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Proposition 4.4.7. As above let \( \mathcal{L}/\mathcal{K} \) be an extension of \( \mathfrak{p} \) -adic fields of degree \( n \), let \( a \in \mathcal{L} \) be such that \( \mathcal{L} = \mathcal{K}\left( a\right) \), and let \( P\left( X\right) \) be the minimal monic polynomial of a over \( \mathcal{K} \) . There ... | Proof. For any monic \( Q \in \mathcal{K}\left\lbrack X\right\rbrack \), let \( Q = \prod \left( {X - {b}_{i}}\right) \) be the factorization of \( Q \) in an algebraic closure of \( \mathcal{K} \) . We thus have \( \prod \left( {a - {b}_{i}}\right) = Q\left( a\right) = Q\left( a\right) - P\left( a\right) \) , so if we... | Yes |
Corollary 4.4.8. Let \( P \in \mathcal{K}\left\lbrack X\right\rbrack \) be a monic irreducible polynomial, let a be a root of \( P \) in an algebraic closure of \( \mathcal{K} \), and let \( \left( {Q}_{i}\right) \) be a sequence of monic polynomials in \( \mathcal{K}\left\lbrack X\right\rbrack \) of the same degree as... | Proof. As soon as \( \begin{Vmatrix}{{Q}_{i} - P}\end{Vmatrix} < \varepsilon \) we can apply the above proposition, which shows that \( \left| {a - {b}_{i}}\right| \) is small for at least one root \( {b}_{i} \) of \( {Q}_{i} \) belonging to \( \mathcal{K}\left( a\right) \) , more precisely that \( \left| {a - {b}_{i}}... | Yes |
Proposition 4.4.9. Any finite extension \( \mathcal{L} \) of \( {K}_{\mathfrak{p}} \) is isomorphic to \( {L}_{\mathfrak{P}} \) for some finite extension \( L \) of \( K \) such that \( \left\lbrack {L : K}\right\rbrack = \left\lbrack {\mathcal{L} : {K}_{\mathfrak{p}}}\right\rbrack \) and some prime ideal \( \mathfrak{... | Proof. Set \( n = \left\lbrack {\mathcal{L} : {K}_{\mathfrak{p}}}\right\rbrack \) . By the primitive element theorem, there exists \( x \in \mathcal{L} \) such that \( \mathcal{L} = {K}_{\mathfrak{p}}\left( x\right) \), and we may of course assume that \( x \) is integral, i.e., that \( v\left( x\right) \geq 0 \), wher... | Yes |
Theorem 4.4.11. Let \( a \in l \) be a class. There exists a representative \( \alpha \in {\mathbb{Z}}_{\mathfrak{P}} \) of \( \bar{\alpha } \) such that \( \left\lbrack {\mathcal{K}\left( \alpha \right) : \mathcal{K}}\right\rbrack = \left\lbrack {k\left( a\right) : k}\right\rbrack \) . Furthermore, the field \( \mathc... | Proof. Let \( \phi \left( T\right) \in k\left\lbrack T\right\rbrack \) be the minimal monic polynomial of \( a \) over \( k \) . Since \( k \) is a perfect field, \( \phi \) is separable, so \( {\phi }^{\prime }\left( a\right) \neq 0 \) . Let \( \Phi \) be any monic lift of \( \phi \) to \( {\mathbb{Z}}_{\mathfrak{p}}\... | Yes |
Corollary 4.4.12. With the same assumptions, there is a bijection between the unramified subextensions \( \mathcal{M}/\mathcal{K} \) of \( \mathcal{L}/\mathcal{K} \) and the fields \( m \) such that \( k \subset \) \( m \subset l \) . The field \( m \) corresponding to \( \mathcal{M} \) is \( \mathcal{M} \cap {\mathbb{... | Proof. Since an extension \( m/k \) is an extension of finite fields, it is separable; hence by the primitive element theorem we have \( m = k\left( a\right) \) for some \( a \in m \) . The corollary then follows immediately from the theorem. | No |
Corollary 4.4.13. There exists a field \( \mathcal{M} \) such that \( \mathcal{K} \subset \mathcal{M} \subset \mathcal{L} \) such that \( \mathcal{M}/\mathcal{K} \) is unramified and such that every \( {\mathcal{M}}^{\prime } \subset \mathcal{L} \) that is unramified over \( \mathcal{K} \) is contained in \( \mathcal{M... | Proof. We simply let \( \mathcal{M} \) be the field corresponding to \( l \) in the previous corollary. | No |
Corollary 4.4.14. Recall that \( \mathcal{K} = {K}_{\mathfrak{p}} \) and \( \mathcal{L} = {L}_{\mathfrak{P}} \) . There exists a unique unramified subextension \( \mathcal{M} \) of \( \mathcal{L} \) that contains all unramified subextensions of \( \mathcal{L}/\mathcal{K} \), and that is such that \( \mathcal{L}/\mathca... | Proof. The first assertion is a restatement of the above corollary. For the second, note that since \( \mathcal{M}/\mathcal{K} \) is unramified, we have \( e\left( {\mathcal{M}/\mathcal{K}}\right) = 1 \) , and since \( \mathcal{L}/\mathcal{M} \) is totally ramified we have \( e\left( {\mathcal{L}/\mathcal{M}}\right) = ... | Yes |
Corollary 4.4.15. The residue field of the algebraic closure \( \overline{\mathcal{K}} \) of \( \mathcal{K} \) is equal to \( \bar{k} \), the algebraic closure of \( k \) . There exists a (unique) subfield \( {\mathcal{K}}^{u} \) of \( \overline{\mathcal{K}} \) such that a finite extension \( \mathcal{L}/\mathcal{K} \)... | Proof. If \( \phi \left( T\right) \in k\left\lbrack T\right\rbrack \) is monic and irreducible and \( \Phi \left( T\right) \in \mathcal{K}\left\lbrack T\right\rbrack \) is any monic lift, then \( \overline{\mathcal{K}} \) contains all the roots of \( \Phi \left( T\right) \) ; hence its residue field contains all the ro... | No |
Proposition 4.4.19. Let \( 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 \) be an exact sequence of \( G \) -modules. If two of the quotients \( q\left( A\right), q\left( B\right), q\left( C\right) \) are defined, so is the third, and we have \( q\left( A\right) q\left( C\right) = q\left( B\right) \) . | Proof. By the exact hexagon lemma (Proposition 4.4.17) and the standard kernel-image formula we have\n\n\[ \left| {{H}^{1}\left( C\right) }\right| = \left| {\operatorname{Ker}\left( {\delta }_{1}\right) }\right| \left| {\operatorname{Im}\left( {\delta }_{1}\right) }\right| = \left| {\operatorname{Im}\left( {g}_{1}\righ... | Yes |
Corollary 4.4.20. If \( A \subset B \) are \( G \) -modules and the index \( \left\lbrack {B : A}\right\rbrack \) is finite then \( q\left( A\right) = q\left( B\right) \) whenever either one is defined. | Proof. If we set \( C = B/A \), by the proposition we must show that \( q\left( C\right) = 1 \) when \( C \) is finite. But in that case\n\n\[ q\left( C\right) = \frac{\left| \operatorname{Ker}\left( {\left. D\right| }_{C}\right) /N\left( C\right) \right| }{\left| \operatorname{Ker}\left( {\left. N\right| }_{C}\right) ... | Yes |
Proposition 4.4.21. Set \( m = n/d \) and assume that \( R/{mR} \) is finite. Then \( q\left( A\right) \) is defined and equal to \( \left| {R/{mR}}\right| \) . In particular, when \( R = \mathbb{Z} \) we have \( q\left( A\right) = m = \left| {G}_{A}\right| . \) | Proof. We have (again with indices modulo \( d \) )\n\n\[ \n{\sigma }^{k}\left( {\mathop{\sum }\limits_{i}{n}_{i}{e}_{i}}\right) = \mathop{\sum }\limits_{i}{n}_{i - k}{e}_{i} \n\] \n\nhence \n\n\[ \nN\left( {\mathop{\sum }\limits_{i}{n}_{i}{e}_{i}}\right) = \mathop{\sum }\limits_{{0 \leq i < n}}{n}_{i}\mathop{\sum }\li... | Yes |
Corollary 4.4.23. If \( {L}_{\mathfrak{P}}/{K}_{\mathfrak{p}} \) is unramified then the norm is surjective on units. | Proof. Clear since \( {L}_{\mathfrak{P}}/{K}_{\mathfrak{p}} \) is cyclic in that case. | No |
