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Proposition 3. Let \( L/K \) be a (not necessarily finite) Galois extension. Then the intermediate fields of \( L/K \) correspond bijectively to the closed subgroups of \( \operatorname{Gal}\left( {L/K}\right) \) . More precisely, the assertions of the fundamental theorem \( {4.1}/6 \) remain valid if we restrict ourse... | The main work for proving the proposition was already done in Section 4.1; see \( {4.1}/7 \) . It remains only to verify for intermediate fields \( E \) of \( L/K \) that the corresponding Galois group \( \operatorname{Gal}\left( {L/E}\right) \) is a closed subgroup of \( \operatorname{Gal}\left( {L/K}\right) \), and t... | Yes |
Lemma 4. Let \( H \subset \operatorname{Gal}\left( {L/K}\right) \) be a subgroup and let \( {L}^{H} \subset L \) be the corresponding fixed field. Then \( \operatorname{Gal}\left( {L/{L}^{H}}\right) \), viewed as a subgroup of \( \operatorname{Gal}\left( {L/K}\right) \), equals the closure of \( H \) in \( \operatornam... | Proof. As before, we consider the system \( {\left( {L}_{i}\right) }_{i \in I} \) of all intermediate fields of \( L/K \) such that \( {L}_{i}/K \) is a finite Galois extension, together with the restriction maps \( {f}_{i} : \operatorname{Gal}\left( {L/K}\right) \rightarrow \operatorname{Gal}\left( {{L}_{i}/K}\right) ... | Yes |
Corollary 5. Let \( L/K \) be a Galois extension and \( H \) a subgroup of \( \operatorname{Gal}\left( {L/K}\right) \) . Then the following assertions are equivalent:\n\n(i) \( H \) is open in \( \operatorname{Gal}\left( {L/K}\right) \) .\n\n(ii) \( H \) is closed in \( \operatorname{Gal}\left( {L/K}\right) \) and the ... | Proof. First assume that \( H \) is open in \( \operatorname{Gal}\left( {L/K}\right) \) . Then \( H \) is closed in \( \operatorname{Gal}\left( {L/K}\right) \) as well, since all its left (resp. right) cosets in \( \operatorname{Gal}\left( {L/K}\right) \) are open, and hence the complement of \( H \) is open in \( \ope... | Yes |
Proposition 7. The restriction maps \( {f}_{i} : \operatorname{Gal}\left( {L/K}\right) \rightarrow \operatorname{Gal}\left( {{L}_{i}/K}\right) \) define \( \operatorname{Gal}\left( {L/K}\right) \) as the projective limit of the system \( \left( {\operatorname{Gal}\left( {{L}_{i}/K}\right) ,{f}_{ij}}\right) \), i.e., \[... | Proof. It is enough to check the defining universal property of a projective limit in terms of ordinary groups, since the topology given on \( \operatorname{Gal}\left( {L/K}\right) \) coincides by its definition with the projective limit of the topologies on the groups \( \operatorname{Gal}\left( {{L}_{i}/K}\right) \) ... | Yes |
Theorem 11. Let \( \\mathbb{F} \) be a finite field and \( \\overline{\\mathbb{F}} \) an algebraic closure. Then there exists a canonical isomorphism of topological groups\n\n\[\\operatorname{Gal}\\left( {\\mathbb{F}/\\mathbb{F}}\\right) \\simeq \\mathop{\\prod }\\limits_{{\\ell \\text{ prime }}}{\\mathbb{Z}}_{\\ell }\... | In particular, we thereby see that the free cyclic subgroup \( \\mathbb{Z} \\subset \\operatorname{Gal}\\left( {\\overline{\\mathbb{F}}/\\mathbb{F}}\\right) \) that is generated by the relative Frobenius homomorphism \( \\sigma \) is significantly \ | No |
Proposition 1. Let \( f \in K\left\lbrack X\right\rbrack \) be a separable polynomial of degree \( n > 0 \) with splitting field \( L \) over \( K \), and let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in L \) be the zeros of \( f \) . Then\n\n\[ \varphi : \operatorname{Gal}\left( {L/K}\right) \rightarrow S\left( \left\{ ... | Proof. Consider an automorphism \( \sigma \in \operatorname{Gal}\left( {L/K}\right) \) . Since \( \sigma \) leaves the coefficients of \( f \) invariant, it maps zeros of \( f \) to zeros of \( f \) . Furthermore, \( \sigma \) is injective and hence induces on \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \)... | Yes |
Proposition 3. Every symmetric rational function in \( k\left( {{T}_{1},\ldots ,{T}_{n}}\right) \) can be uniquely written as a rational function over \( k \) in the elementary symmetric polynomials \( {s}_{1},\ldots ,{s}_{n} \) . In more precise terms:\n\n(i) \( k\left( {{s}_{1},\ldots ,{s}_{n}}\right) = K \) .\n\n(ii... | Proof. To justify (i), observe that\n\n\[ \left\lbrack {L : K}\right\rbrack = \operatorname{ord}{\mathfrak{S}}_{n} = n! \]\n\nand that \( k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \subset K \) . Therefore, it is enough to establish the estimate\n\n\[ \left\lbrack {L : k\left( {{s}_{1},\ldots ,{s}_{n}}\right) }\right\rbr... | Yes |
Proposition 4. The generic polynomial \( p\left( X\right) \in k\left( {{S}_{1},\ldots ,{S}_{n}}\right) \left\lbrack X\right\rbrack \) of degree \( n \) is separable and irreducible. It admits \( {\mathfrak{S}}_{n} \) as its Galois group. | Proof. We consider the rational function field \( L = k\left( {{T}_{1},\ldots ,{T}_{n}}\right) \) in \( n \) variables \( {T}_{1},\ldots ,{T}_{n} \) over \( k \), as well as the fixed field\n\n\[ K = {L}^{{\mathfrak{S}}_{n}} = k\left( {{s}_{1},\ldots ,{s}_{n}}\right) \]\n\nof all symmetric rational functions; cf. Propo... | Yes |
For every symmetric polynomial \( f \in k\left\lbrack {{T}_{1},\ldots ,{T}_{n}}\right\rbrack \), there exists a unique polynomial \( g \in k\left\lbrack {{S}_{1},\ldots ,{S}_{n}}\right\rbrack \) in \( n \) variables \( {S}_{1},\ldots ,{S}_{n} \) such that \( f = g\left( {{s}_{1},\ldots ,{s}_{n}}\right) . | The uniqueness assertion follows directly from the algebraic independence of the polynomials \( {s}_{1},\ldots ,{s}_{n} \) over \( k \), as established in Proposition 3.\n\nTo settle the existence part, consider the lexicographic order on \( {\mathbb{N}}^{n} \), where we write \( \nu < {\nu }^{\prime } \) for two tuple... | Yes |
Theorem 1 (Fundamental theorem on symmetric polynomials). As before, consider the polynomial ring \( R\left\lbrack T\right\rbrack = R\left\lbrack {{T}_{1},\ldots ,{T}_{n}}\right\rbrack \) in \( n \) variables over a ring \( R \) and let \( {s}_{1},\ldots ,{s}_{n} \) be the corresponding elementary symmetric polynomials... | Proof. We conclude by induction on \( n \) . The case \( n = 1 \) is trivial, since in this case, \( {s}_{1} = {T}_{1} \) and every polynomial in \( R\left\lbrack {T}_{1}\right\rbrack \) is symmetric. Therefore, assume \( n > 1 \), and let \( {s}_{0}^{\prime },\ldots ,{s}_{n - 1}^{\prime } \) be the elementary symmetri... | Yes |
Lemma 2. Consider the polynomial ring \( A\left\lbrack X\right\rbrack \) in a variable \( X \) over a ring \( A \) , and let \( h = {c}_{0}{X}^{n} + {c}_{1}{X}^{n - 1} + \ldots + {c}_{n} \) be a polynomial in \( A\left\lbrack X\right\rbrack \) whose leading coefficient \( {c}_{0} \) is a unit in \( A \) . Then every \(... | It remains to supply the proof of Lemma 2. To do this, we have to show that every polynomial \( f \in A\left\lbrack X\right\rbrack \) admits a representation\n\n\[ f = \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}\left( {\mathop{\sum }\limits_{{j \geq 0}}{a}_{ij}{h}^{j}}\right) {X}^{i} = \mathop{\sum }\limits_{{j \geq 0}}\... | No |
