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Theorem 5.23. If \( k \) is a field and \( f\left( x\right) \in k\left\lbrack x\right\rbrack \), then any two splitting fields of \( f\left( x\right) \) over \( k \) are isomorphic. | Proof. Let \( E \) and \( {E}^{\prime } \) be splitting fields of \( f\left( x\right) \) over \( k \) . If \( \varphi \) is the identity, then Proposition 5.22(i) applies at once. | No |
Corollary 5.24. The Galois group \( \operatorname{Gal}\left( {E/k}\right) \) of a polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) with splitting field \( E \) depends only on \( f\left( x\right) \) and \( k \), but not upon the choice of \( E \) . | Proof. If \( \varphi : E \rightarrow {E}^{\prime } \) is an isomorphism fixing \( k \), then there is an isomorphism \( \operatorname{Gal}\left( {E/k}\right) \rightarrow \operatorname{Gal}\left( {{E}^{\prime }/k}\right) \) given by \( \sigma \mapsto {\varphi \sigma }{\varphi }^{-1} \) . | Yes |
Corollary 5.25 (E. H. Moore). Any two finite fields having exactly \( {p}^{n} \) elements are isomorphic. | Proof. If \( E \) is a field with \( q = {p}^{n} \) elements, then Lagrange’s theorem applied to the multiplicative group \( {E}^{ \times } \) shows that \( {a}^{q - 1} = 1 \) for every \( a \in {E}^{ \times } \) . It follows that every element of \( E \), including \( a = 0 \), is a root of \( f\left( x\right) = {x}^{... | Yes |
Theorem 5.26. If \( E/k \) is the splitting field of some polynomial in \( k\left\lbrack x\right\rbrack \), where \( k \) is a field of characteristic 0, then \( \left| {\operatorname{Gal}\left( {E/k}\right) }\right| = \left\lbrack {E : k}\right\rbrack \) . | Proof. This is the special case of Proposition 5.22(ii) when \( k = {k}^{\prime }, E = {E}^{\prime } \) , and \( \varphi = {1}_{k} \) . | Yes |
Corollary 5.27. Let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial of degree \( n \) , where \( k \) is a field of characteristic 0 . If \( E/k \) is a splitting field of \( f\left( x\right) \) over \( k \) , then \( n \) is a divisor of \( \left| {\operatorname{Gal}\left( {E/k}\ri... | Proof. If \( z \in E \) is a root of \( f\left( x\right) \), then \( \left\lbrack {k\left( z\right) : k}\right\rbrack = n \), as in Example 5.14. But \( \left\lbrack {E : k}\right\rbrack = \left\lbrack {E : k\left( z\right) }\right\rbrack \left\lbrack {k\left( z\right) : k}\right\rbrack \), so that \( n \mid \left\lbra... | Yes |
Lemma 5.28. Let \( B/k \) be a splitting field of some polynomial \( g\left( x\right) \in k\left\lbrack x\right\rbrack \) . If \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) is irreducible, and if\n\n\[ p\left( x\right) = {q}_{1}\left( x\right) \cdots {q}_{t}\left( x\right) \]\n\nis the factorization of \( p\l... | Proof. Regard \( p\left( x\right) \) as a polynomial in \( B\left\lbrack x\right\rbrack \) (for \( k \subseteq B \) implies \( k\left\lbrack x\right\rbrack \subseteq B\left\lbrack x\right\rbrack \) ), and let \( E = B\left( {{z}_{1},\ldots ,{z}_{n}}\right) \) be a splitting field of \( p\left( x\right) \), where \( {z}... | Yes |
Theorem 5.29. Let \( E/k \) be a finite field extension. Then \( E/k \) is a splitting field of some polynomial in \( k\left\lbrack x\right\rbrack \) if and only if every irreducible polynomial in \( k\left\lbrack x\right\rbrack \) having a root in \( E \) must split in \( E\left\lbrack x\right\rbrack \) . | Proof. Suppose that \( E/k \) is a splitting field of some polynomial in \( k\left\lbrack x\right\rbrack \) . Let \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) be irreducible, and let \( p\left( x\right) = {q}_{1}\left( x\right) \cdots {q}_{t}\left( x\right) \) be its factorization into irreducibles in \( E\l... | Yes |
Theorem 5.31. Let \( k \subseteq K \subseteq E \) be a tower of fields, where both \( K/k \) and \( E/k \) are normal extensions. Then \( \operatorname{Gal}\left( {E/K}\right) \) is a normal subgroup of \( \operatorname{Gal}\left( {E/k}\right) \), and \[ \operatorname{Gal}\left( {E/k}\right) /\operatorname{Gal}\left( {... | Proof. Since \( K/k \) is a normal extension, it is a splitting field of some polynomial in \( k\left\lbrack x\right\rbrack \), by Theorem 5.29. Hence, if \( \sigma \in \operatorname{Gal}\left( {E/k}\right) \), then \( \sigma \left( K\right) = K \) , by Corollary 5.19. Define \( \rho : \operatorname{Gal}\left( {E/k}\ri... | Yes |
Lemma 5.32. Let \( B \) be a finite extension of a field \( k \). (i) There is a finite extension \( F/B \) with \( F/k \) a normal extension. | Proof. (i) Since \( B \) is a finite extension, \( B = k\left( {{z}_{1},\ldots ,{z}_{\ell }}\right) \) for elements \( {z}_{1},\ldots ,{z}_{\ell } \) . For each \( i \), Theorem 3.116 gives an irreducible polynomial \( {p}_{i}\left( x\right) \in k\left\lbrack x\right\rbrack \) with \( {p}_{i}\left( {z}_{i}\right) = 0 \... | Yes |
Lemma 5.33. Let \( k\left( u\right) /k \) be a pure extension of prime type \( p \) distinct from the characteristic of \( k \) . If \( k \) contains the pth roots of unity and if \( u \notin k \), then \( \operatorname{Gal}\left( {k\left( u\right) /k}\right) \cong {\mathbb{I}}_{p} \) | Proof. Denote \( \operatorname{Gal}\left( {k\left( u\right) /k}\right) \) by \( G \) . Let \( a = {u}^{p} \in k \) . If \( \omega \) is a primitive \( p \) th root of unity, then the roots \( 1,\omega ,\ldots ,{\omega }^{p - 1} \) are distinct [because \( p \neq \operatorname{char}\left( k\right) \) ], and the roots of... | Yes |
Theorem 5.34. Let \( k = {K}_{0} \subseteq {K}_{1} \subseteq {K}_{2} \subseteq \cdots \subseteq {K}_{t} \) be a radical extension of a field \( k \) . Assume, for each \( i \), that each \( {K}_{i} \) is a pure extension of prime type \( {p}_{i} \) over \( {K}_{i - 1} \), where \( {p}_{i} \neq \operatorname{char}\left(... | Proof.\n\n(i) Defining \( {G}_{i} = \operatorname{Gal}\left( {{K}_{t}/{K}_{i}}\right) \) gives a sequence of subgroups of \( \operatorname{Gal}\left( {{K}_{t}/k}\right) \) . Since \( {K}_{1} = k\left( u\right) \), where \( {u}^{{p}_{1}} \in k \), the assumption that \( k \) contains a primitive \( p \) th root of unity... | Yes |
Theorem 5.36. Every quotient \( G/N \) of a solvable group \( G \) is itself a solvable group. | Proof. By the first isomorphism theorem for groups, quotient groups are isomorphic to homomorphic images, and so it suffices to prove that if \( f : G \rightarrow H \) is a surjection (for some group \( H \) ), then \( H \) is a solvable group.\n\nLet \( G = {G}_{0} \geq {G}_{1} \geq {G}_{2} \geq \cdots \geq {G}_{t} = ... | Yes |