Proposition 4.4.24. If \( {L}_{\mathfrak{P}}/{K}_{\mathfrak{p}} \) is a cyclic extension then\n\n\[ \left\lbrack {{K}_{\mathfrak{p}}^{ * } : {\mathcal{N}}_{{L}_{\mathfrak{P}}/{K}_{\mathfrak{p}}}\left( {L}_{\mathfrak{P}}^{ * }\right) }\right\rbrack = \left\lbrack {{L}_{\mathfrak{P}} : {K}_{\mathfrak{p}}}\right\rbrack = ... | Proof. With the above notation we have \( {L}_{\mathfrak{P}}^{ * } = {\pi }^{\mathbb{Z}} \times {U}_{\mathfrak{P}} \), hence \( {L}_{\mathfrak{P}}^{ * }/{U}_{\mathfrak{P}} \) is isomorphic to \( \mathbb{Z} \) and \( G \) acts trivially on it (since \( \sigma \left( \pi \right) /\pi \in {U}_{\mathfrak{P}} \) ). It follo... | Yes |
Lemma 4.4.26. For every \( n \), up to isomorphism there exists exactly one unramified extension \( \mathcal{K} \) of \( {\mathbb{Q}}_{p} \) of degree \( n \), which is the splitting field of \( {X}^{q} - X \) over \( {\mathbb{Q}}_{p} \), where \( q = {p}^{n} \) . | Proof. The residue field of \( {\mathbb{Q}}_{p} \) is the finite field \( {\mathbb{F}}_{p} \simeq \mathbb{Z}/p\mathbb{Z} \) . Up to isomorphism, for every \( n \) there exists one extension \( k \) of \( {\mathbb{F}}_{p} \) of degree \( n \) : it has \( q = {p}^{n} \) elements and is the splitting field of \( {X}^{q} -... | Yes |
Corollary 4.4.27. Let \( \mathcal{K} = {K}_{\mathfrak{p}} \) be any \( \mathfrak{p} \) -adic field, and let \( q = \mathcal{N}\mathfrak{p} \) be the cardinality of its residue field. For every \( n \), up to isomorphism there exists exactly one unramified extension \( \mathcal{L} \) of \( \mathcal{K} \) of degree \( n ... | Proof. Once again by Corollary 4.4.15 there exists exactly one unramified extension \( \mathcal{L} \) of \( \mathcal{K} \) in the algebraic closure of \( \mathcal{K} \) whose residue field is \( {\mathbb{F}}_{Q} \), and \( \mathcal{L} = {L}_{\mathfrak{P}} \) for some \( L \) and \( \mathfrak{P} \) by Proposition 4.4.9.... | Yes |
Corollary 4.4.28. The unramified closure \( {\mathcal{K}}^{u} \) (see Corollary 4.4.15) is obtained by adjoining to \( \mathcal{K} \) the \( m \) th roots of unity for all \( m \) prime to the residue field characteristic \( p \) . | Proof. By the previous corollary, \( {\mathcal{K}}^{u} \) is obtained by adjoining the \( \left( {{q}^{n} - 1}\right) \) st roots of unity for all \( n \geq 1 \) . Since for any \( m \) prime to \( p \), hence to \( q \), there exists \( n \) such that \( {q}^{n} \equiv 1\left( {\;\operatorname{mod}\;m}\right) \) (the ... | Yes |
Corollary 4.4.29. As above let \( \mathcal{K} = {K}_{\mathfrak{p}}, q = \mathcal{N}\mathfrak{p},\mathcal{L} \) the unique unramified extension of \( \mathcal{K} \) of degree \( n \), and \( Q = {q}^{n} \), so that \( \mathcal{L} \) is the splitting field of \( {X}^{Q} - X \) over \( \mathcal{K} \) . If \( \sigma \) is ... | Proof. Recall first that \( \sigma \left( x\right) = {x}^{q} \) only at the level of the residue field. Let \( {\mu }_{Q - 1} \) be the group of \( \left( {Q - 1}\right) \) st roots of unity in \( \mathcal{L} \) . By definition \( \left| {\mu }_{Q - 1}\right| = Q - 1 \) and it is a finite subgroup of \( {\mathcal{L}}^{... | Yes |
Proposition 4.4.30. Keep the above notation, and assume that \( {L}_{\mathfrak{P}}/{K}_{\mathfrak{p}} \) is an unramified extension of \( \mathfrak{p} \) -adic fields. Then any lift of a normal basis \( \bar{\alpha } \) of \( l \) over \( k \) gives a normal basis of \( {\mathbb{Z}}_{\mathfrak{P}} \) over \( {\mathbb{Z... | Proof. We simply apply Nakayama’s Lemma 4.3.15 to the \( {\mathbb{Z}}_{\mathfrak{p}} \) -module \( M = \) \( {\mathbb{Z}}_{\mathfrak{P}}/\left( {{\mathbb{Z}}_{\mathfrak{p}}\left\lbrack G\right\rbrack \alpha }\right) \), where \( \alpha \) is a lift of a normal basis \( \bar{\alpha } \) of \( l \) over \( k \) . Note th... | Yes |
Corollary 4.4.31. Under the same assumptions, the trace map \( {\operatorname{Tr}}_{{L}_{\mathfrak{P}}/{K}_{\mathfrak{p}}} \) is a surjective map from \( {\mathbb{Z}}_{\mathfrak{P}} \) to \( {\mathbb{Z}}_{\mathfrak{p}} \) . | Proof. Again set \( G = \operatorname{Gal}\left( {{L}_{\mathfrak{P}}/{K}_{\mathfrak{p}}}\right) \), and let \( \alpha \in {\mathbb{Z}}_{\mathfrak{P}} \) be a normal basis of \( {\mathbb{Z}}_{\mathfrak{P}} \) over \( {\mathbb{Z}}_{\mathfrak{p}} \) . I claim that under the above isomorphism between \( {\mathbb{Z}}_{\math... | Yes |
Theorem 4.4.32. Let \( {L}_{\mathfrak{P}}/{K}_{\mathfrak{p}} \) be an unramified extension of \( \mathfrak{p} \) -adic fields. Denote by \( {U}_{\mathfrak{p}} \) the group of \( \mathfrak{p} \) -adic units of \( {K}_{\mathfrak{p}} \), and similarly \( {U}_{\mathfrak{P}} \) for \( {L}_{\mathfrak{P}} \) . Then \( {\mathc... | Since we know that unramified extensions are in bijective correspondence with extensions of the residue fields, these surjectivity results on the trace and norm also follow from Propositions 2.4.11 and 2.4.12. | No |
Corollary 4.4.33. Let \( \pi \) be a uniformizer of \( \mathfrak{p} \) . We have\n\n\[ \n{\mathcal{N}}_{{L}_{\mathfrak{P}}/{K}_{\mathfrak{p}}}\left( {L}_{\mathfrak{P}}^{ * }\right) = {\pi }^{f\left( {\mathfrak{P}/\mathfrak{p}}\right) \mathbb{Z}}{U}_{\mathfrak{p}} \n\]\n\nIn particular, the map that sends \( \pi \) to t... | Proof. Since \( {L}_{\mathfrak{P}}/{K}_{\mathfrak{p}} \) is unramified, a uniformizer \( \pi \) of \( \mathfrak{p} \) is also one of \( \mathfrak{P} \) . It follows that \( {L}_{\mathfrak{P}}^{ * } = {\pi }^{\mathbb{Z}}{U}_{\mathfrak{P}} \) ; hence \( {\mathcal{N}}_{{L}_{\mathfrak{P}}/{K}_{\mathfrak{p}}}\left( {L}_{\ma... | Yes |
Proposition 4.4.36. Let \( \mathcal{K} = {\mathbb{Q}}_{p}\left( {\zeta }_{p}\right) = {\mathbb{Q}}_{p}\left( {\left( -p\right) }^{1/\left( {p - 1}\right) }\right) \), and let \( \mathfrak{p} \) be the prime ideal of \( \mathcal{K} \) above \( p \) (generated by \( \zeta - 1 \) or by \( {\left( -p\right) }^{1/\left( {p ... | Proof. In the proof of Theorem 4.3.18 (1) we have seen that if \( \zeta \) is a primitive \( p \) th root of unity there exists \( \pi \) such that \( {\pi }^{p - 1} + p = 0 \) with \( (\zeta - \) \( 1)/\pi \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) \), or equivalently, \( \zeta \equiv 1 + \pi \left( {\... | Yes |
Proposition 4.4.38. (1) The series \( {D}_{\pi }\left( X\right) \) converges at least for \( {v}_{p}\left( x\right) > \) \( - \left( {p - 1}\right) /{p}^{2} \), and in particular has a radius of convergence strictly greater than 1.\n\n(2) More precisely, if we set \( {D}_{\pi }\left( X\right) = \mathop{\sum }\limits_{{... | Proof. (1) and (2). We have\n\n\[ \n{D}_{\pi }\left( {X/\pi }\right) = \exp \left( {X + {X}^{p}/p}\right) = \mathop{\sum }\limits_{{k \geq 0}}{u}_{k}{X}^{k}, \n\]\n\nwhere \( {v}_{p}\left( {u}_{k}\right) \geq - k\left( {{2p} - 1}\right) /\left( {\left( {p - 1}\right) {p}^{2}}\right) \) by Theorem 4.2.22. Since \( {d}_{... | Yes |