Lemma 5. Let \( f, g \in R\left\lbrack X\right\rbrack \) be polynomials as before.\n\n(i) Fix \( {X}^{i - 1},\ldots ,{X}^{0} \) as a free system of generators of \( R{\left\lbrack X\right\rbrack }_{i} \) over \( R \) for each \( i \) . Then the transpose \( S \) of the matrix \( \left( *\right) \) corresponds to the \(... | Proof. Assertion (i) is immediately clear, since the coefficient vectors of the \( \Phi \) -images of\n\n\[ \left( {{X}^{n - 1},0}\right) ,\ldots ,\left( {{X}^{0},0}\right) ,\left( {0,{X}^{m - 1}}\right) ,\ldots ,\left( {0,{X}^{0}}\right) \]\n\ncoincide with the rows of the matrix \( \left( *\right) \) and hence with t... | Yes |
Proposition 6. Let \( f, g \in R\left\lbrack X\right\rbrack \) be polynomials as before and assume that \( m + n \geq 1 \) . Then there are polynomials \( p, q \in R\left\lbrack X\right\rbrack ,\deg p < n,\deg q < m \) , such that \( \operatorname{res}\left( {f, g}\right) = {pf} + {qg} \) . | Proof. We consider the map \( \Phi \) of Lemma 5 and claim that \( \operatorname{res}\left( {f, g}\right) \), viewed as a constant polynomial in \( R\left\lbrack X\right\rbrack \), belongs to the image of \( \Phi \) . To justify this, we use Cramer's rule\n\n\[ S \cdot {S}^{ * } = \left( {\det S}\right) \cdot E \]\n\nw... | Yes |
Proposition 7. Let \( f \in R\left\lbrack X\right\rbrack \) be a monic polynomial of degree \( m \) . View the residue class ring \( A = R\left\lbrack X\right\rbrack /\left( f\right) \) as an \( R \) -module under the canonical map \( R \rightarrow R\left\lbrack X\right\rbrack /\left( f\right) \) and write \( x \) for ... | Proof. Since \( f \) is monic, we can use Euclidean division by \( f \) in \( R\left\lbrack X\right\rbrack \), which is unique; see \( {2.1}/4 \) . Therefore, the projection \( R\left\lbrack X\right\rbrack \rightarrow A \) induces an isomorphism of \( R \) -modules \( R{\left\lbrack X\right\rbrack }_{m} \rightarrow A \... | Yes |
Corollary 8. Let \( f, g \) be nontrivial polynomials with coefficients in a field \( K \) . Furthermore, let \( \deg f = m \) and \( \deg g \leq n \) . Then the following conditions are equivalent:\n\n(i) The resultant \( \operatorname{res}\left( {f, g}\right) \) of formal degree \( \left( {m, n}\right) \) is nonzero.... | Proof. Following Remark 4, we may assume \( f \) to be monic. If \( \operatorname{res}\left( {f, g}\right) \neq 0 \), we see from Proposition 7 that the determinant of the multiplication by \( g\left( x\right) \) is nonzero on \( K\left\lbrack X\right\rbrack /\left( f\right) \) and hence invertible. Then the multiplica... | Yes |
Let\n\n\\[ \n f = \alpha \mathop{\\prod }\\limits_{{i = 1}}^{m}\\left( {X - {\\alpha }_{i}}\\right) ,\\;g = \\beta \\mathop{\\prod }\\limits_{{j = 1}}^{n}\\left( {X - {\\beta }_{j}}\\right) ,\n\\]\n\nbe factorizations of two polynomials \\( f, g \\in R\\left\\lbrack X\\right\\rbrack \\) with constants \\( \\alpha ,\\be... | Proof. Making use of Remark 4, we may assume \\( R = {R}^{\\prime } \\), as well as \\( \\alpha = 1 = \\beta \\) , and hence that \\( f \\) and \\( g \\) are monic polynomials admitting a factorization into linear factors in \\( R\\left\\lbrack X\\right\\rbrack \\) . It is easily checked that \\( \\operatorname{res}\\l... | Yes |
Corollary 10. Let \( f \in R\left\lbrack X\right\rbrack \) be a monic polynomial of degree \( m > 0 \) and \( {f}^{\prime } \) its derivative. Then the discriminant \( {\Delta }_{f} \) is related to the resultant \( \operatorname{res}\left( {f,{f}^{\prime }}\right) \) of formal degree \( \left( {m, m - 1}\right) \) by\... | Proof. The second equation follows from the first one by means of Proposition 7. To derive the first equation, we may replace \( R \) by a suitable extension ring; use the definition of the discriminant in conjunction with Remark 4 (iii). Thereby we may assume that \( f \) decomposes over \( R \) into a product of line... | Yes |
Proposition 5. Let \( n \in \mathbb{N} \) . An element \( \bar{a} \) generates the additive cyclic group \( \mathbb{Z}/n\mathbb{Z} \) if and only if \( \bar{a} \) is a unit in the residue class ring \( \mathbb{Z}/n\mathbb{Z} \) . In particular, if \( n \neq 0 \), then \( \mathbb{Z}/n\mathbb{Z} \) contains precisely \( ... | Proof. Clearly, \( \mathbb{Z}/n\mathbb{Z} \) is generated by \( \bar{a} \) if and only if the residue class \( \overline{1} \) of \( 1 \in \mathbb{Z} \) is contained in the cyclic subgroup generated by \( \bar{a} \) . This is the case if and only if there is some \( r \in \mathbb{Z} \) satisfying \( \overline{1} = r \c... | Yes |
Corollary 6. Let \( K \) be a field and \( n \in \mathbb{N} - \{ 0\} \) an integer such that \( \operatorname{char}K \nmid n \) . Then the group \( {U}_{n} \) of \( n \) th roots of unity contains precisely \( \varphi \left( n\right) \) elements that are primitive. If \( \zeta \in {U}_{n} \) is such a primitive \( n \)... | Proof. Following Proposition 1, the group \( {U}_{n} \) is isomorphic to \( \mathbb{Z}/n\mathbb{Z} \) . Therefore, \( {U}_{n} \) contains precisely \( \varphi \left( n\right) \) primitive \( n \) th roots of unity, as we can read from Proposition 5. Furthermore, for a primitive \( n \) th root of unity \( \zeta \in {U}... | Yes |
Proposition 7. Let \( K \) be a field and \( {\zeta }_{n} \in \bar{K} \) a primitive \( n \) th root of unity, where \( \operatorname{char}K \nmid n \) . Then:\n\n(i) \( K\left( {\zeta }_{n}\right) /K \) is a finite abelian Galois extension of a degree dividing \( \varphi \left( n\right) \) . | Proof. We know already that \( K\left( {\zeta }_{n}\right) /K \) is a finite Galois extension. Using assertions (ii) and (iii), we may view the Galois group \( \operatorname{Gal}\left( {K\left( {\zeta }_{n}\right) /K}\right) \) as a subgroup of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) . In particular, \( \op... | Yes |
Corollary 9. Let \( {\zeta }_{m},{\zeta }_{n} \in \overline{\mathbb{Q}} \) be primitive \( m \) th and \( n \) th roots of unity, where \( \gcd \left( {m, n}\right) = 1 \) . Then\n\n\[ \mathbb{Q}\left( {\zeta }_{m}\right) \cap \mathbb{Q}\left( {\zeta }_{n}\right) = \mathbb{Q} \]\n\nand the map\n\n\[ \operatorname{Gal}\... | Proof. We know from Remark 2 that \( {\zeta }_{mn} = {\zeta }_{m}{\zeta }_{n} \) is a primitive \( {mn} \) th root of unity. Therefore, the composite field of \( \mathbb{Q}\left( {\zeta }_{m}\right) \) and \( \mathbb{Q}\left( {\zeta }_{n}\right) \) in \( \overline{\mathbb{Q}} \) is given by\n\n\[ \mathbb{Q}\left( {\zet... | Yes |
Proposition 11. (i) \( {\Phi }_{n} \) is a monic separable polynomial in \( K\left\lbrack X\right\rbrack \) satisfying \( \deg {\Phi }_{n} = \varphi \left( n\right) \) | Proof. Concerning (i), we have only to show that \( {\Phi }_{n} \), which does not have multiple zeros and hence is separable, admits coefficients in \( K \) . To justify this, observe that \( L = K\left( {\zeta }_{1}\right) = K\left( {{\zeta }_{1},\ldots ,{\zeta }_{\varphi \left( n\right) }}\right) \) is a finite Galo... | Yes |
Proposition 12. Let \( \zeta \in {\overline{\mathbb{F}}}_{q} \) be a primitive \( n \) th root of unity and assume that \( \gcd \left( {n, q}\right) = 1 \), where \( q \) is a prime power.\n\n(i) Look at the injection \( \psi : \operatorname{Gal}\left( {{\mathbb{F}}_{q}\left( \zeta \right) /{\mathbb{F}}_{q}}\right) \ho... | Proof. The relative Frobenius homomorphism over \( {\mathbb{F}}_{q} \) is given on \( {U}_{n} \) by the map \( \zeta \rightarrow {\zeta }^{q} \) and therefore, using the canonical isomorphism \( \operatorname{Aut}\left( {U}_{n}\right) \simeq {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) of Proposition 7 (iii), corre... | Yes |