Theorem 5.38 (Abel-Ruffini). For all \( n \geq 5 \), the general polynomial of degree \( n \) ,\n\n\[ f\left( x\right) = \left( {x - {y}_{1}}\right) \left( {x - {y}_{2}}\right) \cdots \left( {x - {y}_{n}}\right) ,\]\n\nis not solvable by radicals. | Proof. In Example 5.17, we saw that if \( F \) is a field, \( E = F\left( {{y}_{1},\ldots ,{y}_{n}}\right) \) is the field of all rational functions in \( n \) variables \( {y}_{1},\ldots ,{y}_{n} \) with coefficients in \( F \) , and \( k = F\left( {{a}_{0},\ldots ,{a}_{n}}\right) \), where the \( {a}_{i} \) are the c... | Yes |
Lemma 5.40. Let \( E/k \) be a Galois extension with Galois group \( G = \operatorname{Gal}\left( {E/k}\right) \) . Given subgroups \( G \geq H \geq L \), then\n\n\[ \left\lbrack {{E}^{L} : {E}^{H}}\right\rbrack = \left\lbrack {H : L}\right\rbrack \] | Proof. Since \( H \mapsto {E}^{H} \) is order-reversing, there is a tower of fields\n\n\[ k = {E}^{G} \subseteq {E}^{H} \subseteq {E}^{L} \subseteq E \]\n\n(we have \( k = {E}^{G} \) because \( E/k \) is a Galois extension). Theorem 4.31 gives \( \left\lbrack {{E}^{L} : k}\right\rbrack = \left\lbrack {{E}^{L} : {E}^{H}... | Yes |
Theorem 5.41 (Gauss). Let \( p \) be an odd prime. A regular p-gon is constructible if and only if \( p = {2}^{m} + 1 \) for some \( m \geq 0 \) . | Proof. Necessity was proved in Theorem 4.59, where it was shown that \( m \) must be a power of 2 when \( m > 0 \) .\n\nIf \( p \) is a prime, then \( {x}^{p} - 1 = \left( {x - 1}\right) {\Phi }_{p}\left( x\right) \), where \( {\Phi }_{p}\left( x\right) \) is the \( p \) th cyclotomic polynomial. A primitive \( p \) th... | Yes |
Lemma 6.1. If \( S \) and \( T \) are subgroups of an abelian group \( G \), then \( G = S \oplus T \) if and only if \( S + T = G \) and \( S \cap T = \{ 0\} \) . | Proof. Assume that \( G = S \oplus T \) . Every \( g \in G \) has a unique expression of the form \( g = s + t \), where \( s \in S \) and \( t \in T \) ; hence, \( G = S + T \) . If \( x \in S \cap T \) , then \( x \) has two expressions as \( s + t \), namely, \( x = x + 0 \) and \( x = 0 + x \) . Since expressions a... | Yes |
Proposition 6.2. Let \( S \) and \( T \) be subgroups of an abelian group \( G \) with \( G = \) \( S + T \) . If \( G = S \oplus T \) (that is, \( S \cap T = \{ 0\} \) ), then there is an isomorphism \( \varphi : S \oplus T \rightarrow S \times T \) with \( \varphi \left( S\right) = {S}^{ * } \) and \( \varphi \left( ... | Proof. If \( g \in S \oplus T \), then Lemma 6.1 says that \( g \) has a unique expression of the form \( g = s + t \) . Define \( \varphi : S \oplus T \rightarrow S \times T \) by \( \varphi \left( g\right) = \varphi \left( {s + t}\right) = \) \( \left( {s, t}\right) \) . Uniqueness of the expression \( g = s + t \) i... | Yes |
Proposition 6.4. Let \( G = {S}_{1} + {S}_{2} + \cdots + {S}_{n} \), where the \( {S}_{i} \) are subgroups; that is, each \( g \in G \) has an expression of the form\n\n\[ g = {s}_{1} + {s}_{2} + \cdots + {s}_{n} \]\n\nwhere \( {s}_{i} \in {S}_{i} \) for all \( i \) . Then the following conditions are equivalent.\n\n(i... | Proof.\n\n(i) \( \Rightarrow \) (ii) If \( g \in G \) and \( g = {s}_{1} + \cdots + {s}_{n} \), then define \( \varphi : G \rightarrow {S}_{1} \times \cdots \times {S}_{n} \) by \( \varphi \left( g\right) = \varphi \left( {{s}_{1} + \cdots + {s}_{n}}\right) = \left( {{s}_{1},\ldots ,{s}_{n}}\right) \) . Uniqueness of t... | Yes |
Lemma 6.7. If \( p \) is a prime and \( G \neq \{ 0\} \) is a finite \( p \) -primary abelian group, then \( G \) has a nonzero pure cyclic subgroup. | Proof. Let \( G = \left\langle {{x}_{1},\ldots ,{x}_{q}}\right\rangle \) . The order of \( {x}_{i} \) is \( {p}^{{n}_{i}} \) for all \( i \), because \( G \) is \( p \) - primary. If \( x \in G \), then \( x = \mathop{\sum }\limits_{i}{a}_{i}{x}_{i} \), where \( {a}_{i} \in \mathbb{Z} \), so that if \( \ell \) is the l... | Yes |
Proposition 6.8. If \( G \) is an abelian group and \( p \) is a prime, then \( G/{pG} \) is a vector space over \( {\mathbb{F}}_{p} \) which is finite-dimensional when \( G \) is finite. | Proof. If \( \left\lbrack r\right\rbrack \in {\mathbb{F}}_{p} \) and \( a \in G \), define scalar multiplication\n\n\[ \left\lbrack r\right\rbrack \left( {a + {pG}}\right) = {ra} + {pG}. \]\n\nThis formula is well-defined, for if \( k \equiv r{\;\operatorname{mod}\;p} \), then \( k = r + {pm} \) for some integer \( m \... | Yes |
Lemma 6.9. If \( G \) is a p-primary abelian group, then \( d\left( G\right) = 1 \) if and only if \( G \) is cyclic. | Proof. If \( G \) is cyclic, then so is any quotient of \( G \) ; in particular, \( G/{pG} \) is cyclic, and so \( \dim \left( {G/{pG}}\right) = 1 \) .\n\nConversely, if \( G/{pG} = \langle z + {pG}\rangle \), then \( G/{pG} \cong {\mathbb{I}}_{p} \) . Since \( {\mathbb{I}}_{p} \) is a simple group, the correspondence ... | Yes |
Lemma 6.10. Let \( G \) be a finite p-primary abelian group.\n\n(i) If \( S \subseteq G \), then \( d\left( {G/S}\right) \leq d\left( G\right) \) . | Proof.\n\n(i) By the correspondence theorem, \( p\left( {G/S}\right) = \left( {{pG} + S}\right) /S \), so that\n\n\[ \left( {G/S}\right) /p\left( {G/S}\right) = \left( {G/S}\right) /\left\lbrack {\left( {{pG} + S}\right) /S}\right\rbrack \cong G/\left( {{pG} + S}\right) ,\n\]\n\nby the third isomorphism theorem. Since ... | Yes |
Theorem 6.11 (Basis Theorem). Every finite abelian group \( G \) is a direct sum of primary cyclic groups. | Proof. By the primary decomposition, Theorem 6.5, we may assume that \( G \) is \( p \) -primary for some prime \( p \) (for if every primary component is a direct sum of cyclic groups, so is \( G \) ). We prove that \( G \) is a direct sum of cyclic groups by induction on \( d\left( G\right) \geq 1 \) . The base step ... | Yes |
Lemma 6.12. Let \( G \) be a finite \( p \) -primary abelian group, where \( p \) is a prime, and let \( G = {\bigoplus }_{j}{C}_{j} \), where each \( {C}_{j} \) is cyclic. If \( {b}_{n} \) is the number of summands \( {C}_{j} \) having order \( {p}^{n} \), then there is some \( t \geq 1 \) with | Proof. Let \( {B}_{n} \) be the direct sum of all \( {C}_{j} \), if any, with order \( {p}^{n} \) . Thus,\n\n\[ G = {B}_{1} \oplus {B}_{2} \oplus \cdots \oplus {B}_{t} \]\n\nfor some \( t \) . Now\n\n\[ {p}^{n}G = {p}^{n}{B}_{n + 1} \oplus \cdots \oplus {p}^{n}{B}_{t} \]\n\nbecause \( {p}^{n}{B}_{j} = \{ 0\} \) for all... | Yes |
