Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 21. Let \( A \) be an \( n \times n \) matrix over the field \( F \) . Using the three elementary row and column operations above, the \( n \times n \) matrix \( {xI} - A \) with entries from \( F\left\lbrack x\right\rbrack \) can be put into the diagonal form (called the Smith Normal Form for \( A \) ) | Proof: cf. the exercises. | No |
(1) If a matrix \( A \) is similar to a diagonal matrix \( D \), then \( D \) is the Jordan canonical form of \( A \) . | Proof: The first assertion is immediate from the uniqueness of Jordan canonical forms because a diagonal matrix is itself in Jordan form (with Jordan blocks of size 1). | No |
If \( A \) is an \( n \times n \) matrix with entries from \( F \) and \( F \) contains all the eigenvalues of \( A \), then \( A \) is similar to a diagonal matrix over \( F \) if and only if the minimal polynomial of \( A \) has no repeated roots. | Proof: Suppose \( A \) is similar to a diagonal matrix. The minimal polynomial of a diagonal matrix has no repeated roots (its roots are precisely the distinct elements along the diagonal). Since similar matrices have the same minimal polynomial it follows that the minimal polynomial for \( A \) has no repeated roots.\... | Yes |
Proposition 1. The characteristic of a field \( F,\operatorname{ch}\left( F\right) \), is either 0 or a prime \( p \) . If \( \operatorname{ch}\left( F\right) = p \) then for any \( \alpha \in F \) , | Proof: Only the second statement has not been proved, and this follows immediately from the evident equality \( p \cdot \alpha = p \cdot \left( {{1}_{F}\alpha }\right) = \left( {p \cdot {1}_{F}}\right) \left( \alpha \right) \) in \( F \) . | No |
Theorem 3. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then there exists a field \( K \) containing an isomorphic copy of \( F \) in which \( p\left( x\right) \) has a root. Identifying \( F \) with this isomorphic copy shows that there exists an ... | Proof: Consider the quotient\n\n\[ K = F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \]\n\nof the polynomial ring \( F\left\lbrack x\right\rbrack \) by the ideal generated by \( p\left( x\right) \). Since by assumption \( p\left( x\right) \) is an irreducible polynomial in the P.I.D. \( F\left\lbrack ... | Yes |
Theorem 4. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( n \) over the field \( F \) and let \( K \) be the field \( F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) . Let \( \theta = x{\;\operatorname{mod}\;\left( {p\left( x\right) }\right) } \i... | Proof: Let \( a\left( x\right) \in F\left\lbrack x\right\rbrack \) be any polynomial with coefficients in \( F \) . Since \( F\left\lbrack x\right\rbrack \) is a Euclidean Domain (this is Theorem 3 of Chapter 9), we may divide \( a\left( x\right) \) by \( p\left( x\right) \) :\n\n\[ a\left( x\right) = q\left( x\right) ... | Yes |
Corollary 5. Let \( K \) be as in Theorem 4, and let \( a\left( \theta \right), b\left( \theta \right) \in K \) be two polynomials of degree \( < n \) in \( \theta \) . Then addition in \( K \) is defined simply by usual polynomial addition and multiplication in \( K \) is defined by\n\n\[ a\left( \theta \right) b\left... | By the results proved above, this definition of addition and multiplication on the polynomials of degree \( < n \) in \( \theta \) make \( K \) into a field, so that one can also divide by nonzero elements as well, which is not so immediately obvious from the definitions of the operations. | Yes |
Theorem 6. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial. Suppose \( K \) is an extension field of \( F \) containing a root \( \alpha \) of \( p\left( x\right) : p\left( \alpha \right) = 0 \) . Let \( F\left( \alpha \right) \) denote the subfield of... | Proof: There is a natural homomorphism\n\n\[ \varphi : F\left\lbrack x\right\rbrack \rightarrow F\left( \alpha \right) \subseteq K \]\n\n\[ a\left( x\right) \mapsto a\left( \alpha \right) \]\n\nobtained by mapping \( F \) to \( F \) by the identity map and sending \( x \) to \( \alpha \) and then extending so that the ... | Yes |
Theorem 8. Let \( \varphi : F\overset{ \sim }{ \rightarrow }{F}^{\prime } \) be an isomorphism of fields. Let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial and let \( {p}^{\prime }\left( x\right) \in {F}^{\prime }\left\lbrack x\right\rbrack \) be the irreducible polynomial obtaine... | Proof: As noted above, the isomorphism \( \varphi \) induces a natural isomorphism from \( F\left\lbrack x\right\rbrack \) to \( {F}^{\prime }\left\lbrack x\right\rbrack \) which maps the maximal ideal \( \left( {p\left( x\right) }\right) \) to the maximal ideal \( \left( {{p}^{\prime }\left( x\right) }\right) \) . Tak... | Yes |
Proposition 9. Let \( \alpha \) be algebraic over \( F \) . Then there is a unique monic irreducible polynomial \( {m}_{\alpha, F}\left( x\right) \in F\left\lbrack x\right\rbrack \) which has \( \alpha \) as a root. A polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) has \( \alpha \) as a root if and o... | Proof: Let \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) be a polynomial of minimal degree having \( \alpha \) as a root. Multiplying \( g\left( x\right) \) by a constant, we may assume \( g\left( x\right) \) is monic. Suppose \( g\left( x\right) \) were reducible in \( F\left\lbrack x\right\rbrack \), say \(... | Yes |
Corollary 10. If \( L/F \) is an extension of fields and \( \alpha \) is algebraic over both \( F \) and \( L \) , then \( {m}_{\alpha, L}\left( x\right) \) divides \( {m}_{\alpha, F}\left( x\right) \) in \( L\left\lbrack x\right\rbrack \) . | Proof: This is immediate from the second statement in Proposition 9 applied to \( L \) , since \( {m}_{\alpha, F}\left( x\right) \) is a polynomial in \( L\left\lbrack x\right\rbrack \) having \( \alpha \) as a root. | Yes |
Proposition 11. Let \( \alpha \) be algebraic over the field \( F \) and let \( F\left( \alpha \right) \) be the field generated by \( \alpha \) over \( F \) . Then\n\n\[ F\left( \alpha \right) \cong F\left\lbrack x\right\rbrack /\left( {{m}_{\alpha }\left( x\right) }\right) \]\n\nso that in particular\n\n\[ \left\lbra... | Proof: This follows immediately from Theorem 6. | No |
The element \( \alpha \) is algebraic over \( F \) if and only if the simple extension \( F\left( \alpha \right) /F \) is finite. More precisely, if \( \alpha \) is an element of an extension of degree \( n \) over \( F \) then \( \alpha \) satisfies a polynomial of degree at most \( n \) over \( F \) and if \( \alpha ... | Proof: If \( \alpha \) is algebraic over \( F \), then the degree of the extension \( F\left( \alpha \right) /F \) is the degree of the minimal polynomial for \( \alpha \) over \( F \) . Hence the extension is finite, of degree \( \leq n \) if \( \alpha \) satisfies a polynomial of degree \( n \) . Conversely, suppose ... | Yes |