Proposition 2 (E. Artin). Distinct characters \( {\chi }_{1},\ldots ,{\chi }_{n} \) on a group \( G \) with values in a field \( K \) are linearly independent in \( \operatorname{Map}\left( {G, K}\right) \) . | Proof. We proceed indirectly and assume that the assertion of the proposition is false. Then there is a minimal number \( n \in \mathbb{N} \) such that there exists a linearly dependent system of \( K \) -valued characters \( {\chi }_{1},\ldots ,{\chi }_{n} \) on \( G \) . Of course, we must have \( n \geq 2 \), since ... | Yes |
Corollary 3. Let \( L/K \) be a finite separable field extension and \( {x}_{1},\ldots ,{x}_{n} \) a basis of \( L \) as a \( K \) -vector space. Furthermore, let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) denote the \( K \) -homomorphisms of \( L \) to an algebraic closure \( \bar{K} \) of \( K \) . Then the vectors\n\... | Proof. The linear dependence of the \( {\xi }_{i} \) would imply the linear dependence of the \( {\sigma }_{i} \) . However, as we can read from Proposition 2, the \( {\sigma }_{i} \) form a linearly independent system. | Yes |
Lemma 2. Let \( L/K \) be a finite field extension of degree \( n = \left\lbrack {L : K}\right\rbrack \), and consider an element \( a \in L \) . (i) If \( a \in K \), then\n\n\[ \n{\operatorname{tr}}_{L/K}\left( a\right) = {na},\;{\mathrm{\;N}}_{L/K}\left( a\right) = {a}^{n}.\n\]\n\n(ii) If \( L = K\left( a\right) \) ... | Proof. For \( a \in K \) the linear map \( {\varphi }_{a} : L \rightarrow L \) is described by \( a \) times the unit matrix of \( {K}^{n \times n} \) . This justifies the formulas in (i). Furthermore, if \( L = K\left( a\right) \), the minimal polynomial of \( a \) coincides with the minimal polynomial of the endomorp... | Yes |
Lemma 3. Consider an element \( a \in L \) of a finite field extension \( L/K \), and let \( s = \left\lbrack {L : K\left( a\right) }\right\rbrack \) . Then\n\n\[ \n{\operatorname{tr}}_{L/K}\left( a\right) = s \cdot {\operatorname{tr}}_{K\left( a\right) /K}\left( a\right) ,\;{\mathrm{N}}_{L/K}\left( a\right) = {\left( ... | Proof. Choose a \( K \) -basis \( {x}_{1},\ldots ,{x}_{r} \) of \( K\left( a\right) \), as well as a \( K\left( a\right) \) -basis \( {y}_{1},\ldots ,{y}_{s} \) of \( L \) . Then the products \( {x}_{i}{y}_{j} \) form a \( K \) -basis of \( L \) . Let \( A \in {K}^{r \times r} \) be the matrix describing the multiplica... | Yes |
A finite field extension \( L/K \) is separable if and only if the \( K \) -linear map \( {\operatorname{tr}}_{L/K} : L \rightarrow K \) is nontrivial and hence surjective. If \( L/K \) is separable, the symmetric bilinear map\n\n\[ \operatorname{tr} : L \times L \rightarrow K,\;\left( {x, y}\right) \mapsto {\operatorn... | Proof. We assume that \( L/K \) is separable. If \( {\sigma }_{1},\ldots ,{\sigma }_{r} \) are the \( K \) -homomorphisms of \( L \) into an algebraic closure of \( K \), we get\n\n\[ {\operatorname{tr}}_{L/K} = {\sigma }_{1} + \ldots + {\sigma }_{r} \]\n\nby Proposition 4. Furthermore, Proposition 4.6/2 on the linear ... | Yes |
Corollary 8. Let \( L/K \) be a finite separable field extension with a \( K \) -basis \( {x}_{1},\ldots ,{x}_{n} \) of \( L \) . Then there exists a unique \( K \) -basis \( {y}_{1},\ldots ,{y}_{n} \) of \( L \) such that \( {\operatorname{tr}}_{L/K}\left( {{x}_{i}{y}_{j}}\right) = {\delta }_{ij} \) for \( i, j = 1,\l... | Proof. Use the existence and uniqueness of the dual basis of \( {x}_{1},\ldots ,{x}_{n} \) . | No |
Theorem 1 (Hilbert 90). Let \( L/K \) be a finite cyclic Galois extension and let \( \sigma \in \operatorname{Gal}\left( {L/K}\right) \) be a generating element. Then the following conditions are equivalent for elements \( b \in L \) :\n\n(i) \( {\mathrm{N}}_{L/K}\left( b\right) = 1 \) .\n\n(ii) There exists an element... | Proof. If \( b = a \cdot \sigma {\left( a\right) }^{-1} \) for some \( a \in {L}^{ * } \), we conclude from \( {4.7}/6 \) that\n\n\[{\mathrm{N}}_{L/K}\left( b\right) = \frac{{\mathrm{N}}_{L/K}\left( a\right) }{{\mathrm{N}}_{L/K}\left( {\sigma \left( a\right) }\right) } = 1.\]\n\nConversely, consider an element \( b \in... | Yes |
Theorem 2. If \( L/K \) is a finite Galois extension with Galois group \( G \), then \( {H}^{1}\left( {G,{L}^{ * }}\right) = \{ 1\} \), i.e., every 1-cocycle is a 1-coboundary. | Proof. Let \( f : G \rightarrow {L}^{ * } \) be a 1-cocycle. To show that it is a 1-coboundary, look at the Poincaré series\n\n\[ b = \mathop{\sum }\limits_{{{\sigma }^{\prime } \in G}}f\left( {\sigma }^{\prime }\right) \cdot {\sigma }^{\prime }\left( c\right) \]\n\nfor elements \( c \in {L}^{ * } \) . Using the linear... | Yes |
Proposition 3. Let \( L/K \) be a field extension and \( n \) an integer \( > 0 \) such that char \( K \nmid n \) . Moreover, assume that \( K \) contains a primitive \( n \) th root of unity.\n\n(i) Every cyclic Galois extension \( L/K \) of degree \( n \) is of type \( L = K\left( a\right) \) for an element \( a \in ... | Proof. Let \( \zeta \in K \) be a primitive \( n \) th root of unity. If \( L/K \) is a cyclic Galois extension of degree \( n \), then \( {\mathrm{N}}_{L/K}\left( {\zeta }^{-1}\right) = {\zeta }^{-n} = 1 \) by \( {4.7}/2 \) . Furthermore, using Hilbert’s Theorem 90, there exists an element \( a \in {L}^{ * } \) such t... | Yes |
Theorem 4 (Hilbert 90, additive version). Let \( L/K \) be a finite cyclic Galois extension and \( \sigma \in \operatorname{Gal}\left( {L/K}\right) \) a generating element. The following conditions are equivalent for elements \( b \in L \) :\n\n(i) \( {\operatorname{tr}}_{L/K}\left( b\right) = 0 \) .\n\n(ii) There exis... | Proof. We proceed similarly as in the proof of Theorem 1. If \( b = a - \sigma \left( a\right) \) for some element \( a \in L \), then\n\n\[ \n{\operatorname{tr}}_{L/K}\left( b\right) = {\operatorname{tr}}_{L/K}\left( a\right) - {\operatorname{tr}}_{L/K}\left( {\sigma \left( a\right) }\right) = 0 \n\]\n\nby \( {4.7}/6 ... | Yes |
Proposition 1. As before, consider a field \( K \) and an integer \( n > 0 \) such that char \( K \nmid n \) and \( {U}_{n} \subset {K}^{ * } \) . Furthermore, let \( C \subset {K}^{ * } \) be a subgroup containing \( {K}^{*n} \) . Then:\n\n(i) The extension \( K\left( {C}^{1/n}\right) /K \) is Galois and abelian of so... | Proof. Assertion (i) follows from the injectivity of \( {\varphi }_{1} \) in (ii). To show that \( {\varphi }_{1} \) is injective, consider an element \( \sigma \in {G}_{C} \) such that \( \sigma \left( {c}^{1/n}\right) = {c}^{1/n} \) for all \( c \in C \) . Then \( \sigma \left( a\right) = a \) for all \( a \in K\left... | Yes |
For an integer \( n \in \mathbb{N} - \{ 0\} \), consider a finite abelian group \( H \) of some exponent dividing \( n \). Then there exists a (noncanonical) isomorphism \( H \simeq \operatorname{Hom}\left( {H,\mathbb{Z}/n\mathbb{Z}}\right) . | Proof. Since \( \operatorname{Hom}\left( {\cdot ,\mathbb{Z}/n\mathbb{Z}}\right) \) is compatible with finite direct sums, we can apply the fundamental theorem of finitely generated abelian groups \( {2.9}/9 \) and thereby assume that \( H \) is cyclic of some order \( d \), where \( d \mid n \). Then we have to constru... | Yes |