Theorem 4.4.39. With the above notation we have \( {D}_{\pi }\left( 1\right) = {\zeta }_{\pi } \) ; in other words, \( {D}_{\pi }\left( 1\right) \) is equal to the unique pth root of unity congruent to \( 1 + \pi \) modulo \( {\mathfrak{p}}^{2} \) (and in particular is not equal to 1). Furthermore, for any \( a \in {\m... | Proof. Since the radius of convergence of \( {D}_{\pi }\left( X\right) \) is strictly greater than 1, we may compute \( {D}_{\pi }\left( a\right) \) for any \( a \in {\mathbb{Z}}_{p} \) . Furthermore, \( {D}_{\pi }{\left( X\right) }^{p} = \) \( \exp \left( {p\pi X}\right) \exp \left( {-{p\pi }{X}^{p}}\right) \), and by... | Yes |
Proposition 4.4.40. (1) The series \( {D}_{\pi, f}\left( X\right) \) converges at least for \( {v}_{p}\left( x\right) > \) \( - \left( {p - 1}\right) /{p}^{f + 1} \), and in particular has a radius of convergence strictly greater than 1. | Proof. We can write formally\n\n\[ {D}_{\pi, f}\left( X\right) = \mathop{\prod }\limits_{{0 \leq j < f}}\exp \left( {\pi \left( {{X}^{{p}^{j}} - {X}^{{p}^{j + 1}}}\right) }\right) .\n\nBy Proposition 4.4.38, the \( j \) th series converges at least for \( {v}_{p}\left( {x}^{{p}^{j}}\right) > - (p - \) \( 1)/{p}^{2} \) ... | Yes |
Theorem 4.4.41. Let \( L = K\\left( \\theta \\right) \) be an extension of number fields, let \( T\\left( X\\right) \\in K\\left\\lbrack X\\right\\rbrack \) be the minimal monic polynomial of \( \\theta \) over \( K \), let \( \\mathfrak{p} \) be a prime ideal of \( K \), let \( \\mathfrak{p}{\\mathbb{Z}}_{L} = \\matho... | Proof. (1). It is clear that \( {\\left| \\right| }_{{\\mathfrak{P}}_{j}}^{1/\\left( {{e}_{j}{f}_{j}}\\right) } \) is an absolute value and that if \( x \\in {\\mathbb{Z}}_{\\mathfrak{p}} \),\n\nwe have\n\n\[{\\left| x\\right| }_{{\\mathfrak{P}}_{j}} = \\mathcal{N}{\\left( {\\mathfrak{P}}_{j}\\right) }^{-{v}_{{\\mathfr... | Yes |
The algebraic closure \( \overline{{\mathbb{Q}}_{p}} \) of \( {\mathbb{Q}}_{p} \) is not complete for the extension of the natural absolute value given by Corollary 4.4.4. However, the completion \( {\mathbb{C}}_{p} \) of \( \overline{{\mathbb{Q}}_{p}} \) is algebraically closed (and evidently complete), and there exis... | Proof. Denote as usual by \( {\zeta }_{m} \) a primitive \( m \) th root of unity in \( \overline{{\mathbb{Q}}_{p}} \), and\n\nset\n\[ \alpha = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\zeta }_{{n}^{\prime }}{p}^{n} \]\n\nwhere \( {n}^{\prime } = n \) if \( p \nmid n \) and \( {n}^{\prime } = 1 \) (so \( {\zeta }_{{n}... | Yes |
Proposition 4.4.43. Let \( K \) be an algebraically closed field, let \( G \) be an abelian group, and let \( H \) be a subgroup of \( G \) . Any group homomorphism \( \phi \) from \( H \) to \( {K}^{ * } \) can be extended to a homomorphism from \( G \) to \( {K}^{ * } \). | Proof. Consider the set of subgroups \( {H}^{\prime } \) of \( G \) containing \( H \) and on which \( \phi \) can be extended. It is evidently nonempty since \( H \) belongs to it, and is ordered by set inclusion. Thus by Zorn’s lemma there exists a maximal subgroup \( {H}_{m} \) in this set (the use of the axiom of c... | Yes |
Proposition 4.4.44. Denote by \( \mu \) the group of roots of unity of order not divisible by \( p \), and by \( {U}_{1} \) the group of \( x \) such that \( \left| {x - 1}\right| < 1 \) . Then \[ {\mathbb{C}}_{p}^{ * } = {p}^{\mathbb{Q}} \times \mu \times {U}_{1} \] | Proof. Let \( \alpha \in {\mathbb{C}}_{p}^{ * } \) . By definition, we can find a sequence of \( {\alpha }_{i} \in \overline{{\mathbb{Q}}_{p}} \) tending to \( \alpha \) as \( i \rightarrow \infty \) . Thus for \( i \) sufficiently large, \( \left| {{\alpha }_{i} - \alpha }\right| < \left| \alpha \right| \), so that by... | Yes |
There exists a unique extension of \( {\log }_{p}\left( x\right) \) (as defined by the power series for \( \left. {\left| {x - 1}\right| < 1}\right) \) to \( {\mathbb{C}}_{p}^{ * } \) satisfying \( {\log }_{p}\left( p\right) = 0 \) and \( {\log }_{p}\left( {xy}\right) = \) \( {\log }_{p}\left( x\right) + {\log }_{p}\le... | If \( \alpha \in {\mathbb{C}}_{p}^{ * } \) we can write \( \alpha = {p}^{a}\omega \left( x\right) \langle x\rangle \) with \( \left| {\langle x\rangle - 1}\right| < 1 \) . This decomposition is not unique since it depends on the choice of \( {p}^{a} \) (note that here we do not need the homomorphism \( a \mapsto {p}^{a... | Yes |
Proposition 4.4.46. Set \( {q}_{p} = p \) if \( p \geq 3 \) and \( {q}_{2} = 4 \), and let \( x \in {\mathbb{Z}}_{p}^{ * } \) .\n\n(1) We have\n\n\[ \n{\log }_{p}\left( x\right) \equiv \left\{ \begin{array}{ll} 1 - {x}^{p - 1}\left( {{\;\operatorname{mod}\;{p}^{2}}{\mathbb{Z}}_{p}}\right) & \text{ if }p \geq 3 \\ \left... | Proof. (1). Assume first that \( p \geq 3 \) . Note that \( 1/\left( {p - 1}\right) = - 1/\left( {1 - p}\right) \equiv \) \( - \left( {1 + p}\right) \left( {{\;\operatorname{mod}\;{p}^{2}}{\mathbb{Z}}_{p}}\right) \) . Thus by Corollary 4.2.13 we have\n\n\[ \n{\log }_{p}\left( x\right) = \frac{1}{p - 1}{\log }_{p}\left(... | Yes |
Proposition 4.4.47. Assume that \( a \in {\mathbb{C}}_{p} \) and \( x \in {\mathbb{Z}}_{p}^{ * } \) . The series \( \langle x{\rangle }^{a} = \) \( {\exp }_{p}\left( {a{\log }_{p}\left( {\langle x\rangle }\right) }\right) \) converges as a power series in \( \langle x\rangle - 1 \) . | Proof. Immediate from Corollary 4.2.16 since by definition of \( \langle x\rangle \) we have \( {v}_{p}\left( {\langle x\rangle - 1}\right) \geq {v}_{p}\left( {q}_{p}\right) > 1/\left( {p - 1}\right) . | Yes |
Theorem 4.5.1 (Strassmann). Let \( f\left( X\right) = {f}_{0} + {f}_{1}X + \cdots + {f}_{n}{X}^{n} + \cdots \) , and assume that \( {f}_{n} \) tends to 0 (in other words that the domain of convergence contains \( {\mathcal{Z}}_{p} \), so that the radius of convergence is greater than or equal to 1), but that not all th... | Proof. We use induction on \( N \) . First, let \( N = 0 \) and assume that \( f\left( x\right) = 0 \) for some \( x \in {\mathcal{Z}}_{p} \), so that \( {f}_{0} = - \mathop{\sum }\limits_{{n \geq 1}}{f}_{n}{x}^{n} \) . Since \( \left| {f}_{n}\right| < \left| {f}_{0}\right| \) for \( n \geq 1 \) and \( \left| x\right| ... | Yes |
Lemma 4.5.5. Let \( f\left( X\right) \) be a rational function with no poles in \( {\mathcal{Z}}_{p} \) . Then \( f \in W \) . | Proof. Decompose \( f \) into partial fractions: the polynomial part is of course in \( W \) . The polar part is a linear combination of fractions of the form \( 1/(X - \) \( \alpha {)}^{k} \) with \( k \geq 1 \) . By assumption, \( \alpha \notin {\mathcal{Z}}_{p} \), so that\n\n\[ \frac{1}{{\left( X - \alpha \right) }... | Yes |
Lemma 4.5.6. (1) The ring \( W \) is complete for the topology induced by \( \parallel \parallel \) . | Proof. Let \( {f}^{j}\left( X\right) = \mathop{\sum }\limits_{{n \geq 0}}{f}_{n}^{\left( j\right) }{X}^{n} \) be a Cauchy sequence in \( W \) . In partic-\n\nular, for each \( n \) the sequence \( {f}_{n}^{\left( j\right) } \) is a Cauchy sequence in \( {\mathbb{C}}_{p} \), hence has a limit, which we denote by \( {f}_... | Yes |
