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Lemma 3.3. The subset \( {\mathcal{C}}_{\mathbb{R}} \subseteq \mathbb{R} \) of constructible numbers is a subfield of \( \mathbb{R} \). Likewise, \( {\mathcal{C}}_{\mathbb{C}} \) is a subfield of \( \mathbb{C} \), and in fact \( {\mathcal{C}}_{\mathbb{C}} = {\mathcal{C}}_{\mathbb{R}}\left( i\right) \). | Proof. The set \( {\mathcal{C}}_{\mathbb{R}} \subseteq \mathbb{R} \) is nonempty, so in order to show it is a field, we only need to show that it is closed with respect to subtraction and division by a nonzero constructible number (cf. Proposition III6.2).\n\nThe reader will check that \( {\mathcal{C}}_{\mathbb{R}} \) ... | No |
Example 3.5. Regular pentagons are constructible. | Indeed, it suffices to construct the point \( A = \left( {\cos \left( {{2\pi }/5}\right) ,0}\right) \), and it so happens that \( \gamma = \cos \left( {{2\pi }/5}\right) \) satisfies\n\n(*) \n\n\[ \n4{\gamma }^{2} + {2\gamma } - 1 = 0 \n\] \n\n(in fact, \( \gamma \) is half of the inverse of the golden ratio: \( \gamma... | No |
Corollary 3.6. Let \( \gamma \in {\mathcal{C}}_{\mathbb{C}} \) be a constructible number. Then \( \left\lbrack {\mathbb{Q}\left( \gamma \right) : \mathbb{Q}}\right\rbrack \) is a power of 2 . | Proof. By Lemma 3.3 and Theorem 3.4, there exist \( {\delta }_{1}\ldots ,{\delta }_{k} \in \mathbb{R} \) such that\n\n\[ \gamma \in \mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{k}, i}\right) \]\n\nand each \( {\delta }_{j} \) has degree \( \leq 2 \) over \( \mathbb{Q}\left( {{\delta }_{1},\ldots ,{\delta }_{j - 1... | Yes |
The splitting field \( F \) of \( {x}^{8} - 1 \) over \( \mathbb{Q} \) is generated by \( \zeta \mathrel{\text{:=}} {e}^{{2\pi i}/8} \) | indeed, the roots of \( {x}^{8} - 1 \) are all the 8-th roots of 1, and all of them are powers of \( \zeta \) : In fact, \( \zeta \) is a root of the polynomial \( {x}^{4} + 1 \), which is irreducible over \( \mathbb{Q} \) ; therefore \( F = \mathbb{Q}\left( \zeta \right) \) is ’already’ the splitting field of \( {x}^{... | Yes |
Example 4.5. Variation on the theme: \( {x}^{4} - 1 \) . The situation changes if instead of \( {x}^{4} + 1 \) we consider \( {x}^{4} - 1 \) : this polynomial factors over \( \mathbb{Q} \) , | \[ {x}^{4} - 1 = \left( {x - 1}\right) \left( {x + 1}\right) \left( {{x}^{2} + 1}\right) \] and it follows that the splitting field is the same as for \( {x}^{2} + 1 \), that is, just \( \mathbb{Q}\left( i\right) \) . | Yes |
Example 4.6. Variation on the theme: \( {x}^{4} + 2 \) . The overoptimistic reader may now hope that the difference between the splitting fields of \( {x}^{4} + 1 \) vs. \( {x}^{4} - 1 \) over \( \mathbb{Q} \) is just due to the fact that the first polynomial is irreducible over \( \mathbb{Q} \) and the second is not. ... | \[ \sqrt[4]{2}\zeta ,\sqrt[4]{2}{\zeta }^{3},\sqrt[4]{2}{\zeta }^{5},\sqrt[4]{2}{\zeta }^{7} \] Therefore, with \( K = \mathbb{Q}\left( {\sqrt[4]{2}\zeta ,\sqrt[4]{2}{\zeta }^{3},\sqrt[4]{2}{\zeta }^{5},\sqrt[4]{2}{\zeta }^{7}}\right) \) the splitting field of \( {x}^{4} + 2 \) , \[ K \subseteq \mathbb{Q}\left( {\zeta ... | Yes |
If a complex root of an irreducible polynomial \( p\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) may be expressed as a polynomial in \( i \) and \( \sqrt[4]{2} \) with rational coefficients, then all roots of \( p\left( x\right) \) may be expressed likewise in terms of \( i \) and \( \sqrt[4]{2} \) . | Indeed, we have checked (Example 4.6) that \( \mathbb{Q}\left( {i,\sqrt[4]{2}}\right) \) is a splitting field over \( \mathbb{Q} \) ; hence it is a normal extension of \( \mathbb{Q} \) . | Yes |
Let \( p \) be a prime, and consider the field \( {\mathbb{F}}_{p}\left( t\right) \) of rational functions over \( {\mathbb{F}}_{p} \). Then the polynomial \[ {x}^{p} - t \in {\mathbb{F}}_{p}\left( t\right) \left\lbrack x\right\rbrack \] is irreducible. | by Eisenstein’s criterion it is irreducible in \( {\mathbb{F}}_{p}\left\lbrack t\right\rbrack \left\lbrack x\right\rbrack \) (since \( \left( t\right) \) is prime in \( {\mathbb{F}}_{p}\left\lbrack t\right\rbrack \) ), hence in \( {\mathbb{F}}_{p}\left( t\right) \left\lbrack x\right\rbrack \) by Proposition V14.16 Let ... | Yes |
Lemma 4.13. Let \( k \) be a field, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) . Then \( f\left( x\right) \) is separable if and only if \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) are relatively prime. | Proof. First assume that \( f\left( x\right) \) is not separable. Then \( f\left( x\right) \) has a multiple root in a splitting field \( F \) ; that is,\n\n\[ f\left( x\right) = {\left( x - \alpha \right) }^{m}g\left( x\right) \]\n\nfor some \( \alpha \in F, g\left( x\right) \in F\left\lbrack x\right\rbrack \), and \(... | Yes |
Lemma 4.14. Let \( k \) be a field, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an inseparable irreducible polynomial. Then \( {f}^{\prime }\left( x\right) = 0 \) . | Proof. Since \( f\left( x\right) \) is inseparable, \( f\left( x\right) \) and \( {f}^{\prime }\left( x\right) \) have a common irreducible factor \( q\left( x\right) \) by Lemma 4.13 but as \( f\left( x\right) \) is itself irreducible, \( q\left( x\right) \) must be an associate of \( f\left( x\right) \), and in parti... | Yes |
Proposition 4.17. Let \( k \) be a field. Then \( k \) is perfect if and only if all irreducible polynomials in \( k\left\lbrack x\right\rbrack \) are separable. | Proof. I will prove that irreducible polynomials over a perfect field are separable, leaving the other implication to the reader (Exercise 4.12).\n\nWe have already noted that irreducible polynomials are separable over fields of characteristic zero. In positive characteristic \( p \), we have observed that an inseparab... | No |
Corollary 4.18. Finite fields are perfect. Therefore, over finite fields, irreducible polynomials are separable. | Proof. The Frobenius map is injective (because it is a homomorphism of fields), so it is surjective over finite fields, by the pigeon-hole principle. Therefore finite fields are perfect, and the second part of the statement follows from Proposition 4.17. | Yes |