Theorem 6.13. If \( p \) is a prime, then any two decompositions of a finite \( p \) - primary abelian group \( G \) into direct sums of cyclic groups have the same number of cyclic summands of each type. More precisely, for each \( n \geq 0 \), the number of cyclic summands having order \( {p}^{n + 1} \) is \( {U}_{p}... | Proof. By the basis theorem, there exist cyclic subgroups \( {C}_{i} \) with \( G = {\bigoplus }_{i}{C}_{i} \) . The lemma shows, for each \( n \geq 0 \), that the number of \( {C}_{i} \) having order \( {p}^{n + 1} \) is \( {U}_{p}\left( {n, G}\right) \), a number that is defined without any mention of the given decom... | Yes |
Corollary 6.14. If \( G \) and \( {G}^{\prime } \) are finite p-primary abelian groups, then \( G \cong {G}^{\prime } \) if and only if \( {U}_{p}\left( {n, G}\right) = {U}_{p}\left( {n,{G}^{\prime }}\right) \) for all \( n \geq 0 \) . | Proof. If \( \varphi : G \rightarrow {G}^{\prime } \) is an isomorphism, then \( \varphi \left( {{p}^{n}G}\right) = {p}^{n}{G}^{\prime } \) for all \( n \geq \) 0, and hence it induces isomorphisms of the \( {\mathbb{F}}_{p} \) -vector spaces \( {p}^{n}G/{p}^{n + 1}G \cong \) \( {p}^{n}{G}^{\prime }/{p}^{n + 1}{G}^{\pr... | Yes |
Theorem 6.15 (Fundamental Theorem of Finite Abelian Groups). Finite abelian groups \( G \) and \( {G}^{\prime } \) are isomorphic if and only if they have the same elementary divisors; that is, any two decompositions of \( G \) and \( {G}^{\prime } \) as direct sums of primary cyclic groups have the same number of summ... | Proof. By the primary decomposition, Theorem 6.5(ii), \( G \cong {G}^{\prime } \) if and only if, for each prime \( p \), their primary components are isomorphic: \( {G}_{p} \cong {G}_{p}^{\prime } \) . The result now follows from Theorem 6.13. - | Yes |
Proposition 6.17. If \( H \) is a subgroup of a finite group \( G \), then the number of conjugates of \( H \) in \( G \) is \( \left\lbrack {G : {N}_{G}\left( H\right) }\right\rbrack \) . | Proof. This is a special case of Theorem 2.141: the size of the orbit of an element is the index of its stabilizer. | No |
Lemma 6.18. Let \( P \) be a Sylow p-subgroup of a finite group \( G \). (i) Every conjugate of \( P \) is also a Sylow p-subgroup of \( G \). (ii) \( \left| {{N}_{G}\left( P\right) /P}\right| \) is prime to \( p \). (iii) If \( g \in G \) has order some power of \( p \) and if \( {gP}{g}^{-1} = P \), then \( g \in P \... | Proof. (i) If \( g \in G \), then \( {gP}{g}^{-1} \) is a \( p \) -subgroup of \( G \) ; if it is not a maximal such, then there is a \( p \) -subgroup \( Q \) with \( {gP}{g}^{-1} < Q \). Hence, \( P < {g}^{-1}{Qg} \), contradicting the maximality of \( P \). (ii) If \( p \) divides \( \left| {{N}_{G}\left( P\right) /... | Yes |
Corollary 6.20. A finite group \( G \) has a unique Sylow p-subgroup \( P \), for some prime \( p \), if and only if \( P \vartriangleleft G \) . | Proof. Assume that \( P \), a Sylow \( p \) -subgroup of \( G \), is unique. For each \( a \in G \) , the conjugate \( {aP}{a}^{-1} \) is also a Sylow \( p \) -subgroup; by uniqueness, \( {aP}{a}^{-1} = P \) for all \( a \in G \), and so \( P \vartriangleleft G \) .\n\nConversely, assume that \( P \vartriangleleft G \)... | Yes |
Theorem 6.21 (Sylow). If \( G \) is a finite group of order \( {p}^{e}m \), where \( p \) is a prime and \( p \nmid m \), then every Sylow \( p \) -subgroup \( P \) of \( G \) has order \( {p}^{e} \) . | Proof. We first show that \( p \nmid \left\lbrack {G : P}\right\rbrack \) . Now\n\n\[ \left\lbrack {G : P}\right\rbrack = \left\lbrack {G : {N}_{G}\left( P\right) }\right\rbrack \left\lbrack {{N}_{G}\left( P\right) : P}\right\rbrack . \]\n\nThe first factor, \( \left\lbrack {G : {N}_{G}\left( P\right) }\right\rbrack = ... | Yes |
Theorem 6.23 (= Theorem 6.21). If \( G \) is a finite group of order \( {p}^{e}m \), where \( p \) is a prime and \( p \nmid m \), then \( G \) has a subgroup of order \( {p}^{e} \) . | Proof. If \( X \) is the family of all those subsets of \( G \) having exactly \( {p}^{e} \) elements, then \( \left| X\right| = \left( \begin{matrix} n \\ {p}^{e} \end{matrix}\right) \) ; by Exercise 1.66 on page 56, \( p \nmid \left| X\right| \) . Now \( G \) acts on \( X \) : define \( {\alpha }_{g}\left( B\right) =... | Yes |
Lemma 6.24. There is no nonabelian simple group \( G \) of order \( \left| G\right| = {p}^{e}m \) , where \( p \) is prime and \( m > 1, p \nmid m \), and \( {p}^{e} \nmid \left( {m - 1}\right) \)!. | Proof. Suppose that such a simple group \( G \) exists. By Sylow’s theorem, \( G \) contains a subgroup \( P \) of order \( {p}^{e} \), hence of index \( m \) . By Theorem 2.67, the representation of \( G \) on the cosets of \( P \), there exists a homomorphism \( \varphi : G \rightarrow {S}_{m} \) with \( \ker \varphi... | Yes |
Proposition 6.25. There are no nonabelian simple groups of order less than 60. | Proof. If \( p \) is a prime, then Exercise 2.106 on page 204 says that every \( p \) -group \( G \) with \( \left| G\right| > p \) is not simple.\n\nThe reader may check that the only integers \( n \) between 2 and 59, neither a prime power nor having a factorization of the form \( n = {p}^{e}m \) as in the statement ... | Yes |
Proposition 6.26. Let \( G \) be a finite group. If \( p \) is a prime and if \( {p}^{k} \) divides \( \left| G\right| \) , then \( G \) has a subgroup of order \( {p}^{k} \) . | Proof. If \( \left| G\right| = {p}^{e}m \), where \( p \nmid m \), then a Sylow \( p \) -subgroup \( P \) of \( G \) has order \( {p}^{e} \) . Hence, if \( {p}^{k} \) divides \( \left| G\right| \), then \( {p}^{k} \) divides \( \left| P\right| \) . By Proposition 2.150, \( P \) has a subgroup of order \( {p}^{k} \) ; a... | Yes |
Proposition 6.27. \( \mathrm{{UT}}\left( {n, k}\right) \) is a subgroup of \( \mathrm{{GL}}\left( {n, k}\right) \) for every field \( k \) . | Proof. If \( A \in \operatorname{UT}\left( {n, k}\right) \), then \( A = I + N \), where \( N \) is strictly upper triangular; that is, \( N \) is an upper triangular matrix having only 0 ’s on its diagonal. Note that the sum and product of strictly upper triangular matrices is again strictly upper triangular.\n\nLet \... | Yes |
Proposition 6.28. Let \( q = {p}^{e} \), where \( p \) is a prime. For each \( n \geq 2,\operatorname{UT}\left( {n,{\mathbb{F}}_{q}}\right) \) is a p-group of order \( {q}^{n\left( {n - 1}\right) /2} \) . | Proof. The number of entries in an \( n \times n \) unitriangular matrix lying strictly above the diagonal is \( \frac{1}{2}\left( {{n}^{2} - n}\right) = n\left( {n - 1}\right) /2 \) . Since each of these entries can be any element of \( {\mathbb{F}}_{q} \), there are exactly \( {q}^{n\left( {n - 1}\right) /2}n \times ... | Yes |