Lemma 4.5.7. Let \( f \in W \), and assume that \( f \) vanishes in some nonempty open set \( S \subset {\mathcal{Z}}_{p} \). Then \( f = 0 \). Equivalently, let \( f \) and \( g \) be in \( W \) and assume that \( f \) and \( g \) coincide on some nonempty open subset of \( {\mathcal{Z}}_{p} \). Then \( f = g \). | Proof. Assume first that \( 0 \in S \), so that \( f\left( x\right) = 0 \) for all \( x \) such that \( \left| x\right| < \delta \) for some \( \delta > 0 \). Let \( \zeta \in {\mathcal{Z}}_{p} \) be such that \( \zeta \neq 0 \) and \( \left| z\right| < \delta \), and set \( F\left( x\right) = f\left( {\zeta x}\right) ... | Yes |
Lemma 4.5.8. Let \( f \) be a function from \( {\mathcal{Z}}_{p} \) to \( {\mathbb{C}}_{p} \) . A necessary and sufficient condition that \( f \) belong to \( W \) is that it be a uniform limit of rational functions all having their poles outside \( {\mathcal{Z}}_{p} \) . | Proof. Assume first that \( f\left( X\right) = \mathop{\sum }\limits_{{n \geq 0}}{f}_{n}{X}^{n} \in W \) . Then \( f \) is the uniform limit of the polynomials \( {h}^{\left( n\right) }\left( X\right) = \mathop{\sum }\limits_{{0 \leq j \leq n}}{f}_{j}{X}^{j} \), which have no poles (or simply the point at infinity, if ... | Yes |
Lemma 4.5.10. Let \( f \) and \( g \) be two Krasner analytic functions on the set \( \mathcal{D} \) defined above. If \( f \) and \( g \) coincide on some nonempty open subset of \( \mathcal{D} \) then \( f = g \) . | Proof. Consider the map from \( \mathcal{D} \) to \( {\mathbb{C}}_{p} \) sending \( x \) to \( t = 1/\left( {x - 1}\right) \) . Since \( \left| {x - 1}\right| \geq 1 \) for \( x \in \mathcal{D} \), this map is a well-defined map from \( \mathcal{D} \) to \( {\mathcal{Z}}_{p} \), and since \( x = 1 + 1/t \) for \( t \ne... | Yes |
Lemma 4.5.11. Let \( z \in {\mathbb{C}}_{p} \) . The following conditions are equivalent:\n\n(1) \( \left| {z - 1}\right| \geq 1 \) ; in other words, \( z \in \mathcal{D} \) .\n\n(2) \( \left| z\right| \neq 1 \) or \( \left| z\right| = \left| {z - 1}\right| = 1 \) .\n\n(3) \( \left| {z/\left( {1 - z}\right) }\right| \l... | Proof. Left to the reader (Exercise 40). | No |
Theorem 4.5.12 (Weierstrass preparation theorem). Let \( f\left( X\right) = {f}_{0} + {f}_{1}X + \cdots + {f}_{n}{X}^{n} + \cdots \), assume that \( {f}_{n} \) tends to 0, and denote by \( N \) the largest integer such that \( \left| {f}_{N}\right| = \mathop{\max }\limits_{n}\left| {f}_{n}\right| \) . There exists a po... | Proof. We need some lemmas. | No |
Lemma 4.5.13. Let \( R\left( X\right) \in {\mathbb{C}}_{p}\left\lbrack X\right\rbrack \) and let \( G\left( X\right) = {G}_{0} + {G}_{1}X + \cdots + {G}_{N}{X}^{N} \in {\mathbb{C}}_{p}\left\lbrack X\right\rbrack \) be a nonzero polynomial such that \( \left| {G}_{N}\right| = \mathop{\max }\limits_{n}\left| {G}_{n}\righ... | Proof. Denote by \( {R}_{m} \) (respectively \( {L}_{m} \) ) the coefficient of \( {X}^{m} \) in \( R\left( X\right) \) (respectively in \( L\left( X\right) \) ), and let \( d = \deg \left( {R\left( X\right) }\right) \), so that \( d - N = \deg \left( {L\left( X\right) }\right) \). For all \( j \) we have \[ {R}_{d - j... | Yes |
Lemma 4.5.14. Let \( R\left( X\right) \in W \) and let \( G\left( X\right) = {G}_{0} + {G}_{1}X + \cdots + {G}_{N}{X}^{N} \in \) \( {\mathbb{C}}_{p}\left\lbrack X\right\rbrack \) be a nonzero polynomial such that \( \left| {G}_{N}\right| = \mathop{\max }\limits_{n}\left| {G}_{n}\right| \) . There exist \( L\left( X\rig... | Proof. Let \( {R}^{\left( j\right) }\left( X\right) \in {\mathbb{C}}_{p}\left\lbrack X\right\rbrack \) be a sequence of polynomials tending to \( R\left( X\right) \) . If we perform the Euclidean division of \( {R}^{\left( j\right) }\left( X\right) \) by \( G\left( X\right) \), we can find polynomials \( {L}^{\left( j\... | Yes |
Lemma 4.5.15. Assume that for some \( c < 1 \) there exist \( G \in {\mathbb{C}}_{p}\left\lbrack X\right\rbrack \) and \( H \in W \) such that \( \deg \left( G\right) = N,\parallel f - G\parallel /\parallel f\parallel \leq c \), and \( \parallel H - 1\parallel \leq c \) . Then there exist \( {G}_{1} \) and \( {H}_{1} \... | Proof. We have \( f - {GH} = H\left( {f - G}\right) + f\left( {1 - H}\right) \), hence\n\n\[ \parallel f - {GH}\parallel \leq \max \left( {\parallel H\parallel \parallel f - G\parallel ,\parallel f\parallel \parallel 1 - H\parallel }\right) \leq c\parallel f\parallel \]\n\nsince \( \parallel H - 1\parallel \leq c < 1 \... | Yes |
Corollary 4.5.16. Let \( f\left( X\right) = \mathop{\sum }\limits_{{i \geq 0}}{f}_{i}{X}^{i} \) be an entire function, in other words a power series with infinite radius of convergence. Then either \( f \) has a zero in \( {\mathbb{C}}_{p} \) or \( f = {f}_{0} \) is constant. | Proof. Assume that \( f \) has no zeros in \( {\mathbb{C}}_{p} \) (which implies in particular that \( \left. {{f}_{0} \neq 0}\right) \), let \( n > 0 \), and consider \( {f}_{n}\left( X\right) = f\left( {X/{p}^{n}}\right) \) . By the preparation theorem there exist a polynomial \( {g}_{n}\left( X\right) \) and a power... | Yes |
Proposition 4.5.18 (Nagell). Let \( {u}_{n} \) be the sequence defined by \( {u}_{0} = 0 \) , \( {u}_{1} = 1 \), and \( {u}_{n} = {u}_{n - 1} - 2{u}_{n - 2} \) for \( n \geq 2 \) . Then \( {u}_{n} = \pm 1 \) if and only if \( n = 1,2,3,5 \), and 13 . | Proof. Note that the values of \( {u}_{n} \) for \( 0 \leq n \leq {13} \) are respectively \( 0,1,1 \) , \( - 1, - 3, - 1,5,7, - 3, - {17}, - {11},{23},{45}, - 1 \), so that the indicated values of \( n \) indeed give \( {u}_{n} = \pm 1 \) . | No |
Corollary 4.5.19. The only solutions in \( \\mathbb{Z} \) of\n\n\[ \n{x}^{2} + 7 = {2}^{m}\n\]\n\nare \( \\left( {x, m}\\right) = \\left( {\\pm 1,3}\\right) ,\\left( {\\pm 3,4}\\right) ,\\left( {\\pm 5,5}\\right) ,\\left( {\\pm {11},7}\\right) \), and \( \\left( {\\pm {181},{15}}\\right) \) . | Proof. It is clear that \( x \) is odd; hence set \( x = {2y} - 1 \) with \( y \\in \\mathbb{Z} \), so that we obtain \( {y}^{2} - y + 2 = {2}^{m - 2} \). Let \( \\alpha = \\left( {1 + \\sqrt{-7}}\\right) /2 \) in some algebraic closure of \( \\mathbb{Q} \), and \( \\beta = 1 - \\alpha \), so that \( \\alpha \) and \( ... | Yes |
Proposition 5.1.2. Let \( \\left( {V, q}\\right) \) be a nondegenerate quadratic module.\n\n(1) All morphisms from \( \\left( {V, q}\\right) \) into some other quadratic module \( \\left( {{V}^{\\prime },{q}^{\\prime }}\\right) \) are injective. | Proof. (1). If \( f \) is a morphism from \( \\left( {V, q}\\right) \) to \( \\left( {{V}^{\\prime },{q}^{\\prime }}\\right) \) and \( x \) is such that \( f\\left( x\\right) = 0 \), then for all \( y \\in V \) we have \( x \\cdot y = f\\left( x\\right) \\cdot f\\left( y\\right) = 0 \), hence \( x = 0 \) since \( q \) ... | Yes |
