} ... | Yes |
Corollary 61.2. Let \( F \) be a field, \( x \) a transcendental element over \( F \) , and \( u \) an element of \( F\left( x\right) \) . Then \( F\left( u\right) = F\left( x\right) \) if and only if there exist elements \( a, b, c, d \) in \( F \) with \( {ad} - {bc} \) nonzero and \( u = \frac{{ax} + b}{{cx} + d} \)... | [Note the condition on \( a, b, c, d \) insures that \( u \) does not lie in \( F \).] | No |
Proposition 61.3. Let \( F \) be a field and \( x \) a transcendental element over \( F \), then\n\n\( G\left( {F\left( x\right) /F}\right) = \)\n\n\[ \left\{ {f : F\left( x\right) \rightarrow F\left( x\right) \mid x \mapsto \frac{{ax} + b}{{cx} + d}}\right. \text{an}F\text{-homomorphism with} \]\n\n\[ a, b, c, d\text{... | The group \( {\mathrm{{GL}}}_{2}\left( F\right) /Z\left( {{\mathrm{{GL}}}_{2}\left( F\right) }\right) \) is called the (second) projective linear group and denoted by \( {\operatorname{PGL}}_{2}\left( F\right) \) . If \( F = \mathbb{C} \), then \( \mathbb{C} = {\mathbb{C}}^{2} \), so \( {\mathrm{{PGL}}}_{2}\left( \math... | Yes |
Proposition 62.2. The number \( {N}_{p, n} \) of monic irreducible polynomials of degree \( n \) in \( \left( {\mathbb{Z}/p\mathbb{Z}}\right) \left\lbrack t\right\rbrack \), with \( p \) a prime, satisfies | \[ {N}_{p, n} = \frac{1}{n}\mathop{\sum }\limits_{{d \mid n}}\mu \left( \frac{n}{d}\right) {p}^{d} \] | Yes |
Example 62.3. Let \( p \) be a prime not dividing 12, then\n\n\[ \n{\Phi }_{12} = \mathop{\prod }\limits_{{d \mid {12}}}{\left( {t}^{\frac{12}{d}} - 1\right) }^{\mu \left( d\right) }\n\] | \[ \n= {\left( {t}^{12} - 1\right) }^{\mu \left( 1\right) }{\left( {t}^{6} - 1\right) }^{\mu \left( 2\right) }{\left( {t}^{4} - 1\right) }^{\mu \left( 3\right) }{\left( {t}^{3} - 1\right) }^{\mu \left( 4\right) }\n\]\n\n\[ \n{\left( {t}^{2} - 1\right) }^{\mu \left( 6\right) }{\left( t - 1\right) }^{\mu \left( {12}\righ... | Yes |
Lemma 62.4. Let \( F \) be a finite field with \( q \) elements and \( i \) a positive integer. Set\n\n\[ S\left( {\widehat{t}}^{i}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{x \in F}}{x}^{i}\text{ in }F \]\n\nthen\n\n\[ S\left( {\widehat{t}}^{i}\right) = \left\{ \begin{array}{ll} - 1 & \text{ if }q - 1 \mid i ... | Proof. Suppose that \( q - 1 \mid i \), then \( {x}^{i} = 1 \) for all \( x \) in \( {F}^{ \times } \) and \( S\left( \widehat{{t}^{i}}\right) = \mathop{\sum }\limits_{{F}^{ \times }}{x}^{i} = \left| {F}^{q - 1}\right| {1}_{F} = - 1 \) in \( F \) . So we may assume that \( q - 1/i \) . Then there exists an element \( y... | Yes |
Theorem 62.5. (Chevalley-Warning Theorem) Let \( F \) be a finite field of characteristic \( p \) and \( f \) a non-constant polynomial in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) having total degree \( \deg f < n \) . Set \[ V = \left\{ {\underline{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \in {F}... | Proof. Suppose that \( F \) has \( q \) elements. Set \( P = 1 - {f}^{q - 1} \) in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . As \( \left| {F}^{ \times }\right| = q - 1 \), if \( \underline{x} \) lies in \( V \), then \( f\left( \underline{x}\right) = 0 \) in \( F \), so \( P\left( \underline{x}\right... | Yes |
Lemma 63.6. Let \( {\alpha }_{0} < \cdots < {\alpha }_{m} \) be real numbers and \( z\left( x\right) \) an \( m \) times differential real-valued function on the closed interval \( \left\lbrack {{\alpha }_{0},{\alpha }_{m}}\right\rbrack \) . Then there exist a real number \( \alpha \) satisfying\n\n\[ \frac{{z}^{\left(... | Proof. Let \( y\left( t\right) = {a}_{0} + {a}_{1}t + \cdots + {a}_{m}{t}^{m} \) in \( \mathbb{R}\left\lbrack t\right\rbrack \) be the unique polynomial satisfying \( y\left( {\alpha }_{i}\right) = z\left( {\alpha }_{i}\right) \) for \( i = 0,\ldots, m \) given by Lagrange Interpolation. By Cramer's Rule, the system of... | Yes |
Corollary 63.9. Let \( {f}_{1}\left( {x, t}\right) ,\ldots ,{f}_{m}\left( {x, t}\right) \) be irreducible polynomials in \( \mathbb{Q}\left\lbrack {x, t}\right\rbrack \) . Then there exist infinitely many rational numbers \( \alpha \) such that \( {f}_{1}\left( {\alpha, t}\right) ,\ldots ,{f}_{m}\left( {\alpha, t}\righ... | Proof. By the proof of the proposition, we can work with the union of all the \( y\left( x\right) \) ’s that arise from the \( {f}_{i}\left( {x, t}\right) \) ’s with \( N \) the number of all of these. | No |
Theorem 63.12. (Kronecker’s Criterion) Let \( F \) be a field and \( f \) an element in \( {P}_{d}\left( {n, F}\right) \) . Then \( f \) is irreducible in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) if and only if for all non-trivial factorizations (i.e., polynomials of positive degree)\n\n\[{S}_{d}\left... | Proof. \( \left( \Leftarrow \right) \) : If \( f = {f}_{1}{f}_{2} \) with \( {f}_{1},{f}_{2} \in F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \), then we have \( {f}_{1},{f}_{2} \in {P}_{d}\left( {n, f}\right) ,{S}_{d}\left( f\right) = {S}_{d}\left( {{f}_{1}{f}_{2}}\right) = {S}_{d}\left( {f}_{1}\right) {S}_{d}... | Yes |
If \( f = {t}_{1}^{2} + {t}_{2}^{2} \) in \( \mathbb{Q}\left\lbrack {{t}_{1},{t}_{2}}\right\rbrack \), we have \( {S}_{3}\left( f\right) = {t}^{2}\left( {1 + {t}^{4}}\right) \) | with \( g = {t}_{1}^{2},{S}_{3}\left( g\right) = {t}^{2} \) with \( h = 1 + {t}_{1}{t}_{2} \), and \( {gh} = \) \( {t}_{1}^{2} + {t}_{1}^{3}{t}_{2} \) | No |
Corollary 63.14. Let \( F \) be a field and \( f \in F\left\lbrack {{t}_{0},\ldots ,{t}_{n}}\right\rbrack \) irreducible with \( {\deg }_{{t}_{i}} < d \) for \( i = 1,\ldots, n \) . Let \( {S}_{d} : {P}_{d}\left( {n, F\left( {t}_{0}\right) }\right) \rightarrow {P}_{{d}^{n}}\left( {1, F\left( {t}_{0}\right) }\right) \) ... | Proof. As \( F\left\lbrack {t}_{0}\right\rbrack \) is a UFD, all factorizations of \( {S}_{d}\left( f\right) \) into two polynomials in \( F\left\lbrack {{t}_{0}, x}\right\rbrack \) arise from (*) (up to constants in \( F \) ). View \( f \) as an irreducible polynomial in \( F\left( {t}_{0}\right) \left\lbrack {{t}_{1}... | Yes |