Corollary 13. If the extension \( K/F \) is finite, then it is algebraic. | Proof: If \( \alpha \in K \), then the subfield \( F\left( \alpha \right) \) is in particular a subspace of the vector space \( K \) over \( F \) . Hence \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \leq \left\lbrack {K : F}\right\rbrack \) and so \( \alpha \) is algebraic over \( F \) by the proposition. | Yes |
Corollary 15. Suppose \( L/F \) is a finite extension and let \( K \) be any subfield of \( L \) containing \( F, F \subseteq K \subseteq L \) . Then \( \left\lbrack {K : F}\right\rbrack \) divides \( \left\lbrack {L : F}\right\rbrack \) . | Proof: This is immediate. | No |
Lemma 16. \( F\left( {\alpha ,\beta }\right) = \left( {F\left( \alpha \right) }\right) \left( \beta \right) \), i.e., the field generated over \( F \) by \( \alpha \) and \( \beta \) is the field generated by \( \beta \) over the field \( F\left( \alpha \right) \) generated by \( \alpha \) . | Proof: This follows by the minimality of the fields in question. The field \( F\left( {\alpha ,\beta }\right) \) contains \( F \) and \( \alpha \), hence contains the field \( F\left( \alpha \right) \), and since it also contains \( \beta \), we have the inclusion \( \left( {F\left( \alpha \right) }\right) \left( \beta... | Yes |
Theorem 17. The extension \( K/F \) is finite if and only if \( K \) is generated by a finite number of algebraic elements over \( F \) . More precisely, a field generated over \( F \) by a finite number of algebraic elements of degrees \( {n}_{1},{n}_{2},\ldots ,{n}_{k} \) is algebraic of degree \( \leq {n}_{1}{n}_{2}... | Proof: If \( K/F \) is finite of degree \( n \), let \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \) be a basis for \( K \) as a vector space over \( F \) . By Corollary 15, \( \left\lbrack {F\left( {\alpha }_{i}\right) : F}\right\rbrack \) divides \( \left\lbrack {K : F}\right\rbrack = n \) for \( i = 1,2,\ldo... | Yes |
Corollary 18. Suppose \( \alpha \) and \( \beta \) are algebraic over \( F \) . Then \( \alpha \pm \beta ,{\alpha \beta },\alpha /\beta \) (for \( \beta \neq 0 \) ), (in particular \( {\alpha }^{-1} \) for \( \alpha \neq 0 \) ) are all algebraic. | Proof: All of these elements lie in the extension \( F\left( {\alpha ,\beta }\right) \), which is finite over \( F \) by the theorem, hence they are algebraic by Corollary 13. | Yes |
Corollary 19. Let \( L/F \) be an arbitrary extension. Then the collection of elements of \( L \) that are algebraic over \( F \) form a subfield \( K \) of \( L \) . | Proof: This is immediate from the previous corollary. | No |
Theorem 20. If \( K \) is algebraic over \( F \) and \( L \) is algebraic over \( K \), then \( L \) is algebraic over \( F \) . | Proof: Let \( \alpha \) be any element of \( L \) . Then \( \alpha \) is algebraic over \( K \), so \( \alpha \) satisfies some polynomial equation\n\n\[ \n{a}_{n}{\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{1}\alpha + {a}_{0} = 0 \n\]\n\nwhere the coefficients \( {a}_{0},{a}_{1},\ldots ,{a}_{n} \) are ... | Yes |
Proposition 21. Let \( {K}_{1} \) and \( {K}_{2} \) be two finite extensions of a field \( F \) contained in \( K \) . Then\n\n\[ \left\lbrack {{K}_{1}{K}_{2} : F}\right\rbrack \leq \left\lbrack {{K}_{1} : F}\right\rbrack \left\lbrack {{K}_{2} : F}\right\rbrack \]\n\nwith equality if and only if an \( F \) -basis for o... | Proof: From \( {K}_{1}{K}_{2} = F\left( {{\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n},{\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m}}\right) = {K}_{1}\left( {{\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m}}\right) , \) we see as above that \( {\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m} \) span \( {K}_{1}{K}_{2} ... | Yes |
Corollary 22. Suppose that \( \left\lbrack {{K}_{1} : F}\right\rbrack = n,\left\lbrack {{K}_{2} : F}\right\rbrack = m \) in Proposition 21, where \( n \) and \( m \) are relatively prime: \( \left( {n, m}\right) = 1 \) . Then \( \left\lbrack {{K}_{1}{K}_{2} : F}\right\rbrack = \left\lbrack {{K}_{1} : F}\right\rbrack \l... | Proof: In general the extension degree \( \left\lbrack {{K}_{1}{K}_{2} : F}\right\rbrack \) is divisible by both \( n \) and \( m \) since \( {K}_{1} \) and \( {K}_{2} \) are subfields of \( {K}_{1}{K}_{2} \), hence is divisible by their least common multiple. In this case, since \( \left( {n, m}\right) = 1 \), this me... | Yes |
Theorem 24. None of the classical Greek problems: (I) Doubling the Cube, (II) Trisecting an Angle, and (III) Squaring the Circle, is possible. | Proof: (I) Doubling the cube amounts to constructing \( \sqrt[3]{2} \) in the reals starting with the unit 1 . Since \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) is not a power of 2, this is impossible.\n\n(II) If an angle \( \theta \) can be constructed, then determining the p... | Yes |
Theorem 25. For any field \( F \), if \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) then there exists an extension \( K \) of \( F \) which is a splitting field for \( f\left( x\right) \) . | Proof: We first show that there is an extension \( E \) of \( F \) over which \( f\left( x\right) \) splits completely into linear factors by induction on the degree \( n \) of \( f\left( x\right) \) . If \( n = 1 \), then take \( E = F \) . Suppose now that \( n > 1 \) . If the irreducible factors of \( f\left( x\righ... | Yes |
Corollary 28. (Uniqueness of Splitting Fields) Any two splitting fields for a polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) over a field \( F \) are isomorphic. | Proof: Take \( \varphi \) to be the identity mapping from \( F \) to itself and \( E \) and \( {E}^{\prime } \) to be two splitting fields for \( f\left( x\right) \left( { = {f}^{\prime }\left( x\right) }\right) \) . | No |
Proposition 29. Let \( \overline{F} \) be an algebraic closure of \( F. \) Then \( \overline{F} \) is algebraically closed. | Proof: Let \( f\left( x\right) \) be a polynomial in \( \bar{F}\left\lbrack x\right\rbrack \) and let \( \alpha \) be a root of \( f\left( x\right) \) . Then \( \alpha \) generates an algebraic extension \( \bar{F}\left( \alpha \right) \) of \( \bar{F} \), and \( \bar{F} \) is algebraic over \( F \) . By Theorem \( {20... | Yes |