Theorem 3. Let \( K \) be a field and \( n > 0 \) an integer such that \( \operatorname{char}K \nmid n \) as well as \( {U}_{n} \subset {K}^{ * } \) . The maps\n\n\[ \n\\left\\{ \\begin{matrix} \\text{ subgroups }C \\subset {K}^{ * } \\\\ \\text{ such that }{K}^{*n} \\subset C \\end{matrix}\\right\\} \\xrightarrow[{*\\... | Proof. Due to Proposition 1 and Lemma 2, it remains only to show that the maps \( \\Phi ,\\Psi \) are bijective and mutually inverse to each other. Starting with the relation \( \\Psi \\circ \\Phi = \\mathrm{{id}} \), we consider a subgroup \( C \\subset {K}^{ * } \) satisfying \( C \\supset {K}^{*n} \), where in a fir... | Yes |
Theorem 1. Let \( K \) be a field and \( G \) its absolute Galois group. Consider a continuous \( G \) -module \( A \) together with a surjective \( G \) -homomorphism \( \wp : A \rightarrow A \) , whose kernel \( {\mu }_{n} \) is a finite cyclic subgroup in \( {A}_{K} \) of order \( n \) . Assume for every cyclic Galo... | Proof. Similarly as in the proof of \( {4.9}/1 \), we start by showing that \( {\varphi }_{1} \) and \( {\varphi }_{2} \) are injective. To do this, consider an element \( \sigma \in {G}_{C} \) such that \( \langle \sigma ,\bar{c}\rangle = 0 \) for all \( c \in C \) . Then we see that \( \sigma \left( a\right) = a \) f... | Yes |
Lemma 2. The substitution endomorphism\n\n\\[ \n{\\omega }_{n} : \\mathbb{Z}\\left\\lbrack \\frac{1}{p}\\right\\rbrack \\left\\lbrack {{X}_{0},\\ldots ,{X}_{n}}\\right\\rbrack \\rightarrow \\mathbb{Z}\\left\\lbrack \\frac{1}{p}\\right\\rbrack \\left\\lbrack {{X}_{0},\\ldots ,{X}_{n}}\\right\\rbrack \n\\]\n\n\\[ \nf\\le... | Proof. Indeed, \\( {\\omega }_{n} \\) is surjective, since each of the variables \\( {X}_{0},\\ldots ,{X}_{n} \\) can be written as a polynomial in \\( {W}_{0},\\ldots ,{W}_{n} \\) . But then \\( {\\omega }_{n} \\) is injective as well for general reasons; for example, extend coefficients from \\( \\mathbb{Z}\\left\\lb... | Yes |
Lemma 3. Assume that \( p \) is invertible in \( R \) . Then the map\n\n\[ \nw : W\left( R\right) = {R}^{\mathbb{N}} \rightarrow {R}^{\mathbb{N}},\;x \mapsto {\left( {W}_{n}\left( x\right) \right) }_{n \in \mathbb{N}}, \]\n\nis bijective. | Proof. The homomorphisms \( {\omega }_{n} \) and \( \omega \) of Lemma 2 are substitution homomorphisms, and the same is true for their inverses \( {\omega }_{n}^{-1} \) and \( {\omega }^{-1} \), due to the universal property of polynomial rings dealt with in \( {2.5}/5 \), resp. \( {2.5}/1 \) . Therefore, there exist ... | Yes |
Lemma 4. Let \( R \) be a ring such that \( p = p \cdot 1 \) is not a zero divisor in \( R \) . The following conditions are equivalent for elements \( {a}_{0},\ldots ,{a}_{n},{b}_{0},\ldots ,{b}_{n} \in R \) and \( r \in \mathbb{N} - \{ 0\} \) :\n\n(i) \( {a}_{i} \equiv {b}_{i}{\;\operatorname{mod}\;\left( {p}^{r}\rig... | Proof. We proceed by induction on \( n \), the case \( n = 0 \) being trivial. Therefore, assume \( n > 0 \) . Conditions (i) and (ii) are equivalent for \( n - 1 \) in place of \( n \) by the induction hypothesis. Therefore, if one of the conditions (i) and (ii) holds, we may in either case assume that both conditions... | Yes |
For \( a \in W\left( R\right) \), let \( \left( {p \cdot a}\right) \) be the \( p \) -fold sum of \( a \) in \( W\left( R\right) \) and \( {\left( p \cdot a\right) }_{n} \) its associated component of index \( n \) . Likewise, let \( {\left( \left( V \circ F\right) \left( a\right) \right) }_{n} \) be the component of \... | Proof. Using a similar argument to that applied in the proof of Proposition 6, we may assume that \( p \) is not a zero divisor in \( R \) . Then, by Lemma 4, the stated congruences are equivalent to\n\n\[ \n{W}_{n}\left( {\left( {V \circ F}\right) \left( a\right) }\right) \equiv {W}_{n}\left( {p \cdot a}\right) \;{\;\... | Yes |
Theorem 8. Assume \( \operatorname{char}K = p > 0 \) . Then \( A = {W}_{r}\left( {K}_{s}\right) \), viewed as a G-module, together with the G-homomorphism\n\n\[ \wp : A \rightarrow A,\;a \mapsto F\left( a\right) - a, \]\n\nsatisfies the conditions of Theorem 1 for Kummer theory of exponent \( {p}^{r} \) over \( K \) . | We divide the proof into several steps.\n\nLemma 9. \( \wp : {W}_{r} | No |
Lemma 9. \( \wp : {W}_{r}\left( {K}_{s}\right) \rightarrow {W}_{r}\left( {K}_{s}\right) \) is surjective. | Proof. For \( r = 1 \), we get \( A = {W}_{1}\left( {K}_{s}\right) = {K}_{s} \), and it has to be shown that the map\n\n\[ \wp : {K}_{s} \rightarrow {K}_{s},\;\alpha \mapsto {\alpha }^{p} - \alpha ,\]\n\nis surjective. However, this is clear, since polynomials of type \( {X}^{p} - X - c \), where \( c \in {K}_{s} \), a... | Yes |
Lemma 10. The kernel of \( \wp : {W}_{r}\left( {K}_{s}\right) \rightarrow {W}_{r}\left( {K}_{s}\right) \) satisfies \( \ker \wp = {W}_{r}\left( {\mathbb{F}}_{p}\right) \) . This group is cyclic of order \( {p}^{r} \) and generated by the unit element \( e \in {W}_{r}\left( {\mathbb{F}}_{p}\right) \) . | Proof. The solutions of the equation \( {x}^{p} = x \) in \( {K}_{s} \) consist precisely of the elements of the prime subfield \( {\mathbb{F}}_{p} \subset {K}_{s} \) . Therefore, we get ker \( \wp = {W}_{r}\left( {\mathbb{F}}_{p}\right) \), due to the definition of \( \wp \) . Thus, ker \( \wp \) is a group of order \... | Yes |
Proposition 3. Let \( {K}^{\prime }/K \) be a field extension such that \( K \) is the fixed field of a subgroup \( G \subset \operatorname{Aut}\left( {K}^{\prime }\right) \) . Furthermore, consider a \( {K}^{\prime } \) -vector space \( {V}^{\prime } \), together with a \( K \) -form \( V \) and its corresponding cano... | Proof. Assertion (i) is easy to obtain. Fix a \( K \) -basis \( {\left( {v}_{i}\right) }_{i \in I} \) of \( V \) and write \( v = \mathop{\sum }\limits_{i}{a}_{i}{v}_{i} \) with coefficients \( {a}_{i} \in {K}^{\prime } \) . Since \( {f}_{\sigma }\left( v\right) = \mathop{\sum }\limits_{i}\sigma \left( {a}_{i}\right) {... | Yes |
Proposition 4. Let \( {K}^{\prime }/K \) be a field extension such that \( K \) is the fixed field of a subgroup \( G \subset \operatorname{Aut}\left( {K}^{\prime }\right) \) . Furthermore, consider a \( {K}^{\prime } \) -vector space \( {V}^{\prime } \) . For each \( \sigma \in G \), let \( {f}_{\sigma } : {V}^{\prime... | Proof. Of course, \( V \) is a \( K \) -form of \( {K}^{\prime }{ \otimes }_{K}V \) for trivial reasons. Let \( h \) be the canonical action of \( G \) on \( {K}^{\prime }{ \otimes }_{K}V \), where \( {h}_{\sigma } : {K}^{\prime }{ \otimes }_{K}V \rightarrow {K}^{\prime }{ \otimes }_{K}V \) is characterized by \( a \ot... | Yes |
Proposition 9 (Class equation). Let \( G \) be a finite group with center \( Z \) . Furthermore, consider the conjugation action on \( G \) and let \( {x}_{1},\ldots ,{x}_{n} \) be a system of representatives of the orbits contained in \( G - Z \) . Then\n\n\[ \operatorname{ord}G = \operatorname{ord}Z + \mathop{\sum }\... | Proof. The orbit of an element \( x \in Z \) consists only of the element \( x \) itself. On the other hand, we can identify the orbit of an element \( x \in G - Z \) with \( G/{Z}_{\{ x\} } \) ; cf. Remark 6. Therefore, the assertion follows from the orbit equation in Proposition 7. | Yes |