Theorem 63.15. (Hilbert Irreducibility Theorem) Let \( f\left( {{x}_{1},\ldots ,{x}_{n}, t}\right) \) be an irreducible polynomial in \( \mathbb{Q}\left\lbrack {{x}_{1},\ldots ,{x}_{n}, t}\right\rbrack \) . Then there exist infinitely many \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) in \( {\mathbb{Q}}^{n}... | Proof. We induct on \( n \) . The case of \( n = 1 \) is Proposition 63.8 . We shall apply the previous corollary and its notation to complete the proof. If \( f \in {P}_{d}\left( {n,\mathbb{Q}\left( {x}_{1}\right) }\right) \) and \( {S}_{d} : {P}_{d}\left( {n,\mathbb{Q}\left( {x}_{1}\right) }\right) \rightarrow {P}_{1... | Yes |
Corollary 63.18. Let \( m \geq 2 \) . Then there exists an element algebraic of degree \( {2}^{m} \) over \( \mathbb{Q} \) that is not constructible. | Proof. Let \( f \) be an irreducible polynomial of degree \( n \) in \( \mathbb{Q}\left\lbrack t\right\rbrack \) with Galois group \( {S}_{n} \) with \( n = {2}^{m} \) and \( K = F\left( \alpha \right) \) a splitting field of \( f \) over \( \mathbb{Q} \) . If \( \alpha \) was constructible (from \( z = 0, z = 1 \) ), ... | Yes |
Proposition 64.4. Let \( \alpha \) be a real number. Then \( \alpha \) is a Liouville number if and only if for every positive integer \( N \) there exist integers \( p \) and \( q \), with \( q \geq 2 \) satisfying\n\n(*) \n\n\[ \n0 < \left| {\frac{p}{q} - \alpha }\right| < \frac{1}{{q}^{N}} \n\] | Proof. \( \left( \Rightarrow \right) \) is the remark above.\n\n\( \left( \Leftarrow \right) \) : Let \( c > 0 \) and \( n \geq 2 \) be given. Choose an integer \( m \) so that \( \frac{1}{{2}^{m}} < c \) and set \( N = n + m \) . By assumption, there exist integers \( p \) and \( q \), with \( q \geq 2 \) satisfying\n... | Yes |
Let \( \alpha = \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{{10}^{k!}} \), then \( \alpha \) is a Liouville number: | Let \( \mathop{\sum }\limits_{{k = 1}}^{N}\frac{1}{{10}^{k!}} = \frac{{P}_{N}}{{10}^{N!}} \), so \( {P}_{N} \) and \( {10}^{N!} \) are positive integers satisfying\n\n\[ 0 < \left| {\alpha - \frac{{P}_{N}}{{10}^{N!}}}\right| = \mathop{\sum }\limits_{{k = N + 1}}^{\infty }\frac{1}{{10}^{k!}} \leq \frac{1}{{10}^{\left( N... | Yes |
The real number \( e \) is transcendental over \( \mathbb{Q} \) . | Suppose that there exist integers \( {a}_{0},\ldots ,{a}_{m} \) not all zero satisfying\n\n\[ \n{a}_{m}{e}^{m} + {a}_{m - 1}{e}^{m - 1} + \cdots + {a}_{1}e + {a}_{0} = 0, \n\] \n\ni.e., \( e \) is a root of the nonzero polynomial \( \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{t}^{i} \) in \( \mathbb{Z}\left\lbrack t\rig... | Yes |
Lemma 65.5. Let \( h\left( x\right) = \frac{{x}^{n}g\left( x\right) }{n!} \) with \( g \) a polynomial in \( \mathbb{Z}\left\lbrack t\right\rbrack \) . Then\n\n(1) The number \( {h}^{\left( j\right) }\left( 0\right) \) is an integer for all \( j \) .\n\n(2) The integer \( n + 1 \) satisfies \( n + 1 \mid {h}^{\left( j\... | Proof. Let \( {t}^{n}g\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{c}_{j}{t}^{j} \) in \( \mathbb{Z}\left\lbrack t\right\rbrack \) (almost all \( {c}_{j} = 0 \) ). So we have \( {c}_{j} \) are integers and zero if \( j < n \) . Since \( h\left( x\right) = \frac{{x}^{n}g\left( x\right) }{n!} \), we see th... | Yes |
Theorem 66.4. (Fundamental Theorem of Symmetric Functions) Let \( R \) be a commutative ring, \( {s}_{i} = {s}_{i}\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) in \( R\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . Then \( R\left\lbrack {{s}_{1},\ldots ,{s}_{n}}\right\rbrack = R{\left\lbrack {t}_{1},\ldots ,{t}_{... | Proof. Existence: Let \( f \in R\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) be symmetric in \( {t}_{1},\ldots ,{t}_{n} \) of total degree \( k \) and leading term \( a{t}_{1}^{{i}_{1}}\ldots {t}_{n}^{{t}_{n}} \) in the lexicographical ordering. In particular, \( k = {i}_{1} + \cdots + {i}_{n} \) . As \( f \)... | Yes |
Corollary 66.6. Let \( F \) be a field and \( f \) in \( F\left\lbrack t\right\rbrack \) a polynomial of degree \( n \) with roots \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) in \( F \) . Suppose that \( P \in F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) is symmetric in \( {t}_{1},\ldots ,{t}_{n} \), then \( P... | Proof. Using Observation 66.3, we know if \( f = {a}_{n}{t}^{n} + \cdots + {a}_{0} \) in \( F\left\lbrack t\right\rbrack \), then \( \pm {s}_{i}\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = {a}_{i}/{a}_{n} \) lies in \( F \) . The result follows. | No |
Problem 67.6. (Hilbert Seventh Problem) Let \( \alpha \) and \( \beta \) be algebraic integers with \( \alpha \) not zero or one and \( \beta \) not rational, is it true that \( {\alpha }^{\beta } \) is transcendental over \( \mathbb{Q} \), e.g., \( {\sqrt{2}}^{\sqrt{2}} \) is transcendental over \( \mathbb{Q} \) ? | This was solved in the affirmative independently by Gelfand and Schneider. They showed | No |
Lemma 68.4. Let \( R \) be a commutative ring, \( P \in \mathcal{Y}\left( R\right) \). (1) If \( a, b \in R \) satisfy \( {ab} \in P \) then either \( P + {aP} \in \mathcal{Y}\left( R\right) \) or \( P - {bP} \in \mathcal{Y}\left( R\right) \). | Proof. (1): We need only show that -1 cannot lie in both \( P + {aP} \) and \( P - {bP} \). If it does then there exist \( x, y, z, w \in P \) satisfying \( - 1 = \) \( x + {ay} = z - {bw} \), so \[ \left( {ay}\right) \left( {-{bw}}\right) = \left( {-1 - x}\right) \left( {-1 - z}\right) = 1 + x + z + {xz}, \] hence \( ... | Yes |
Proposition 68.5. Let \( F \) be a formally real field and \( P \in \mathcal{Y}\left( F\right) \) . Then\n\n\[ P = \mathop{\bigcap }\limits_{{P \subset \widetilde{P} \in \mathcal{X}\left( F\right) }}\widetilde{P} \] | Proof. The inclusion \( P \subset \mathop{\bigcap }\limits_{{P \subset \widetilde{P} \in \mathcal{X}\left( F\right) }}\widetilde{P} \) is clear. Conversely, suppose that \( x \notin P \) . By definition, \( - 1 \notin P \) . We show that \( P - {xP} \in \mathcal{Y}\left( F\right) \) . To do this, it suffices to show th... | Yes |
Corollary 68.6. (Artin-Schreier) Suppose that \( F \) is formally real. Then \( \sum {F}^{2} = \mathop{\bigcap }\limits_{{\mathcal{X}\left( F\right) }}P \) . In particular, \( \mathcal{X}\left( F\right) \neq \varnothing \) . | Proof. As \( F \) is formally real, \( \sum {F}^{2} \in \mathcal{Y}\left( F\right) \) . | No |