Proposition 31. Let \( K \) be an algebraically closed field and let \( F \) be a subfield of \( K \) . Then the collection of elements \( \bar{F} \) of \( K \) that are algebraic over \( F \) is an algebraic closure of \( F \) . An algebraic closure of \( F \) is unique up to isomorphism. | Proof: By definition, \( \bar{F} \) is an algebraic extension of \( F \) . Every polynomial \( f\left( x\right) \in \) \( F\left\lbrack x\right\rbrack \) splits completely over \( K \) into linear factors \( x - \alpha \) (the same is true for every polynomial even in \( K\left\lbrack x\right\rbrack \) ). But each \( \... | Yes |
A polynomial \( f\left( x\right) \) has a multiple root \( \alpha \) if and only if \( \alpha \) is also a root of \( {D}_{x}f\left( x\right) \), i.e., \( f\left( x\right) \) and \( {D}_{x}f\left( x\right) \) are both divisible by the minimal polynomial for \( \alpha \) . In particular, \( f\left( x\right) \) is separa... | Proof: Suppose first that \( \alpha \) is a multiple root of \( f\left( x\right) \) . Then over a splitting field,\n\n\[ f\left( x\right) = {\left( x - \alpha \right) }^{n}g\left( x\right) \]\n\nfor some integer \( n \geq 2 \) and some polynomial \( g\left( x\right) \) . Taking derivatives we obtain\n\n\[ {D}_{x}f\left... | Yes |
Every irreducible polynomial over a field of characteristic 0 (for example, \( \mathbb{Q} \)) is separable. A polynomial over such a field is separable if and only if it is the product of distinct irreducible polynomials. | Proof: Suppose \( F \) is a field of characteristic 0 and \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) is irreducible of degree \( n \). Then the derivative \( {D}_{x}p\left( x\right) \) is a polynomial of degree \( n - 1 \). Up to constant factors the only factors of \( p\left( x\right) \) in \( F\left\lbra... | Yes |
Proposition 35. Let \( F \) be a field of characteristic \( p \) . Then for any \( a, b \in F \) , \[ {\left( a + b\right) }^{p} = {a}^{p} + {b}^{p},\;\text{ and }\;{\left( ab\right) }^{p} = {a}^{p}{b}^{p}. \] Put another way, the \( {p}^{\text{th }} \) -power map defined by \( \varphi \left( a\right) = {a}^{p} \) is a... | Proof: The Binomial Theorem for expanding \( {\left( a + b\right) }^{n} \) for any positive integer \( n \) holds (by the standard induction proof) over any commutative ring: \[ {\left( a + b\right) }^{n} = {a}^{n} + \left( \begin{array}{l} n \\ 1 \end{array}\right) {a}^{n - 1}b + \cdots + \left( \begin{array}{l} n \\ ... | Yes |
Corollary 36. Suppose that \( \mathbb{F} \) is a finite field of characteristic \( p \) . Then every element of \( \mathbb{F} \) is a \( {p}^{\text{th }} \) power in \( \mathbb{F} \) (notationally, \( \mathbb{F} = {\mathbb{F}}^{p} \) ). | Proof: The injectivity of the Frobenius endomorphism of \( \mathbb{F} \) implies that it is also surjective when \( \mathbb{F} \) is finite, which is the statement of the corollary. | Yes |
Proposition 37. Every irreducible polynomial over a finite field \( \mathbb{F} \) is separable. | The important part of the proof of this result is the fact that every element in the characteristic \( p \) field \( \mathbb{F} \) was a \( {p}^{\text{th }} \) power in \( \mathbb{F} \) . This suggests the following definition:\n\nDefinition. A field \( K \) of characteristic \( p \) is called perfect if every element ... | No |
Corollary 42. The degree over \( \mathbb{Q} \) of the cyclotomic field of \( {n}^{\text{th }} \) roots of unity is \( \varphi \left( n\right) \) : \n\n\[ \n\left\lbrack {\mathbb{Q}\left( {\zeta }_{n}\right) : \mathbb{Q}}\right\rbrack = \varphi \left( n\right) \n\] | Proof: By the theorem, \( {\Phi }_{n}\left( x\right) \) is the minimal polynomial for any primitive \( {n}^{\text{th }} \) root of unity \( {\zeta }_{n} \) . | Yes |
Proposition 1. Aut \( \left( K\right) \) is a group under composition and Aut \( \left( {K/F}\right) \) is a subgroup. | Proof: It is clear that \( \operatorname{Aut}\left( K\right) \) is a group. If \( \sigma \) and \( \tau \) are automorphisms of \( K \) which fix \( F \) then also \( {\sigma \tau } \) and \( {\sigma }^{-1} \) are the identity on \( F \), which shows that \( \operatorname{Aut}\left( {K/F}\right) \) is a subgroup. | Yes |
Proposition 2. Let \( K/F \) be a field extension and let \( \alpha \in K \) be algebraic over \( F \) . Then for any \( \sigma \in \operatorname{Aut}\left( {K/F}\right) ,{\sigma \alpha } \) is a root of the minimal polynomial for \( \alpha \) over \( F \) i.e., Aut \( \left( {K/F}\right) \) permutes the roots of irred... | Proof: Suppose \( \alpha \) satisfies the equation\n\n\[ {\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{1}\alpha + {a}_{0} = 0 \]\n\nwhere \( {a}_{0},{a}_{1},\ldots ,{a}_{n - 1} \) are elements of \( F \) . Applying the automorphism \( \sigma \) we obtain (using the fact that \( \sigma \) is an additive h... | Yes |
Proposition 3. Let \( H \leq \operatorname{Aut}\left( K\right) \) be a subgroup of the group of automorphisms of \( K \) . Then the collection \( F \) of elements of \( K \) fixed by all the elements of \( H \) is a subfield of \( K \) . | Proof: Let \( h \in H \) and let \( a, b \in F \) . Then by definition \( h\left( a\right) = a, h\left( b\right) = b \) so that \( h\left( {a \pm b}\right) = h\left( a\right) \pm h\left( b\right) = a \pm b, h\left( {ab}\right) = h\left( a\right) h\left( b\right) = {ab} \) and \( h\left( {a}^{-1}\right) = h{\left( a\rig... | Yes |
Proposition 4. The association of groups to fields and fields to groups defined above is inclusion reversing, namely\n\n(1) if \( {F}_{1} \subseteq {F}_{2} \subseteq K \) are two subfields of \( K \) then \( \operatorname{Aut}\left( {K/{F}_{2}}\right) \leq \operatorname{Aut}\left( {K/{F}_{1}}\right) \), and\n\n(2) if \... | Proof: Any automorphism of \( K \) that fixes \( {F}_{2} \) also fixes its subfield \( {F}_{1} \), which gives (1). The second assertion is proved similarly. | No |
Theorem 7. (Linear Independence of Characters) If \( {\chi }_{1},{\chi }_{2},\ldots ,{\chi }_{n} \) are distinct characters of \( G \) with values in \( L \) then they are linearly independent over \( L \) . | Proof: Suppose the characters were linearly dependent. Among all the linear dependence relations (2) above, choose one with the minimal number \( m \) of nonzero coefficients \( {a}_{i} \) . We may suppose (by renumbering, if necessary) that the \( m \) nonzero coefficients are \( {a}_{1},{a}_{2},\ldots ,{a}_{m} \) :\n... | Yes |
Corollary 10. Let \( K/F \) be any finite extension. Then\n\n\[ \left| {\operatorname{Aut}\left( {K/F}\right) }\right| \leq \left\lbrack {K : F}\right\rbrack \]\n\nwith equality if and only if \( F \) is the fixed field of \( \operatorname{Aut}\left( {K/F}\right) \) . Put another way, \( K/F \) is Galois if and only if... | Proof: Let \( {F}_{1} \) be the fixed field of \( \operatorname{Aut}\left( {K/F}\right) \), so that\n\n\[ F \subseteq {F}_{1} \subseteq K \]\n\nBy Theorem 9, \( \left\lbrack {K : {F}_{1}}\right\rbrack = \left| {\operatorname{Aut}\left( {K/F}\right) }\right| \) . Hence \( \left\lbrack {K : F}\right\rbrack = \left| {\ope... | Yes |