Lemma 4.22. Let \( k \subseteq k\left( \alpha \right) \) be a simple algebraic extension. Then \( {\left\lbrack k\left( \alpha \right) : k\right\rbrack }_{s} \) equals the number of distinct roots in \( \bar{k} \) of the minimal polynomial of \( \alpha \) . In particular, \( {\left\lbrack k\left( \alpha \right) : k\rig... | Proof. The proof is essentially (and not by coincidence) a rehash of the proof of Corollary 1.7. Associate with each \( \iota : k\left( \alpha \right) \rightarrow \bar{k} \) extending \( {\operatorname{id}}_{k} \) the image \( \iota \left( \alpha \right) \), which must be a root of the minimal polynomial of \( \alpha \... | Yes |
Lemma 4.23. Let \( k \subseteq E \subseteq F \) be algebraic extensions. Then \( {\left\lbrack F : k\right\rbrack }_{s} \) is finite if and only if both \( {\left\lbrack F : E\right\rbrack }_{s},{\left\lbrack E : k\right\rbrack }_{s} \) are finite, and in this case \[ {\left\lbrack F : k\right\rbrack }_{s} = {\left\lbr... | Proof. Different embeddings of \( E \) into \( \bar{k} \) extend to different embeddings of \( F \) into \( \bar{k} \) , by Lemma 2.8, and embeddings of \( F \) into \( \bar{E} = \bar{k} \) extending the identity on \( E \) extend a fortiori the identity on \( k \) . Therefore, if any of \( {\left\lbrack F : E\right\rb... | Yes |
Proposition 4.24. Let \( k \subseteq F \) be a finite extension. Then \( {\left\lbrack F : k\right\rbrack }_{s} \leq \left\lbrack {F : k}\right\rbrack \), and the following are equivalent:\n\n(i) \( F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \), where each \( {\alpha }_{i} \) is separable over \( k \) ;\n... | Proof. Since \( F \) is finite over \( k \), it is finitely generated. Let \( F = k\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right) \) . Then using Lemma 4.23, Lemma 4.22, and Proposition 1.10,\n\n\[{\left\lbrack F : k\right\rbrack }_{s} = {\left\lbrack k\left( {\alpha }_{1},\ldots ,{\alpha }_{r - 1}\right) \left( {... | Yes |
Let \( q = {p}^{d} \) be a power of a prime integer \( p \) . Then the polynomial \( {x}^{q} - x \) is separable over \( {\mathbb{F}}_{p} \), and the splitting field of the polynomial \( {x}^{q} - x \) over \( {\mathbb{F}}_{p} \) is a field with precisely \( q \) elements. Conversely, let \( F \) be a field with exactl... | Proof. Let \( F \) be the splitting field of \( {x}^{q} - x \) over \( {\mathbb{F}}_{p} \) . Let \( E \) be the set of roots of \( f\left( x\right) = {x}^{q} - x \) in \( F \) . Since \( {f}^{\prime }\left( x\right) = q{x}^{q - 1} - 1 = - 1 \) (as \( q = 0 \) in characteristic \( p \) ), we have \( \left( {f\left( x\ri... | Yes |
Corollary 5.2. For every prime power \( q \) there exists one and only one finite field of order \( q \), up to isomorphism. | Proof. This follows immediately from Theorem 5.1 and the uniqueness of splitting fields (Lemma 4.2). | Yes |
Let \( p \) be a prime integer. Then I claim that the polynomial \( {x}^{4} + 1 \) is reducible over \( {\mathbb{F}}_{p} \) (and therefore over every finite field). | Since \( {x}^{4} + 1 = {\left( x + 1\right) }^{4} \) in \( {\mathbb{F}}_{2}\left\lbrack x\right\rbrack \), the statement holds for \( p = 2 \) . Thus, we may assume that \( p \) is an odd prime. Then I claim that \( {x}^{4} + 1 \) divides \( {x}^{{p}^{2}} - x \) . Indeed, the square of every odd number is congruent to ... | No |
Corollary 5.4. Let \( p \) be a prime, and let \( d \leq e \) be positive integers. Then there is an extension \( {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{e}} \) if and only if \( d \mid e \) . Further, if \( d \mid e \), then there is exactly one such extension, in the sense that \( {\mathbb{F}}_{{p}^{e}} \... | Proof. If there is an extension as stated, then \( {\mathbb{F}}_{p} \subseteq {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{e}} \) ; hence \( \left\lbrack {{\mathbb{F}}_{{p}^{d}} : {\mathbb{F}}_{p}}\right\rbrack \) divides \( \left\lbrack {{\mathbb{F}}_{{p}^{e}} : {\mathbb{F}}_{p}}\right\rbrack \) by Corollary 1.... | Yes |
Corollary 5.5. Let \( F \) be a finite field. Then for all integers \( n \geq 1 \) there exist irreducible polynomials of degree \( n \) in \( F\left\lbrack x\right\rbrack \) . | Proof. We know \( F = {\mathbb{F}}_{{p}^{d}} \) for some prime \( p \) and some \( d \geq 1 \) . By Corollary 5.4 there is an extension \( {\mathbb{F}}_{{p}^{d}} \subseteq {\mathbb{F}}_{{p}^{dn}} \), generated by an element \( \alpha \) . Then \( \left\lbrack {{\mathbb{F}}_{{p}^{dn}} : {\mathbb{F}}_{{p}^{d}}}\right\rbr... | Yes |
Corollary 5.6. Let \( F = {\mathbb{F}}_{q} \) be a finite field, and let \( n \) be a positive integer. Then the factorization of \( {x}^{{q}^{n}} - x \) in \( F\left\lbrack x\right\rbrack \) consists of all irreducible monic polynomials of degree \( d \), as \( d \) ranges over the positive divisors of \( n \) . In pa... | Proof. By Theorem 5.1, \( {\mathbb{F}}_{{q}^{n}} \) is the splitting field of \( {x}^{{q}^{n}} - x \) over \( {\mathbb{F}}_{p} \), and hence over \( {\mathbb{F}}_{q} = F \) . If \( f\left( x\right) \) is a monic irreducible polynomial of degree \( d \), then \( F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\r... | Yes |
Let’s contemplate the case \( q = 2 : {\mathbb{F}}_{2} = \mathbb{Z}/2\mathbb{Z} \). | - \( n = 1 \) : the polynomial \( {x}^{2} - x \) factors as the product of \( x \) and \( \left( {x - 1}\right) \) (which we could write as \( \left( {x + 1}\right) \) just as well, since we are working over \( {\mathbb{F}}_{2} \) ). These are all the irreducible polynomials of degree 1 over \( {\mathbb{F}}_{2} \).\n\n... | Yes |
Proposition 5.8. \( {\operatorname{Aut}}_{{\mathbb{F}}_{p}}\left( {\mathbb{F}}_{{p}^{d}}\right) \) is cyclic, generated by the Frobenius isomorphism. | Proof. Let \( \varphi \) be the Frobenius homomorphism \( {\mathbb{F}}_{{p}^{d}} \rightarrow {\mathbb{F}}_{{p}^{d}} : \varphi \left( x\right) = {x}^{p} \) . The Frobenius homomorphism is an isomorphism on a finite field (Corollary 4.18) and restricts to the identity on \( {\mathbb{F}}_{p} \) (Exercise 4.11), so \( \var... | Yes |
If \( n = p \) is prime, then every nonidentity element of \( {\mu }_{p} \cong {C}_{p} \) is a generator: every \( p \) -th root of 1 is primitive except 1 itself. | \[ {\Phi }_{p}\left( x\right) = \frac{{x}^{p} - 1}{x - 1} = {x}^{p - 1} + \cdots + 1 \] is the particular case encountered in Example V15.19, where we proved that \( {\Phi }_{p}\left( x\right) \) is indeed irreducible. | No |