Proposition 6.29. If \( p \) is an odd prime, then there exists a nonabelian group \( G \) of order \( {p}^{3} \) with \( {x}^{p} = 1 \) for all \( x \in G \) . | Proof. If \( G = \operatorname{UT}\left( {3,{\mathbb{F}}_{p}}\right) \), then \( \left| G\right| = {p}^{3} \) . If \( A \in G \), then \( A = I + N \), where \( {N}^{3} = 0 \) ; hence \( {N}^{p} = 0 \) because \( p \geq 3 \) . Since \( {IN} = N = {NI} \), the binomial theorem gives \( {A}^{p} = {\left( I + N\right) }^{... | Yes |
Theorem 6.30. Let \( {\mathbb{F}}_{q} \) denote the finite field with \( q \) elements. Then \[ \left| {\mathrm{{GL}}\left( {n,{\mathbb{F}}_{q}}\right) }\right| = \left( {{q}^{n} - 1}\right) \left( {{q}^{n} - q}\right) \left( {{q}^{n} - {q}^{2}}\right) \cdots \left( {{q}^{n} - {q}^{n - 1}}\right) . | Proof. Let \( V \) be an \( n \) -dimensional vector space over \( {\mathbb{F}}_{q} \) . We show first that there is a bijection \( \Phi : \mathrm{{GL}}\left( {n,{\mathbb{F}}_{q}}\right) \rightarrow \mathcal{B} \), where \( \mathcal{B} \) is the set of all bases of \( V \) . Choose, once for all, a basis \( {e}_{1},\ld... | Yes |
Corollary 6.31. \( \left| {\mathrm{{GL}}\left( {n,{\mathbb{F}}_{q}}\right) }\right| = {q}^{n\left( {n - 1}\right) /2}\left( {{q}^{n} - 1}\right) \left( {{q}^{n - 1} - 1}\right) \cdots \left( {{q}^{2} - 1}\right) \left( {q - 1}\right) \) . | Proof. The number of powers of \( q \) in the formula\n\n\[ \left| {\mathrm{{GL}}\left( {n,{\mathbb{F}}_{q}}\right) }\right| = \left( {{q}^{n} - 1}\right) \left( {{q}^{n} - q}\right) \left( {{q}^{n} - {q}^{2}}\right) \cdots \left( {{q}^{n} - {q}^{n - 1}}\right) \]\n\nis \( {q}^{1 + 2 + \cdots + \left( {n - 1}\right) } ... | Yes |
Theorem 6.32. If \( p \) is a prime and \( q = {p}^{m} \), then the unitriangular group \( \operatorname{UT}\left( {n,{\mathbb{F}}_{q}}\right) \) is a Sylow p-subgroup of \( \operatorname{GL}\left( {n,{\mathbb{F}}_{q}}\right) \) . | \( \begin{array}{ll} \text{ Proof. } & \operatorname{Now}\left| {\mathrm{{UT}}\left( {n,{\mathbb{F}}_{q}}\right) }\right| = {q}^{n\left( {n - 1}\right) /2}\left( {{q}^{n} - 1}\right) \left( {{q}^{n - 1} - 1}\right) \cdots \left( {{q}^{2} - 1}\right) \left( {q - 1}\right) ,\mathrm{{by}} \end{array} \) Corollary 6.31, so... | Yes |
Lemma 6.35. Let \( A, P, Q \) be distinct points in the plane, let \( C = C\left\lbrack {P;{PA}}\right\rbrack \) be the circle with center \( P \) and radius \( {PA} \), and let \( {C}^{\prime } = C\left\lbrack {Q;{QA}}\right\rbrack \) be the circle with center \( Q \) and radius \( {QA} \) . Then \( C \cap {C}^{\prime... | Proof. We use analytic geometry. Draw \( P \) and \( Q \) as the points 0 and 1 on the \( x \) - axis, and let \( A = \left( {a, b}\right) \) . The equation of \( C \) is \( {x}^{2} + {y}^{2} = {\left| PA\right| }^{2} = {a}^{2} + {b}^{2} \), and the equation of \( {C}^{\prime } \) is \( {\left( x - 1\right) }^{2} + {y}... | Yes |
Proposition 6.36. Let \( \varphi : \mathbb{C} \rightarrow \mathbb{C} \) be an isometry fixing 0 .\n\n(i) There is some \( \theta \) with \( \varphi \left( 1\right) = {e}^{i\theta } \) . If \( \varphi \left( 1\right) = 1 \), then \( \varphi \) fixes the \( x \) -axis pointwise, and \( \varphi \) is either the identity o... | Proof.\n\n(i) Let \( z \in \mathbb{R} \) be distinct from 0, and let \( {C}_{\left| z\right| } \) be the circle with center 0 and radius \( \left| z\right| = \left| {0z}\right| \) . Since \( \varphi \) is an isometry fixing 0, we have \( \varphi \left( {C}_{\left| z\right| }\right) = {C}_{\left| z\right| } \), for isom... | Yes |
Corollary 6.37. If \( \varphi \) is an isometry with \( \varphi \left( 0\right) = c \), then there is some \( \theta \) so that\n\n\[ \varphi \left( z\right) = {e}^{i\theta }z + c\;\text{ or }\;\varphi \left( z\right) = {e}^{i\theta }\bar{z} + c. \] | Proof. If \( \varphi \) is a translation, say, \( \varphi : z \mapsto z + c \), then \( \varphi \) has the formula \( \varphi \left( z\right) = \) \( {e}^{i\theta }z + c \) with \( \theta = 0 \) . In general, given \( \varphi : z \mapsto {e}^{i\theta }z + c \), define \( \tau \) to be translation by \( c = \varphi \lef... | Yes |
Proposition 6.40. If \( \varphi : z \mapsto {e}^{i\theta }\bar{z} + c \) is not a reflection, then \( \varphi = {\tau \rho } \), where \( \rho \) is a reflection, say, with axis \( \ell \), and \( \tau \) is a translation \( z \mapsto z + \frac{1}{2}w \), where \( w \) has direction that of \( \ell \) . | Proof. As in the proof of (i) \( \Rightarrow \) (ii) in Proposition 6.38, we have \( {\varphi }^{2}\left( z\right) = \) \( z + {e}^{i\theta }\bar{c} + c \) . We define \( w = {e}^{i\theta }\bar{c} + c \), so that\n\n\[{\varphi }^{2} : z \mapsto z + w\]\n\n(1)\n\nNow define\n\n\[\tau : z \mapsto z + \frac{1}{2}w\]\n\nso... | Yes |
Theorem 6.42. Every isometry is either a translation, a rotation, a reflection, or a glide-reflection. | Proof. The theorem follows from Proposition 6.36(ii), Corollary 6.37, Proposition 6.38, and Proposition 6.40. - | Yes |
Proposition 6.43. Let \( \varphi \in \operatorname{Isom}\left( {\mathbb{R}}^{2}\right) \).\n\n(i) If \( \varphi \) has no fixed points, then \( \varphi \) is either a translation or a glide-reflection.\n\n(ii) If \( \varphi \) has only one fixed point, then \( \varphi \) is a rotation.\n\n(iii) If \( \varphi \) has mor... | Proof. There are only four types of isometry, by Theorem 6.42: translations, which have no fixed points; rotations, which have exactly one fixed point; reflections, which have infinitely many fixed points, namely, every point on their axes; glide-reflections. It suffices to show that a glide-reflection \( \varphi \) ha... | Yes |
Lemma 6.45. Let \( \varphi \in \operatorname{Isom}\left( {\mathbb{R}}^{2}\right) \), and let \( {z}_{1},\ldots ,{z}_{n} \) be distinct points in \( \mathbb{C} \) . Then \( \varphi \left( u\right) = {u}^{\prime } \), where \( u \) is the center of gravity of \( {z}_{1},\ldots ,{z}_{n} \) and \( {u}^{\prime } \) is the c... | Proof. By Theorem 6.42, \( \varphi \) is either a translation, a rotation, a reflection, or a glide-reflection. A rotation about \( c \) is a composite \( {\tau \rho } \), where \( \tau \) is the translation \( z \mapsto z + c \) and \( \rho \) is a rotation around 0 . Proposition 6.40 shows that a glide-reflection is ... | Yes |
Lemma 6.46. If \( G \leq \operatorname{Isom}\left( {\mathbb{R}}^{2}\right) \) is a finite subgroup, then there is \( u \in \mathbb{C} \) with \( \varphi \left( u\right) = u \) for all \( \varphi \in G \) . | Proof. Choose \( z \in \mathbb{C} \), and let \( \mathcal{O} \) be its orbit:\n\n\[ \mathcal{O} = \{ \varphi \left( z\right) : \varphi \in G\} \]\n\nSince \( G \) is finite, \( \mathcal{O} \) is finite: \( \mathcal{O} = \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \), where \( {z}_{1} = z \) . Now \( G \) acts on \( \math... | Yes |