Let \( G \) be a finite subgroup of automorphisms of a field \( K \) and let \( F \) be the fixed field. Then every automorphism of \( K \) fixing \( F \) is contained in \( G \), i.e., \( \operatorname{Aut}\left( {K/F}\right) = G \), so that \( K/F \) is Galois, with Galois group \( G \). | Proof: By definition \( F \) is fixed by all the elements of \( G \) so we have \( G \leq \operatorname{Aut}\left( {K/F}\right) \) (and the question is whether there are any automorphisms of \( K \) fixing \( F \) not in \( G \) i.e., whether this containment is proper). Hence \( \left| G\right| \leq \left| {\operatorn... | Yes |
Corollary 12. If \( {G}_{1} \neq {G}_{2} \) are distinct finite subgroups of automorphisms of a field \( K \) then their fixed fields are also distinct. | Proof: Suppose \( {F}_{1} \) is the fixed field of \( {G}_{1} \) and \( {F}_{2} \) is the fixed field of \( {G}_{2} \) . If \( {F}_{1} = {F}_{2} \) then by definition \( {F}_{1} \) is fixed by \( {G}_{2} \) . By the previous corollary any automorphism fixing \( {F}_{1} \) is contained in \( {G}_{1} \), hence \( {G}_{2}... | Yes |
Proposition 15. Any finite field is isomorphic to \( {\mathbb{F}}_{{p}^{n}} \) for some prime \( p \) and some integer \( n \geq 1 \). | The field \( {\mathbb{F}}_{{p}^{n}} \) is the splitting field over \( {\mathbb{F}}_{p} \) of the polynomial \( {x}^{{p}^{n}} - x \), with cyclic Galois group of order \( n \) generated by the Frobenius automorphism \( {\sigma }_{p} \). The subfields of \( {\mathbb{F}}_{{p}^{n}} \) are all Galois over \( {\mathbb{F}}_{p... | No |
Corollary 16. The irreducible polynomial \( {x}^{4} + 1 \in \mathbb{Z}\left\lbrack x\right\rbrack \) is reducible modulo every prime \( p \) . | Proof: Consider the polynomial \( {x}^{4} + 1 \) over \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) for the prime \( p \) . If \( p = 2 \) we have \( {x}^{4} + 1 = {\left( x + 1\right) }^{4} \) and the polynomial is reducible. Assume now that \( p \) is odd. Then \( {p}^{2} - 1 \) is divisible by 8 since \( p \) is... | Yes |
Proposition 17. The finite field \( {\mathbb{F}}_{{p}^{n}} \) is simple. In particular, there exists an irreducible polynomial of degree \( n \) over \( {\mathbb{F}}_{p} \) for every \( n \geq 1 \) . | We have described the finite fields \( {\mathbb{F}}_{{p}^{n}} \) above as the splitting fields of the polynomials \( {x}^{{p}^{n}} - x \) . By the previous proposition, this field can also be described as a quotient of \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \), namely by the minimal polynomial for \( \theta \) ... | Yes |
Proposition 19. Suppose \( K/F \) is a Galois extension and \( {F}^{\prime }/F \) is any extension. Then \( K{F}^{\prime }/{F}^{\prime } \) is a Galois extension, with Galois group\n\n\[ \operatorname{Gal}\left( {K{F}^{\prime }/{F}^{\prime }}\right) \cong \operatorname{Gal}\left( {K/K \cap {F}^{\prime }}\right) \]\n\ni... | Proof: If \( K/F \) is Galois, then \( K \) is the splitting field of some separable polynomial \( f\left( x\right) \) in \( F\left\lbrack x\right\rbrack \) . Then \( K{F}^{\prime }/{F}^{\prime } \) is the splitting field of \( f\left( x\right) \) viewed as a polynomial in\n\n\( {F}^{\prime }\left\lbrack x\right\rbrack... | Yes |
Corollary 20. Suppose \( K/F \) is a Galois extension and \( {F}^{\prime }/F \) is any finite extension.\n\nThen\n\[ \left\lbrack {K{F}^{\prime } : F}\right\rbrack = \frac{\left\lbrack {K : F}\right\rbrack \left\lbrack {{F}^{\prime } : F}\right\rbrack }{\left\lbrack K \cap {F}^{\prime } : F\right\rbrack }.\] | Proof: This follows by the proposition from the equality \( \left\lbrack {K{F}^{\prime } : {F}^{\prime }}\right\rbrack = \left\lbrack {K : K \cap {F}^{\prime }}\right\rbrack \) given by the orders of the Galois groups in the proposition. | No |
Proposition 21. Let \( {K}_{1} \) and \( {K}_{2} \) be Galois extensions of a field \( F \) . Then\n\n(1) The intersection \( {K}_{1} \cap {K}_{2} \) is Galois over \( F \) .\n\n(2) The composite \( {K}_{1}{K}_{2} \) is Galois over \( F \) . The Galois group is isomorphic to the subgroup\n\n\[ H = \left\{ {\left( {\sig... | Proof: (1) Suppose \( p\left( x\right) \) is an irreducible polynomial in \( F\left\lbrack x\right\rbrack \) with a root \( \alpha \) in \( {K}_{1} \cap {K}_{2} \) . Since \( \alpha \in {K}_{1} \) and \( {K}_{1}/F \) is Galois, all the roots of \( p\left( x\right) \) lie in \( {K}_{1} \) . Similarly all the roots lie i... | Yes |
Corollary 22. Let \( {K}_{1} \) and \( {K}_{2} \) be Galois extensions of a field \( F \) with \( {K}_{1} \cap {K}_{2} = F \) . Then\n\n\[ \operatorname{Gal}\left( {{K}_{1}{K}_{2}/F}\right) \cong \operatorname{Gal}\left( {{K}_{1}/F}\right) \times \operatorname{Gal}\left( {{K}_{2}/F}\right) . \] | Proof: The first part follows immediately from the proposition. For the second, let \( {K}_{1} \) be the fixed field of \( {G}_{1} \subset G \) and let \( {K}_{2} \) be the fixed field of \( {G}_{2} \subset G \) . Then \( {K}_{1} \cap {K}_{2} \) is the field corresponding to the subgroup \( {G}_{1}{G}_{2} \), which is ... | No |
Corollary 23. Let \( E/F \) be any finite separable extension. Then \( E \) is contained in an extension \( K \) which is Galois over \( F \) and is minimal in the sense that in a fixed algebraic closure of \( K \) any other Galois extension of \( F \) containing \( E \) contains \( K \) . | Proof: There exists a Galois extension of \( F \) containing \( E \), for example the composite of the splitting fields of the minimal polynomials for a basis for \( E \) over \( F \) (which are all separable since \( E \) is separable over \( F \) ). Then the intersection of all the Galois extensions of \( F \) contai... | Yes |
Proposition 24. Let \( K/F \) be a finite extension. Then \( K = F\left( \theta \right) \) if and only if there exist only finitely many subfields of \( K \) containing \( F \). | Proof: Suppose first that \( K = F\left( \theta \right) \) is simple. Let \( E \) be a subfield of \( K \) containing \( F : F \subseteq E \subseteq K \) . Let \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) be the minimal polynomial for \( \theta \) over \( F \) and let \( g\left( x\right) \in E\left\lbrack x\... | Yes |