Lemma 5.11. For all positive integers \( n \) , \n\n\[ \n{x}^{n} - 1 = \mathop{\prod }\limits_{{1 \leq d \mid n}}{\Phi }_{d}\left( x\right) \n\] | Proof. If \( n = {de} \), then every \( d \) -th root \( \zeta \) of 1 is an \( n \) -th root of 1, because \( {\zeta }^{n} = \) \( {\zeta }^{de} = {\left( {\zeta }^{d}\right) }^{e} = 1 \) . In particular, every primitive \( d \) -th root \( \zeta \) of 1 is an \( n \) -th root of 1 . \n\nOn the other hand, every \( \z... | Yes |
Corollary 5.12. The cyclotomic polynomials \( {\Phi }_{n}\left( x\right) \) have integer coefficients. | Proof. Use induction on \( n \) . Note that \( {\Phi }_{1}\left( x\right) = x - 1 \), and assume we have shown that all \( {\Phi }_{m}\left( x\right) \) have integer coefficients for \( m < n \) . In particular, \( f\left( x\right) \mathrel{\text{:=}} \) \( \mathop{\prod }\limits_{{1 \leq d \mid n, d < n}}{\Phi }_{d}\l... | Yes |
The reader can spend some quality time computing explicitly the cyclotomic polynomials \( {\Phi }_{n}\left( x\right) \) for several nonprime numbers \( n \), working inductively and capitalizing on the fact that we know explicitly \( {\Phi }_{p}\left( x\right) \) for prime \( p \) . | For example, \( {x}^{4} - 1 = {\Phi }_{1}\left( x\right) {\Phi }_{2}\left( x\right) {\Phi }_{4}\left( x\right) \) ; therefore\n\n\[ \n{\Phi }_{4}\left( x\right) = \frac{{x}^{4} - 1}{{x}^{2} - 1} = {x}^{2} + 1 \n\]\n\nSince \( {x}^{6} - 1 = {\Phi }_{1}\left( x\right) {\Phi }_{2}\left( x\right) {\Phi }_{3}\left( x\right)... | Yes |
Proposition 5.14. For all positive \( n,{\Phi }_{n}\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) is irreducible over \( \mathbb{Q} \) . | (Since \( {\Phi }_{n}\left( x\right) \) is monic, irreducibility in \( \mathbb{Q}\left\lbrack x\right\rbrack \) is equivalent to irreducibility in \( \mathbb{Z}\left\lbrack x\right\rbrack \) ; cf. Corollary V 4.17\n\nProof. Arguing by contradiction, assume \( {\Phi }_{n}\left( x\right) \) is reducible. Then its roots \... | Yes |
Proposition 5.16. \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) }\right) \) is isomorphic to the group of units in \( \mathbb{Z}/n\mathbb{Z} \) . | Proof. We know that \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) }\right) \) has cardinality \( \phi \left( n\right) \) (Corollary 1.7) the roots are distinct since \( {\Phi }_{n}\left( x\right) \) is separable), so all we need to do is exhibit an injective homomorphism\n\n\[ j : {\l... | Yes |
Proposition 5.19. Every finite separable extension is simple. | Proof. Arguing inductively as in the proof of Proposition 5.18 we may assume \( F = k\left( {\alpha ,\beta }\right) \), with \( \alpha \) and \( \beta \) separable (and in particular algebraic) over \( k \), and we may assume \( k \) is an infinite field.\n\nConsider the set \( I \) of embeddings \( \iota : F \hookrigh... | Yes |
Corollary 5.20. Let \( k \subseteq F \) be a finite, separable extension. Then\n\n\[ \left| {{\operatorname{Aut}}_{k}\left( F\right) }\right| \leq \left\lbrack {F : k}\right\rbrack \]\n\nwith equality if and only if \( k \subseteq F \) is a normal extension. | Proof. Since \( k \subseteq F \) is finite and separable, Proposition 5.19 implies it is simple: \( F = k\left( \alpha \right) \) for some \( \alpha \in F \) . The inequality follows immediately from Corollary 1.7, and equality holds if and only if the minimal polynomial \( f\left( x\right) \) of \( \alpha \) factors i... | Yes |
Lemma 6.3. The Galois correspondence is inclusion-reversing. Further, for all subgroups \( G \) of \( {\operatorname{Aut}}_{k}\left( F\right) \) and all intermediate fields \( k \subseteq E \subseteq F \) :\n\n\[ \n\text{-}E \subseteq {F}^{{\operatorname{Aut}}_{E}\left( F\right) }\text{;}\n\]\n\n\[ \n\text{-}G \subsete... | ## Proof. Exercise 6.1 | No |
Consider the extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( \sqrt[3]{2}\right) \). Since \( \left\lbrack {\mathbb{Q}\left( \sqrt[3]{2}\right) : \mathbb{Q}}\right\rbrack = 3 \) is prime, the only intermediate fields are \( \mathbb{Q} \) and \( \mathbb{Q}\left( \sqrt[3]{2}\right) \) (by Corollary 1.11). Concerning \(... | Thus, in this example the Galois correspondence acts between a set with two elements and a singleton: \( \{ \mathbb{Q},\mathbb{Q}\left( \sqrt[3]{2}\right) \} \rightleftarrows \left\{ {{\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\left( \sqrt[3]{2}\right) }\right) }\right\} = \{ e\} \). | Yes |
Proposition 6.5. Let \( k \subseteq F \) be a finite extension, and let \( G \) be a subgroup of \( {\operatorname{Aut}}_{k}\left( F\right) \) . Then \( \left| G\right| = \left\lbrack {F : {F}^{G}}\right\rbrack \), and | \[ G = {\operatorname{Aut}}_{{F}^{G}}\left( F\right) \] | No |
Lemma 6.6. Let \( k \subseteq F \) be a finite extension, and let \( G \) be a subgroup of \( {\operatorname{Aut}}_{k}\left( F\right) \) . Then \( {F}^{G} \subseteq F \) is a finite, simple, normal, separable extension. | Proof of Lemma 6.6. The extension \( {F}^{G} \subseteq F \) is finite because \( k \subseteq F \) is finite.\n\nLet \( \alpha \in F \) ; by the remark following the statement of the lemma, \( \alpha \) is a root of a separable polynomial \( {q}_{\alpha }\left( t\right) \) with coefficients in \( {F}^{G} \) . It follows... | Yes |
Theorem 6.9. Let \( k \subseteq F \) be a finite field extension. Then the following are equivalent:\n\n(1) \( F \) is the splitting field of a separable polynomial \( f\left( t\right) \in k\left\lbrack t\right\rbrack \) over \( k \) ;\n\n(2) \( k \subseteq F \) is normal and separable;\n\n(3) \( \left| {{\operatorname... | Proof. Most of the needed implications have been proven along the way.\n\n\( \left( 1\right) \Leftrightarrow \left( 2\right) \) by Theorem 4.8. \( \left( 2\right) \Rightarrow \left( 3\right) \) by Corollary 5.20. \( \left( 3\right) \Leftrightarrow \left( 4\right) \) follows from Proposition 6.5, applied to the extensio... | Yes |
Theorem 6.12. Let \( k \subseteq F \) be a Galois extension. The Galois correspondence is an inclusion-reversing isomorphism of the lattice of intermediate subfields of \( k \subseteq F \) with the lattice of subgroups of \( {\operatorname{Aut}}_{k}\left( F\right) \) . That is (with notation as in Lemma 6.3), if \( {E}... | Proof. This follows immediately from Theorem 6.9 and Lemma 6.3, which gives \[ {\operatorname{Aut}}_{{E}_{1}{E}_{2}}\left( F\right) = {G}_{1} \cap {G}_{2},\;{F}^{\left\langle {G}_{1},{G}_{2}\right\rangle } = {E}_{1} \cap {E}_{2} \] as needed. | Yes |