Theorem 69.2. Let \( \left( {F, P}\right) \) be an ordered field.\n\n(1) Let \( d \in F \) and \( K = F\left( \sqrt{d}\right) \) . Then there exists an extension of\n\nP to \( K \) if and only if \( d \in P \) . | Proof. (1): Suppose \( d \in P \) . We may assume that \( d \in P \smallsetminus {F}^{2} \) . Let\n\n\[ S = \left\{ {\sum {x}_{i}{y}_{i}^{2} \mid {x}_{i} \in P,{y}_{i} \in K\text{ for all }i}\right\} .\n\]\n\nWe show that \( S \in \mathcal{Y}\left( K\right) \) . Certainly, \( S \) is closed under addition and multiplic... | Yes |
Corollary 69.4. \( \left( {F, P}\right) \) is real closed \( \left( {\operatorname{rel}P}\right) \) if and only if \( F \) is real closed. Moreover, if this is the case, then \( P = {F}^{2} \) . | Proof. Suppose that \( \left( {F, P}\right) \) is real closed. Let \( d \in P \) . By the theorem, \( P \) extends to \( F\left( \sqrt{d}\right) \) so \( d \in {F}^{2} \) by hypothesis. As \( \left( {F, P}\right) \) is real closed, we must have \( P = {F}^{2} \in \mathcal{X}\left( F\right) \) . Since \( {F}^{2} \subset... | Yes |
Proposition 69.5. Every ordered field \( \left( {F, P}\right) \) has a real closure \( \left( {\widetilde{F},{\widetilde{F}}^{2}}\right) \) . | Proof. Let \( \widehat{F} \) be an algebraic closure of \( F \) . Let \( P \in \mathcal{X}\left( F\right) \) . The statement follows from a Zorn's Lemma argument on \[ \{ \widehat{F}/K/F \mid \left( {K, Q}\right) /\left( {F, P}\right) \text{is an extension}\} \text{.} \] Because of the last results, if \( \left( {K, P}... | No |
Theorem 69.7. (Fundamental Theorem of Algebra) The following are equivalent:\n\n(1) \( F \) is real closed.\n\n(2) \( F \) is euclidean and every polynomial \( f \in F\\left\\lbrack t\\right\\rbrack \) of odd degree has a root in \( F \).\n\n(3) \( F \) is not algebraically closed but \( F\\left( \\sqrt{-1}\\right) \) ... | Proof. \( \\left( 1\\right) \\Rightarrow \\left( 2\\right) \) : We have seen the hypothesis implies that \( F \) is euclidean. Let \( p \\in F\\left\\lbrack t\\right\\rbrack \) be irreducible of odd degree. As \( K = F\\left\\lbrack t\\right\\rbrack /\\left( p\\right) \) is an extension of odd degree any ordering of \(... | Yes |
Lemma 69.9. (Sylvester) Let \( \\left( {F, P}\\right) \) be an ordered field and \( K \) a real closure of \( \\left( {F, P}\\right) \) . Suppose that \( f \) is a non-constant polynomial in \( F\\left\\lbrack t\\right\\rbrack \) and \( A = F\\left\\lbrack t\\right\\rbrack /\\left( f\\right) \) . Let \( \\varphi : A \\... | Proof. Let \( \\varphi \) be the trace form on \( A = F\\left\\lbrack t\\right\\rbrack /\\left( f\\right) \) and \( f = u{p}_{1}^{{e}_{1}}\\cdots {p}_{r}^{{e}_{r}} \) a factorization of \( f \) into monic irreducibles in \( K\\left\\lbrack t\\right\\rbrack \) with \( u \\in {K}^{ \\times } \) . By the Chinese Remainder... | Yes |
Lemma 69.10. Let \( \left( {F, P}\right) \) be an ordered field and \( K \) a real closure of \( \left( {F, P}\right) \) . Suppose that \( E = F\left( \alpha \right) \) is an algebraic extension of \( F \) with \( \varphi = {\varphi }_{E} : E \times E \rightarrow F \) the trace form. Suppose that \( r = \) \( {\operato... | Proof. Every \( F \) -embedding of \( E \) into an algebraic closure of \( K \) must take \( \alpha \) to a root of \( f \) . By Lemma 69.9, the polynomial \( f \) has \( r \) distinct roots in \( K \) . Clearly, \( {\widetilde{P}}_{i} = {\sigma }_{i}^{-1}\left( {K}^{2}\right) \supset P \) . Suppose that \( Q \in \math... | Yes |
Let \( \left( {F, P}\right) \) be an ordered field.\n\n(1) Let \( \left( {E, Q}\right) /\left( {F, P}\right) \) be an algebraic extension of ordered fields and \( K \) a real closure of \( \left( {F, P}\right) \) . Then there exists an order preserving \( F \) -homomorphism \( E \rightarrow F \) . | Proof. (1): Apply Zorn's Lemma to\n\n\( \{ M \mid E/M/F \) and there exists an order preserving\n\n\[ F\text{-homomorphism}\left. {{\psi }_{M} : \left( {M, Q \cap M}\right) \rightarrow \left( {K,{K}^{2}}\right) }\right\} \]\n\nto obtain a maximal such \( {M}_{0} \) . Suppose \( {M}_{0} \neq E \) and \( x \in E \smallse... | Yes |
Corollary 70.9. Let \( F \) be a field. If the absolute Galois group of \( F \) contains a nontrivial element of finite order, then \( F \) is formally real and the element is an involution. | Proof. By Theorem 70.2 the result follows if \( F \) is perfect, so we may assume that \( \operatorname{char}F = p > 0 \) . Let \( C \) be an algebraic closure of \( F \) and \( {F}_{sep} \) the separable closure of \( F \) in \( C \) . Suppose that \( C/K/{F}_{sep} \) is an intermediate field and \( \sigma : K \righta... | Yes |
Lemma 71.1. Let \( F \) be a field of characteristic zero and \( K/E \) be a finite field extension of degree \( n \) . Then there exists an \( E \) -basis \( \mathcal{B} \) for \( K \) such that the matrix representation of the trace form \( \varphi : K \times K \rightarrow E \) in the basis \( \mathcal{B} \) is diago... | Proof. We must produce an \( E \) -basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) for \( K \) satisfying \( \varphi \left( {{x}_{i}{x}_{j}}\right) = {\delta }_{ij} \) for all \( i, j = 1,\ldots, n \) As \( K \) is a field and \( E \) perfect, the trace form is just the non-degenerate field trace \( {\operatornam... | No |
Lemma 71.4. Let \( R \) be a commutative ring. Then the following are equivalent\n\n(1) \( R \) is semi-real.\n\n(2) There exists \( P \in \mathcal{Y}\left( R\right) \) such that \( P \cap - P \) is a prime ideal.\n\n(3) There exists a formally real prime ideal \( \mathfrak{p} \) in \( R \) .\n\n(4) There exists a ring... | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) : As \( R \) is semi-real, we have \( \sum {R}^{2} \in \mathcal{Y}\left( R\right) \) . Thus (2) follows by Lemma 68.4.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) : The prime ideal \( P \cap - P \) in \( \left( 2\right) \) is clearly semi-real and exclud... | Yes |
Corollary 71.5. Let \( F \) be a real closed field and \( R \) a domain that is an affine \( F \) -algebra. If \( R \) is semi-real the there exists an \( F \) -algebra homomorphism \( R \rightarrow F \) . | Proof. There exists a formally real prime ideal \( \mathfrak{p} \) in \( R \) . Then \( R/\mathfrak{p} \) is a real \( F \) -affine ring that is a domain so there exists an \( F \) -algebra homomorphism \( R/\mathfrak{p} \rightarrow R \) by the Lang Homomorphism Theorem 71.2. The composition \( R \rightarrow R/\mathfra... | No |