Theorem 25. (The Primitive Element Theorem) If \( K/F \) is finite and separable, then \( K/F \) is simple. In particular, any finite extension of fields of characteristic 0 is simple. | Proof: Let \( L \) be the Galois closure of \( K \) over \( F \) . Then any subfield of \( K \) containing \( F \) corresponds to a subgroup of the Galois group \( \operatorname{Gal}\left( {L/F}\right) \) by the Fundamental Theorem. Since there are only finitely many such subgroups, the previous proposition shows that ... | No |
Theorem 26. The Galois group of the cyclotomic field \( \mathbb{Q}\left( {\zeta }_{n}\right) \) of \( {n}^{\text{th }} \) roots of unity is isomorphic to the multiplicative group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \) . The isomorphism is given explicitly by the map\n\n\[ \n{\left( \mathbb{Z}/n\mathbb... | Proof: The discussion above shows that \( {\sigma }_{a} \) is an automorphism for any \( a\left( {\;\operatorname{mod}\;n}\right) \) , so the map above is well defined. It is a homomorphism since\n\n\[ \n\left( {{\sigma }_{a}{\sigma }_{b}}\right) \left( {\zeta }_{n}\right) = {\sigma }_{a}\left( {\zeta }_{n}^{b}\right) ... | Yes |
Corollary 27. Let \( n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{k}^{{a}_{k}} \) be the decomposition of the positive integer \( n \) into distinct prime powers. Then the cyclotomic fields \( \mathbb{Q}\left( {\zeta }_{{p}_{i}^{{a}_{i}}}\right), i = 1,2,\ldots, k \) intersect only in the field \( \mathbb{Q} \) an... | Proof: The only statement which has not been proved is the identification of the isomorphism of Galois groups with the statement of the Chinese Remainder Theorem on the group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \), which is quite simple and is left for the exercises. | No |
Proposition 29. The regular \( n \) -gon can be constructed by straightedge and compass if and only if \( n = {2}^{k}{p}_{1}\cdots {p}_{r} \) is the product of a power of 2 and distinct Fermat primes. | The proof above actually indicates a procedure for constructing the regular \( n \) -gon as a succession of square roots. For example, the construction of the regular 17-gon (solved by Gauss in 1796 at age 19) requires the construction of the subfields of degrees \( 2,4,8 \) and 16 in \( \mathbb{Q}\left( {\zeta }_{17}\... | No |
Corollary 31. (Fundamental Theorem on Symmetric Functions) Any symmetric function in the variables \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) is a rational function in the elementary symmetric functions \( {s}_{1},{s}_{2},\ldots ,{s}_{n} \) . | Proof: A symmetric function lies in the fixed field of \( {S}_{n} \) above, hence is a rational function in \( {s}_{1},\ldots ,{s}_{n} \) . | No |
Proposition 34. The Galois group of \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) is a subgroup of \( {A}_{n} \) if and only if the discriminant \( D \in F \) is the square of an element of \( F \) . | Proof: This is a restatement of Proposition 33 in this case. The Galois group is contained in \( {A}_{n} \) if and only if every element of the Galois group fixes\n\n\[ \sqrt{D} = \mathop{\prod }\limits_{{i < j}}\left( {{\alpha }_{i} - {\alpha }_{j}}\right) \]\n\ni.e., if and only if \( \sqrt{D} \in F \) . | No |
Proposition 36. Let \( F \) be a field of characteristic not dividing \( n \) which contains the \( {n}^{\text{th }} \) roots of unity. Then the extension \( F\left( \sqrt[n]{a}\right) \) for \( a \in F \) is cyclic over \( F \) of degree dividing \( n \) . | Proof: The extension \( K = F\left( \sqrt[n]{a}\right) \) is Galois over \( F \) if \( F \) contains the \( {n}^{\text{th }} \) roots of unity since it is the splitting field for \( {x}^{n} - a \) . For any \( \sigma \in \operatorname{Gal}\left( {K/F}\right) ,\sigma \left( \sqrt[n]{a}\right) \) is another root of this ... | Yes |
Lemma 38. If \( \alpha \) is contained in a root extension \( K \) as in (21) above, then \( \alpha \) is contained in a root extension which is Galois over \( F \) and where each extension \( {K}_{i + 1}/{K}_{i} \) is cyclic. | Proof: Let \( L \) be the Galois closure of \( K \) over \( F \) . For any \( \sigma \in \operatorname{Gal}\left( {L/F}\right) \) we have the chain of subfields\n\n\[ F = \sigma {K}_{0} \subset \sigma {K}_{1} \subset \cdots \subset \sigma {K}_{i} \subset \sigma {K}_{i + 1} \subset \cdots \subset \sigma {K}_{s} = {\sigm... | Yes |
Theorem 39. The polynomial \( f\left( x\right) \) can be solved by radicals if and only if its Galois group is a solvable group. | Proof: Suppose first that \( f\left( x\right) \) can be solved by radicals. Then each root of \( f\left( x\right) \) is contained in an extension as in the lemma. The composite \( L \) of such extensions is\n\nagain of the same type by Proposition 21. Let \( {G}_{i} \) be the subgroups corresponding to the subfields \(... | Yes |
Corollary 41. For any prime \( p \) not dividing the discriminant of \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \), the Galois group of \( f\left( x\right) \) over \( \mathbb{Q} \) contains an element with cycle decomposition \( \left( {{n}_{1},{n}_{2},\ldots ,{n}_{k}}\right) \) where \( {n}_{1},{n}_... | ## Example\n\nConsider the polynomial \( {x}^{5} - x - 1 \) . The discriminant of this polynomial is \( {2869} = {19} \cdot {151} \) so we reduce at primes \( \neq {19},{151} \) . Reducing mod 2 the polynomial \( {x}^{5} - x - 1 \) factors as \( \left( {{x}^{2} + x + 1}\right) \left( {{x}^{3} + {x}^{2} + 1}\right) \lef... | Yes |
Proposition 1. Let \( I \) be a nonempty countable set and for each \( i \in I \) let \( {A}_{i} \) be a set. The cardinality of the Cartesian product is the product of the cardinalities of the sets \( {A}_{i} \), i.e., \[ \left| {\mathop{\prod }\limits_{{i \in I}}{A}_{i}}\right| = \mathop{\prod }\limits_{{i \in I}}\le... | Proof: In order to count the number of choice functions note that each \( i \in I \) may be mapped to any of the \( \left| {A}_{i}\right| \) elements of \( {A}_{i} \) and for \( i \neq j \) the values of choice functions at \( i \) and \( j \) may be chosen completely independently. Thus the number of choice functions ... | Yes |
Theorem 2. Assuming the usual (Zermelo-Fraenkel) axioms of set theory, the following are equivalent: (1) Zorn's Lemma (2) the Axiom of Choice (3) the Well Ordering Principle. | Proof: This follows from elementary set theory. We refer the reader to Real and Abstract Analysis by Hewitt and Stromberg, Springer-Verlag, 1965, Section 3 for these equivalences and some others. | No |
Proposition 0.1. Let \( f : A \rightarrow B \) .\n\n(1) The map \( f \) is injective if and only if \( f \) has a left inverse. | Proof. (a) Suppose \( f \) is injective so that \( {f}^{-1}\left( b\right) \) contains a single element for \( b \in f\left( A\right) \) . Thus \( g\left( b\right) = {f}^{-1}\left( b\right) \) is well-defined for \( b \in f\left( A\right) \) . Hence \( g\left( {f\left( a\right) }\right) = a \) for all \( a \in A \) . N... | Yes |