The extension \( \mathbb{Q}\left( {\sqrt{2},\sqrt{3}}\right) = \mathbb{Q}\left( {\sqrt{2} + \sqrt{3}}\right) \) studied in Example 1.19 is the splitting field of the polynomial \( {t}^{4} - {10}{t}^{2} + 1 \), so it is Galois. | We found that its Galois group is \( \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \) ; the lattice of this group has no mysteries for us: and therefore the lattice of intermediate fields is just as transparent: (The intermediate fields are determined by recalling the generators of the corresponding subgroups, a... | Yes |
Proposition 6.17. Suppose \( k \subseteq F \) is a Galois extension and \( k \subseteq K \) is any finite extension. Then \( K \subseteq {KF} \) is a Galois extension, and \( {\operatorname{Aut}}_{K}\left( {KF}\right) \cong {\operatorname{Aut}}_{F \cap K}\left( F\right) \) . | Proof. As \( k \subseteq F \) is Galois, it is the splitting field of a separable polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \subseteq K\left\lbrack x\right\rbrack \) . The roots of \( f\left( x\right) \) generate \( F \) over \( k \), so they generate \( {KF} \) over \( K \) ; in other words, \( {... | Yes |
We have studied cyclotomic fields \( \mathbb{Q}\left( {\zeta }_{n}\right) \) as extensions of \( \mathbb{Q};\mathbb{Q}\left( {\zeta }_{n}\right) \) is the splitting field of \( {x}^{n} - 1 \), so these extensions are Galois; we have proved (Proposition 5.16) that \( {\operatorname{Aut}}_{\mathbb{Q}}\left( {\mathbb{Q}\l... | Now let \( k \) be any field of characteristic zero. The splitting field of \( {x}^{n} - 1 \) over \( k \) is the composite \( k\left( \zeta \right) \) of \( k \) and \( \mathbb{Q}\left( \zeta \right) \) . By Proposition 6.17 the extension \( k \subseteq k\left( \zeta \right) \) is Galois, and \( {\operatorname{Aut}}_{... | Yes |
Proposition 6.19. Let \( k \subseteq F \) be an extension of degree \( m \) . Assume that \( k \) contains a primitive \( m \) -th root of 1 and char \( k \) does not divide 24 \( m \) . Then \( k \subseteq F \) is Galois and cyclic if and only if \( F = k\left( \delta \right) \), with \( {\delta }^{m} \in k \) . | Proof. Let \( \zeta \in k \) be a primitive \( m \) -th root of 1 . First assume that \( F = k\left( \delta \right) \), with \( {\delta }^{m} = c \in k \) . Then all \( m \) roots of the polynomial \( {x}^{m} - c \) , \[ \delta ,{\zeta \delta },{\zeta }^{2}\delta ,\cdots ,{\zeta }^{m - 1}\delta \] are in \( F \), and \... | Yes |
Theorem 7.1. \( \mathbb{C} \) is algebraically closed. | Proof. Let \( f\left( x\right) \in \mathbb{C}\left\lbrack x\right\rbrack \) be a nonconstant polynomial; we have to prove that \( f\left( x\right) \) has roots in \( \mathbb{C} \) . Note that if \( f\left( x\right) \) has no roots in \( \mathbb{C} \), then neither does \( f\left( x\right) \overline{f\left( x\right) } \... | No |
Proposition 7.2. Let \( k \subseteq F \) be a Galois extension, and assume \( \left\lbrack {F : k}\right\rbrack = {p}^{r} \) for some prime \( p \) and \( r \geq 0 \) . Then there exist intermediate fields\n\n\[ k = {E}_{0} \subseteq {E}_{1} \subseteq {E}_{2} \subseteq \cdots \subseteq {E}_{r} = F \]\n\nsuch that \( \l... | Proof. As the Galois correspondence is bijective for Galois extensions (Theorem 6.9, part (5)), this statement follows immediately from the fact that a group of order \( {p}^{r} \), with \( p \) prime, has a complete series of \( p \) -subgroups; cf. for example the discussion following the statement of Theorem IV 2.8 | No |
Theorem 7.3. The regular n-gon is constructible by straightedge and compass if and only if \( \phi \left( n\right) \) is a power of 2 . | Proof. As recalled above, we have already established the \( \Rightarrow \) direction.\n\nFor the converse, assume \( \phi \left( n\right) = {2}^{r} \) for some \( r \) . The extension \( \mathbb{Q} \subseteq \mathbb{Q}\left( {\zeta }_{n}\right) \) is Galois (it is the splitting field of \( {\Phi }_{n}\left( x\right) \... | Yes |
Theorem 7.4 (Fundamental theorem on symmetric functions). Let \( K \) be a field, and let \( \varphi \in K\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) . Then \( \varphi \) is symmetric if and only if it is a rational function (with coefficients in \( K \) ) of the elementary symmetric functions \( {s}_{1},\ldots ,{s}_{n}... | Proof. Let \( F = K\left( {{t}_{1},\ldots ,{t}_{n}}\right) \), and let \( k = K\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) be the subfield generated by the elementary symmetric functions over \( K \) . Then \( F \) is a splitting field of the separable polynomial \( {P}_{n}\left( x\right) \) over \( k \) . In particular... | Yes |
Corollary 7.6. Let \( G \) be a finite group. Then there exists a Galois extension \( k \subseteq F \) such that \( {\operatorname{Aut}}_{k}\left( F\right) \cong G \) . | ## Proof. Exercise 7.4 | No |
Lemma 7.10. Every separable radical extension is contained in a Galois radical extension. | Proof. Let \( k \subseteq F \) be a separable radical extension. In particular \( k \subseteq F \) is finite and separable, so \( F = k\left( \alpha \right) \) for some \( \alpha \in F \) (Proposition 5.19). Let \( p\left( x\right) \) be the minimal polynomial of \( \alpha \) over \( k \) . The splitting field \( L \) ... | Yes |
Let \( k \) be a field of characteristic 0, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then there exists a formula solving \( f\left( x\right) \) by radicals if and only if the splitting field of \( f\left( x\right) \) is contained in a Galois radical extension. | Proof. If the splitting field of \( f\left( x\right) \) is contained in a radical extension (Galois or not), then the roots may be written as combinations of field operations and radicals, as needed.\n\nFor the converse, assume \( f\left( x\right) \) is solvable by radicals. A formula for a root \( {t}_{1} \) can be tu... | Yes |
Lemma 7.13. Let \( k \subseteq F \) be a Galois extension, with \( \operatorname{char}k = 0 \) . Provided it has enough roots of \( 1, k \subseteq F \) is radical if and only if it is solvable. | Proof. Modulo the fundamental theorem of Galois theory, this is a straightforward generalization of Proposition 6.19\n\nIndeed, assume that \( k \subseteq F \) is radical:\n\n\[ k \subseteq k\left( {\delta }_{1}\right) \subseteq \cdots \subseteq k\left( {{\delta }_{1},\ldots ,{\delta }_{r}}\right) = F \]\n\nwith \( {\d... | Yes |