Corollary 71.6. Let \( F \) be a real closed field and \( R \) a real affine \( F \) - algebra that is a domain. Let \( {f}_{1},\ldots ,{f}_{n} \in R \smallsetminus \{ 0\} \) . Then there exists an \( F \) -algebra homomorphism \( \varphi : R \rightarrow F \) so that \( \varphi \left( {f}_{i}\right) \neq 0 \) for all \... | Proof. \( R\left\lbrack {{f}_{1}^{-1},\ldots {f}_{n}^{-1}}\right\rbrack \) is a real affine \( F \) -algebra that is a domain. | No |
Corollary 71.7. Let \( F \) be a real closed field and \( R \) an affine \( F \) -algebra. Let \( {f}_{1},\ldots ,{f}_{r},{g}_{1},\ldots ,{g}_{s} \) lie in \( R \) . Suppose that there exists a maximal preordering \( P \in \mathcal{Y}\left( R\right) \) such that \( 0 \neq {f}_{i} \in P \) for all \( i \) and \( {g}_{j}... | Proof. Let \( \mathfrak{p} = P \cap - P \), a prime ideal in \( R \) . Replacing \( R \) by \( R/\mathfrak{p} \) we may assume that \( P \cap - P = \mathfrak{p} = 0 \) . In particular, \( R \) is a domain and \( P \) is an ordering on \( R \) . Clearly, \( P \) extends to an ordering of the quotient field \( K \) of \(... | Yes |
Corollary 71.8. (Weak Real Nullstellensatz) If \( F \) is real closed and \( \mathfrak{A} \) is an ideal in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \), then \( \mathfrak{A} \) is semi-real if and only if \( {Z}_{F}\left( \mathfrak{A}\right) \neq \varnothing \) . | Proof. Let \( {A}_{F}\left( \mathfrak{A}\right) = F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . If \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {Z}_{F}\left( \mathfrak{A}\right) \), then \( f : {A}_{F}\left( \mathfrak{A}\right) \rightarrow F \) by \( {x}_{i} \rightarrow {a}_{i} \) defines an \( F \) -alge... | Yes |
Theorem 71.9. (Artin) Let \( F \) be a real closed field and \( \mathfrak{p} \) a formally real prime ideal in \( F\left\lbrack {{t}_{1},\ldots ,{t}_{n}}\right\rbrack \) . Let \( K \) be the quotient field of \( {A}_{F}\left( \mathfrak{p}\right) \) . Let \( f \in K \) . If \( f \) is positive, then \( f \) is a sum of ... | Proof. Suppose that \( f \) is not a sum of squares in \( K \) . Then there exists \( P \in \mathcal{X}\left( K\right) \) such that \( f{ < }_{P}0 \) by Corollary 68.6. Thus \( P \) extends to \( E = K\left( \sqrt{-f}\right) \) by Theorem 69.2. Let \( f = g/h \), with \( g, h \in {A}_{F}\left( \mathfrak{p}\right) \) an... | Yes |
Corollary 71.10. (Hilbert's 17th Problem) Any positive semi-definite function in \( \mathbf{R}\left( {{t}_{1},\ldots {t}_{n}}\right) \) is a sum of squares in \( \mathbf{R}\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) . | The Motzkin polynomial \( {t}_{1}^{4}{t}_{2}^{2} + {t}_{1}^{2}{t}_{2}^{4} - 3{t}_{1}^{2}{t}_{2}^{2} + 1 \) in \( \mathbf{R}\left\lbrack {{t}_{1},{t}_{2}}\right\rbrack \) is positive semi-definite but not a sum of squares in \( \mathbf{R}\left\lbrack {{t}_{1},{t}_{2}}\right\rbrack \) . (It is a sum of four squares, but ... | No |
Proposition 72.3. Let \( B/A \) be an extension of nonzero commutative rings and \( x \) an element in \( B \) . Then \( x \) is integral over \( A \) if and only if \( A\left\lbrack x\right\rbrack \) is a finitely generated \( A \) -module. | Proof. ( \( \Rightarrow \) ): Let \( x \) be a root of the monic polynomial \( f \) in \( A\left\lbrack t\right\rbrack \) . By the General Division Algorithm 31.4, we can write\n\n\[ g = {fq} + r\text{with}q, r \in A\left\lbrack t\right\rbrack \text{and}r = 0\text{or}\deg r < \deg f.\text{ }\]\n\nTherefore, if \( n = \... | Yes |
Corollary 72.5. Let \( B/A \) be an extension of nonzero commutative rings, \( {x}_{1},\ldots ,{x}_{n} \) elements of \( B \) . Then \( {x}_{1},\ldots ,{x}_{n} \) are all integral over \( A \) if and only if \( A\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is a finitely generated \( A \) -module. | Proof. \( \left( \Rightarrow \right) \) : The case for \( n = 1 \) follows from the proposition. By induction, \( A\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \) is a finitely generated \( A \) -module, say with generating set \( S \) . As \( {x}_{n} \) is integral over \( A \), it is a root of a monic poly... | Yes |
Corollary 72.7. Let \( B/A \) be an extension of nonzero commutative rings, \( x \) and \( y \) elements of \( B \) both integral over \( A \) . Then \( x \pm y \), and \( {xy} \) are integral over \( A \) . In particular, \[ \{ x \in B \mid x\text{ integral over }A\} \] is a subring of \( B \) . | Proof. If \( z \) is any element in \( A\left\lbrack {x, y}\right\rbrack \), then \( A\left\lbrack {x, y, z}\right\rbrack = A\left\lbrack {x, y}\right\rbrack \) is a finitely generated \( A \) -module, so \( z \) is integral over \( A \) . | No |
Corollary 72.9. Let \( C/B \) and \( B/A \) be extensions of nonzero commutative rings. Then \( C/B \) and \( B/A \) are integral extensions if and only if \( C/A \) is an integral extension. | Proof. We need only show if \( C/B \) and \( B/A \) are integral extensions so is \( C/A \) . Let \( c \) be an element in \( C \) . As \( c \) is integral over \( B \), it satisfies an equation \( {c}^{n} + {b}_{n - 1}{c}^{n - 1} + \cdots + {b}_{0} = 0 \) in \( B \) for some \( {b}_{0},\ldots ,{b}_{n - 1} \) in \( B \... | No |
Proposition 72.13. A UFD is integrally closed. In particular, any PID is integrally closed. | Proof. Let \( A \) be a UFD and \( a \) and \( b \) be elements in \( A \) with \( b \) nonzero. Suppose that \( a/b \) is integral over \( A \) . As \( A \) is a UFD, we may assume that \( a \) and \( b \) are relatively prime. By definition, \( a/b \) satisfies an equation\n\n\[{\left( \frac{a}{b}\right) }^{n} + {c}_... | Yes |
Lemma 73.1. Let \( K/F \) be a finite separable extension of fields of degree \( n,{K}^{ * } \mathrel{\text{:=}} {\operatorname{Hom}}_{F}\left( {K, F}\right) \), the \( F \) -linear dual space of \( K \) . Then the map \( T : K \rightarrow {K}^{ * } \) defined by \( x \mapsto {\operatorname{tr}}_{x} : y \mapsto {\opera... | Proof. By Dedekind’s Lemma 51.3, the map \( {\operatorname{Tr}}_{K/F} : K \rightarrow F \) is nontrivial, so there exists an element \( z \) in \( K \) with \( {\operatorname{Tr}}_{K/F}\left( z\right) \) nonzero. Therefore, \( {\operatorname{Tr}}_{K/F}\left( {x{x}^{-1}z}\right) \) is not zero for every nonzero \( x \) ... | Yes |