Lemma 5.1.5. Let \( x \neq 0 \) be an isotropic vector of a nondegenerate quadratic module \( \left( {V, q}\right) \). There exists a subspace \( U \) of \( V \) containing \( x \) that is a hyperbolic plane. | Proof. Since \( q \) is nondegenerate, there exists \( z \in V \) such that \( x \cdot z \neq 0 \). The element \( y = {2z} - \left( {z \cdot z/x \cdot z}\right) x \) is clearly isotropic and \( x \cdot y \neq 0 \), so the subspace \( U \) generated by \( x \) and \( y \) is a hyperbolic plane containing \( x \) . | Yes |
Corollary 5.1.6. If \( \left( {V, q}\right) \) is nondegenerate and contains a nonzero isotropic vector then \( q\left( V\right) = K \) . In another language (which we will soon use systematically), if \( q \) represents 0 nontrivially then \( q \) represents every element of \( K \) . | Proof. Thanks to the lemma we know that there exists an isotropic vector \( y \) such that \( x \cdot y = 1 \) . Thus if \( a \in K \) it is clear that \( a = q\left( {x + \left( {a/2}\right) y}\right) \), proving the corollary. | No |
Proposition 5.1.7. Every quadratic module has an orthogonal basis. | Proof. We prove this by induction on \( n = \dim \left( V\right) \), the case \( n = 0 \) being trivial. If \( V \) is isotropic, then all bases are orthogonal. Otherwise, choose some nonisotropic element \( {e}_{1} \in V \), and let \( H = {e}_{1}^{ \bot } \) . This is a subspace of \( V \) of dimension exactly \( n -... | Yes |
Theorem 5.1.9. Let \( \left( {V, q}\right) \) be a nondegenerate quadratic module of dimension at least 3 and let \( \mathcal{B} \) and \( {\mathcal{B}}^{\prime } \) be two orthogonal bases of \( V \) . There exists a finite sequence \( {\left( {\mathcal{B}}^{\left( i\right) }\right) }_{0 \leq i \leq m} \) of orthogona... | Proof. Let \( \mathcal{B} = \left( {{e}_{1},\ldots ,{e}_{n}}\right) \) and \( {\mathcal{B}}^{\prime } = \left( {{e}_{1}^{\prime },\ldots ,{e}_{n}^{\prime }}\right) \) . We consider three different cases.\n\nCase 1: \( {\left( {e}_{1} \cdot {e}_{1}^{\prime }\right) }^{2} \neq \left( {{e}_{1} \cdot {e}_{1}}\right) \left(... | Yes |
Lemma 5.1.10. There exists \( a \in K \) such that \( {e}_{a} = {e}_{1}^{\prime } + a{e}_{2}^{\prime } \) is not isotropic and generates with \( {e}_{1} \) a nondegenerate plane. | Proof. We enumerate the forbidden values of \( a \) . The vector \( {e}_{a} \) is not isotropic if and only if \( {a}^{2} \neq - \left( {{e}_{1}^{\prime } \cdot {e}_{1}^{\prime }}\right) /\left( {{e}_{2}^{\prime } \cdot {e}_{2}^{\prime }}\right) \) . The subspace \( K{e}_{1} + K{e}_{a} \) is a nondegenerate plane if an... | Yes |
Lemma 5.1.11. With the above assumptions, if \( \left( {U, q}\right) \) is degenerate we can extend \( s \) to an injective morphism \( {s}_{1} \) from \( {U}_{1} \) to \( {V}^{\prime } \), where \( {U}_{1} \) is a subspace of \( V \) containing \( U \) as a hyperplane. | Proof. Let \( x \in \operatorname{rad}\left( U\right) \) be nonzero. There exists a linear form \( k \) on \( U \) such that \( k\left( x\right) = 1 \) . Since \( V \) is nondegenerate, as we have seen in the proof of Proposition 5.1.2 the canonical map \( {q}_{U} \) from \( V \) to \( {U}^{ * } \) is surjective; hence... | Yes |
Corollary 5.1.13. Two isomorphic subspaces of a nondegenerate quadratic module have isomorphic orthogonal supplements. | Proof. Indeed, we simply extend the isomorphism between the subspaces to an automorphism of the quadratic module, and then restrict to the orthogonal supplements. | Yes |
Proposition 5.1.16. If \( q \) is nondegenerate and represents 0, then there exists a hyperbolic form \( h \) and a nondegenerate form \( {q}^{\prime } \) such that \( q \sim h \oplus {q}^{\prime } \) . Furthermore, \( q \) represents all elements of \( K \) . | Proof. This is Lemma 5.1.5 and Corollary 5.1.6. The fact that \( {q}^{\prime } \) is nondegenerate follows from Proposition 5.1.2. | No |
Corollary 5.1.17. Let \( q \) be a nondegenerate quadratic form in \( n \) variables and let \( c \in {K}^{ * } \) . The following three conditions are equivalent:\n\n(1) The form \( q \) represents \( c \) .\n\n(2) There exists a quadratic form \( {q}_{1} \) in \( n - 1 \) variables such that \( q \sim c{x}_{0}^{2} \o... | Proof. (2) implies (1) is trivial by taking \( {x}_{0} = 1 \) and the other variables equal to 0 . Conversely, if \( q \) represents \( c \), there exists \( x \in {K}^{n} \) such that \( q\left( x\right) = \) \( x \cdot x = c \) . Since \( q \) is nondegenerate, if \( H = {x}^{ \bot } \) we have \( {K}^{n} = H{ \oplus... | Yes |
Corollary 5.1.18. Let \( {q}_{1} \) and \( {q}_{2} \) be two nonzero nondegenerate quadratic forms and let \( q = {q}_{1} \ominus {q}_{2} \) as defined above. The following properties are equivalent:\n\n(1) The form \( q \) represents 0 .\n\n(2) There exists \( c \in {K}^{ * } \) represented by both \( {q}_{1} \) and \... | Proof. The equivalence of (2) and (3) follows from the above corollary, and (2) implies (1) is trivial. Let us prove that (1) implies (2). If \( q = {q}_{1} \ominus {q}_{2} \) represents 0 there exist \( x \) and \( y \) in the corresponding quadratic modules such that \( {q}_{1}\left( x\right) = {q}_{2}\left( y\right)... | Yes |
Theorem 5.1.19. Let \( q \) be a quadratic form in \( n \) variables. There exists an equivalent form that is a diagonal quadratic form; in other words, there exist \( {a}_{i} \in K \) such that \( q \sim \mathop{\sum }\limits_{{1 \leq i \leq n}}{a}_{i}{x}_{i}^{2} \) . | Proof. This is a translation of Proposition 5.1.7. As already mentioned, this can be proved computationally using Gauss's reduction of quadratic forms into sums of squares, which gives an explicit algorithm for finding the \( {a}_{i} \) and the linear equivalence from \( q \) to the diagonal form. | No |
Theorem 5.1.20 (Witt). Let \( q = {q}_{1} \oplus {q}_{2} \) and \( {q}^{\prime } = {q}_{1}^{\prime } \oplus {q}_{2}^{\prime } \) be two nondegenerate quadratic forms. If \( q \sim {q}^{\prime } \) and \( {q}_{1} \sim {q}_{1}^{\prime } \), then \( {q}_{2} \sim {q}_{2}^{\prime } \). | Proof. This is the translation of Corollary 5.1.13, the corollary to Witt's theorem. It is this theorem to which one usually refers when talking about Witt's theorem. | No |
Corollary 5.1.21. If \( q \) is nondegenerate there exist hyperbolic forms \( {h}_{i} \) for \( 1 \leq i \leq m \) and a form \( {q}^{\prime } \) that does not represent 0 such that\n\n\[ q \sim {h}_{1} \oplus \cdots \oplus {h}_{m} \oplus {q}^{\prime } \]\n\nand this decomposition is unique up to equivalence. | Proof. Existence follows from Proposition 5.1.16. Let us prove uniqueness. With evident notation let\n\n\[ q \sim \mathop{\sum }\limits_{{1 \leq i \leq m}}{h}_{i} \oplus {q}_{1} \sim \mathop{\sum }\limits_{{1 \leq i \leq {m}^{\prime }}}{h}_{i}^{\prime } \oplus {q}_{2}, \]\n\nand assume for instance that \( {m}^{\prime ... | Yes |
Proposition 5.2.1. A quadratic form over \( {\mathbb{F}}_{q} \) of rank \( n \geq 2 \) represents all elements of \( {\mathbb{F}}_{q}^{ * } \), and a quadratic form of rank \( n \geq 3 \) represents all elements of \( {\mathbb{F}}_{q} \) . | Proof. Corollary 5.1.17 tells us that both statements are equivalent. To prove the first we could apply the Chevalley-Warning Theorem 2.5.2, but we give a direct proof that is a direct application of the so-called pigeonhole principle (\ | No |