Theorem 6.47 (Leonardo). If \( G \leq \operatorname{Isom}\left( {\mathbb{R}}^{2}\right) \) is a finite subgroup, then either \( G \cong {\mathbb{I}}_{m} \) for some \( m \) or \( G \cong {D}_{2n} \) for some \( n \) . | Proof. By Lemma 6.46, there is \( c \in \mathbb{C} \) with \( \varphi \left( c\right) = c \) for all \( \varphi \in G \) . If \( \tau : z \mapsto z - c \), then \( {\tau \varphi }{\tau }^{-1}\left( 0\right) = {\tau \varphi }\left( c\right) = \tau \left( c\right) = 0 \) . Since \( {\tau G}{\tau }^{-1} \cong G \), we may... | Yes |
Lemma 6.48. If \( \varphi \in G \), where \( G \) is a frieze group, then there is some real number \( c \) such that one of the following holds:\n\n(i) If \( \varphi \) is a translation, then \( \varphi \left( z\right) = z + c \) .\n\n(ii) If \( \varphi \) is a rotation, then \( \varphi \) is a half-turn: \( \varphi \... | Proof. We know that \( \varphi : z \mapsto {e}^{i\theta }z + c \) or \( \varphi : z \mapsto {e}^{i\theta }\bar{z} + c \) . Since \( \varphi \left( \mathbb{R}\right) = \mathbb{R} \) , we have \( c = \varphi \left( 0\right) \in \mathbb{R} \) and \( \varphi \left( 1\right) = {e}^{i\theta } + c \in \mathbb{R} \) . Therefor... | Yes |
Corollary 6.49. The point group \( \pi \left( G\right) \) of a frieze group \( G \) is a subgroup of \( \operatorname{im}\pi = \{ 1, f, g, h\} \leq {O}_{2}\left( \mathbb{R}\right) \) (which is isomorphic to the four-group \( \mathbf{V} \) ), where \( f\left( z\right) = - z, g\left( z\right) = - \bar{z} \), and \( h\lef... | Proof. By Lemma 6.48, we have \( \operatorname{im}\pi = \{ 1, f, g, h\} \) . | Yes |
Corollary 6.52. To isomorphism, there are 4 frieze groups, namely, \( \mathbb{Z},{D}_{\infty } \) , \( {\mathbb{I}}_{2} \times \mathbb{Z} \), and \( {\mathbb{I}}_{2} \times {D}_{\infty } \) . | Proof. As stated in the proof of Theorem 6.50, \( \sum \left( {F}_{1}\right) \) and \( \sum \left( {F}_{5}\right) \) are isomorphic to \( \mathbb{Z},\sum \left( {F}_{2}\right) ,\sum \left( {F}_{3}\right) \) and \( \sum \left( {F}_{7}\right) \) are isomorphic to \( {D}_{\infty },\sum \left( {F}_{4}\right) \) is isomorph... | Yes |
Proposition 7.1 (Correspondence Theorem for Rings). If \( I \) is a proper ideal in a commutative ring \( R \), then the natural map \( \pi : R \rightarrow R/I \) induces an inclusion-preserving bijection \( {\pi }^{\prime } \) from the set of all intermediate ideals \( J \) (that is, \( I \subseteq J \subseteq R \) ),... | Proof. If one forgets its multiplication, the commutative ring \( R \) is an additive abelian group and its ideals \( I \) are (normal) subgroups. The correspondence theorem for groups, Theorem 2.121, now applies, and it gives an inclusion-preserving bijection\n\n\[ \n{\pi }_{ * } : \{ \text{all subgroups of}R\text{con... | Yes |
Proposition 7.4. If \( k \) is a field, then a nonzero polynomial \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) is irreducible if and only if \( \left( {p\left( x\right) }\right) \) is a prime ideal. | Proof. Suppose that \( p\left( x\right) \) is irreducible. First, \( \left( p\right) \) is a proper ideal; otherwise, \( R = \left( p\right) \) and hence \( 1 \in \left( p\right) \), so there is a polynomial \( f\left( x\right) \) with \( 1 = p\left( x\right) f\left( x\right) \) . But \( p\left( x\right) \) has degree ... | Yes |
Proposition 7.5. A proper ideal \( I \) in a commutative ring \( R \) is a prime ideal if and only if \( R/I \) is a domain. | Proof. Let \( I \) be a prime ideal. Since \( I \) is a proper ideal, we have \( 1 \notin I \) and so \( 1 + I \neq 0 + I \) in \( R/I \) . If \( 0 = \left( {a + I}\right) \left( {b + I}\right) = {ab} + I \), then \( {ab} \in I \) . Since \( I \) is a prime ideal, either \( a \in I \) or \( b \in I \) ; that is, either... | Yes |
Lemma 7.6. The ideal \( \{ 0\} \) is a maximal ideal in a commutative ring \( R \) if and only if \( R \) is a field. | Proof. It is shown in Proposition 3.43 that every nonzero ideal \( I \) in \( R \) is equal to \( R \) itself if and only if every nonzero element in \( R \) is a unit. That is, \( \{ 0\} \) is a maximal ideal if and only if \( R \) is a field. \( \; \bullet \) | Yes |
Proposition 7.7. A proper ideal \( I \) in a commutative ring \( R \) is a maximal ideal if and only if \( R/I \) is a field. | Proof. The correspondence theorem for rings shows that \( I \) is a maximal ideal if and only if \( R/I \) has no ideals other than \( \{ 0\} \) and \( R/I \) itself; Lemma 7.6 shows that this property holds if and only if \( R/I \) is a field. | Yes |
Every maximal ideal \( I \) in a commutative ring \( R \) is a prime ideal. | Proof. If \( I \) is a maximal ideal, then \( R/I \) is a field. Since every field is a domain, \( R/I \) is a domain, and so \( I \) is a prime ideal. | Yes |
Corollary 7.10. If \( k \) is a field, then \( \left( {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right) \) is a maximal ideal in \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), where \( {a}_{i} \in k \) for \( i = 1,\ldots, n \) . | Proof. By Proposition 3.33, there is a unique homomorphism\n\n\[ \varphi : k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \]\n\nwith \( \varphi \left( c\right) = c \) for all \( c \in k \) and with \( \varphi \left( {x}_{i}\right) = {x}_{i} - {a}_{i... | Yes |
Theorem 7.11. If \( R \) is a PID, then every nonzero prime ideal \( I \) is a maximal ideal. | Proof. Assume there is a proper ideal \( J \) with \( I \subseteq J \) . Since \( R \) is a PID, \( I = \left( a\right) \) and \( J = \left( b\right) \) for some \( a, b \in R \) . Now \( a \in J \) implies that \( a = {rb} \) for some \( r \in R \) , and so \( {rb} \in I \) ; but \( I \) is a prime ideal, so that \( r... | No |
Corollary 7.12. If \( k \) is a field and \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) is irreducible, then the quotient ring \( k\left\lbrack x\right\rbrack /\left( {p\left( x\right) }\right) \) is a field. | Proof. Since \( p\left( x\right) \) is irreducible, Proposition 7.4 says that the principal ideal \( I = \left( {p\left( x\right) }\right) \) is a nonzero prime ideal; since \( k\left\lbrack x\right\rbrack \) is a PID, \( I \) is a maximal ideal, and so \( k\left\lbrack x\right\rbrack /I \) is a field. | Yes |
Proposition 7.13. Let \( R \) be a domain and let \( a, b \in R \). (ii) The principal ideals (a) and (b) are equal if and only if a and b are associates. | Proof. (ii) If \( \left( a\right) = \left( b\right) \), then \( \left( a\right) \subseteq \left( b\right) \) and \( \left( b\right) \subseteq \left( a\right) \) ; hence, \( a \in \left( b\right) \) and \( b \in \left( a\right) \) . Thus, \( a \mid b \) and \( b \mid a \) ; by part (i), \( a \) and \( b \) are associate... | No |