Corollary 7.16. Let \( k \) be a field of characteristic 0, and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial. Then \( f\left( x\right) \) is solvable by radicals if and only if its Galois group is solvable. | ## Proof. This is an immediate consequence of Lemma 7.11 and Proposition 7.14, | Yes |
Example 7.19. Lemma 7.18 and a discriminant computation are all that is needed to compute the Galois group of an irreducible cubic polynomial \[ f\left( x\right) = {x}^{3} + a{x}^{2} + {bx} + c. \] | It would be futile to try to remember the discriminant \[ D = {a}^{2}{b}^{2} - 4{a}^{3}c - 4{b}^{3} + {18abc} - {27}{c}^{2}; \] but one may remember the trick of shifting \( x \) by \( a/3 \) (in characteristic \( \neq 3 \) ), with the effect of killing the coefficient of \( {x}^{2} \) : \[ f\left( {x - \frac{a}{3}}\ri... | Yes |
Let \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( p \), where \( p \) is prime. Assume that \( f\left( x\right) \) has \( p - 2 \) real roots and 2 nonreal, complex roots. Then the Galois group of \( f\left( x\right) \) is \( {S}_{p} \). | Indeed, complex conjugation induces an automorphism of the splitting field and acts by interchanging the two nonreal roots, so the Galois group \( G \), as a subgroup of \( {S}_{p} \), contains a transposition. On the other hand, the degree of the splitting field (and hence \( \left| G\right| \) ) is divisible by \( p ... | Yes |
The operation \( {\operatorname{Aut}}_{k}\left( \_ \right) \) from Galois field extensions to groups is contravariantly functorial. | Indeed, if \( k \subseteq E \subseteq F \) is viewed as a morphism of two Galois extensions \( \left( {k \subseteq E\text{to}k \subseteq F}\right) \), we have a corresponding group homomorphism\n\n\[ \n{\operatorname{Aut}}_{k}\left( F\right) \rightarrow {\operatorname{Aut}}_{k}\left( E\right)\n\]\n\ndefined by restrict... | Yes |
In [111]4.3] I have defined the spectrum of a commutative ring \( R \) , Spec \( R \), as the set of prime ideals of \( R \) . If \( R, S \) are commutative rings and \( \varphi : R \rightarrow S \) is a ring homomorphism, then the inverse image \( {\varphi }^{-1}\left( \mathfrak{p}\right) \) of a prime ideal \( \mathf... | \[ {\varphi }^{ * } : \operatorname{Spec}\left( S\right) \rightarrow \operatorname{Spec}\left( R\right) \] This assignment is clearly functorial, so we can view Spec as a contravariant functor from the category of commutative rings to Set. | Yes |
Let \( \mathrm{C} \) be a category, and let \( X \) be an object of \( \mathrm{C} \). Then the assignments\n\n\[ \nA \mapsto {\operatorname{Hom}}_{\mathsf{C}}\left( {X, A}\right) \;,\;A \mapsto {\operatorname{Hom}}_{\mathsf{C}}\left( {A, X}\right) \n\]\n\ndefine, respectively, covariant and contravariant functors \( \m... | For example, if\n\n\[ \nA\xrightarrow[]{\alpha }B\xrightarrow[]{\beta }C \n\]\n\nis a diagram in \( \mathrm{C} \), stare at\n\n\n\nEvery \( \alpha : A \rightarrow B \) determines a function\n\n\[ \n{\operatorname{Hom... | No |
Construct a category by taking the objects to be nonnegative integers and \( \operatorname{Hom}\left( {m, n}\right) \) to be the set of \( n \times m \) matrices with entries in a field \( k \), with composition defined by product of matrices (and suitable care concerning matrices with no rows or columns). | The resulting category is equivalent to the category of finite-dimensional \( k \) -vector spaces. Indeed, we obtain a functor from the former to the latter by sending \( n \) to the vector space \( {k}^{n} \) (endowed with the standard basis) and each matrix to the corresponding linear map. This functor is clearly ful... | Yes |
Recall (Definition VII 2.19) that the coordinate ring of an affine algebraic set over a field \( K \) is a reduced, commutative, finite-type \( K \) -algebra. We can define the category \( K \) -Aff of affine \( K \) -algebraic sets by prescribing that the objects be algebraic subsets of some affine \( K \) -space and ... | Thus, \( K \) -Aff is defined in such a way that the functor \( K \) -Aff \( {}^{op} \rightarrow K \) -Alg that maps an affine algebraic set \( S \) to its coordinate ring \( K\left\lbrack S\right\rbrack \) is an equivalence of the opposite category of \( K \) -Aff with the subcategory of reduced, commutative, finite-t... | Yes |
Claim 1.13. The limit \( \underset{i}{\overline{\lim }}{A}_{i} \) exists in \( R \) -Mod. | Proof. The product \( \mathop{\prod }\limits_{i}{A}_{i} \) consists of arbitrary sequences \( {\left( {a}_{i}\right) }_{i > 0} \) of elements \( {a}_{i} \in \) \( {A}_{i} \) . Say that a sequence \( {\left( {a}_{i}\right) }_{i > 0} \) is coherent if for all \( i > 0 \) we have \( {a}_{i} = {\varphi }_{i, i + 1}\left( {... | No |
If \( \mathrm{C} = \) Set and all the \( {\psi }_{ij} \) are injective, we are talking about a 'nested sequence of sets':\n\n\[ \n{A}_{1} \subseteq {A}_{2} \subseteq {A}_{3} \subseteq {A}_{4} \subseteq \cdots \n\] \n\nthe direct limit of this sequence would be the ’infinite union’ \( \mathop{\bigcup }\limits_{i}{A}_{i}... | More formally, \( \mathop{\bigcup }\limits_{i}{A}_{i} \) consists of equivalence classes of pairs \( \left( {i,{a}_{i}}\right) \), where \( {a}_{i} \in {A}_{i} \) and \( \left( {i,{a}_{i}}\right) \) is equivalent to \( \left( {j,{a}_{j}}\right) \) for \( i \leq j \) if \( {a}_{j} = {\psi }_{ij}\left( {a}_{i}\right) \) ... | Yes |
Lemma 1.17. Right-adjoint functors commute with limits.\n\nThat is, if \( \mathcal{G} : \mathrm{D} \rightarrow \mathrm{C} \) has a left-adjoint \( \mathcal{F} : \mathrm{C} \rightarrow \mathrm{D} \) and \( \mathcal{A} : \mathrm{I} \rightarrow \mathrm{D} \) is another functor, then there is a canonical isomorphism\n\n\[ ... | Assume \( \mathcal{G} : \mathrm{D} \rightarrow \mathrm{C} \) is right-adjoint to \( \mathcal{F} : \mathrm{C} \rightarrow \mathrm{D} \) and \( \mathcal{A} : \mathrm{I} \rightarrow \mathrm{D} \) is a given functor. As we have seen, the limit of \( \mathcal{A} \) is final subject to fitting in commutative diagrams\n\n\n\nwhere \( \otimes \left( {r, n}\right) = {rn} \) . By the uniqueness property of universal ob... | Yes |
For all \( R \) -modules \( M, N, P \), there is an isomorphism of \( R \) -modules\n\n\[{\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{R}\left( {N, P}\right) }\right) \cong {\operatorname{Hom}}_{R}\left( {M{ \otimes }_{R}N, P}\right) . | Proof. As noted before the statement, every \( \alpha \in {\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{R}\left( {N, P}\right) }\right) \) determines an \( R \) -bilinear map \( \varphi : M \times N \rightarrow P \), by\n\n\[ \left( {m, n}\right) \mapsto \alpha \left( m\right) \left( n\right) .\n\nBy the univ... | No |