Proposition 5.2.2. Let \( c \in {\mathbb{F}}_{q}^{ * } \) that is not a square in \( {\mathbb{F}}_{q}^{ * } \) . A nondegenerate quadratic form over \( {\mathbb{F}}_{q} \) is equivalent to \( {x}_{1}^{2} + \cdots + {x}_{n - 1}^{2} + a{x}_{n}^{2} \) with \( a = 1 \) if its discriminant is a square, and with \( a = c \) ... | Proof. Since \( q \) is odd, the map \( x \mapsto {x}^{2} \) is a group homomorphism from \( {\mathbb{F}}_{q}^{ * } \) to itself with kernel \( \{ \pm 1\} \) ; hence \( {\mathbb{F}}_{q}^{ * }/{\mathbb{F}}_{q}^{*2} \) has order 2, generated by the class modulo squares of \( c \) . Thus if \( n = 1 \) the result is true.... | Yes |
Corollary 5.2.3. Two nondegenerate quadratic forms over \( {\mathbb{F}}_{q} \) are equivalent if and only if they have the same rank and the same discriminant in \( {\mathbb{F}}_{q}^{ * }/{\mathbb{F}}_{q}^{*2} \) . | Proof. Clear from the above proposition. | No |
Proposition 5.2.5. Let \( a \) and \( b \) be in \( {\mathcal{K}}^{ * } \) . We have \( \left( {a, b}\right) = 1 \) if and only if \( a \in \mathcal{N}\left( {\mathcal{K}{\left( \sqrt{b}\right) }^{ * }}\right) \), i.e., if and only if \( a \) is the norm of an element of \( \mathcal{K}{\left( \sqrt{b}\right) }^{ * } \)... | Proof. If \( b \) is a square, then \( \left( {0,1,\sqrt{b}}\right) \) is evidently a solution to our equation, and \( \mathcal{K}\left( \sqrt{b}\right) = \mathcal{K} \), so the proposition is clear in this case. Otherwise the elements of the quadratic extension \( \mathcal{K}\left( \sqrt{b}\right) \) have the form \( ... | Yes |
Proposition 5.2.6. We have the following formulas, where all the elements that occur are assumed to be nonzero:\n\n(1) \( \left( {a, b}\right) = \left( {b, a}\right) \) and \( \left( {a,{c}^{2}}\right) = 1 \) .\n\n(2) \( \left( {a, - a}\right) = 1 \) and \( \left( {a,1 - a}\right) = 1 \) .\n\n(3) \( \left( {a, b}\right... | Proof. (1) is clear. When \( b = - a \) ,(respectively \( b = 1 - a \) ), then \( \left( {1,1,0}\right) \) (respectively \( \left( {1,1,1}\right) \) ) is a nontrivial solution to \( a{x}^{2} + b{y}^{2} = {z}^{2} \), proving (2). For (3), by the preceding proposition if \( \left( {a, b}\right) = 1 \) then \( a \in \math... | Yes |
Theorem 5.2.7. (1) For \( \mathcal{K} = \mathbb{R} \), we have \( \left( {a, b}\right) = - 1 \) if \( a < 0 \) and \( b < 0 \) , and \( \left( {a, b}\right) = 1 \) if \( a \) or \( b \) is positive. | Proof. (1) is trivial, so we assume that \( \mathcal{K} = {\mathbb{Q}}_{p} \) . Note first the following lemma. | No |
Lemma 5.2.8. Assume that \( b \in {U}_{p} \) . Then if the equation \( p{x}^{2} + b{y}^{2} = {z}^{2} \) has a nontrivial solution in \( {\mathbb{Q}}_{p} \), it has one such that \( x \in {\mathbb{Z}}_{p} \) and \( y \) and \( z \) are in \( {U}_{p} \) . | Proof. Let \( \left( {x, y, z}\right) \) be a nontrivial solution. Dividing by \( {p}^{v} \) with \( v = \) \( \min \left( {{v}_{p}\left( x\right) ,{v}_{p}\left( y\right) ,{v}_{p}\left( z\right) }\right) \), we may assume that \( x, y \), and \( z \) are in \( {\mathbb{Z}}_{p} \) with at least one of them in \( {U}_{p}... | Yes |
Corollary 5.2.9. The Hilbert symbol is a nondegenerate bilinear form on the \( {\mathbb{F}}_{2} \) -vector space \( {\mathcal{K}}^{ * }/{{\mathcal{K}}^{ * }}^{2} \) . | Proof. When \( \mathcal{K} = \mathbb{R},{\mathcal{K}}^{ * }/{{\mathcal{K}}^{ * }}^{2} \) is a vector space of dimension 1 over \( {\mathbb{F}}_{2} \) having \( \{ 1, - 1\} \) as representatives, and the result is trivial. When \( K = {\mathbb{Q}}_{p} \) , the bilinearity comes from the multiplicativity of the Legendre-... | Yes |
Corollary 5.2.10. If \( b \) is not a square in \( {\mathcal{K}}^{ * } \), then \( \mathcal{N}\left( {K{\left( \sqrt{b}\right) }^{ * }}\right) \) is a subgroup of index 2 in \( {\mathcal{K}}^{ * } \) . | Proof. Clear since the map \( a \mapsto \left( {a, b}\right) \) from \( {\mathcal{K}}^{ * } \) to \( \{ \pm 1\} \) has kernel \( \mathcal{N}\left( {K{\left( \sqrt{b}\right) }^{ * }}\right) \) by Proposition 5.2.5, and is surjective since the Hilbert symbol is nondegenerate. | No |
Proposition 5.2.11. Let \( q\left( {x, y, z}\right) = a{x}^{2} + b{y}^{2} + c{z}^{2} \) be a nondegenerate quadratic form in three variables with coefficients in \( {\mathbb{Q}}_{p} \) (including \( p = \infty \) ). Set \( \varepsilon = \varepsilon \left( q\right) = \left( {a, b}\right) \left( {a, c}\right) \left( {b, ... | Proof. The form \( q \) represents 0 if and only if the form \( - {cq} \) does, hence if and only if \( - {ac}{x}^{2} - {bc}{y}^{2} = {z}^{2} \) has a nontrivial solution, in other words by definition \( \left( {-{ac}, - {bc}}\right) = 1 \) . By bilinearity this condition is\n\n\[ 1 = \left( {-{ac}, - {bc}}\right) = \l... | Yes |
Corollary 5.2.12. Let \( c \in {\mathbb{Q}}_{p}^{ * } \), and let \( q\left( {x, y}\right) = a{x}^{2} + b{y}^{2} \) be a nondegenerate quadratic form in two variables. Then \( q \) represents \( c \) in \( {\mathbb{Q}}_{p} \) if and only if \( \left( {c, - {ab}}\right) = \left( {a, b}\right) \) . | Proof. By Corollary 5.1.17, \( q \) represents \( c \) if and only if the form \( {q}^{\prime }\left( {x, y, z}\right) = \) \( a{x}^{2} + b{y}^{2} - c{z}^{2} \) represents 0 nontrivially. By the above proposition, this is true if and only if \( \left( {-1,{abc}}\right) = \left( {a, b}\right) \left( {a, - c}\right) \lef... | Yes |
Corollary 5.2.13. Let \( c \in {\mathbb{Q}}_{p}^{ * } \) . Then \( c \) is a sum of two squares of elements of \( {\mathbb{Q}}_{p} \) if and only if one of the following holds:\n\n(1) \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \).\n\n(2) \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) and \( {v}_{p}\l... | Proof. By the preceding corollary \( c \) is a sum of two squares if and only if \( \left( {c, - 1}\right) = 1 \) . By Theorem 5.2.7 we have \( \left( {c, - 1}\right) = {\left( \frac{-1}{p}\right) }^{{v}_{p}\left( c\right) } \) for \( p \neq 2 \) and \( \left( {c, - 1}\right) = {\left( -1\right) }^{\left( {c/{v}_{2}\le... | Yes |
Lemma 5.2.14. Let \( \mathcal{K} = {\mathbb{Q}}_{p} \) with \( p \neq \infty \) .\n\n(1) We have \( \left| {{\mathcal{K}}^{ * }/{{\mathcal{K}}^{ * }}^{2}}\right| = {2}^{r} \) with \( r = 2 \) for \( p \neq 2 \) and \( r = 3 \) for \( p = 2 \) . | Proof. (1) has been proved in the preceding chapter (Proposition 4.3.20). | Yes |
Theorem 5.2.15. The value of \( \varepsilon \left( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \right) \) is independent of the linear change of variables that transforms \( q \) into diagonal form, hence is an invariant of the quadratic form itself, which we will denote by \( \varepsilon \left( q\right) \) . | Proof. We use once again the language of quadratic modules, which is more suitable for this kind of proof. Let \( V = {K}^{n} \) and let \( \left( {V, q}\right) \) be the quadratic module associated with \( q \) . The theorem states that \( \varepsilon \) does not depend on the orthogonal basis \( \mathcal{B} \) of \( ... | Yes |