Lemma 73.3. Let \( K/F \) be a finite separable extension of fields. Then \( {\mathrm{N}}_{K/F} = {\mathrm{N}}_{E/F} \circ {\mathrm{N}}_{K/E} \) and \( {\operatorname{Tr}}_{K/F} = {\operatorname{Tr}}_{E/F} \circ {\operatorname{Tr}}_{K/E} \) for any intermediate field \( K/E/F \) . | Proof. Let \( L/F \) be a finite Galois extension with \( L/K \) . Let \( {\sigma }_{1},\ldots ,{\sigma }_{n} : E \rightarrow L \) denote all the \( F \) -homomorphisms and \( {\tau }_{1},\ldots ,{\tau }_{m} \) : \( K \rightarrow L \) all the \( E \) -homomorphisms. Extend each \( {\sigma }_{i} \) to \( {\widehat{\sigm... | Yes |
Proposition 73.4. Let \( A \) be a domain with quotient field \( F \) and \( K/F \) a finite separable extension. If \( x \) in \( K \) is integral over \( A \), then the minimal polynomial \( {m}_{F}\left( x\right) \) of \( x \) lies in \( {A}_{F}\left\lbrack t\right\rbrack \) and both \( {\operatorname{Tr}}_{K/F}\lef... | Proof. Let \( L/F \) be a finite Galois extension satisfying \( L/K \) and \( {\sigma }_{1},\ldots ,{\sigma }_{n} : K \rightarrow L \) all the distinct \( F \) -homomorphisms. If \( x \) is an element of \( K \), then any elementary symmetric function in \( {\sigma }_{1}\left( x\right) ,\ldots ,{\sigma }_{n}\left( x\ri... | Yes |
Theorem 73.5. Let \( A \) be an integrally closed Noetherian domain with quotient field \( F \) . Let \( K/F \) be a finite separable extension. Then the integral closure of \( A \) in \( K \) is a finitely generated \( A \) -module. In particular, \( {A}_{K} \) is also an integrally closed Noetherian domain. | Proof. By Theorem 37.6, we know that any finitely generated \( A \) -module is a Noetherian \( A \) -module, since \( A \) is a Noetherian ring. In particular, it suffices to show there exists a finitely generated \( A \) - module \( M \) with \( {A}_{K} \) a submodule of \( M \) . Let \( n = \left\lbrack {K : F}\right... | Yes |
Corollary 73.8. Let \( A \) be a PID with quotient field \( F \) and \( K/F \) a finite separable field extension. Then \( {A}_{K} \) is a finitely generated free A-module of rank \( \left\lbrack {K : F}\right\rbrack \) . | Proof. By the theorem, \( {A}_{K} \) is a finitely generated \( A \) -module. Since \( {A}_{K} \) is a domain, it is a torsion-free \( A \) -module, hence \( A \) -free by Corollary 41.16 to the Fundamental Theorem of Finite Generated Modules over a PID. Therefore, we need only compute the rank of \( {A}_{K} \) . We sh... | Yes |
Lemma 74.1. Let \( \varphi : A \rightarrow B \) be a ring homomorphism of commutative rings. If \( \mathfrak{P} \) is a prime ideal in \( B \), then \( {\varphi }^{-1}\left( \mathfrak{P}\right) \) is a prime ideal in A. In particular, if \( \varphi \) is the inclusion of rings, then \( \mathfrak{P} \cap A \) is a prime... | Proof. The homomorphism \( \varphi \) induces a monomorphism of rings \( \bar{\varphi } : A/{\varphi }^{-1}\left( \mathfrak{P}\right) \rightarrow B/\mathfrak{P} \) . As \( B/\mathfrak{P} \) is a domain so is \( A/{\varphi }^{-1}\left( \mathfrak{P}\right) \) and the result follows. | Yes |
Theorem 74.5. Let \( A \) be a Dedekind domain with quotient field \( F \) and \( K/F \) a finite separable field extension. Then \( {A}_{K} \) is a Dedekind domain. | Proof. \( {A}_{K} \) is integrally closed by definition and Noetherian by Theorem 73.5. So it suffices to show if \( \mathfrak{P} \) is a nonzero prime ideal in \( {A}_{K} \), then it is a maximal ideal, equivalently, \( {A}_{K}/\mathfrak{P} \) is a field. By the lemma, \( \mathfrak{P} \cap A \) is a prime ideal. If \(... | Yes |
Corollary 74.11. Let \( A \) be a Dedekind domain. Then every fractional ideal of a Dedekind domain \( A \) is invertible. Moreover, \( {I}_{A} \) is a free abelian group on basis the set of maximal ideals in \( A \) . | Proof. Let \( \mathfrak{A} \) be a fractional ideal, then there exists a nonzero element \( a \) in \( A \) satisfying \( a\mathfrak{A} \subset A \) . By the theorem, we have factorizations say \( a\mathfrak{A} = {\mathfrak{P}}_{1}^{{e}_{1}}\cdots {\mathfrak{P}}_{r}^{{e}_{r}} \) and \( \left( a\right) = {\mathfrak{p}}_... | Yes |
Corollary 74.13. Let \( A \) be a Dedekind domain. Then \( A \) is a UFD if and only if \( A \) is a PID if and only if \( C{l}_{A} \) is trivial. | Proof. We know that a domain is a UFD if and only if every nonzero prime ideal contains a prime element by Kaplansky's Theorem 28.1. As \( A \) is a Dedekind domain, this is equivalent to every maximal ideal being principal which, by Theorem 74.9, is equivalent to every ideal being principal. | Yes |
Theorem 75.4. (Kummer-Dedekind) Let \( A \) be a Dedekind domain with quotient field \( F \) and \( \mathfrak{p} \) a nonzero prime ideal in \( A \). Suppose that \( K/F \) is a finite separable field extension with \( K = F\left( \alpha \right) ,\alpha \in \) \( {A}_{K} \), and \( \mathfrak{p} \) relatively prime to t... | Proof. Step 1. \( {A}_{K}/\mathfrak{p}{A}_{K} \cong A\left\lbrack \alpha \right\rbrack /\mathfrak{p}A\left\lbrack \alpha \right\rbrack \) : Since \( \mathfrak{p}{A}_{K} \) and \( \mathfrak{f} \) are relatively prime, \( {A}_{K} = \mathfrak{p}{A}_{K} + \mathfrak{f} \) . But \( \mathfrak{f} \subset A\left\lbrack \alpha \... | Yes |
Proposition 76.1. Let \( A \) be a Dedekind domain with quotient field \( F \) and \( \mathfrak{p} \) a nonzero prime ideal in \( A \). If \( K/F \) is a finite Galois extension, then the Galois group \( G\left( {K/F}\right) \) acts transitively on the set of prime ideals in \( {A}_{K} \) lying over \( \mathfrak{p} \). | Proof. Let \( {\mathfrak{P}}_{i} \mid \mathfrak{p} \) be prime ideals in \( {A}_{K} \) for \( i = 1,2 \). Suppose that \( {\mathfrak{P}}_{2} \neq \sigma \left( {\mathfrak{P}}_{1}\right) \) for any \( \sigma \in G\left( {K/F}\right) \). By the Chinese Remainder Theorem, there exists an element \( x \) in \( {A}_{K} \) s... | Yes |
Proposition 76.2. Let \( A \) be a Dedekind domain with \( F \) its quotient field and \( \mathfrak{p} \) a nonzero prime ideal in \( A \) . If \( K/F \) is a finite Galois extension, then \( e\left( {\mathfrak{P}/\mathfrak{p}}\right) = e\left( {{\mathfrak{P}}^{\prime }/\mathfrak{p}}\right) \) and \( f\left( {\mathfrak... | Proof. By the last proposition, there exists an element \( \sigma \) in \( G\left( {K/F}\right) \) satisfying \( {\mathfrak{P}}^{\prime } = \sigma \left( \mathfrak{P}\right) \), hence we have an isomorphism \( {A}_{K}/\mathfrak{P} \rightarrow {A}_{K}/\sigma \left( \mathfrak{P}\right) \) given by \( x + \mathfrak{P} \ma... | Yes |