Lemma 7.14. Let \( R \) be a PID.\n\n(i) There is no infinite strictly ascending chain of ideals\n\n\[ {I}_{1} \subsetneq {I}_{2} \subsetneq \cdots \subsetneq {I}_{n} \subsetneq {I}_{n + 1} \subsetneq \cdots . \] | (i) If, on the contrary, an infinite strictly ascending chain exists, then define \( J = \) \( \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{I}_{n} \) . We claim that \( J \) is an ideal. If \( a \in J \), then \( a \in {I}_{n} \) for some \( n \) ; if \( r \in R \) , then \( {ra} \in {I}_{n} \), because \( {I}_{n} \) i... | Yes |
Proposition 7.15. Let \( R \) be a domain in which every \( r \in R \), neither 0 nor a unit, is a product of irreducibles. Then \( R \) is a UFD if and only if, for every irrreducible element \( p \in R \), the principal ideal \( \left( p\right) \) is a prime ideal in \( R \) . | Proof. Assume that \( R \) is a UFD. If \( a, b \in R \) and \( {ab} \in \left( p\right) \), then there is \( r \in R \) with\n\n\[ \n{ab} = {rp}\text{.}\n\]\n\nFactor each of \( a, b \), and \( r \) into irreducibles; by unique factorization, the left side of the equation must involve an associate of \( p \) . This as... | Yes |
Theorem 7.16. If \( R \) is a PID, then \( R \) is a UFD. In particular, every euclidean ring is a UFD. | Proof. In view of the last two results, it suffices to prove that \( \left( p\right) \) is a prime ideal whenever \( p \) is irreducible. Suppose that \( p \mid {ab} \) ; we must show that \( p \mid b \) or \( p \mid a \) . The subset\n\n\[ I = \{ {sb} + {tp} : s, t \in R\} \]\n\nis an ideal in \( R \) and, hence, \( I... | Yes |
Proposition 7.17. If \( R \) is a UFD, then the gcd of any finite set of elements \( {a}_{1},\ldots ,{a}_{n} \) exists. | Proof. It suffices to prove that the gcd of two elements \( a \) and \( b \) exists, for an easy inductive proof then shows that a gcd of any finite number of elements exists.\n\nThere are units \( u \) and \( v \) and distinct irreducibles \( {p}_{1},\ldots ,{p}_{t} \) with\n\n\[ a = u{p}_{1}^{{e}_{1}}{p}_{2}^{{e}_{2}... | Yes |
Lemma 7.19. If \( R \) is a UFD and \( f\left( x\right), g\left( x\right) \in R\left\lbrack x\right\rbrack \) are both primitive, then their product \( f\left( x\right) g\left( x\right) \) is also primitive. | Proof. If \( \pi : R \rightarrow R/\left( p\right) \) is the natural map \( \pi : a \mapsto a + \left( p\right) \), then Proposition 3.33 shows that the function \( \widetilde{\pi } : R\left\lbrack x\right\rbrack \rightarrow \left( {R/\left( p\right) }\right) \left\lbrack x\right\rbrack \), which replaces each coeffici... | Yes |
Lemma 7.20. Let \( R \) be a UFD.\n\n(i) Every nonzero \( f\\left( x\\right) \\in R\\left\\lbrack x\\right\\rbrack \) has a factorization\n\n\[ f\\left( x\\right) = c\\left( f\\right) {f}^{ * }\\left( x\\right) \]\n\nwhere \( c\\left( f\\right) \\in R \) and \( {f}^{ * }\\left( x\\right) \\in R\\left\\lbrack x\\right\\... | Proof.\n\n(i) If \( f\\left( x\\right) = {a}_{n}{x}^{n} + \\cdots + {a}_{1}x + {a}_{0} \) and \( c\\left( f\\right) \) is the content of \( f \), then there are factorizations \( {a}_{i} = c\\left( f\\right) {b}_{i} \) in \( R \) for \( i = 0,1,\\ldots, n \) ; if we define \( {f}^{ * }\\left( x\\right) = \) \( {b}_{n}{... | Yes |
Theorem 7.21 (Gauss). If \( R \) is a UFD, then \( R\left\lbrack x\right\rbrack \) is also a UFD. | Proof. We show first, by induction on \( \deg \left( f\right) \), that every \( f\left( x\right) \in R\left\lbrack x\right\rbrack \), neither zero nor a unit, is a product of irreducibles. If \( \deg \left( f\right) = 0 \), then \( f\left( x\right) \) is a constant, hence lies in \( R \) . Since \( R \) is a UFD, \( f ... | Yes |
Corollary 7.22. If \( k \) is a field, then \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is a UFD. | Proof. The proof is by induction on \( n \geq 1 \) . We proved, in Chapter 3, that the polynomial ring \( k\left\lbrack {x}_{1}\right\rbrack \) in one variable is a UFD. For the inductive step, recall that \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n},{x}_{n + 1}}\right\rbrack = R\left\lbrack {x}_{n + 1}\right\rbrack \), w... | Yes |
Corollary 7.23. Let \( R \) be a UFD, let \( Q = \operatorname{Frac}\left( R\right) \), and let \( f\left( x\right) \in R\left\lbrack x\right\rbrack \) . If\n\n\[ f\left( x\right) = G\left( x\right) H\left( x\right) \text{ in }Q\left\lbrack x\right\rbrack ,\] \n\nthen there is a factorization\n\n\[ f\left( x\right) = g... | Proof. By Lemma 7.20, there is a factorization\n\n\[ f\left( x\right) = c\left( G\right) c\left( H\right) {G}^{ * }\left( x\right) {H}^{ * }\left( x\right) \text{ in }Q\left\lbrack x\right\rbrack ,\] \n\nwhere \( {G}^{ * }\left( x\right) ,{H}^{ * }\left( x\right) \in R\left\lbrack x\right\rbrack \) are primitive polyno... | Yes |
Corollary 7.24. Let \( k \) be a field and let \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) be a primitive polynomial in \( R\left\lbrack {x}_{n}\right\rbrack \), where \( R = k\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \) . If \( f \) cannot be factored into two polynomials of lower degree in \( R\left\... | Proof. Let us write \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) = F\left( {x}_{n}\right) \), for we wish to view \( f \) as a polynomial in \( R\left\lbrack {x}_{n}\right\rbrack \) ; that is, we view \( f \) as a polynomial in \( {x}_{n} \) having coefficients in \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbr... | Yes |
Proposition 7.26. The following conditions are equivalent for a commutative ring \( R \) .\n\n(i) \( R \) has the ACC.\n\n(ii) \( R \) satisfies the maximum condition: every nonempty family \( \mathcal{F} \) of ideals in \( R \) has a maximal element; that is, there is some \( {I}_{0} \in \mathcal{F} \) for which there... | Proof. (i) \( \Rightarrow \) (ii): Let \( \mathcal{F} \) be a family of ideals in \( R \), and assume that \( \mathcal{F} \) has no maximal element. Choose \( {I}_{1} \in \mathcal{F} \) . Since \( {I}_{1} \) is not a maximal element, there is \( {I}_{2} \in \mathcal{F} \) with \( {I}_{1} \subsetneq {I}_{2} \) . Now \( ... | Yes |
Corollary 7.27. If \( I \) is an ideal in a nonzero noetherian ring \( R \), then there exists a maximal ideal \( M \) in \( R \) containing \( I \) . In particular, every noetherian ring has maximal ideals. | Proof. Let \( \mathcal{F} \) be the family of all those proper ideals in \( R \) which contain \( I \) ; note that \( \mathcal{F} \neq \varnothing \) because \( I \in \mathcal{F} \) . Since \( R \) is noetherian, the maximum condition gives a maximal element \( M \) in \( \mathcal{F} \) . We must still show that \( M \... | Yes |