For every \( R \) -module \( N \), the functor \( {}_{ - }{ \otimes }_{R}N \) is left-adjoint to the functor \( {\operatorname{Hom}}_{R}\left( {N,}\right) \) . | Proof. The claim is that the isomorphism found in Lemma 2.4 is natural in the sense hinted at, but not fully explained, in [1.5] the interested reader should have no problems checking this naturality. | No |
Corollary 2.7. For any two sets \( A, B \) :\n\n\[ \n{R}^{\oplus A}{ \otimes }_{R}{R}^{\oplus B} \cong {R}^{\oplus A \times B}.\n\] | Indeed, 'distributing' the direct sum identifies the left-hand side with the direct sum \( {\left( {R}^{\oplus A}\right) }^{\oplus B} \), which is isomorphic to the right-hand side (Exercise III 6.5). For finitely generated free modules, this simply says that \( {R}^{\oplus m} \otimes {R}^{\oplus n} \cong {R}^{\oplus {... | No |
Corollary 2.8. For all \( R \) -modules \( N \) and all ideals \( I \) of \( R \) , \[ \frac{R}{I}{ \otimes }_{R}N \cong \frac{N}{IN} \] | Indeed, \( {}_{ - }{ \otimes }_{R}N \) is right-exact; thus, the exact sequence \[ 0 \rightarrow I \rightarrow R \rightarrow \frac{R}{I} \rightarrow 0 \] induces an exact sequence \[ I{ \otimes }_{R}N \rightarrow R{ \otimes }_{R}N \rightarrow \frac{R}{I}{ \otimes }_{R}N \rightarrow 0. \] The image of \( I{ \otimes }_{R... | Yes |
Corollary 2.9. For all ideals \( I, J \) of \( R \) ,\n\n\[ \frac{R}{I}{ \otimes }_{R}\frac{R}{J} \cong \frac{R}{I + J} \] | This follows immediately from Corollary 2.8 and the 'third isomorphism theorem', Proposition III 5.17. Indeed, \( {IR}/J = \left( {I + J}\right) /J \) . | Yes |
Multiplication by 2 gives an inclusion\n\n\[ \n{\mathbb{Z}}^{ \subset \cdot 2} \rightarrow \mathbb{Z} \]\n\nidentifying the first copy of \( \mathbb{Z} \) with the ideal (2) in the second copy. Tensoring by \( \mathbb{Z}/2\mathbb{Z} \) over \( \mathbb{Z} \) (and keeping in mind that \( R{ \otimes }_{R}N \cong N \) ), w... | This is the zero-morphism, and in particular it is not injective. | Yes |
Consider the affine algebraic set \( \mathcal{V}\left( {xy}\right) \) in the plane \( {\mathbb{A}}^{2} \) (over a fixed field \( k \) ) and the ’projection on the first coordinate’ \( \mathcal{V}\left( {xy}\right) \rightarrow {\mathbb{A}}^{1},\left( {x, y}\right) \mapsto x \) : | In terms of coordinate rings (cf. SVII 2.3), this map corresponds to the homomorphism of \( k \) -algebras:\n\n\[ k\left\lbrack x\right\rbrack \rightarrow \frac{k\left\lbrack {x, y}\right\rbrack }{\left( xy\right) } \]\n\ndefined by mapping \( x \) to the coset \( x + \left( {xy}\right) \) (this will be completely clea... | Yes |
Lemma 3.2. Let \( R \) be a commutative ring; let \( M, N \) be \( R \) -modules, and let \( G \) be an abelian group. Then every \( \mathbb{Z} \) -bilinear, \( R \) -balanced map \( \varphi : M \times N \rightarrow G \) factors through \( M{ \otimes }_{R}N \) ; that is, there exists a unique group homomorphism \( \bar... | The universal property explored in [2.1] is recovered as the statement that if \( G \) is an \( R \) -module and \( \varphi \) is \( R \) -bilinear, then the induced group homomorphism \( M{ \otimes }_{R}N \rightarrow G \) is in fact an \( R \) -linear map. | No |
Lemma 3.5. Suppose \( M \) is an \( R \) -module, \( N \) is an \( \left( {R, S}\right) \) -bimodule, and \( P \) is an \( S \) -module. Then there is a canonical isomorphism of abelian groups\n\n\[ \n{\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{S}\left( {N, P}\right) }\right) \cong {\operatorname{Hom}}_{S}\... | Proof. Every element \( \alpha \in {\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{S}\left( {N, P}\right) }\right) \) determines a map\n\n\[ \n\varphi : M \times N \rightarrow P \n\]\n\nvia \( \varphi \left( {m,\_ }\right) \mathrel{\text{:=}} \alpha \left( m\right) ;\varphi \) is clearly \( \mathbb{Z} \) -bilin... | Yes |
Proposition 3.6. Let \( f : R \rightarrow S \) be a homomorphism of commutative rings. Then, with notation as above, \( {f}_{ * } \) is right-adjoint to \( {f}^{ * } \) and left-adjoint to \( {f}^{!} \) . In particular, \( {f}_{ * } \) is exact, \( {f}^{ * } \) is right-exact, and \( {f}^{!} \) is left-exact. | Proof. Let \( M \), resp., \( N \), be an \( R \) -module, resp., an \( S \) -module. Note that, trivially, \( {\operatorname{Hom}}_{S}\left( {S, N}\right) \) is canonically isomorphic to \( N \) (as an \( S \) -module) and to \( {f}_{ * }\left( N\right) \) (as an \( R \) -module). Thus 15\n\n\[ \n{\operatorname{Hom}}_... | Yes |
Lemma 4.2. Let \( \varphi : {M}^{\ell } \rightarrow P \) be an \( R \) -multilinear function.\n\nIf \( \varphi \) is alternating, then for all \( \sigma \in {S}_{\ell } \), and all \( {m}_{1},\ldots ,{m}_{\ell } \), \n\n\[ \varphi \left( {{m}_{\sigma \left( 1\right) },\ldots ,{m}_{\sigma \left( \ell \right) }}\right) =... | Proof. For the first statement, it suffices to show that interchanging any two factors switches the sign of an alternating function (since transpositions generate the symmetric group). Since the other factors have no effect on this operation, this reduces the question to the case \( \ell = 2 \) . Therefore, we only hav... | Yes |
Lemma 4.3. Let \( R \) be a commutative ring, and let \( M \) be a free \( R \)-module of rank \( r \). Then \( {\Lambda }_{R}^{\ell }\left( M\right) \) is a free \( R \)-module of rank \( \left( \begin{array}{l} r \\ \ell \end{array}\right) \). | Proof. There are \( \left( \begin{array}{l} r \\ \ell \end{array}\right) \) sequences of indices \( {i}_{1},\ldots ,{i}_{\ell } \) satisfying \( 1 \leq {i}_{1} < \cdots < {i}_{\ell } \leq r \) , so we just need to show that the generators \( {e}_{{i}_{1}} \land \cdots \land {e}_{{i}_{\ell }} \) are linearly independent... | Yes |
For \( V = {k}^{4},{\Lambda }_{k}^{2}\left( V\right) \) has dimension \( \left( \begin{array}{l} 4 \\ 2 \end{array}\right) = 6 \) . On ’pure wedges’ \( {a}_{1} \land {a}_{2} \), the isomorphism \( {\mathbb{A}}_{k}^{2}\left( V\right) \rightarrow {k}^{6} \) works as follows. View the vectors \( {a}_{1},{a}_{2} \) as the ... | \[ A = \left( \begin{array}{ll} {a}_{1}^{1} & {a}_{2}^{1} \\ {a}_{1}^{2} & {a}_{2}^{2} \\ {a}_{1}^{3} & {a}_{2}^{3} \\ {a}_{1}^{4} & {a}_{2}^{4} \end{array}\right) \mapsto \left( \begin{array}{l} {a}_{1}^{1}{a}_{2}^{2} - {a}_{1}^{2}{a}_{2}^{1} \\ {a}_{1}^{1}{a}_{2}^{3} - {a}_{1}^{3}{a}_{2}^{1} \\ {a}_{1}^{1}{a}_{2}^{4}... | Yes |