Proposition 76.3. Let \( A \) be a Dedekind domain with quotient field \( F \) and \( K/F \) a finite Galois extension, \( \mathfrak{P} \) a prime ideal in \( {A}_{K} \) with \( \mathfrak{P} \) lying over \( \mathfrak{p} \), then\n\n(1) \( \mathfrak{P} \) is the only prime ideal over \( \mathfrak{P} \cap {Z}_{\mathfrak... | Proof. By Proposition 76.2, we have \( \left\lbrack {K : {Z}_{\mathfrak{P}}}\right\rbrack = {e}^{\prime }{f}^{\prime }{r}^{\prime } \) where \( {e}^{\prime } \) is the ramification index of each prime ideal in \( {A}_{K} \) lying over \( \mathfrak{P} \cap {Z}_{\mathfrak{P}} \) , \( {f}^{\prime } \) its inertia index, a... | Yes |
Proposition 76.5. Let \( A \) be a Dedekind domain with \( F \) its quotient field and \( \mathfrak{p} \) a nonzero prime ideal in \( A \). If \( K/F \) is a finite Galois extension and \( \mathfrak{P} \) a prime in \( {A}_{K} \) lying over \( \mathfrak{p} \), then \( \left( {{A}_{K}/\mathfrak{P}}\right) /\left( {A/\ma... | Proof. Let \( - : {A}_{K}\left\lbrack t\right\rbrack \rightarrow \left( {{A}_{K}/\mathfrak{P}}\right) \left\lbrack t\right\rbrack \) be the natural epimorphism. Since \( f\left( {\mathfrak{P} \cap {A}_{{Z}_{\mathfrak{P}}}/\mathfrak{p}}\right) = 1 \), we have \( A/\mathfrak{p} = {A}_{{Z}_{\mathfrak{P}}}/\left( {\mathfra... | Yes |
Proposition 77.2. Let \( K/\mathbb{Q} \) be a finite field extension of degree \( n \) and \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) a subset of \( {\mathbb{Z}}_{K} \) forming a \( \mathbb{Q} \) -basis for \( K \) . Then we have \[ \mathbb{Z}\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbr... | Proof. The first inclusion is trivial. Let \( \Delta = \Delta \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) and \( \delta = \det \left( {\left( {{\sigma }_{i}\left( {\alpha }_{j}\right) }\right) \in {\mathbb{Z}}_{K}}\right. \), where \( {\sigma }_{1},\ldots ,{\sigma }_{n} : K \rightarrow \mathbb{C} \) are the ... | Yes |
Proposition 77.4. Let \( A \) be a Dedekind domain with quotient field \( F \) and \( K/F \) a finite separable field extension. Suppose that \( K = F\left( \alpha \right) \) with \( \alpha \) an element of \( {A}_{K} \) . If \( \mathfrak{p} \) is a nonzero prime ideal of \( A \) relatively prime to both \( {\mathfrak{... | Proof. Let \( - : {A}_{K}\left\lbrack t\right\rbrack \rightarrow \left( {{A}_{K}/\mathfrak{p}{A}_{K}}\right) \left\lbrack t\right\rbrack \) be the canonical epimorphism and\n\n\[ \overline{{m}_{F}\left( \alpha \right) } = {\overline{{p}_{1}}}^{{e}_{1}}\cdots {\overline{{p}_{r}}}^{{e}_{r}} \]\n\nbe a factorization in \(... | Yes |
Proposition 77.5. Let \( K \) be a number field and \( p \) a prime integer. If (p) ramifies in \( {\mathbb{Z}}_{K} \), then \( p \mid {d}_{K} \) . | Proof. Let \( L/K \) be a finite extension with \( L/\mathbb{Q} \) a finite Galois extension. Suppose that \( p/{d}_{K} \) and \( \left( p\right) \) ramifies in \( {\mathbb{Z}}_{K} \), say \( p{\mathbb{Z}}_{K} = \) \( {\mathfrak{P}}_{1}^{{e}_{1}}\cdots {\mathfrak{P}}_{r}^{{e}_{r}} \) is a factorization of \( p{\mathbb{... | No |
Proposition 77.6. Let \( p \) be an odd prime, \( \zeta \) a primitive pth root of unity in \( \mathbb{C} \), and \( K = \mathbb{Q}\left( \zeta \right) \). Then \( {\mathbb{Z}}_{K} = \mathbb{Z}\left\lbrack \zeta \right\rbrack \). In particular, \( {\Delta }_{K} = \) \( {\left( -1\right) }^{\frac{p - 1}{2}}{p}^{p - 2} \... | Proof. By Example 60.7, we know that \( \Delta \left( \zeta \right) = {\left( -1\right) }^{\frac{p - 1}{2}}{p}^{p - 2} \) and that \( \mathrm{N}\left( \zeta \right) = \mathrm{N}\left( {1 - \zeta }\right) \). Certainly, \( \mathbb{Z}\left\lbrack \zeta \right\rbrack = \mathbb{Z}\left\lbrack {1 - \zeta }\right\rbrack \). ... | Yes |
Corollary 78.5. Let \( F = \mathbb{Q}\left( \alpha \right) \) with \( \alpha \) integral over \( \mathbb{Z} \) satisfying \( g\left( \alpha \right) = 0 \) with \( g \in \mathbb{Z}\left\lbrack t\right\rbrack \) monic and \( p \) a prime in \( \mathbb{Z} \). If \( p/{\mathrm{N}}_{F/\mathbb{Q}}\left( {\left( {{g}^{\prime ... | Proof. Let \( f = {m}_{\mathbb{Q}}\left( \alpha \right) \). By Proposition 60.6, we know that \( {\Delta }_{F/\mathbb{Q}}\left( \alpha \right) = \pm {\mathrm{N}}_{F/\mathbb{Q}}\left( {{f}^{\prime }\left( \alpha \right) }\right) \). Let \( g = {fh} \) in \( \mathbb{Z}\left\lbrack t\right\rbrack \), then \( {g}^{\prime }... | Yes |
Proposition 79.1. Let \( p \) and \( l \) be odd rational primes, \( L = \mathbb{Q}\left( \zeta \right) \) with \( \zeta \) a primitive \( l \) th root of unity, and \( K = \mathbb{Q}\left( \sqrt{\left( \frac{-1}{l}\right) l}\right) \) . Then \( \left( p\right) \) splits completely in \( {\mathbb{Z}}_{K} \) if and only... | Proof. By Proposition 56.14, we have \( K \) is a subfield of \( L \) . Suppose that \( p \) splits completely in \( {\mathbb{Z}}_{K} \), say \( p{\mathbb{Z}}_{K} = {\mathfrak{P}}_{1}{\mathfrak{P}}_{2} \) is its factorization. There exists an automorphism \( \sigma \in G\left( {L/F}\right) \) satisfying \( \sigma \left... | Yes |
Corollary 80.3. Let \( A \) be a noetherian domain, not a field. Then \( A \) is a valuation ring if and only if \( A \) is a discrete valuation ring. | Proof. \( \left( \Rightarrow \right) \) : By the Properties 80.2, \( A \) is a local PID so every nonzero prime ideal is maximal and integrally closed. Since it is Noetherian, it is a local Dedekind domain, hence a discrete valuation ring. \( \left( \Leftarrow \right) \) : Is Exercise 74.14(13). A key to the proof is i... | No |
Lemma 80.5. Let \( A \) be a domain with quotient field \( F \) and \( M, N \) be \( A \) -submodules of \( F \) satisfying \( {MN} = A \) . Then \( N = {M}^{-1} \) and \( M \) is a finitely generated \( A \) -module. In particular, \( M \in {I}_{A} \) . | Proof. By definition \( N \subset {M}^{-1} \), so \( N \subset {M}^{-1}{MN} \subset {AN} = N \) . It follows that \( N = {M}^{-1} \) . If \( M \) is finitely generated then \( M \in {I}_{A} \) by Remark 2 above. In any case, since \( A = M{M}^{-1} \), we have an equation \( 1 = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i... | Yes |