Corollary 7.28. If \( R \) is a noetherian ring and \( J \) is an ideal in \( R \), then \( R/J \) is also noetherian. | Proof. If \( A \) is an ideal in \( R/I \), then the correspondence theorem provides an ideal \( J \) in \( R \) with \( J/I = A \) . Since \( R \) is noetherian, the ideal \( J \) is finitely generated, say, \( J = \left( {{b}_{1},\ldots ,{b}_{n}}\right) \), and so \( A = J/I \) is also finitely generated (by the cose... | Yes |
Lemma 7.29. A commutative ring \( R \) is noetherian if and only if, for every sequence \( {a}_{1},\ldots ,{a}_{n},\ldots \) of elements in \( R \), there exists \( m \geq 1 \) and \( {r}_{1},\ldots ,{r}_{m} \in R \) with \( {a}_{m + 1} = {r}_{1}{a}_{1} + \cdots + {r}_{m}{a}_{m} \) . | Proof. Assume that \( R \) is noetherian and that \( {a}_{1},\ldots ,{a}_{n},\ldots \) is a sequence of elements in \( R \) . If \( {I}_{n} = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \), then there is an ascending chain of ideals, \( {I}_{1} \subseteq {I}_{2} \subseteq \cdots \) . By the ACC, there exists \( m \geq 2 \)... | Yes |
Theorem 7.30 (Hilbert Basis Theorem). If \( R \) is a commutative noetherian ring, then \( R\left\lbrack x\right\rbrack \) is also noetherian. | Proof. Assume that \( I \) is an ideal in \( R\left\lbrack x\right\rbrack \) that is not finitely generated; of course, \( I \neq \{ 0\} \) . Define \( {f}_{0}\left( x\right) \) to be a polynomial in \( I \) of minimal degree and define, inductively, \( {f}_{n + 1}\left( x\right) \) to be a polynomial of minimal degree... | Yes |
(i) If \( k \) is a field, then \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is noetherian. | Proof. The proofs of the first two items are by induction on \( n \geq 1 \), using the theorem, while the proof of item (iii) follows from Corollary 7.28. | No |
Proposition 7.32. Let \( k \) be an infinite field and let \( k\left\lbrack X\right\rbrack = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . If \( f\left( X\right), g\left( X\right) \in k\left\lbrack X\right\rbrack \) satisfy \( {f}^{\mathrm{b}} = {g}^{\mathrm{b}} \), then \( f\left( {{x}_{1},\ldots ,{x}_{n}}... | Proof. The proof is by induction on \( n \geq 1 \) ; the base step is Corollary 3.52. For the inductive step, write\n\n\[ f\left( {X, y}\right) = \mathop{\sum }\limits_{i}{p}_{i}\left( X\right) {y}^{i}\text{ and }g\left( {X, y}\right) = \mathop{\sum }\limits_{i}{q}_{i}\left( X\right) {y}^{i}, \]\n\nwhere \( X \) denote... | Yes |
Proposition 7.33. If \( k \) is an algebraically closed field and \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) is not a constant, then \( f\left( X\right) \) has a zero. | Proof. We prove the result by induction on \( n \geq 1 \), where \( X = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . The base step follows at once from our assuming that \( {k}^{1} = k \) is algebraically closed. As in the proof of Proposition 7.32, write\n\n\[ f\left( {X, y}\right) = \mathop{\sum }\limits_{i}{g}_{i}\l... | Yes |
Proposition 7.36. Let \( k \) be a field.\n\n(i) \( \operatorname{Var}\left( {{x}_{1},{x}_{1} - 1}\right) = \varnothing \) and \( \operatorname{Var}\left( 0\right) = {k}^{n} \), where 0 is the zero polynomial.\n\n(ii) If \( I \) and \( J \) are ideals in \( k\left\lbrack X\right\rbrack \), then\n\n\[ \operatorname{Var}... | Proof.\n\n(i) If \( a = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in \operatorname{Var}\left( {{x}_{1},{x}_{1} - 1}\right) \), then \( {a}_{1} = 0 \) and \( {a}_{1} = 1 \) ; plainly, there are no such points \( a \), and so \( \operatorname{Var}\left( {{x}_{1},{x}_{1} - 1}\right) = \varnothing \) . That \( \operatorname... | Yes |
Proposition 7.37. If \( A \subseteq {k}^{n} \), where \( k \) is a field, then there is an isomorphism\n\n\[ k\left\lbrack X\right\rbrack /\operatorname{Id}\left( A\right) \cong k\left\lbrack A\right\rbrack \] | Proof. The restriction map res : \( k\left\lbrack X\right\rbrack \rightarrow k\left\lbrack A\right\rbrack \) is a surjection with kernel \( \operatorname{Id}\left( A\right) \), and so the result follows from the first isomorphism theorem. Note that two polynomials agreeing on \( A \) lie in the same coset of \( \operat... | Yes |
Proposition 7.38. Let \( k \) be a field.\n\n(i) \( \operatorname{Id}\left( \varnothing \right) = k\left\lbrack X\right\rbrack \) and, if \( k \) is algebraically closed, \( \operatorname{Id}\left( {k}^{n}\right) = \{ 0\} \) .\n\n(ii) If \( A \subseteq B \) are subsets of \( {k}^{n} \), then \( \operatorname{Id}\left( ... | Proof.\n\n(i) If \( f\left( X\right) \in \operatorname{Id}\left( A\right) \) for some subset \( A \subseteq {k}^{n} \), then \( f\left( a\right) = 0 \) for all \( a \in A \) ; hence, if \( f\left( X\right) \notin \operatorname{Id}\left( A\right) \), then there exists \( a \in A \) with \( f\left( a\right) \neq 0 \) . I... | Yes |
Proposition 7.39. If an ideal \( I = \operatorname{Id}\left( A\right) \) for some \( A \subseteq {k}^{n} \), where \( k \) is a field, then it is a radical ideal. Hence, the coordinate ring \( k\left\lbrack A\right\rbrack \) has no nilpotent elements. | Proof. Since \( I \subseteq \sqrt{I} \) is always true, it suffices to check the reverse inclusion. By hypothesis, \( I = \operatorname{Id}\left( A\right) \) for some \( A \subseteq {k}^{n} \) ; hence, if \( f \in \sqrt{I} \), then \( {f}^{m} \in \operatorname{Id}\left( A\right) \) ; that is, \( f{\left( a\right) }^{m}... | Yes |
Lemma 7.41. Let \( k \) be a field and let \( \varphi : k\left\lbrack X\right\rbrack \rightarrow k \) be a surjective ring homomorphism which fixes \( k \) pointwise. If \( J = \ker \varphi \), then \( \operatorname{Var}\left( J\right) \neq \varnothing \) . | Proof. For each \( i \), we have \( {x}_{i} \in k\left\lbrack X\right\rbrack \) ; let \( \varphi \left( {x}_{i}\right) = {a}_{i} \in k \) and let \( a = \) \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {k}^{n} \) . If \( f\left( X\right) = \mathop{\sum }\limits_{{{e}_{1},\ldots ,{e}_{n}}}{c}_{{e}_{1},\ldots ,{e}_{n}}{... | Yes |
Corollary 7.43. If \( I \) is a proper ideal in \( \mathbb{C}\left\lbrack X\right\rbrack \), then there is an element \( a = \) \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {\mathbb{C}}^{n} \) with \( f\left( a\right) = 0 \) for all \( f \in I \) . | Proof. Choose \( a \) to be any element in \( \operatorname{Var}\left( I\right) \) . | No |