Proposition 3. Let \( G \) be a p-group of order \( {p}^{k} \), for a prime number \( p \) and an exponent \( k \geq 1 \) . Then \( p \) divides the order of the center \( Z \) of \( G \), so that \( Z \neq \{ 1\} \) . | Proof. Look at the class equation \( {5.1}/9 \) for the conjugation action of \( G \) on itself\n\n\[ \n\operatorname{ord}G = \operatorname{ord}Z + \mathop{\sum }\limits_{{i = 1}}^{n}\left( {G : {Z}_{\left\{ {x}_{i}\right\} }}\right)\n\]\n\nwhere \( {x}_{1},\ldots ,{x}_{n} \) is a system of representatives of the \( G ... | Yes |
Corollary 4. Let \( G \) be a p-group of order \( {p}^{k} \), for a prime number \( p \) . Then there is a descending chain of subgroups\n\n\[ G = {G}_{k} \supset {G}_{k - 1} \supset \ldots \supset {G}_{0} = \{ 1\} \]\n\nsuch that \( \operatorname{ord}{G}_{\ell } = {p}^{\ell } \) and \( {G}_{\ell - 1} \) is a normal su... | Proof. We conclude by induction on \( k \), the case \( k = 0 \) being trivial. Therefore, assume \( k > 0 \) . Applying Proposition 3, the center \( Z \subset G \) is nontrivial and there\n\n\( {}^{1} \) The quotients \( {G}_{\ell }/{G}_{\ell - 1} \) are of order \( p \) and hence cyclic as well as abelian. Thereby it... | Yes |
For a prime number \( p \), let \( G \) be a group of order \( {p}^{2} \) . Then \( G \) is abelian. More precisely, we have\n\n\[ G \simeq \mathbb{Z}/{p}^{2}\mathbb{Z}\;\text{ or }\;G \simeq \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}. \] | Proof. To start with we show that \( G \) is abelian. From Proposition 3 we conclude that \( p \mid \operatorname{ord}Z \) for the center \( Z \) of \( G \) and hence that \( Z \) is of order \( p \) or \( {p}^{2} \) . If ord \( Z = {p}^{2} \), then \( G = Z \) and \( G \) is abelian. On the other hand, if \( \operator... | Yes |
Lemma 7. Let \( G \) be a finite group of order \( n = {p}^{k}m \), where \( p \) is prime, but not necessarily relatively prime to \( m \) . Then the number \( s \) of p-subgroups \( H \subset G \) having order \( \operatorname{ord}H = {p}^{k} \) satisfies the relation\n\n\[ s \equiv \left( \begin{matrix} n - 1 \\ {p}... | Proof. We write \( X \) for the set of all subsets in \( G \) that consist of precisely \( {p}^{k} \) elements. Then\n\n\[ \operatorname{ord}X = \left( \begin{matrix} n \\ {p}^{k} \end{matrix}\right) \]\n\nand \( G \) acts on \( X \) by \ | No |
Lemma 9. Let \( G \) be a finite group, \( H \subset G \) a p-subgroup, and \( S \subset G \) a p-Sylow subgroup. Then there is an element \( g \in G \) such that \( H \subset {gS}{g}^{-1} \) . | Proof. On \( G/S \), the set of left cosets of \( S \) in \( G \), we consider the \( H \) -action\n\n\[ H \times G/S \rightarrow G/S,\;\left( {h,{gS}}\right) \mapsto \left( {hg}\right) S, \]\n\nand apply the theorem of Lagrange \( {1.2}/3 \), in conjunction with the orbit-stabilizer lemma \( {5.1}/6 \) as well as the ... | Yes |
Lemma 10. Let \( G \) be a finite group and \( S \) a p-Sylow subgroup in \( G \) . Writing \( {N}_{S} \) for the normalizer of \( S \) in \( G \), the index \( \left( {G : {N}_{S}}\right) \) equals the number of p-Sylow subgroups in \( G \) . | Proof. Let \( X \) be the set of \( p \) -Sylow subgroups in \( G \) . Since all \( p \) -Sylow subgroups are conjugate in \( G \), the conjugation action\n\n\[ G \times X \rightarrow X,\;\left( {g,{S}^{\prime }}\right) \mapsto g{S}^{\prime }{g}^{-1}, \]\n\nis transitive. In particular, the orbit-stabilizer lemma \( {5... | No |
Corollary 11. Let \( G \) be a finite group and \( p \) a prime number. Then:\n\n(i) If \( p \mid \operatorname{ord}G \), then \( G \) admits an element of order \( p \) . | Proof. Assertion (i) follows from Proposition 8, or alternatively, from Theorem 6 (i), in conjunction with Corollary 4. | No |
Proposition 12. Let \( p, q \) be prime numbers such that \( p < q \) and \( p \nmid \left( {q - 1}\right) \) . Then every group \( G \) of order \( {pq} \) is cyclic. | Proof. Let \( s \) be the number of \( p \) -Sylow subgroups in \( G \) . Then, by Theorem 6 (iii), we have \( s \mid \) ord \( G \), i.e., \( s \mid {pq} \), as well as \( s \equiv 1\left( p\right) \) . This implies \( p \nmid s \) and hence \( s \mid q \) . Since \( q = s \equiv 1\left( p\right) \) is excluded by the... | Yes |
Proposition 1. Let \( n \geq 2 \) .\n\n(i) If \( {\pi }_{1},{\pi }_{2} \in {\mathfrak{S}}_{n} \) are disjoint cycles, then \( {\pi }_{1} \circ {\pi }_{2} = {\pi }_{2} \circ {\pi }_{1} \) . | Proof. Assertion (i) is trivial. | No |
Proposition 3. For \( n \geq 3 \), the group \( {\mathfrak{A}}_{n} \) consists of all permutations \( \pi \in {\mathfrak{S}}_{n} \) that can be written as a product of 3-cycles. | Proof. Consider elements \( {x}_{1},{x}_{2},{x}_{3},{x}_{4} \in \{ 1,\ldots, n\} \) . If \( {x}_{1},{x}_{2},{x}_{3} \) are distinct, then the following formula holds:\n\n\[ \left( {{x}_{1},{x}_{2}}\right) \circ \left( {{x}_{2},{x}_{3}}\right) = \left( {{x}_{1},{x}_{2},{x}_{3}}\right) . \]\n\nFurthermore, if \( {x}_{1},... | Yes |
Proposition 4. A group \( G \) is solvable if and only if there is an integer \( n \in \mathbb{N} \) such that \( {D}^{n}G = \{ 1\} \) . | Proof. First, assume that \( G \) is solvable and let\n\n\[ G = {G}_{0} \supset {G}_{1} \supset \ldots \supset {G}_{n} = \{ 1\} \]\n\nbe a normal series with abelian factors. We show by induction that \( {D}^{i}G \subset {G}_{i} \) for \( i = 0,\ldots, n \) . For \( i = 0 \) the inclusion holds for trivial reasons. Now... | Yes |
Proposition 7. Let \( G \) be a finite solvable group. Then every strictly decreasing normal series in \( G \) with abelian factors can be refined to a normal series whose factors are cyclic of prime order. | Proof. Let \( G = {G}_{0} \supset \ldots \supset {G}_{n} \) be a strictly decreasing normal series with abelian factors. If one of the factors, say \( {G}_{i}/{G}_{i + 1} \), is not cyclic of prime order, choose a nontrivial element \( \bar{a} \in {G}_{i}/{G}_{i + 1} \), where replacing \( \bar{a} \) by a suitable powe... | Yes |
Proposition 8. Let \( G \) be a group and \( H \subset G \) a subgroup. If \( G \) is solvable, the same is true for \( H \) as well. If \( H \) is normal in \( G \), then \( G \) is solvable if and only if \( H \) and \( G/H \) are solvable. | Proof. First, assume that \( G \) is solvable. Then, since \( {D}^{i}H \subset {D}^{i}G \), we see that \( H \) is solvable as well. Furthermore, if \( H \) is normal in \( G \), we can consider the canonical epimorphism \( \pi : G \rightarrow G/H \) . Since \( {D}^{i}\left( {\pi \left( G\right) }\right) = \pi \left( {... | Yes |
Corollary 9. The Cartesian product \( \mathop{\prod }\limits_{{i = 1}}^{n}{G}_{i} \) of finitely many groups \( {G}_{1},\ldots ,{G}_{n} \) is solvable if and only if all groups \( {G}_{i} \) are solvable. | Proof. Conclude by induction, and for \( n = 2 \), apply Proposition 8 to the projection \( {G}_{1} \times {G}_{2} \rightarrow {G}_{2} \), which admits \( {G}_{1} \) as its kernel. | Yes |
Lemma 4. Given a chain of finite field extensions \( K \subset L \subset M \), the extension \( M/K \) is solvable (resp. solvable by radicals) if and only if \( M/L \) and \( L/K \) are solvable (resp. solvable by radicals). | Proof of Lemma 4. Again we start by considering the property \ | No |