Lemma 80.6. If \( A \) is a semi-local domain, then \( \operatorname{Inv}\left( A\right) = {P}_{A} \) . | Proof. Let \( {\mathfrak{m}}_{1},\ldots ,{\mathfrak{m}}_{n} \) be the maximal ideals in \( A \) and suppose that \( M \in \operatorname{Inv}\left( A\right) \), so \( M{M}^{-1} = A \) . For each \( i = 1,\ldots, n \), choose \( {a}_{i} \in M \) and \( {b}_{i} \in {M}^{-1} \) such that \( {a}_{i}{b}_{i} \notin {\mathfrak... | Yes |
Proposition 80.8. Let \( A \) be a domain with quotient field \( F \) and \( M \) an A-submodule of \( F \) . Then the following are equivalent:\n\n(1) \( M \in \operatorname{Inv}\left( A\right) \).\n\n(2) \( M \) is a finitely generated \( A \) -module and \( {M}_{\mathfrak{p}} \in \operatorname{Inv}\left( A\right) \)... | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) follows from lemmas 80.7 and 80.5, \( \left( 2\right) \Rightarrow \left( 3\right) \) is immediate, and (3) \( \Rightarrow \) (4) follows from Lemma 80.6.\n\n\( \left( 4\right) \Rightarrow \left( 1\right) \) : Suppose that \( M{M}^{-1} < A \) . Then there exist a ... | Yes |
Theorem 80.11. Let \( A \) be a domain. Then the following are equivalent:\n\n1. A is a Püfer domain.\n\n2. \( {A}_{\mathfrak{p}} \) is a valuation ring for all prime ideals \( \mathfrak{p} \) in \( A \) .\n\n3. \( {A}_{\mathfrak{m}} \) is a valuation ring for all maximal ideals \( \mathfrak{m} \) in \( A \) . | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) : Let \( \mathfrak{A} \) be a finitely generated ideal in \( {A}_{\mathfrak{p}} \) with \( \mathfrak{p} \) a prime ideal in \( A \) . Then there exist \( {a}_{i} \) in \( R \smallsetminus \mathfrak{p} \) and \( {s}_{i} \) in \( S, i = 1,\ldots, n \) , some \( n \... | No |
Lemma 80.13. (Chevalley) Let \( A \) be a subring of the commutative ring \( B \) and \( \mathfrak{A} < A \) an ideal. If \( u \) is a unit in \( B \), then \( \mathfrak{A} \) survives in either \( A\left\lbrack u\right\rbrack \) or in \( A\left\lbrack {u}^{-1}\right\rbrack \) . | Proof. Suppose this is false. Then there exist elements \( {a}_{1},\ldots ,{a}_{n} \) and \( {b}_{1},\ldots ,{b}_{m} \) in \( \mathfrak{A} \) satisfying:\n\n(i)\n\n\[ 1 = {a}_{0} + {a}_{1}u + \cdots + {a}_{n}{u}^{n} \]\n\n(ii)\n\n\[ 1 = {b}_{0} + {b}_{1}u + \cdots + {b}_{m}{u}^{-m} \]\n\nin \( B \), where we may assume... | Yes |
Theorem 80.14. Let \( A \) be a domain with quotient field \( F \) with \( K/F \) a field extension, and \( \mathfrak{A} < A \) an ideal. Then there exists a valuation ring \( B \) of \( K \) containing \( A \) with \( \mathfrak{A} \) surviving in \( B \) . | Proof. Let\n\n\[ \mathcal{S} = \left\{ {\left( {{A}_{\alpha },{\mathfrak{A}}_{\alpha }}\right) \mid A \subset {A}_{\alpha } \subset K}\right. \text{subrings,}\]\n\n\[ \left. {\mathfrak{A} \subset {\mathfrak{A}}_{\alpha } < {A}_{\alpha }\text{ with }{\mathfrak{A}}_{\alpha }\text{ an ideal in }{A}_{\alpha }}\right\}\]\n\... | Yes |
Corollary 80.16. Let \( A \) be a local domain with its quotient field \( F \) lying in a field \( K \) . Set\n\n\[ \n{\mathcal{D}}_{K} = \{ B\text{ a local ring } \mid B \subset K\}\n\]\n\npartially ordered by domination. Then\n\n(1) There exists a valuation ring \( V \) in \( {\mathcal{D}}_{K} \) dominating \( A \) .... | Proof. (1) follows from Theorem 80.14 with \( \mathfrak{A} \) the maximal ideal of \( A \) .\n\n(2): The proof of Theorem 80.14 implies the if statement. Conversely, suppose that \( A \subset B \subset F \) are valuation rings of \( F \) with maximal ideals \( {\mathfrak{m}}_{A} \) and \( {\mathfrak{m}}_{B} \), respect... | Yes |
Corollary 80.17. Let \( R \) be a Prüfer domain with quotient field \( F \) . Suppose that \( A \) is a valuation ring in \( F \) containing \( R \) . Then there exists a prime ideal \( \mathfrak{p} \) in \( R \) such that \( A = {R}_{\mathfrak{p}} \) . | Proof. Let \( \mathfrak{m} \) be the maximal ideal of \( A \) and \( \mathfrak{p} = \mathfrak{m} \cap R \), a prime ideal in \( R \) . Let \( s \in A \smallsetminus \mathfrak{p} \) . Then \( {s}^{-1} \) lies in \( A \), lest \( s \in \mathfrak{m} \cap R \) . Therefore, \( {R}_{\mathfrak{p}} \subset A \) . By Theorem 80... | Yes |
Corollary 80.18. Let \( F \) be a field and \( A \) a valuation ring satisfying \( F \subset A \subset F\left( t\right) \) . Then there exists an irreducible polynomial \( f \) in \( F\left\lbrack t\right\rbrack \) satisfying \( A = F{\left\lbrack t\right\rbrack }_{\left( f\right) } \) or \( A = F{\left\lbrack {t}^{-1}... | Proof. If \( t \) lies in \( A \), then \( F\left\lbrack t\right\rbrack \subset A \) . As \( F\left\lbrack t\right\rbrack \) is a PID, it is Prüfer, so the result follows from Corollary 80.17. So we may assume that \( t \notin A \) . As \( A \) is a valuation ring \( {t}^{-1} \) lies in \( A \) . By Corollary 80.17, \(... | Yes |
Lemma 80.19. Let \( A \) be an integrally closed local domain with quotient field \( F \) and \( a \in F \) . Suppose that there exists a polynomial \( f \) in \( A\left\lbrack t\right\rbrack \) with \( f\left( a\right) = 0 \) and at least one coefficient of \( f \) is a unit in \( A \) . Then either a or \( {a}^{-1} \... | Proof. Suppose that \( b{a}^{n} + c{a}^{n - 1} + g\left( a\right) = f\left( a\right) = 0 \) with \( b, c \in A \) and \( g \in A\left\lbrack t\right\rbrack \) satisfying \( \deg g < n - 1 \) . If \( b \) is a unit in \( A \), then we are done as \( A \) is integrally closed, so we may assume that \( b \notin {A}^{ \tim... | Yes |
Theorem 80.20. Let \( A \) be a Prüfer domain with quotient field \( F \) . Let \( L/F \) be an algebraic (possibly infinite) algebraic field extension. Then \( {A}_{L} \) is Prüfer. | Proof. Let \( \mathfrak{m} \) be a maximal ideal in \( {A}_{L} \) and \( \mathfrak{p} = A \cap \mathfrak{m} \) . We need to show that \( {B}_{\mathfrak{m}} \) is a valuation ring. Since \( A \) is a Prüfer domain, \( {A}_{\mathfrak{p}} \) is a valuation ring by Theorem 80.11. Let \( a \) be an element in \( L \) and \(... | Yes |
Theorem 80.23. Let \( F \) be an algebraically closed field and \( A \) a domain with \( R \subset A \) a subring. Suppose that \( A \) is a finitely generated \( R \) -algebra and \( a \in A \) nonzero. Then there exists an element \( 0 \neq r \in R \) satisfying the following condition: whenever \( \varphi : A \right... | Proof. By induction it suffices to assume that \( A = F\left\lbrack u\right\rbrack \) . \n\nCase 1. The element \( u \) is transcendental over \( R \) .\n\nLet \( a = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{u}^{i} \) in \( F\left\lbrack u\right\rbrack \) with \( {a}_{0} \) nonzero. We show \( r = {a}_{0} \) works. S... | Yes |