Corollary 5.2.17. Let \( c \in {\mathbb{Q}}_{p}^{ * }/{\mathbb{Q}}_{p}^{*2} \) . A nondegenerate form \( q \) in \( n \) variables with invariants \( d \) and \( \varepsilon \) represents \( c \) if and only if one of the following holds:\n\n(1) \( n = 1 \) and \( c = d \) .\n\n(2) \( n = 2 \) and \( \left( {c, - d}\ri... | Proof. By Corollary 5.1.17, \( q \) represents \( c \) if and only if the nondegenerate form in \( n + 1 \) variables \( {q}_{1} = - c{x}_{0}^{2} + q \) represents 0 nontrivially. We have\n\n\( d\left( {q}_{1}\right) = - {cd}\left( q\right) \) and \( \varepsilon \left( {q}_{1}\right) = \left( {-c, d\left( q\right) }\ri... | Yes |
Proposition 5.2.18. Two quadratic forms over \( {\mathbb{Q}}_{p} \) are equivalent if and only if they have the same rank, discriminant, and invariant \( \varepsilon \left( q\right) \) . | Proof. Left to the reader (Exercise 5). See Exercises 6, 7, and 8 for further properties of quadratic forms over \( {\mathbb{Q}}_{p} \) . | No |
Theorem 5.3.1 (Product formula). If \( a \) and \( b \) are in \( {\mathbb{Q}}^{ * } \) then \( {\left( a, b\right) }_{v} = 1 \) for almost all \( v \in P \) (in other words for all but a finite number), and we have the product formula \[ \mathop{\prod }\limits_{{v \in P}}{\left( a, b\right) }_{v} = 1 \] | Proof. By bilinearity, it is sufficient to prove the theorem when \( a \) and \( b \) are equal to -1 or to a prime number. In these cases, Theorem 5.2.7 gives the answer:\n\n- If \( a = - 1 \) and \( b = - 1 \), then \( {\left( -1, - 1\right) }_{\infty } = {\left( -1, - 1\right) }_{2} = - 1 \), and \( {\left( -1, - 1\... | Yes |
Theorem 5.3.3 (Hasse-Minkowski). Let \( K \) be a number field and let \( q \) be a quadratic form in \( n \) variables with coefficients in \( K \) . Then \( q \) represents 0 in \( K \) if and only if it represents 0 in every completion of \( K \) . | The proof of this theorem for a general number field is outside the scope of this book. Even for \( K = \mathbb{Q} \), the proof is not short, and we will prove the theorem only in this case, as well as stronger statements. | No |
Over any field \( K \) of characteristic different from 2 the form \( a{x}^{2} + {bxy} + c{y}^{2} \) represents 0 nontrivially if and only if \( {b}^{2} - {4ac} \) is a square in \( K \) . | Proof. We note the formal identity\n\n\[ \n{\left( 2ax + by\right) }^{2} - {y}^{2}\left( {{b}^{2} - {4ac}}\right) = {4a}\left( {a{x}^{2} + {bxy} + c{y}^{2}}\right) .\n\]\n\nWe consider two cases. If \( a \neq 0 \), then if \( \left( {x, y}\right) \neq \left( {0,0}\right) \) is such that \( a{x}^{2} + \) \( {bxy} + c{y}... | Yes |
Proposition 5.3.5. Let \( q\left( {x, y, z}\right) \) be a quadratic form in three variables, and assume that \( q\left( {x, y, z}\right) = 0 \) has a nontrivial solution in every completion of \( \mathbb{Q} \) except perhaps in one. Then it has a nontrivial solution in \( \mathbb{Q} \), hence in all places. | Proof. As usual we may assume that \( q \) is nondegenerate, since otherwise the result is trivial, and by changing \( q \) into an equivalent form, that \( q\left( {x, y, z}\right) = a{x}^{2} + b{y}^{2} + c{z}^{2} \) is a diagonal form. By Proposition 5.2.11, \( q \) represents 0 in \( {\mathbb{Q}}_{v} \) if and only ... | Yes |
Theorem 5.3.6. Let \( q \) be a quadratic form in four variables such that \( q = 0 \) has a nontrivial solution in \( \mathbb{R} \) and every \( {\mathbb{Q}}_{p} \) . Then \( q = 0 \) has a nontrivial solution in \( \mathbb{Q} \) . | Proof. We may assume that \( q = {a}_{1}{x}_{1}^{2} + {a}_{2}{x}_{2}^{2} - {a}_{3}{x}_{3}^{2} - {a}_{4}{x}_{4}^{2} \) . Let \( v \) be a place of \( \mathbb{Q} \) . Since \( q \) represents 0 in \( {\mathbb{Q}}_{v} \), Corollary 5.1.18 tells us that there exists \( {c}_{v} \in {\mathbb{Q}}_{v}^{ * } \) that is represen... | Yes |
Theorem 5.3.7. Let \( q \) be a quadratic form in five variables, and assume that \( q = 0 \) has a real solution, in other words that \( q \) is indefinite or singular. Then \( q = 0 \) has a solution in \( \mathbb{Q} \) . | Proof. The proof is very similar to the case \( n = 4 \), so we only give a sketch. We may assume \( q \) nonsingular and diagonal, and with evident notation we write \( g = {a}_{1}{x}_{1}^{2} + {a}_{2}{x}_{2}^{2}, h = - {a}_{3}{x}_{3}^{2} - {a}_{4}{x}_{4}^{2} - {a}_{5}{x}_{5}^{2} \), and we assume \( {a}_{1} > 0 \) an... | Yes |
Corollary 5.3.8. Let \( q \) be a quadratic form in \( n \geq 5 \) variables, and assume that \( q \) represents 0 in \( \mathbb{R} \), in other words that \( q \) is indefinite or singular. Then \( q \) represents 0 in \( \mathbb{Q} \) . | Proof. We may assume \( q \) nonsingular and diagonal, and that \( {a}_{1} > 0 \) , \( {a}_{5} < 0 \) for instance. Then \( q = {q}_{1} + {q}_{2} \) with \( {q}_{1} = \mathop{\sum }\limits_{{1 \leq i \leq 5}}{a}_{i}{x}_{i}^{2} \) . By the above theorem \( {q}_{1} = 0 \) has a nontrivial solution, and we choose \( {x}_{... | Yes |
Proposition 5.4.1. A nondegenerate quadratic form with coefficients in \( \mathbb{Q} \) represents \( c \in {\mathbb{Q}}^{ * } \) if and only if it represents \( c \) in \( \mathbb{R} \) and in every \( {\mathbb{Q}}_{p} \) . | Proof. Clear from Corollary 5.1.17 and the Hasse-Minkowski theorem. | No |
Proposition 5.4.2. Let \( q \) be a nondegenerate quadratic form in \( n \) variables with coefficients in \( \mathbb{Q} \), and let \( d\left( q\right) \in {\mathbb{Q}}^{ * }/{\mathbb{Q}}^{*2} \) be its discriminant. Then \( q \) represents 0 in \( \mathbb{Q} \) if and only one of the following holds:\n\n(1) \( n = 2 ... | Proof. Clearly the given conditions do not change by an invertible linear change of variables, since the discriminant is well defined modulo squares. We may thus assume that \( q \) is in diagonal form. For \( n = 2,{a}_{1}{x}_{1}^{2} + {a}_{2}{x}_{2}^{2} \) represents 0 in \( \mathbb{Q} \) if and only if \( - {a}_{1}/... | Yes |
Corollary 5.4.3. Let \( q \) be a nondegenerate quadratic form in \( n \) variables. A number \( c \in {\mathbb{Q}}^{ * } \) is represented by \( q \) in \( \mathbb{Q} \) if and only if one of the following holds:\n\n(1) \( n = 1 \) and \( c/d\left( q\right) = 1 \in {\mathbb{Q}}^{ * }/{{\mathbb{Q}}^{ * }}^{2} \) .\n\n(... | Proof. Clear from Corollary 5.1.17 and the above proposition. | No |
Proposition 5.4.4. Let \( q \) be a positive definite matrix-integral quadratic form in \( k \) variables. Assume that for every \( x \in {\mathbb{Q}}^{k} \) there exists \( y \in {\mathbb{Z}}^{k} \) such that \( q\left( {x - y}\right) < 1 \) . Then if \( n \in \mathbb{Z} \) is represented by \( q \) in \( \mathbb{Q} \... | Proof. If \( x \) and \( y \) in \( {\mathbb{Q}}^{k} \) are represented by column vectors \( X \) and \( Y \), and if \( Q \) is the matrix of \( q \) with respect to the canonical basis of \( {\mathbb{Q}}^{k} \), we denote by \( x \cdot y \) their scalar product with respect to the quadratic form \( q \) ; in other wo... | Yes |