Corollary 6. Let \( L/K \) be a separable field extension of degree \( \leq 4 \) . Then \( L/K \) is solvable, and in particular, solvable by radicals. | Proof. By the primitive element theorem \( {3.6}/{12} \), the extension \( L/K \) is simple, say \( L = K\left( a\right) \) . Let \( f \in K\left\lbrack X\right\rbrack \) be the minimal polynomial of \( a \) over \( K \) and let \( {L}^{\prime } \) be a splitting field of \( f \) over \( K \) . Then \( \deg f = \left\l... | No |
There exist finite separable field extensions that are not solvable by radicals. For example, the generic equation of degree \( n \) is not solvable by radicals for \( n \geq 5 \) . | It is enough to know that the generic equation of degree \( n \) admits the full permutation group \( {\mathfrak{S}}_{n} \) as its Galois group for \( n \geq 2 \) ; cf. Section 4.3, Example (4). Since \( {\mathfrak{S}}_{n} \) is not solvable for \( n \geq 5 \) by \( {5.4}/5 \), we see from Theorem 5 that the correspond... | Yes |
Lemma 9. In the setting of Lemma 8, let \( G \) be a solvable group and consider an element \( \sigma \in G \) . If \( \sigma \), as a bijective self-map on \( \{ 1,\ldots, p\} \), admits two different fixed points, then \( \sigma = \mathrm{{id}} \) . | Proof. Following Lemma 8, there is a normal subgroup \( H \) of order \( p \) in \( G \), and \( H \) is necessarily cyclic of order \( p \), say generated by some element \( \pi \in G \subset {\mathfrak{S}}_{p} \) . Factoring \( \pi \) into a product of disjoint cycles, see \( {5.3}/1 \) (ii), and using ord \( \pi = p... | Yes |
Proposition 10. Let \( K \) be a field and \( f \in K\left\lbrack X\right\rbrack \) an irreducible separable polynomial of prime degree \( p \) with splitting field \( L \) over \( K \) . Assume that the corresponding Galois group \( \operatorname{Gal}\left( {L/K}\right) \) is solvable. Then \( L = K\left( {\alpha ,\be... | Proof. Every element \( \sigma \in G = \operatorname{Gal}\left( {L/K}\right) \) induces a permutation of the zeros \( {\alpha }_{1},\ldots ,{\alpha }_{p} \) of \( f \), and we may view \( G \) as a subgroup of the permutation group \( {\mathfrak{S}}_{p} \) ; see \( {4.3}/1 \) . Given two zeros \( \alpha ,\beta \in L \)... | Yes |
Proposition 1 (Cardano’s formulas). Let \( K \) be a field satisfying \( \operatorname{char}K \neq 2,3 \) . For coefficients \( p, q \in K \), the solutions of the algebraic equation \( {x}^{3} + {px} + q = 0 \) are given by\n\n\[ \n{x}_{1} = u + v,\;{x}_{2} = {\zeta }^{2}u + {\zeta v},\;{x}_{3} = {\zeta u} + {\zeta }^... | Proof. If we replace in the above expressions \( u, v \) by \( {\zeta u},{\zeta }^{2}v \), resp. \( {\zeta }^{2}u,{\zeta v} \), this only produces a permutation of \( {x}_{1},{x}_{2},{x}_{3} \) . Therefore, we may assume without loss of generality that \( u = \frac{1}{3}\left( {\zeta, x}\right) \), as well as \( v = \f... | Yes |
Proposition 3. For an integer \( n \geq 3 \), the regular \( n \) -gon is constructible if and only if \( \varphi \left( n\right) \) is a power of 2, where \( \varphi \) is Euler’s \( \varphi \) -function (cf. \( {4.5}/3 \) ). | Proof. Let \( {\zeta }_{n} \) be a primitive \( n \) th root of unity over \( \mathbb{Q} \) . Then we know from \( {4.5}/8 \) that \( \mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q} \) is an abelian Galois extension of degree \( \varphi \left( n\right) \) . Assuming that the regular \( n \) -gon is constructible, i.e.... | Yes |
Proposition 5. The following conditions are equivalent for \( n \geq 2 \) :\n\n(i) \( \varphi \left( n\right) \) is a power of 2 .\n\n(ii) There exist distinct Fermat primes \( {p}_{1},\ldots ,{p}_{r} \) and an integer \( m \in \mathbb{N} \) such that \( n = {2}^{m}{p}_{1}\ldots {p}_{r} \) . | Proof. Given a prime \( p \), the expression \( \varphi \left( {p}^{m}\right) = \left( {p - 1}\right) {p}^{m - 1} \) is a power of 2 if and only if \( p = 2 \) or if \( {p}^{m - 1} = 1 \), i.e., \( m = 1 \), and \( p - 1 \) is a power of 2 . Therefore, using the multiplicativity of the \( \varphi \) -function, the asse... | No |
Lemma 6. A prime number \( p \geq 3 \) is a Fermat prime if and only if \( p - 1 \) is a power of 2 . | Proof. From the definition of Fermat numbers \( p \) we see that \( p - 1 \) is a power of 2 . Conversely, assume for a prime \( p \) that \( p - 1 \) is a power of 2, say \( p = {\left( {2}^{{2}^{\ell }}\right) }^{r} + 1 \) for an odd exponent \( r \) . Then, if \( r > 1 \), we can factorize \( p \) according to the f... | Yes |
Proposition 3. Let \( L/K \) be a field extension. A system \( \mathfrak{X} \) of elements in \( L \) is a transcendence basis of \( L/K \) if and only if \( \mathfrak{X} \) is a maximal system in \( L \) that is algebraically independent over \( K \) . In particular, every field extension \( L/K \) admits a transcende... | Proof. First assume that \( \mathfrak{X} \) is a maximal algebraically independent system of \( L/K \) . Then, by the maximality of \( \mathfrak{X} \), every element of \( L \) is algebraic over \( K\left( \mathfrak{X}\right) \) , so that \( \mathfrak{X} \) is a transcendence basis of \( L/K \) . Indeed, for \( x \in L... | Yes |
Lemma 4. Consider a field extension \( L/K \) and a system \( \mathfrak{Y} \) of elements in \( L \) such that \( L \) is algebraic over \( K\left( \mathfrak{Y}\right) \) . Furthermore, let \( {\mathfrak{X}}^{\prime } \subset L \) be a system that is algebraically independent over \( K \) . Then \( {\mathfrak{X}}^{\pri... | Proof. Using Zorn’s lemma \( {3.4}/5 \), we choose a maximal subsystem \( {\mathfrak{X}}^{\prime \prime } \subset \mathfrak{Y} \) such that the composite system \( \mathfrak{X} = {\mathfrak{X}}^{\prime } \cup {\mathfrak{X}}^{\prime \prime } \) is algebraically independent over \( K \) . Similarly as in the proof of Pro... | Yes |
Lemma 6 (Schröder-Bernstein theorem). Assume for two sets \( M \) and \( N \) that there are injections \( \sigma : M \hookrightarrow N \) and \( \tau : N \hookrightarrow M \) . Then there is a bijection \( \rho : M \rightarrow N \) . | Proof. We denote by \( {M}^{\prime } \subset M \) the set of all elements \( x \in M \) satisfying for every \( n \in \mathbb{N} \) the implication\n\n\[ x \in {\left( \tau \circ \sigma \right) }^{n}\left( M\right) \; \Rightarrow \;x \in {\left( \tau \circ \sigma \right) }^{n} \circ \tau \left( N\right) . \]\n\nIn othe... | Yes |
Lemma 7. Every infinite set \( M \) is a disjoint union of sets that are countably infinite. | Proof. Consider the set \( X \) of all pairs \( \left( {A, Z}\right) \), where \( A \) is an infinite subset of \( M \) , and where \( Z \) is a disjoint decomposition of \( A \) into countably infinite subsets. In other words, \( Z \) is a system of countably infinite disjoint subsets of \( A \) whose union equals \( ... | Yes |
Corollary 8. Two purely transcendental field extensions \( L/K \) and \( {L}^{\prime }/K \) admit a \( K \) -isomorphism \( L \rightarrow {L}^{\prime } \) if and only if \( L \) and \( {L}^{\prime } \) are of the same transcendence degree over \( K \) . | In particular, this implies that there cannot exist a \( K \) -isomorphism between polynomial rings \( K\left\lbrack {{X}_{1},\ldots ,{X}_{m}}\right\rbrack \) and \( K\left\lbrack {{Y}_{1},\ldots ,{Y}_{n}}\right\rbrack \) having different numbers of variables \( m \) and \( n \) . Otherwise, the corresponding fields of... | Yes |