Zariski’s Lemma Let \( K/F \) be an extension of fields. Suppose that \( K \) is a finitely generated \( F \) -algebra. Then \( K/F \) is a finite field extension. | Proof. Let \( \widetilde{F} \) be an algebraic closure of \( F \) . Then apply the theorem with \( R = F, A = \widetilde{F} \), and \( a = 1 \) . By the Theorem, if \( x \) in \( \widetilde{F} \) is transcendental over \( F \), the homomorphism \( \widetilde{F} \rightarrow \widetilde{F} \) sending \( x \rightarrow 0 \)... | No |
Theorem 80.25. (Krull) Let \( A \) be a domain with quotient field \( F \) and \( \mathcal{S} = \{ B \mid A \) is a subring of \( B \) and \( B \) is a valuation ring \( \} \). The \( A \) is integrally closed if and only if \( A = \mathop{\bigcap }\limits_{\mathcal{S}}B \) . | Proof. ( \( \Leftarrow \) ): As valuation rings are integrally closed, this follows by Exercise \( {80.27}\left( 5\right) \) . \( \left( \Rightarrow \right) \) : Certainly, \( A = \mathop{\bigcap }\limits_{\mathcal{S}}B \), so assume that there exists an element \( u \) in \( \left( {\mathop{\bigcap }\limits_{\mathcal{... | No |
Theorem 80.26. Let \( A \) be an integrally closed domain. Then \( A\left\lbrack t\right\rbrack \) is integrally closed. | Proof. We have\n\n(*)\n\n\[ \left( {\mathop{\bigcap }\limits_{\substack{{A \subset B \subset F} \\ {B\text{ a valuation ring }} }}B}\right) \left\lbrack t\right\rbrack = \left( {\mathop{\bigcap }\limits_{\substack{{A \subset B \subset F} \\ {A \subset B \subset F} }}B\left\lbrack t\right\rbrack }\right) .\n\]\n\n(Why?)... | No |
Lemma 81.1. Let \( R \) be a commutative ring. Then\n\n(1) If \( T \) is a subset of \( R \), then \( {V}_{R}\left( T\right) = {V}_{R}\left( {\langle T\rangle }\right) \) . | Proof. This essentially was Exercise 38.15(5). | No |
Lemma 81.2. Let \( V \) be a subset of \( \operatorname{Spec}\left( R\right) \) and \( \bar{V} \) its closure in \( \operatorname{Spec}\left( R\right) \). If \( \mathfrak{A} = \mathop{\bigcap }\limits_{{\mathfrak{p} \in V}}\mathfrak{p} \), then \( \bar{V} = V\left( \mathfrak{A}\right) \). | Proof. (C:) As \( \mathfrak{A} = \mathop{\bigcap }\limits_{V}\mathfrak{p} \subset \mathfrak{P} \) for all \( \mathfrak{P} \in V \), we have \( V \subset V\left( \mathfrak{A}\right) \), hence \( \overline{V} \subset V\left( \mathfrak{A}\right) \).\n\n(D:) Let \( \bar{V} = V\left( \mathfrak{B}\right) \), with \( \mathfra... | Yes |
Corollary 81.3. Let \( R \) be a commutative ring. Then\n\n(1) \( \operatorname{Max}\left( R\right) \subset \operatorname{Spec}\left( R\right) \) is the set of closed points in \( \operatorname{Spec}\left( R\right) \) . | (1): Let \( \mathfrak{p} \in \operatorname{Spec}\left( R\right) \) . Then we have \( \mathfrak{p} \) is a closed point if and only if \( \overline{\mathfrak{p}} = \{ \mathfrak{q} \in \operatorname{Spec}\left( R\right) \mid \mathfrak{p} \subset \mathfrak{q}\} = \{ \mathfrak{p}\} \) if and only if \( \mathfrak{p} \in \op... | Yes |
Corollary 81.6. Let \( \mathfrak{A} \) and \( \mathfrak{B} \) be ideals in \( R \), then \( V\left( \mathfrak{A}\right) = V\left( \sqrt{\mathfrak{A}}\right) \) and \( V\left( \mathfrak{A}\right) \subset V\left( \mathfrak{B}\right) \text{if and only if}\sqrt{\mathfrak{A}} \supset \sqrt{\mathfrak{B}}\text{.} | Proof. Clearly, \( V\left( \mathfrak{A}\right) = V\left( \sqrt{\mathfrak{A}}\right) \) and \( V\left( \mathfrak{A}\right) \subset V\left( \mathfrak{B}\right) \) implies \( \sqrt{\mathfrak{B}} \subset \) \( \mathop{\bigcap }\limits_{{V\left( \mathfrak{A}\right) }}\mathfrak{p} = \sqrt{\mathfrak{A}}. \) | Yes |
Proposition 81.10. Let \( X \) be a nonempty topological space. Then the following are equivalent:\n\n(1) \( X \) is irreducible.\n\n(2) If \( {W}_{1},{W}_{2} < X \) are closed, then \( {W}_{1} \cup {W}_{2} < X \) .\n\n(3) If \( \left\{ {{W}_{1},\ldots ,{W}_{n}}\right\} \) is a finite closed cover of \( X \), then \( X... | We leave the proof as an exercise. | No |
Lemma 81.12. \( \operatorname{Spec}\left( R\right) \) is irreducible if and only if \( \operatorname{nil}\left( R\right) \) is a prime ideal if and only if \( \left| {\operatorname{Min}\left( R\right) }\right| = 1 \) . In particular, if \( R \) is a domain, then \( \operatorname{Spec}\left( R\right) \) is irreducible. | Proof. We know that \( \operatorname{nil}\left( R\right) = \mathop{\bigcap }\limits_{{\operatorname{Spec}\left( R\right) }}\mathfrak{p} = \mathop{\bigcap }\limits_{{\operatorname{Min}\left( R\right) }}\mathfrak{p} \subset \mathfrak{P} \) for every prime ideal \( \mathfrak{P} \) and \( V\left( 0\right) = V\left( \sqrt{0... | Yes |
Lemma 81.15. Let \( \varphi : A \rightarrow B \) be a ring homomorphism, then \( {}^{a}\varphi \left( \mathfrak{B}\right) \) is a prime ideal of \( A \) for every prime ideal \( \mathfrak{P} \) in \( B \), i.e., \[ {}^{a}\varphi : \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \] is d... | Proof. If \( \mathfrak{P} < B \) is a prime ideal, then the ideal \( {\varphi }^{-1}\left( \mathfrak{P}\right) < A \) is an ideal and \( \varphi \) induces a monomorphism \( A/{\varphi }^{-1}\left( \mathfrak{P}\right) \rightarrow B/\mathfrak{P} \) with \( B/\mathfrak{B} \) a domain. Consequently, \( A/{\varphi }^{-1}\l... | Yes |
Lemma 81.20. Let \( X \) be a topological space and \( Y \) a subset of \( X \). (1) \( Y \) is irreducible if and only if \( \bar{Y} \) in \( X \) is irreducible. | Proof. (1): Let \( U \subset X \) be an open subset, then \( Y \subset \bar{U} \) if and only if \( \bar{Y} \subset \bar{U} \). By definition, the closure of \( U \cap \bar{Y} \) in \( \bar{Y} \) is \( \bar{U} \cap \bar{Y} \), so \( U \cap Y \) is dense in \( Y \) if and only if \( U \cap \bar{Y} \) is dense in \( \bar... | Yes |
Proposition 81.26. Let \( X \) be a Noetherian space. Then there exist finitely many irreducible components of \( X \) and \( X \) is their union. | Proof. This proof is a typical proof when we have a Noetherian condition. Let\n\n\[ \mathcal{F}\left( X\right) \mathrel{\text{:=}} \{ V \mid V \subset X\text{ closed }\} \]\n\n\[ \mathcal{C}\left( X\right) \mathrel{\text{:=}} \{ C \mid C \subset \mathcal{F}\left( X\right) , \]\n\n\( C \) is a finite union of closed irr... | Yes |
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