The polynomial ring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) over any ring \( R \) carries a natural grading, given by the (ordinary) degree of polynomials. | We may write\n\n\[\nR\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack = R \oplus \left\langle {{x}_{1},\ldots ,{x}_{n}}\right\rangle \oplus \left\langle {{x}_{1}^{2},{x}_{1}{x}_{2},\ldots ,{x}_{n}^{2}}\right\rangle \oplus \cdots .\n\] | No |
Lemma 4.9. Let \( S = {\bigoplus }_{i}{S}_{i} \) be a graded ring and let \( I \subseteq S \) be an ideal of \( S \) . Then the following are equivalent:\n\n(i) I is homogeneous;\n\n(ii) if \( s \in S \) and \( s = \mathop{\sum }\limits_{i}{s}_{i} \) is the decomposition of \( s \) into homogeneous elements \( {s}_{i} ... | Proof. (i) \( \Leftrightarrow \) (ii) is the very definition of homogeneous ideal; (ii) \( \Leftrightarrow \) (iii) is left to the reader (Exercise 4.10).\n\n(ii) \( \Rightarrow \) (iv): Assuming (ii) holds, define a grading on \( S/I \) by letting the piece of degree \( i \) consist of (0 and) the cosets of the elemen... | No |
The ideal \( I = \left( {y - {x}^{2}}\right) \) is not homogeneous in the ring \( k\left\lbrack {x, y}\right\rbrack \) , if this is given the grading by the usual degree. | indeed, \( y - {x}^{2} \in I \) while \( y \notin I \) , contradicting condition (ii) of Lemma 4.9. | Yes |
Lemma 4.12. Let \( {I}_{\mathbb{S}},{I}_{\mathbb{A}} \subseteq {\mathbb{T}}_{R}^{ * }\left( M\right) \) be the ideals respectively generated by all elements of the form \( \left( {m \otimes n - n \otimes m}\right) \) as \( m, n \in M \) and by elements of the form \( m \otimes m \) as \( m \in M \) . Then\n\n\[ \n{\mat... | Proof. We have observed in [4.3] that the kernel of a graded homomorphism is the direct sum of the kernels of the induced homomorphisms in each degree; the statement of the lemma then follows easily from the explicit descriptions of the kernels of the canonical projections from the tensor powers to the symmetric and ex... | No |
Proposition 4.16. Let \( R \) be a commutative ring, and let \( M \) be an \( R \) -module. Then for every \( R \) -algebra \( A \) and every \( R \) -module homomorphism \( \lambda : M \rightarrow A \) such that \( \lambda {\left( m\right) }^{2} = 0\forall m \in M \), there exists a unique homomorphism of \( R \) -alg... | Details may now safely be left to the reader (who may for example establish the first proposition from the universal property of tensor powers and then deduce the second and third from Lemma 4.12). | No |
The free case is particularly easy to understand. For instance,\n\n\[ \n{\mathbb{S}}_{R}^{ * }\left( {R}^{\oplus r}\right) \cong R\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \n\] | Indeed, the polynomial ring satisfies the appropriate universal property with respect to mapping to commutative rings (cf. [111,2.2] and [111,6.4]). Likewise, \( {\mathbb{T}}_{R}^{ * }\left( {R}^{\oplus r}\right) \) should be thought of as a ’noncommutative’ polynomial ring, in which the \( r \) inde-terminates do not ... | No |
Proposition 5.2. For every \( R \) -module \( N \), the functor \( {\operatorname{Hom}}_{R}\left( {\_, N}\right) \) is right-adjoint to itself. | Proof. Let \( L, M, N \) denote \( R \) -modules. Recall (cf. the considerations preceding Lemma 2.4) that \( R \) -bilinear maps\n\n\[ \varphi : L \times M \rightarrow N \]\n\nmay be identified with \( R \) -linear maps\n\n\[ L \rightarrow {\operatorname{Hom}}_{R}\left( {M, N}\right) \]\n\nBy the same token, they may ... | Yes |
Proposition 5.5. Let \( M \) be any \( R \) -module, and let \( F \) be a free \( R \) -module of finite rank. Then\n\n\[ \n{\operatorname{Hom}}_{R}\left( {M, F}\right) \cong {M}^{ \vee }{ \otimes }_{R}F.\n\] | Proof. By hypothesis \( F \cong {R}^{\oplus n} \cong {R}^{n} \) ; hence\n\n\( {\operatorname{Hom}}_{R}\left( {M, F}\right) \cong {\operatorname{Hom}}_{R}\left( {M,{R}^{n}}\right) \cong {\operatorname{Hom}}_{R}{\left( M, R\right) }^{n} \cong {\operatorname{Hom}}_{R}\left( {M, R}\right) { \otimes }_{R}{R}^{n} \cong {M}^{... | Yes |
Corollary 5.7. The dual of a free module is isomorphic to a product of copies of \( R \) :\n\n\[ \n{\left( {R}^{\oplus S}\right) }^{ \vee } \cong {R}^{S} \n\]\n\nIn particular, \( {\left( {R}^{n}\right) }^{ \vee } \cong {R}^{n} \) : if \( F \) is a free \( R \) -module of finite rank, then \( {F}^{ \vee } \cong F \) . | Proof. This follows from Lemma 5.6 and the fact that \( {R}^{ \vee } = {\operatorname{Hom}}_{R}\left( {R, R}\right) \) is isomorphic to \( R \) . | No |
To see that these isomorphisms do depend on the choice of the basis, consider the standard basis \( \left( {{\mathbf{e}}_{1},{\mathbf{e}}_{2}}\right) \) of \( {R}^{2} \) and the corresponding dual basis \( \left( {{\check{\mathbf{e}}}_{1},{\check{\mathbf{e}}}_{2}}\right) \) of \( {\left( {R}^{2}\right) }^{ \vee } \), a... | By definition\n\n\[ \n{\check{\mathbf{e}}}_{1}^{\prime }\left( {\mathbf{e}}_{2}^{\prime }\right) = 0 \n\]\n\nwhile\n\n\[ \n{\check{\mathbf{e}}}_{1}\left( {\mathbf{e}}_{2}^{\prime }\right) = {\check{\mathbf{e}}}_{1}\left( {{\mathbf{e}}_{1} + {\mathbf{e}}_{2}}\right) = 1 + 0 = 1. \n\]\n\nTherefore, \( {\check{\mathbf{e}}... | Yes |
Lemma 5.12. The duality functor is left-exact: every exact sequence\n\n\\[ \nL \rightarrow M \rightarrow N \rightarrow 0 \n\\]\n\nof \\( R \\) -modules induces an exact sequence\n\n\\[ \n0 \rightarrow {N}^{ \\vee } \rightarrow {M}^{ \\vee } \rightarrow {L}^{ \\vee } \n\\] | Proof. This is an immediate consequence of the left-exactness of Hom. | No |
Proposition 5.13. Let\n\n\[ \n0 \rightarrow M\xrightarrow[]{\;\mu \;}N\xrightarrow[]{\;\nu \;}P \rightarrow 0 \n\] \n\nbe an exact sequence of \( R \) -modules, with \( P \) free. Then the induced sequence \n\n\[ \n0 \rightarrow {P}^{ \vee }\xrightarrow[]{{\nu }^{ \vee }}{N}^{ \vee }\xrightarrow[]{{\mu }^{ \vee }}{M}^{... | Proof. Lemma 5.12 takes care of all but the surjectivity of the map \( {N}^{ \vee } \rightarrow {M}^{ \vee } \) induced from \( M \rightarrow N \) : \n\nThe question is whether every \( R \) -linear \( f : M \rightarrow R \) can be extended to an \( R \) -linear map \( g : N \rightarrow R \) so that \( f = g \circ \mu ... | Yes |