Corollary 9. Let \( \varphi : K\left\lbrack {{X}_{1},\ldots ,{X}_{m}}\right\rbrack \rightarrow K\left\lbrack {{Y}_{1},\ldots ,{Y}_{n}}\right\rbrack \) be a \( K \) -homomorphism between polynomial rings such that every variable \( Y \in \left\{ {{Y}_{1},\ldots ,{Y}_{n}}\right\} \) satisfies an equation of type\n\n\[ \n... | Proof. Let \( R \) be the image of \( \varphi \) . Being a subring of \( K\left\lbrack {{Y}_{1},\ldots ,{Y}_{n}}\right\rbrack \), it is an integral domain, and we can view \( K\left( {{Y}_{1},\ldots ,{Y}_{n}}\right) \) as an extension field of the field of fractions \( Q\left( R\right) \) . Since \( K\left( {{Y}_{1},\l... | Yes |
Proposition 2. The tensor product \( T = M{ \otimes }_{R}N \) exists for arbitrary \( R \) -modules \( M \) and \( N \) . | Proof. The basic construction idea is quite simple. We start with \( {R}^{\left( M \times N\right) } \), the free \( R \) -module generated by all pairs \( \left( {x, y}\right) \in M \times N \), and divide out the smallest submodule \( Q \) such that the residue classes of elements of type \( \left( {x, y}\right) \) a... | Yes |
Proposition 8. Let \( S \subset R \) be a multiplicative system.\n\n(i) The canonical map \( R \rightarrow {R}_{S} \) is flat, i.e., \( {R}_{S} \) is a flat \( R \) -module under this map.\n\n(ii) For every \( R \) -module \( M \), there is a canonical isomorphism of \( R \) -modules, resp. \( {R}_{S} \) -modules,\n\n\... | Proof. We start with assertion (ii). The map\n\n\[ M \times {R}_{S} \rightarrow {M}_{S},\;\left( {x,\frac{a}{s}}\right) \mapsto \frac{ax}{s}, \]\n\nis well defined, as is easily checked, and \( R \) -bilinear. Hence, it gives rise to a unique \( R \) -linear map \( \varphi : M{ \otimes }_{R}{R}_{S} \rightarrow {M}_{S} ... | Yes |
Lemma 9. The above maps \( {\sigma }^{\prime } : {R}^{\prime } \rightarrow {R}^{\prime }{ \otimes }_{R}{R}^{\prime \prime },\;{\sigma }^{\prime \prime } : {R}^{\prime \prime } \rightarrow {R}^{\prime }{ \otimes }_{R}{R}^{\prime \prime } \) admit the following universal property: Given two R-algebra homomorphisms \( {\v... | Proof. To show that \( \varphi \) is unique, look at a tensor \( {a}^{\prime } \otimes {a}^{\prime \prime } \in {R}^{\prime }{ \otimes }_{R}{R}^{\prime \prime } \) . Then we have \[ \varphi \left( {{a}^{\prime } \otimes {a}^{\prime \prime }}\right) = \varphi \left( {\left( {{a}^{\prime } \otimes 1}\right) \cdot \left( ... | Yes |
Proposition 10. Let \( {R}^{\prime } \) be an \( R \) -algebra and \( \mathfrak{X} \) a system of variables, as well as \( \mathfrak{a} \subset R\left\lbrack \mathfrak{X}\right\rbrack \) an ideal. Then there are canonical isomorphisms\n\n\[ R\left\lbrack \mathfrak{X}\right\rbrack { \otimes }_{R}{R}^{\prime } \rightarro... | Proof. The canonical \( R \) -algebra homomorphisms \( {\varphi }^{\prime } : R\left\lbrack \mathfrak{X}\right\rbrack \rightarrow {R}^{\prime }\left\lbrack \mathfrak{X}\right\rbrack \) and \( {\varphi }^{\prime \prime } : {R}^{\prime } \rightarrow {R}^{\prime }\left\lbrack \mathfrak{X}\right\rbrack \) give rise, due to... | Yes |
Lemma 13. Let \( A \) and \( {A}^{\prime } \) be algebras over a field \( K \) and observe for subalgebras \( {A}_{0} \subset A \) and \( {A}_{0}^{\prime } \subset {A}^{\prime } \) that \( {A}_{0}{ \otimes }_{K}{A}_{0}^{\prime } \) is canonically a subalgebra in \( A{ \otimes }_{K}{A}^{\prime } \) . Furthermore, let \(... | Proof. By the flatness of \( K \) -algebras, the inclusions \( {A}_{0} \hookrightarrow A \) and \( {A}_{0}^{\prime } \hookrightarrow {A}^{\prime } \) give rise to injections\n\n\[ {A}_{0}{ \otimes }_{K}{A}_{0}^{\prime } \hookrightarrow {A}_{0}{ \otimes }_{K}{A}^{\prime } \hookrightarrow A{ \otimes }_{K}{A}^{\prime } \]... | Yes |
Proposition 4. Let \( M/K \) be a field extension.\n\n(i) If \( M/K \) is separable and \( L \) is an intermediate field of \( M/K \), then the extension \( L/K \) is separable as well. \( {}^{2} \)\n\n(ii) \( M/K \) is separable if and only if \( L/K \) is separable for every intermediate field \( L \) of \( M/K \) th... | Proof. Consider an intermediate field \( L \) of \( M/K \) and assume that \( M/K \) is separable. Then, for any field extension \( {K}^{\prime }/K \), the inclusion \( L \hookrightarrow M \) gives rise to an inclusion \( L{ \otimes }_{K}{K}^{\prime } \hookrightarrow M{ \otimes }_{K}{K}^{\prime } \), since \( {K}^{\pri... | Yes |
Proposition 7. For a field \( K \) of characteristic \( p > 0 \) and an extension field \( L \) of \( K \), the following conditions are equivalent:\n\n(i) \( L/K \) is separable.\n\n(ii) \( L{ \otimes }_{K}{K}^{{p}^{-\infty }} \) is reduced.\n\n(iii) For every finite extension \( {K}^{\prime }/K \) such that \( {K}^{\... | Proof. The implication (i) \( \Rightarrow \) (ii) is trivial. Furthermore, the implication (ii) \( \Rightarrow \) (iii) follows from the flatness of \( L/K \), since every \( {K}^{\prime } \) in the situation of (iii) is a subfield of \( {K}^{{p}^{-\infty }} \), so that \( L{ \otimes }_{K}{K}^{\prime } \) is a subring ... | No |
Proposition 13. The following conditions are equivalent for a field extension \( L/K \) :\n\n(i) \( L/K \) is primary.\n\n(ii) \( L{ \otimes }_{K}{K}^{\prime } \) is irreducible for every finite separable extension \( {K}^{\prime }/K \) .\n\n(iii) \( K \) is separably closed in \( L \), i.e., every element \( a \in L \... | Proof. The implication (i) \( \Rightarrow \) (ii) is trivial. Therefore, assume condition (ii) and let \( a \in L \) be separable algebraic over \( K \), with corresponding minimal polynomial \( f \in K\left\lbrack X\right\rbrack \) of \( a \) over \( K \) . This polynomial factorizes over \( L \) into a product of irr... | Yes |
Proposition 14. The following conditions are equivalent for a field extension \( L/K \) :\n\n(i) \( L/K \) is regular.\n\n(ii) \( L{ \otimes }_{K}{K}^{\prime } \) is an integral domain for every finite extension \( {K}^{\prime }/K \) .\n\n(iii) \( L/K \) is separable and \( K \) is algebraically closed in \( L \) . | Proof. A ring \( R \) is an integral domain if and only if the zero ideal \( 0 \subset R \) is prime. This is equivalent to the fact that the radical \( \operatorname{rad}R \) is both prime and zero. This justifies the equivalence of (i) and (ii) if we use Propositions 7 and 13.\n\nTo derive the equivalence of (i) and ... | Yes |
Proposition 2. Let \( A \) be an \( R \) -algebra. Then there exists an \( A \) -module \( {\Omega }_{A/R}^{1} \) together with an \( R \) -derivation \( {d}_{A/R} : A \rightarrow {\Omega }_{A/R}^{1} \) such that \( \left( {{\Omega }_{A/R}^{1},{d}_{A/R}}\right) \) admits the following universal property:\n\nFor every \... | Proof. We start with the case \( A = R\left\lbrack \mathfrak{X}\right\rbrack \), for a system \( \mathfrak{X} \) of (arbitrarily many) variables \( {X}_{i}, i \in I \) . Then define \( {\Omega }_{A/R}^{1} = {A}^{\left( I\right) } \) as the free \( A \) -module generated by \( I \) . Writing \( d{X}_{i} \) for the basis... | Yes |