Theorem 0.3. The operations of addition and multiplication on \( \mathbb{Z}/n\mathbb{Z} \) are 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 \( \overline{{a}_{1}}... | \[ {a}_{1} \equiv {b}_{1}{\;\operatorname{mod}\;n}\;\text{ and }\;{a}_{2} \equiv {b}_{2}{\;\operatorname{mod}\;n} \] then \[ {a}_{1} + {a}_{2} \equiv {b}_{1} + {b}_{2}{\;\operatorname{mod}\;n}\;\text{ and }\;{a}_{1}{a}_{2} \equiv {b}_{1}{b}_{2}{\;\operatorname{mod}\;n}. \] | Yes |
2.4.8. Prove that \( {S}_{4} = \langle \left( {1234}\right) ,\left( {1243}\right) \rangle \) . | Proof. First note that \( \left| {S}_{4}\right| = 4! = {24} \) . Since \( \langle \left( {1234}\right) ,\left( {1243}\right) \rangle \leq {S}_{4} \), Lagrange’s theorem states we need to find 13 elements generated by this set so that \( \langle \left( {1234}\right) ,\left( {1243}\right) \rangle = {S}_{4} \) . So we hav... | Yes |
3.2.8. Prove that if \( H \) and \( K \) are finite subgroups of \( G \) whose orders are relatively prime then \( H \cap K = 1 \) . | Proof. Let \( \left| H\right| = m \) and \( \left| K\right| = n \) with \( \left( {m, n}\right) = 1 \) . Now take \( x \in H \cap K \) so \( \left| x\right| \left| {m\text{and}}\right| x\left| \right| n \) so \( x = 1 \) . | Yes |
Theorem 3.19 (The Third Isomorphism Theorem). Let \( G \) be a group and let \( H \) and \( K \) be normal subgroups of \( G \) with \( H \leq K \) . Then \( K/H \trianglelefteq G/H \) and | \[ \left( {G/H}\right) /\left( {K/H}\right) \cong G/K \] | Yes |
8.1.3. Let \( R \) be a Euclidean Domain. Let \( m \) be the minimum integer in the set of norms of nonzero elements of \( R \) . Prove that every nonzero element of \( R \) of norm \( m \) is a unit. Deduce that a nonzero element of norm zero (if such an element exists) is a unit. | Proof. Let \( a \in R \) with \( N\left( a\right) = m \) . Since \( R \) is a Euclidean Domain, there exists \( q, r \in R \) such that \( 1 = {qa} + r \) with \( N\left( r\right) < N\left( a\right) \) or \( r = 0 \) . Since \( a \) is chosen to be minimum norm, \( r = 0 \) . Thus \( a \) is a unit. | No |
9.2.4. Let \( F \) be a finite field. Prove that \( F\left\lbrack x\right\rbrack \) contains infinitely many primes. | Proof. Suppose on the other hand that \( {p}_{1}\left( x\right) ,\ldots ,{p}_{n}\left( x\right) \) are all of the primes in \( F\left\lbrack x\right\rbrack \) . Let \( p\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}{p}_{i}\left( x\right) \) and notice \( 1 + p\left( x\right) \in F\left\lbrack x\right\rbrack \) ... | Yes |
Lemma 9.25. Suppose \( {f}_{1},\ldots {f}_{m} \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) are polynomials with the same multidegree \( \alpha \) and that the linear combination \( h = {a}_{1}{f}_{1} + \cdots + {a}_{m}{f}_{m} \) with constants \( {a}_{i} \in F \) has strictly smaller multidegree. Then | \[ h = \mathop{\sum }\limits_{{i = 2}}^{m}{b}_{i}S\left( {{f}_{i - 1},{f}_{i}}\right) ,\;\text{ for some constants }{b}_{i} \in F. \] | No |
Proposition 13.1. The characteristic of a field \( F,\operatorname{ch}\left( F\right) \), is either 0 or a prime \( p \) . If \( \operatorname{ch}\left( F\right) = p \) then for any \( \alpha \in F \) , | \[ p \cdot \alpha = \alpha + \alpha + \cdots + \alpha = 0. \] | No |
Theorem 13.6. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) be an irreducible polynomial. Suppose \( K \) is an extension field of \( F \) containing a root \( \alpha \) of \( p\left( x\right) : p\left( \alpha \right) = 0 \) . Let \( F\left( \alpha \right) \) denote the subfield... | \[ F\left( \alpha \right) \cong F\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) . \] | No |
13.2.1. Let \( \\mathbb{F} \) be a finite field of characteristic \( p \) . Prove that \( \\left| \\mathbb{F}\\right| = {p}^{n} \) for some positive integer \( n \) . | Proof. Let \( \\varphi : \\mathbb{Z} \\rightarrow \\mathbb{F} \) be the ring map given by \( \\varphi \\left( n\\right) = n \\cdot {1}_{\\mathbb{F}} \) . Since \( p\\mathbb{Z} \\subseteq \\ker \\varphi \) then \( \\bar{\\varphi } : {\\mathbb{F}}_{p} \\rightarrow \\mathbb{F} \) is well defined. But \( \\bar{\\varphi } \... | Yes |
Corollary 13.42. The degree over \( \mathbb{Q} \) of the cyclotomic field of \( {n}^{th} \) roots of unity is \( \varphi \left( n\right) \) : | \[ \left\lbrack {\mathbb{Q}\left( {\zeta }_{n}\right) : \mathbb{Q}}\right\rbrack = \varphi \left( n\right) \] | Yes |
Proposition 3.4.1. For all distinct primes \( r \) and \( \ell \) and positive integers \( n \) ,\n\n\[ \n{c}_{\sigma ,\tau, r,\ell, n} = {\left( \sqrt[{\ell n}]{r}{\sigma }^{{\sigma \tau } - \sigma }\right) }^{\left( {{\log }_{\ell }{\chi }_{\ell }\left( \sigma \right) }\right) /{\log }_{\ell }r}\left( {\sigma ,\tau \... | Proof: Set\n\n\[ \n{g}_{\sigma } = : {g}_{{\Gamma }_{\sigma }, r,\ell ,{\tau }_{\ell }} \in {M}_{r,\ell }\;\left( {\sigma \in G\left( \mathbb{Q}\right) }\right) ,\n\]\n\n(the RHS is as defined in \( §{1.10} \) and the embedding \( {\tau }_{\ell } \) is part of the normalization data) and abridge the notation of \( §2 \... | Yes |
Proposition 3.5.2. If \( p \neq r \) and \( p \neq \ell \), then the extension \( {\bar{G}}_{r,\ell, n, p} \) is trivial and hence, equivalently, \( {h}_{p}\left( \left\lbrack {\bar{G}}_{r,\ell, n}\right\rbrack \right) = 0 \) . | Proof: Firstly, consider the case \( p = \infty \) . Then, by [A2, Theorem 1(VI) of \( §{1.6}\rbrack ,{B}_{\rho } = 1 \) ( \( \rho = \) : complex conjugation), hence \( {\Gamma }_{\rho } = 1 \) . Therefore \( {c}_{\sigma ,\tau } = 1 \) for all \( \sigma ,\tau \in {G}_{\infty } \), and a fortiori the extension \( {\bar{... | Yes |
Proposition 3.6.4. The Hasse invariant \( {h}_{r}\left( {\widetilde{c}}_{r,\ell, n}\right) \) is nonvanishing and independent of \( r \) . | Proof: It is well known [AT] that there exists a unique isomorphism \( t \) : \( {H}^{2}\left( {G\left( \mathbb{Q}\right) ,{C}_{\overline{\mathbb{Q}}}}\right) \widetilde{ \rightarrow }\mathbb{Q}/\mathbb{Z} \) rendering the diagram below commutative:\n\n\[ \left. {{C}_{\mathbb{Q}} = {H}^{0}\left( {G\left( \mathbb{Q}\rig... | No |
Proposition 3.6.5. The restrictions to \( {G}_{r} \) of the 2-cocycles \( {c}_{\sigma ,\tau, r,\ell, n} \) and \( {\widetilde{c}}_{\sigma ,\tau, r,\ell, n} \) are cohomologous. | Proof: Let \( \sigma ,\tau \in {G}_{r} \) be arbitrary. Put\n\n\[ f\left( \sigma \right) = \exp \left( {a\left( \sigma \right) \log r}\right), g\left( \sigma \right) = f{\left( \sigma \right) }^{\sigma },\]\n\nwhere the logarithm \( \log r \) is taken positive. Then\n\n\[ {\widetilde{c}}_{\sigma ,\tau, r,\ell, n} = f\l... | Yes |