Theorem 5.4.7 (Bhargava). For any \( S \subset {\mathbb{Z}}_{ \geq 0} \) there exists a minimal witness \( T = \Phi \left( S\right) \) for \( S \), in other words such that \( {T}^{\prime } \) is a witness for \( S \) if and only if \( T \subset {T}^{\prime } \subset S \) . This minimal witness is clearly unique and is... | Note that we also have the trivial equality \( \Phi \left( S\right) = {\Phi }_{s}\left( {2S}\right) /2 \), with evident notation, so if desired we can consider only the map \( \Phi \), or only the map \( {\Phi }_{s} \). | Yes |
Proposition 5.4.9 (Fermat). Let \( n \) be a positive integer. The following three conditions are equivalent:\n\n(1) The integer \( n \) is a sum of two squares of elements of \( \mathbb{Z} \).\n\n(2) The integer \( n \) is a sum of two squares of elements of \( \mathbb{Q} \).\n\n(3) For every prime \( p \mid n \) such... | Proof. By Corollary 5.2.13, \( n \) is a sum of two squares in every \( {\mathbb{Q}}_{p} \) if and only if \( n > 0,2 \mid {v}_{p}\left( n\right) \) for every \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), and \( n/{2}^{{v}_{2}\left( n\right) } \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . However, thi... | Yes |
Proposition 5.4.10 (Landau). As \( X \rightarrow \infty \) the number \( {N}_{2}\left( X\right) \) of \( n \leq X \) that are sums of two squares satisfies\n\n\[ \n{N}_{2}\left( X\right) \sim C\frac{X}{\sqrt{\log \left( X\right) }}, \n\]\n\nwith\n\n\[ \nC = \frac{1}{\sqrt{2}}\mathop{\prod }\limits_{{p \equiv 3\left( {\... | The value of \( C \) is computed using the methods explained in Section 10.3.6; see Exercise 53 of Chapter 10. | No |
Lemma 5.4.11. A number \( c \in {\mathbb{Q}}^{ * } \) is represented over \( \mathbb{Q} \) by the quadratic form \( {x}^{2} + {y}^{2} + {z}^{2} \) if and only if \( c > 0 \) and \( - c \) is not a square in \( {\mathbb{Q}}_{2} \) . | Proof. Let \( q \) be the quadratic form \( {x}^{2} + {y}^{2} + {z}^{2} \) . We have trivially \( d\left( q\right) = \varepsilon \left( q\right) = 1 \) . Thus by Corollary 5.2.17, for \( p \neq \infty, q \) represents \( c \) in \( {\mathbb{Q}}_{p} \) if and only if either \( - c \) is not a square in \( {\mathbb{Q}}_{... | Yes |
Theorem 5.4.12 (Gauss). Let \( n \) be a positive integer. The following three conditions are equivalent:\n\n(1) The integer \( n \) is a sum of three squares of elements of \( \mathbb{Z} \).\n\n(2) The integer \( n \) is a sum of three squares of elements of \( \mathbb{Q} \).\n\n(3) The integer \( n \) is not of the f... | Proof. By Proposition 4.3.23, the condition that \( n \) has the form \( {4}^{a}\left( {{8k} - 1}\right) \) is equivalent to \( - n \) being a square in \( {\mathbb{Q}}_{2} \). Thus by the above lemma the positive integer \( n \) is a sum of three squares in \( \mathbb{Q} \) if and only if \( n \) is not of the form \(... | Yes |
Corollary 5.4.13. As \( X \rightarrow \infty \) the number of integers \( n \leq X \) that are sums of three squares is asymptotic to \( {5X}/6 \) . | Proof. Immediate from the proposition and left to the reader (Exercise 15). | No |
Corollary 5.4.14 (Lagrange). Every positive integer \( n \) is a sum of four squares of integers. | Proof. We write \( n = {4}^{a}\left( {{8k} + m}\right) \) with \( a \geq 0,4 \nmid m \), and \( 1 \leq m \leq 7 \) . By the above theorem, if \( m \neq 7 \), then \( n \) is a sum of three squares. On the other hand, if \( m = 7, n - {4}^{a} = {4}^{a}\left( {{8k} + m - 1}\right) \) is a sum of three squares, so that \(... | No |
Theorem 5.4.15. Let \( n \geq 1 \) be an integer. We have\n\n\[ \n{r}_{2}\left( n\right) = 4\mathop{\sum }\limits_{{d \mid n}}\left( \frac{-4}{d}\right) \n\] | Remarks. (1) The formula for \( {r}_{2}\left( n\right) \) can easily be obtained from Proposition 5.4.9; see Exercise 18. | No |
Theorem 5.6.2. Let \( K \) be any field of characteristic zero, let \( p \) be a prime number, let \( e \geq 1 \) be an integer, and finally let \( L = K\left( {\zeta }_{{p}^{e}}\right) \). If \( x \in {K}^{ * } \), then \( x \) is a \( {p}^{e} \) th power in \( {L}^{ * } \) if and only if it is a \( {p}^{e} \) th powe... | Proof. If \( x \) is a \( {p}^{e} \) th power in \( K \) (or for the exceptional cases in a quadratic extension of \( K \) contained in \( L \) ), then evidently \( x \) is a \( {p}^{e} \) th power in \( L \) . Conversely, let \( x \in {K}^{ * } \) and assume that \( x = {z}^{{p}^{e}} \) for some \( z \in {L}^{ * } \) ... | Yes |
Theorem 5.6.3. Let \( K \) be a number field, let \( p \) be a prime number, let \( e \geq 1 \) be an integer, and let \( S \) be a finite set of places of \( K \) . Let \( x \in {K}^{ * } \), and assume that for all places \( v \notin S \) the element \( x \) considered as an element of the completion \( {K}_{v} \) is... | Proof. We set \( L = K\left( {\zeta }_{{p}^{e}}\right) \), and consider the Kummer extension \( M \) of \( L \) defined by \( M = L\left( {x}^{1/{p}^{e}}\right) \) . This extension is evidently a cyclic extension with Galois group canonically isomorphic to a subgroup of \( \mathbb{Z}/{p}^{e}\mathbb{Z} \) . By the Cebot... | Yes |
Proposition 5.7.1. The equation\n\n\[ \left( {{x}^{2} - 2}\right) \left( {{x}^{2} - {17}}\right) \left( {{x}^{2} - {34}}\right) = 0 \] \n\nhas solutions everywhere locally but not globally. | Proof. It is clear that the equation has a solution in \( \mathbb{R} \) . It has a solution in \( {\mathbb{Q}}_{2} \) since 17 is a square in \( {\mathbb{Q}}_{2} \) by Proposition 4.3.23, and 2 is a square in \n\n\( {\mathbb{Q}}_{17} \) by Proposition 4.3.22, so it has a solution in \( {\mathbb{Q}}_{17} \) . If \( p \)... | Yes |
Proposition 5.7.2. The equation\n\n\[ \n{y}^{2} + {z}^{2} = \left( {3 - {x}^{2}}\right) \left( {{x}^{2} - 2}\right) \n\]\n\nhas solutions everywhere locally but not globally. | Proof. It is clear that the equation has a solution in \( \mathbb{R} \) . For \( p = 3 \), we take \( x = 1 \), and by Corollary 5.2.13, \( - 2 \) is a sum of two squares in \( {\mathbb{Q}}_{3} \) . For all other \( p \), we choose \( x = 0 \) . Since \( {v}_{p}\left( {-6}\right) = 0 \) for \( p \neq 2 \) and 3, Coroll... | Yes |
Proposition 5.7.3 (Lind). The equation \( 2{y}^{2} = {x}^{4} - {17}{z}^{4} \) has nontrivial solutions everywhere locally, but not globally. | Proof. For the local solubility, we prove the stronger statement that the equation \( 2{y}^{2} = {x}^{4} - {17} \) is everywhere locally soluble. The advantage of this equation is that like any hyperelliptic quartic, it represents a curve of genus 1 , which is immediately seen to be nonsingular outside 2,17, and \( \in... | No |
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