Lemma 5.15. Let \( A \) be the matrix representing a linear map \( \alpha : {R}^{n} \rightarrow {R}^{m} \) with respect to the standard bases. Then the dual map \( {\alpha }^{ \vee } : {\left( {R}^{m}\right) }^{ \vee } \rightarrow {\left( {R}^{n}\right) }^{ \vee } \) is represented by the transpose of \( A \) with resp... | The (easy) verification of this fact is left to the reader (Exercise 5.10). | No |
Proposition 5.16. Let \( R \) be an integral domain, and let \( M \) be an \( R \)-module. Then \( {M}^{ \vee } \) is torsion-free. | Proof. There is a surjection \( {R}^{\oplus S} \rightarrow M \), thus, an exact sequence\n\n\[ \n{R}^{\oplus T} \rightarrow {R}^{\oplus S} \rightarrow M \rightarrow 0.\n\]\n\nDualizing, \( {M}^{ \vee } \) is realized as the kernel of the induced map \( {R}^{S} \rightarrow {R}^{T} \) ; hence \( {M}^{ \vee } \) may be id... | No |
Lemma 6.2. An R-module \( P \) is projective if and only if for all epimorphisms of \( R \) -modules \( \mu : M \rightarrow N \), every \( R \) -linear map \( p : P \rightarrow N \) lifts to an \( R \) -linear map \( \widehat{p} : P \rightarrow M \). | Proof. This is straightforward. Since \( {\operatorname{Hom}}_{R}\left( {\_, Q}\right) \) is left-exact for all \( Q, Q \) is injective if and only if whenever a sequence\n\n\[ 0 \rightarrow L \rightarrow M \]\n\nis exact, then so is the induced sequence\n\n\[ {\operatorname{Hom}}_{R}\left( {M, Q}\right) \rightarrow {\... | No |
For example, assume that \( P \) is projective; then I claim that every exact sequence\n\n\[ 0 \rightarrow L\xrightarrow[]{\lambda }M\xrightarrow[]{\mu }P \rightarrow 0 \]\n\nsplits, in the sense that there is a submodule \( {P}^{\prime } \) of \( M \) such that \( \mu \) restricts to an isomorphism \( {P}^{\prime } \r... | Indeed, since \( P \) is projective, then the identity \( P\overset{ \equiv }{ \rightarrow }P \) lifts to a homomorphism \( \rho : P \rightarrow M \), and the reader can then verify that \( {P}^{\prime } = \rho \left( P\right) \) fits the requirement. Loosely speaking, in this situation we can simply replace \( M \) by... | No |
Proposition 6.4. An R-module \( P \) is projective if and only if it is a direct summand of a free module, that is, if and only if there exists a free \( R \) -module \( F \), an \( R \) - module \( K \), and an isomorphism \( K \oplus P \cong F \) . | Proof. Any set \( S \) of generators of \( P \) determines a surjection of the free module \( F = {R}^{\oplus S} \) onto \( P \) and hence an exact sequence\n\n\[ 0 \rightarrow K \rightarrow F \rightarrow P \rightarrow 0. \]\n\nAs observed above, such a sequence necessarily splits if \( P \) is projective; thus \( F \c... | No |
Let \( {P}_{1},{P}_{2} \) be projective \( R \) -modules. Then \( {P}_{1} \oplus {P}_{2} \) and \( {P}_{1}{ \otimes }_{R}{P}_{2} \) are projective. Projective modules are flat. | Proof. These statements follow easily from Proposition 6.4, the fact that \( \otimes \) is distributive with respect to \( \oplus \), and the fact that free modules are flat. | No |
Theorem 6.6. An R-module \( Q \) is injective if and only if every \( R \) -linear map \( f \) : \( I \rightarrow Q \), with \( I \) an ideal of \( R \), extends to an \( R \) -linear map \( \widehat{f} : R \rightarrow Q \) . | Proof. The 'only if' part of the statement is immediate from the definition of injective. To verify the ’if’ part, assume \( Q \) satisfies the stated extension condition, let \( L \subseteq M \) be any inclusion of \( R \) -modules, and let \( q : L \rightarrow Q \) be a given \( R \) -linear map:\n\n![23387543-548b-4... | Yes |
Corollary 6.7. Let \( R \) be a PID. Then an \( R \) -module \( Q \) is injective if and only if it is divisible. | ## Proof. Exercise 6.14. | No |
Viewed as abelian groups (i.e., \( \mathbb{Z} \) -modules), \( \mathbb{Q} \) and \( \mathbb{Q}/\mathbb{Z} \) are injective. More generally, if \( D \) is any divisible abelian group and \( K \subseteq D \), then \( D/K \) is injective. | Indeed, it is trivially divisible! | No |
Lemma 6.9. Let \( f : S \rightarrow R \) be a homomorphism of commutative rings, and let \( Q \) be an injective \( S \) -module. Then \( {f}^{!}\left( Q\right) \) is an injective \( R \) -module. | Proof. By adjunction (Lemma 3.5),\n\n\[ \n{\operatorname{Hom}}_{R}\left( {\_ ,{f}^{!}\left( Q\right) }\right) \cong {\operatorname{Hom}}_{S}\left( {{f}_{ * }\left( \_ \right), Q}\right) \n\]\n\nas functors \( R \) -Mod \( \rightarrow \mathrm{{Ab}} \) . Since \( {f}_{ * } \) is exact (Proposition 3.6) and \( {\operatorn... | Yes |
Corollary 6.12. Let \( M \) be an \( R \) -module. Then \( M \) can be identified with a submodule of an injective \( R \) -module. | Proof. I claim that it suffices to show that \( \mathbb{Z} \) -Mod has enough injectives, Indeed, this will show that there exists a divisible abelian group \( D \) such that \( M \subseteq D(M \) is in particular an abelian group); since \( R \) -linear maps are in particular \( \mathbb{Z} \) -linear,\n\n\[ M \cong {\... | Yes |
An \( R \) -module \( P \) is projective if and only if \( {\operatorname{Ext}}_{R}^{1}\left( {P,\_ }\right) = 0 \), if and only if \( {\operatorname{Ext}}_{R}^{i}\left( {P,\_ }\right) = 0 \) for all \( i > 0 \) . | The first assertion is proven similarly, using the first definition given above for Ext. | No |
Lemma 1.3. A morphism \( \varphi : A \rightarrow B \) in an additive category is a monomorphism if and only if for all \( \zeta : Z \rightarrow A \) ,\n\n\[ \varphi \circ \zeta = 0 \Rightarrow \zeta = 0. \]\n\nIt is an epimorphism if and only if for all \( \beta : B \rightarrow Z \) ,\n\n\[ \beta \circ \varphi = 0 \Rig... | Proof. This is simply because two morphisms with the same source and target are equal if and only if their difference in the corresponding Hom-set (which is an abelian group by hypothesis) is 0. | Yes |
Lemma 1.4. In any additive category, kernels are monomorphisms and cokernels are epimorphisms. | Proof. Let \( \varphi : A \rightarrow B \) be a morphism in an additive category \( \mathrm{A} \), and let \( \operatorname{coker}\varphi \) : \( B \rightarrow C \) be its cokernel. Let \( \gamma : C \rightarrow Z \) be a morphism such that \( \gamma \circ \operatorname{coker}\varphi = 0 \) . The composition \( \left( ... | No |
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