Proposition 3.7.1. Assumptions and notation as above\n\n\[ \n{g}_{\sigma }\left( {\xi /{\ell }^{n}}\right) \equiv \left( {{u}_{\sigma }\xi \left( {1 - {\log }_{\ell }\xi }\right) /{\log }_{\ell }r + {v}_{\sigma }{n\xi }}\right) /{\ell }^{n}\;\left( {\sigma \in G\left( \mathbb{Q}\right) ,\xi \in {{\mathbf{Z}}_{\ell }}^{... | Proof: In view of Theorem \( 2,{g}_{\sigma } \) is uniquely determined by \( {u}_{\sigma },{v}_{\sigma } \) and the condition \( \bar{g}\left( 1\right) = {u}_{\sigma }/{\log }_{\ell }r \) following from the normalization condition (3.1.1). The explicit formula for \( {g}_{\sigma } \) comes from the discussions of \( §§... | Yes |
\[ {H}^{2}\left( {{G}_{S},{\mathbb{Q}}_{\ell }/{\mathbb{Z}}_{\ell }\left( n\right) }\right) = 0\text{ for }\begin{array}{ll} \text{ a) } & 1 < n,\text{ or } \\ \text{ b) } & 1 > n, \end{array} \] | Case a) has been proved by Soulé [Sou1] via higher Chern classes on algebraic \( K \) -theory; see the discussion in \( §2 \) . | No |
If \( E \) is an elliptic curve over \( k \), then \( {H}^{0}\left( {\bar{E},{\mathbf{Z}}_{\ell }}\right) = {\mathbf{Z}}_{\ell } \) and \( {H}^{2}\left( {\bar{E},{\mathbf{Z}}_{\ell }}\right) = {\mathbf{Z}}_{\ell }\left( {-1}\right) \) are covered by the above. For the remaining case \( i = 1 \), we may use the Kummer s... | \[ {H}^{1}\left( {\bar{E},\mathbf{Z}/{\ell }^{\nu }\left( 1\right) }\right) = {H}^{1}\left( {\bar{E},{\mu }_{{\ell }^{\nu }} = \operatorname{Pic}{\left( \bar{E}\right) }_{{\ell }^{\nu }} = E{\left( \bar{k}\right) }_{{\ell }^{\nu }}.}\right. | Yes |
Consider \( X = \operatorname{Spec}k \) : It follows from results of Borel and Soulé that \( {K}_{2n}\left( {\mathcal{O}}_{k}\right) \) is finite for \( n > 0 \) and that the \( \ell \) -adic Chern class\n\n\[ \n{c}_{1, n} : {K}_{{2n} - 1}\left( {\mathcal{O}}_{k}\right) \otimes {\mathbb{Z}}_{\ell } \rightarrow {H}_{\te... | Now \( {H}_{\text{ét }}^{1}\left( {\operatorname{Spec}\left( {{\mathcal{O}}_{k}\left\lbrack \frac{1}{\ell }\right\rbrack }\right) ,{\mathbb{Z}}_{\ell }\left( n\right) }\right) = {H}^{1}\left( {{G}_{S},{\mathbb{Z}}_{\ell }\left( n\right) }\right) \) for \( S = {S}_{\ell } \) (cf. (16) below), and \( {K}_{r}\left( {\math... | Yes |
Example 4 Let \( k \) be an imaginary quadratic field and let \( X = E \) be an elliptic curve with complex multiplication by \( k \) . Assume that \( \ell \) splits in \( k \) , \( \ell = \mathfrak{p}{\mathfrak{p}}^{ * } \), and that \( E \) has good reduction at \( \ell \) . In [Sou5] Soulé considers a map \[ {r}^{\p... | On the other hand, it follows from the results of \( \mathrm{K} \) . Wingberg mentioned in Example 2 that \( {\alpha }_{\mathfrak{p}} \) has a finite kernel for regular \( \ell \) . Hence \( {r}^{\prime } \) has a finite cokernel, in support of Conjecture 2. | Yes |
For \( \mathfrak{p} \mid p \) one has \[ {\left( {H}^{i}\left( \bar{X},{\mathbb{Q}}_{p}\left( m\right) \right) \mathop{\bigotimes }\limits_{{\mathbb{Q}}_{p}}{\mathbb{C}}_{p}\right) }^{{G}_{p}} = 0\text{ for }\;\begin{array}{ll} \text{ a) }m < 0 & \left( { \Leftrightarrow i + 1 < n}\right) ,\text{ or } \\ \text{ b’) }m ... | In particular, the same vanishing holds for \( {H}^{i}{\left( \bar{X},{\mathbb{Q}}_{p}\left( m\right) \right) }^{{G}_{\mathfrak{p}}} \subseteq \left( {{H}^{i}(\bar{X}}\right. \) , \( {\left. {\mathbb{Q}}_{p}\left( m\right) ) \otimes {\mathbb{C}}_{p}\right) }^{{G}_{\mathfrak{p}}} \), and, by Lemma 11, \( {H}^{2}\left( {... | No |
Let \( G \) be the symmetric group \( {S}_{m} \) with \( m \geq 3 \) and \( \mathfrak{C} = \left( {{2A},\left( {m - 1}\right) A,{mA}}\right) \) the class structure of \( {S}_{m} \) consisting of the transpositions, the \( \left( {m - 1}\right) \) -cycles and the \( m \) -cycles. Then obviously \( \mathfrak{C} \) coinci... | By conjugating the transposition \( {\sigma }_{1} = \left( {ij}\right) \) by a suitable power of \( {\sigma }_{3} \) we get \( {\sigma }_{1} = \left( {1k}\right) \) with the additional property \( 1 \leq k \leq \frac{m + 2}{2} \) . Since \( {\sigma }_{3}{\sigma }_{1} \) has to be a \( \left( {m - 1}\right) \) -cycle, w... | Yes |
Let \( G \) be a cyclic group \( {Z}_{m} \) of order \( m \) generated by \( \sigma \) . Then \( \mathfrak{C} \mathrel{\text{:=}} \left( {\left\lbrack \sigma \right\rbrack ,\left\lbrack {\sigma }^{-1}\right\rbrack }\right) \) is a class structure of \( G \) with \( {\ell }^{i}\left( \mathfrak{C}\right) = 1 \) and \( {\... | If \( \widehat{\mathbb{S}} \) is chosen to be the set consisting of the numerator and the denominator of the principal divisor \( \left( t\right) \) of \( t \) in \( \mathbb{Q}\left( t\right) /\mathbb{Q} \), then by Remark 1 the field \( k \) in Theorem 1 coincides with \( {k}_{\mathfrak{C}} = {\mathbf{Q}}^{\left( m\ri... | Yes |
Proposition 2. For every class structure \( \mathfrak{C} \) of a finite group \( G \) we have \( {\ell }^{i}\left( \mathfrak{C}\right) \leq n\left( \mathfrak{C}\right) \) . Moreover \( {\ell }^{i}\left( \mathfrak{C}\right) = n\left( \mathfrak{C}\right) \) if and only if \( \bar{\sum }\left( \mathfrak{C}\right) = \sum \... | This proposition is proved in [22] and [37]. | No |
Proposition 3. The group of admissible topological automorphisms \( {H}_{\widehat{\mathbf{S}}}\left( \mathfrak{C}\right) \) (and \( {H}_{\widehat{S}}\left( {\mathfrak{C}}^{ * }\right) \) respectively) operates in an effectively computable way on the set \( {\sum }^{i}\left( \mathfrak{C}\right) \) (or \( {\sum }^{i}\lef... | This statement arises from the topological origin of the \( s \) -generator \( \underline{\sigma } = \) \( \left( {{\sigma }_{1},\ldots ,{\sigma }_{s}}\right) \) of \( G \) . According to section 1, the \( {\sigma }_{j} \) are homomorphic images of the homotopy classes \( {\alpha }_{j} \) of loops around \( {P}_{j} \in... | Yes |
Let \( G \) be the group \( {PS}{L}_{2}\left( p\right) \) and let \( {\mathfrak{C}}_{2} \) be the class structure \( \left( {{2A},{pA},{pB}}\right) \) consisting of the class of involutions and the two different classes of elements of order \( p \) in \( G \) . Then the ramification structure spanned by \( {\mathfrak{C... | Moreover we have \( \operatorname{Sym}\left( {\mathfrak{C}}^{ * }\right) = \langle \left( {23}\right) \rangle \) and\n\n\[ \n{\ell }^{i}\left( {\mathfrak{C}}_{2}^{ * }\right) = 2,\;{\widetilde{\ell }}^{i}\left( {\mathfrak{C}}_{2}^{ * }\right) = 1\text{ for }\left( \frac{2}{p}\right) = - 1. \n\] | No |