For an R-algebra \( A \) and an ideal \( \mathfrak{a} \subset A \), write \( B = A/\mathfrak{a} \). Furthermore, let \( \left( {{\Omega }_{A/R}^{1},{d}_{A/R}}\right) \) be the module of relative differential forms of \( A \) over \( R \). Then \[ \Omega = {\Omega }_{A/R}^{1}/\left( {\mathfrak{a}{\Omega }_{A/R}^{1} + A{... | Proof. First observe that \( \Omega \) is indeed a \( B \)-module. Further, since \( {d}_{A/R} \) admits the properties of an \( R \)-derivation, the same is true for \( d \). To establish the universal property for \( d \), consider an \( R \)-derivation \( \bar{\delta } : B \rightarrow M \) to a \( B \)-module \( M \... | Yes |
For every homomorphism of \( R \) -algebras \( \tau : A \rightarrow B \), the sequence\n\n\[ \n{\Omega }_{A/R}^{1}{ \otimes }_{A}B\overset{\alpha }{ \rightarrow }{\Omega }_{B/R}^{1}\overset{\beta }{ \rightarrow }{\Omega }_{B/A}^{1} \rightarrow 0 \n\]\n\ngiven by \( {d}_{A/R}\left( f\right) \otimes b\overset{\alpha }{ \... | Proof. Since \( {\Omega }_{B/A}^{1} \) is generated by the elements of type \( {d}_{B/A}\left( g\right), g \in B \), and since \( \beta \left( {{d}_{B/R}\left( g\right) }\right) = {d}_{B/A}\left( g\right) \), we see that \( \beta \) is surjective. Furthermore, we have \( \beta \circ \alpha = 0 \), which implies \( \ope... | Yes |
Corollary 8. Let \( L/K \) be a purely transcendental field extension that is generated by a transcendence basis \( {\left( {x}_{j}\right) }_{j \in J} \). Then \( {\left( {d}_{L/K}\left( {x}_{j}\right) \right) }_{j \in J} \) is a basis of the \( L \) -vector space \( {\Omega }_{L/K}^{1} \). | Proof. Use Propositions 4 and 6. Alternatively, one can rely on the assertion of Proposition 7, at least if the transcendence basis \( {\left( {x}_{j}\right) }_{j \in J} \) is finite. | No |
Corollary 9. Let \( L/K \) be a separable algebraic field extension. Then, for every derivation \( \delta : K \rightarrow V \) to an \( L \) -vector space \( V \), there exists a unique extension as a derivation \( {\delta }^{\prime } : L \rightarrow V \) . In particular, we get \( {\Omega }_{L/K}^{1} = 0 \) . | Proof. Let \( \delta : K \rightarrow V \) be a derivation to an \( L \) -vector space \( V \), and let \( {L}^{\prime } \) be an intermediate field of \( L/K \) such that \( {L}^{\prime }/K \) is finite. As we know, \( {L}^{\prime }/K \) is simple by the primitive element theorem \( {3.6}/{12} \), say \( {L}^{\prime } ... | Yes |
Corollary 10. Let \( K \) be a field of characteristic \( p > 0 \) and \( L/K \) a purely inseparable field extension of degree \( p \), say \( L = K\left( x\right) \) with minimal polynomial of \( x \) over \( K \) given by \( f = {X}^{p} - c \in K\left\lbrack X\right\rbrack \) . Furthermore, let \( \delta : K \righta... | Proof. According to Proposition 7, the derivation \( \delta \) admits an extension to a derivation \( {\delta }^{\prime } : L \rightarrow V \) satisfying \( {\delta }^{\prime }\left( x\right) = v \) if and only if the equation\n\n\[ \n- \delta \left( c\right) + p{x}^{p - 1} \cdot v = 0 \n\]\n\nis satisfied, hence if an... | Yes |
Proposition 1. Let \( f : A \rightarrow B \) .\n\n(1) The map \( f \) is injective if and only if \( f \) has a left inverse.\n\n(2) The map \( f \) is surjective if and only if \( f \) has a right inverse.\n\n(3) The map \( f \) is a bijection if and only if there exists \( g : B \rightarrow A \) such that \( f \circ ... | Proof: Exercise. | No |
If \( \sim \) defines an equivalence relation on \( A \) then the set of equivalence classes of \( \sim \) form a partition of \( A \) . | Proof: Omitted. | No |
Theorem 3. The operations of addition and multiplication on \( \mathbb{Z}/n\mathbb{Z} \) defined above are both well defined, that is, they do not depend on the choices of representatives for the classes involved. More precisely, if \( {a}_{1},{a}_{2} \in \mathbb{Z} \) and \( {b}_{1},{b}_{2} \in \mathbb{Z} \) with \( \... | Proof: Suppose \( {a}_{1} \equiv {b}_{1}\left( {\;\operatorname{mod}\;n}\right) \), i.e., \( {a}_{1} - {b}_{1} \) is divisible by \( n \) . Then \( {a}_{1} = {b}_{1} + {sn} \) for some integer \( s \) . Similarly, \( {a}_{2} \equiv {b}_{2}\left( {\;\operatorname{mod}\;n}\right) \) means \( {a}_{2} = {b}_{2} + {tn} \) f... | Yes |
Proposition 4. \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } = \{ \bar{a} \in \mathbb{Z}/n\mathbb{Z} \mid \left( {a, n}\right) = 1\} \) . | It is easy to see that if any representative of \( \bar{a} \) is relatively prime to \( n \) then all representatives are relatively prime to \( n \) so that the set on the right in the proposition is well defined. | No |
If \( G \) is a group under the operation \( \star \), then (1) the identity of \( G \) is unique | If \( f \) and \( g \) are both identities, then by axiom (ii) of the definition of a group \( f \star g = f \) (take \( a = f \) and \( e = g \) ). By the same axiom \( f \star g = g \) (take \( a = g \) and \( e = f \) ). Thus \( f = g \), and the identity is unique. | Yes |
Proposition 2. Let \( G \) be a group and let \( a, b \in G \) . The equations \( {ax} = b \) and \( {ya} = b \) have unique solutions for \( x, y \in G \) . In particular, the left and right cancellation laws hold in \( G \), i.e.,\n\n(1) if \( {au} = {av} \), then \( u = v \), and\n\n(2) if \( {ub} = {vb} \), then \(... | Proof: We can solve \( {ax} = b \) by multiplying both sides on the left by \( {a}^{-1} \) and simplifying to get \( x = {a}^{-1}b \) . The uniqueness of \( x \) follows because \( {a}^{-1} \) is unique. Similarly, if \( {ya} = b, y = b{a}^{-1} \) . If \( {au} = {av} \), multiply both sides on the left by \( {a}^{-1} \... | Yes |
Proposition 1. (The Subgroup Criterion) A subset \( H \) of a group \( G \) is a subgroup if and only if\n\n(1) \( H \neq \varnothing \), and\n\n(2) for all \( x, y \in H, x{y}^{-1} \in H \) . | Proof: If \( H \) is a subgroup of \( G \), then certainly (1) and (2) hold because \( H \) contains the identity of \( G \) and the inverse of each of its elements and because \( H \) is closed under multiplication.\n\nIt remains to show conversely that if \( H \) satisfies both (1) and (2), then \( H \leq G \) . Let ... | Yes |
Proposition 2. If \( H = \langle x\rangle \), then \( \left| H\right| = \left| x\right| \) (where if one side of this equality is infinite, so is the other). More specifically\n\n(1) if \( \left| H\right| = n < \infty \), then \( {x}^{n} = 1 \) and \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are all the distinct elements of... | Proof: Let \( \left| x\right| = n \) and first consider the case when \( n < \infty \) . The elements \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are distinct because if \( {x}^{a} = {x}^{b} \), with, say, \( 0 \leq a < b < n \), then \( {x}^{b - a} = {x}^{0} = 1 \), contrary to \( n \) being the smallest positive power of ... | Yes |
Proposition 3. Let \( G \) be an arbitrary group, \( x \in G \) and let \( m, n \in \mathbb{Z} \) . If \( {x}^{n} = 1 \) and \( {x}^{m} = 1 \), then \( {x}^{d} = 1 \), where \( d = \left( {m, n}\right) \) . In particular, if \( {x}^{m} = 1 \) for some \( m \in \mathbb{Z} \), then \( \left| x\right| \) divides \( m \) . | Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers \( r \) and \( s \) such that \( d = {mr} + {ns} \), where \( d \) is the g.c.d. of \( m \) and \( n \) . Thus\n\n\[ \n{x}^{d} = {x}^{{mr} + {ns}} = {\left( {x}^{m}\right) }^{r}{\left( {x}^{n}\right) }^{s} = {1}^{r}{1}^{s} = 1.\n\]\n\nThis pro... | Yes |
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