Let \( G \) be the Mathieu group \( {M}_{12} \) and \( \mathfrak{C} = \left( {{4A},{4A},{10A}}\right) \) the class structure consisting of the class \( {4A} \) of elements with permutation type \( {\left( 1\right) }^{4}{\left( 4\right) }^{2} \) and the class \( {10A} \) of elements of order 10, which have type (2)(10).... | Now there exist only three conjugacy classes of maximal subgroups of \( G \) with elements of order 10, two of type \( {M}_{10} \rtimes {Z}_{2} \) and one of the type \( {S}_{5} \times {Z}_{2} \) (see for example [6]). As can be seen easily (see for example [27]), no \( \left( {{\sigma }_{1},{\sigma }_{2},{\sigma }_{3}... | Yes |
Proposition 4. The profinite pure Hurwitz braid group \( {B}^{s} \) acts in an effectively computable way on the set of \( s \) -generators \( {\sum }^{i}\left( \mathfrak{C}\right) \) and \( {\sum }^{i}\left( {\mathfrak{C}}^{ * }\right) \) . | Now we use \( \Delta \mathrel{\text{:=}} \operatorname{Aut}\left( {\widehat{K}/\mathbb{Q}\left( \underline{t}\right) }\right), B \mathrel{\text{:=}} {B}^{s} \) and for \( \left\lbrack \underline{\sigma }\right\rbrack \in {\sum }_{s}^{i}\left( \mathfrak{C}\right) \)\n\n\[ \n{\Delta }_{\underline{\sigma }}^{i} \mathrel{\... | Yes |
Let \( G \) be the symmetric group \( {S}_{m} \) for \( m \geq 4 \) and \( \mathfrak{C} \) the class structure \( \left( {{2A},{2A},\left( {m - 2}\right) A,{mA}}\right) \) consisting of the class of transpositions \( {2A} \) and of the classes of \( \left( {m - 2}\right) \) -cycles and \( m \) -cycles. Then we get \( {... | The permutation types of \( {\beta }_{1j} \) on \( Z \) are\n\n\[ \operatorname{Typ}\left( {\beta }_{12}\right) = {\left( 1\right) }^{\left( m - 3\right) }\left( 3\right) \]\n\n\[ \operatorname{Typ}\left( {\beta }_{13}\right) = \operatorname{Typ}\left( {\beta }_{14}\right) = \left( 1\right) \left( {m - 1}\right) . \]\n... | Yes |
Let \( G \) be the group \( {PG}{L}_{2}\left( 7\right) \) and \( \mathfrak{C} \) the class structure \( \left( {{2A},{2A},{2B},{6A}}\right) \), which consists of the classes of involutions \( {2A} \) of even sign and \( {2B} \) of odd sign in the natural permutation representation on \( {\mathbb{P}}^{1}\left( {\mathbb{... | For the permutation types of \( {\beta }_{1j} \) on \( Z \) we get\n\n\[ \operatorname{Typ}\left( {\beta }_{12}\right) = {\left( 1\right) }^{2}{\left( 2\right) }^{2}{\left( 3\right) }^{2} \]\n\n\[ \operatorname{Typ}\left( {\beta }_{13}\right) = \operatorname{Typ}\left( {\beta }_{14}\right) = \left( \underline{\underlin... | Yes |
Let \( G \) be the Mathieu group \( {M}_{24} \) and \( \mathfrak{C} \) the class structure \( \left( {{12B},{2A},{2A},{2A}}\right) \) consisting of the classes of permutation type \( {\left( {12}\right) }^{2} \) and \( {\left( 1\right) }^{8}{\left( 2\right) }^{8} \) in the natural permutation representation. Then obvio... | The computer has to work for the results: The set \( {\sum }^{i}\left( {\mathfrak{C}}^{ * }\right) \) is an exceptional \( {\Pi }_{3} \) -orbit \( Z \) of length 144 and\n\n\[ \operatorname{Typ}\left( {\beta }_{1j}\right) = {\left( 2\right) }^{6}{\left( 3\right) }^{39}{\left( 5\right) }^{3}\text{ for }j = 2,3,4. \]\n\n... | Yes |
Proposition 2. We have the inequality:\n\n\[ \operatorname{Krull}\dim \left( {R/{pR}}\right) \geq \delta = {d}^{1} - {d}^{2}. \]\n\nIf \( {d}^{2} = 0 \) (i.e., the lifting problem for \( \bar{\rho } \) is unobstructed) then we have equality above, and moreover \( R \) is a formal power series ring in \( {d}^{1} \) para... | Proof: The ring \( R/{pR} \) is the universal deformation ring for characteristic \( p \) deformations of \( \bar{\rho } \) . Let \( F \) be a power series ring in \( {d}^{1} \) variables over \( k \), and let \( F \rightarrow R/{pR} \) be a continuous homomorphism which induces an isomorphism on Zariski tangent spaces... | Yes |
Proposition 3. Let \( \bar{\rho } \) be an ordinary, absolutely irreducible,2-dimensional residual representation. Then a universal ordinary deformation of \( \bar{\rho } \) exists. That is, there is a local ring\n\n\[ \n{R}^{ \circ } = {R}^{ \circ }\left( {\Pi, k,\bar{\rho }}\right) \in \mathcal{C} \]\n\nand an ordina... | Proof: Similar to the Proposition of \( §2 \) . Note that the notion of ordinariness is, in general, destroyed by twisting by a one-dimensional character. | No |
Proposition 4. Let\n\n\[ \bar{\rho } : \Pi \rightarrow {\mathrm{{GL}}}_{N}\left( k\right) \]\n\nbe a residual representation (absolutely irreducible). Let \( H \subset {\mathrm{{GL}}}_{N}\left( k\right) \) be the image of \( \Pi \) under \( \bar{\rho } \) . Suppose that\n\n\[ {H}^{1}\left( {H,{\operatorname{Ad}}_{H}^{0... | Proof: It suffices to show surjectivity of reduced tangent spaces. That is, it suffices to show that there are no non-constant deformations of \( \bar{\rho } \) to \( k\left\lbrack \epsilon \right\rbrack \left( {{\epsilon }^{2} = 0}\right) \) with traces lying in \( k \subset k\left\lbrack \epsilon \right\rbrack \) .\n... | Yes |
Proposition 6. Let \( \bar{\rho } \) be as above, and suppose that \( p \geq 7 \) . Then one of the two conditions below holds:\n\n(A) There is a Galois extension \( M/\mathbb{Q} \) unramified outside \( S \), containing \( L/\mathbb{Q} \), with Galois group isomorphic to \( {\mathrm{{SL}}}_{2}\left( {{\mathbb{F}}_{p}\... | Proof: In remark (2) at the end of \( §{10} \) we see that, viewing \( \bar{\rho } \) as a representation into \( {\mathrm{{GL}}}_{2}\left( {\mathbb{F}}_{p}\right) \), we have \( \delta \geq 1 \) with equality if \( {d}^{2} = 0 \) . Thus either \( {d}^{2} = 0 \) and \( {d}^{1} = 1 \) which gives (B) or \( {d}^{1} > 1 \... | Yes |
Proposition 7. If \( \bar{\rho } \) is a neat residual representation whose image has order prime to \( p \), then \( {d}^{2} = 0 \) . The universal deformation ring \( R \) of \( \bar{\rho } \) is isomorphic to a power series ring over \( W\left( k\right) \) in \( \delta = {d}^{1} \) parameters. | Proof: Let \( {S}^{\prime } \) denote the set of places of \( L \) lying above the places \( S \) of \( K \) . Let \( {Y}_{L} = \operatorname{Spec}\left( {\mathcal{O}}_{L}\right) \) where \( {\mathcal{O}}_{L} \) is the ring of integers in \( L \) and let\n\n\( {Y}_{L,{S}^{\prime }} \subset {Y}_{L} \) denote the open su... | Yes |
For each prime number \( p \) of the form \( {27} + 4{a}^{3} \) with \( a \in \mathbb{Z} \) there is a unique special \( {S}_{3} \) -representation (up to equivalence), | \[ \bar{\rho } : {G}_{\mathbb{Q},\{ p,\infty \} } \rightarrow {\mathrm{{GL}}}_{2}\left( {\mathbb{F}}_{p}\right) \] The special \( {S}_{3} \) -representations are neat (in the sense of \( §{12} \) above). If \( R = \) \( R\left( \bar{\rho }\right) \) denotes the universal deformation ring of a special \( {S}_{3} \) -rep... | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.