Theorem 7.45. Every maximal ideal \( M \) in \( \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) has the form\n\n\[ M = \left( {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right) \]\n\nwhere \( a = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {\mathbb{C}}^{n} \), and so there is a bijection between \( {... | Proof. Since \( M \) is a proper ideal, we have \( \operatorname{Var}\left( M\right) \neq \varnothing \), by Theorem 7.42; that is, there is \( a = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {k}^{n} \) with \( f\left( a\right) = 0 \) for all \( f \in M \) . Since \( \operatorname{Var}\left( M\right) = \left\{ {b \in {... | Yes |
Proposition 7.46. Let \( k \) be a field.\n\n(i) For every subset \( A \subseteq {k}^{n} \), \n\n\[ \operatorname{Var}\left( {\operatorname{Id}\left( A\right) }\right) \supseteq A\text{.} \]\n\n(ii) For every ideal \( I \subseteq k\left\lbrack X\right\rbrack \). \n\n\[ \operatorname{Id}\left( {\operatorname{Var}\left( ... | Proof.\n\n(i) This result is almost a tautology. If \( a \in A \), then \( f\left( a\right) = 0 \) for all \( f\left( X\right) \in \) \( \operatorname{Id}\left( A\right) \) . But every \( f\left( X\right) \in \operatorname{Id}\left( A\right) \) annihilates \( A \), by definition of \( \operatorname{Id}\left( A\right) \... | Yes |
Theorem 7.48. The functions \( V \mapsto \operatorname{Id}\left( V\right) \) and \( I \mapsto \operatorname{Var}\left( I\right) \) are inverse order-reversing bijections\n\n\[ \left\{ {\text{algebraic sets} \subseteq {\mathbb{C}}^{n}}\right\} \rightleftarrows \left\{ {\text{radical ideals} \subseteq \mathbb{C}\left\lbr... | Proof. By Proposition 7.46, we have \( \operatorname{Var}\left( {\operatorname{Id}\left( V\right) }\right) = V \) for every algebraic set \( V \), while Theorem 7.44 gives \( \operatorname{Id}\left( {\operatorname{Var}\left( I\right) }\right) = \sqrt{I} \) for every ideal \( I \) . | Yes |
Proposition 7.49. Every algebraic set \( V \) in \( {k}^{n} \) is a union of finitely many varieties: | Proof. Call an algebraic set \( W \in {k}^{n} \) good if it is irreducible or a union of finitely many varieties; otherwise, call \( W \) bad. We must show that there are no bad algebraic sets. If \( W \) is bad, it is not irreducible, and so \( W = {W}^{\prime } \cup {W}^{\prime \prime } \), where both \( {W}^{\prime ... | Yes |
An algebraic set \( V \) in \( {k}^{n} \) is a variety if and only if \( \operatorname{Id}\left( V\right) \) is a prime ideal in \( k\left\lbrack X\right\rbrack \) . Hence, the coordinate ring \( k\left\lbrack V\right\rbrack \) of a variety \( V \) is a domain. | Proof. Assume that \( V \) is a variety. It suffices to show that if \( {f}_{1}\left( X\right) ,{f}_{2}\left( X\right) \notin \) \( \operatorname{Id}\left( V\right) \), then \( {f}_{1}\left( X\right) {f}_{2}\left( X\right) \notin \operatorname{Id}\left( V\right) \) . Define, for \( i = 1,2 \) ,\n\n\[ \n{W}_{i} = V \cap... | Yes |
Proposition 7.51. Every algebraic set \( V \) is an irredundant union of varieties | Proof. By Proposition 7.49, \( V \) is a union of finitely many varieties; say, \( V = \) \( {V}_{1} \cup \cdots \cup {V}_{m} \) . If \( m \) is chosen minimal, then this union must be irredundant.\n\nWe now prove uniqueness. Suppose that \( V = {W}_{1} \cup \cdots \cup {W}_{s} \) is an irredundant union of varieties. ... | Yes |
Every radical ideal \( J \) in \( k\left\lbrack X\right\rbrack \) is an irredundant intersection of prime ideals, moreover, the prime ideals \( {P}_{i} \) are uniquely determined by \( J \) . | Proof. Since \( J \) is a radical ideal, there is a variety \( V \) with \( J = \operatorname{Id}\left( V\right) \) . Now \( V \) is an irredundant union of irreducible algebraic subsubsets, so that By Proposition 7.50, \( {V}_{i} \) irreducible implies \( \operatorname{Id}\left( {V}_{i}\right) \) is prime, and so \( J... | Yes |
Proposition 7.53. Let \( X \) be a well-ordered set.\n\n(i) If \( x, y \in X \), then either \( x \preccurlyeq y \) or \( y \preccurlyeq x \).\n\n(ii) Every strictly decreasing sequence is finite. | Proof.\n\n(i) The subset \( S = \{ x, y\} \) has a smallest element, which must be either \( x \) or \( y \) . In the first case, \( x \preccurlyeq y \) ; in the second case, \( y \preccurlyeq x \). (ii) Assume that there is an infinite strictly decreasing sequence, say,\n\n\[ \n{x}_{1} \succ {x}_{2} \succ {x}_{3} \suc... | Yes |
Corollary 7.55. If \( X \) is a well-ordered set, then \( \mathcal{W}\left( X\right) \) is well-ordered in the lexicographic order (which we also denote by \( { \preccurlyeq }_{\text{lex }} \) ). | Proof. We will only give a careful definition of the lexicographic order here; the proof that it is a well-order is left to the reader. First, define \( 1{ \preccurlyeq }_{\text{lex }}w \) for all \( w \in \mathcal{W}\left( X\right) \) . Next, given words \( u = {x}_{1}\cdots {x}_{p} \) and \( v = {y}_{1}\cdots {y}_{q}... | No |
Proposition 7.57. The degree-lexicographic order \( { \preccurlyeq }_{\text{dlex }} \) is a monomial order on \( {\mathbb{N}}^{n} \) . | Proof. It is routine to show that \( { \preccurlyeq }_{\text{dlex }} \) is a partial order on \( {\mathbb{N}}^{n} \) . To see that it is a well-order, let \( S \) be a nonempty subset of \( {\mathbb{N}}^{n} \) . The total degrees of elements in \( S \) form a nonempty subset of \( \mathbb{N} \), and so there is a small... | Yes |
Proposition 7.58. Let \( \preccurlyeq \) be a monomial order on \( {\mathbb{N}}^{n} \), and let \( f\left( X\right), g\left( X\right) \), \( h\left( X\right) \in k\left\lbrack X\right\rbrack = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . (i) If \( \operatorname{DEG}\left( f\right) = \operatorname{DEG}\left... | (i) If \( \operatorname{DEG}\left( f\right) = \alpha = \operatorname{DEG}\left( g\right) \), then \( \operatorname{LT}\left( f\right) = c{X}^{\alpha } \) and \( \operatorname{LT}\left( g\right) = d{X}^{\alpha } \) . Hence, \( \operatorname{LT}\left( g\right) \mid \operatorname{LT}\left( f\right) \) [and also \( \operat... | Yes |
Proposition 7.59. Let \( \preccurlyeq \) be a monomial order on \( {\mathbb{N}}^{n} \), let \( f\left( X\right), g\left( X\right) \in k\left\lbrack X\right\rbrack = \) \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), and assume that \( f\overset{g}{ \rightarrow }h \) ; that is, there is a nonzero term \( {c}... | Proof. Let us write \[ f\left( X\right) = \operatorname{LT}\left( f\right) + {c}_{\kappa }{X}^{\kappa } + \text{ lower terms; } \] since \( {c}_{\beta }{X}^{\beta } \) is a term of \( f\left( X\right) \), we have \( \beta \preccurlyeq \operatorname{DEG}\left( f\right) \) . If \( \operatorname{LT}\left( g\right) = {a}_{... | No |
Theorem 7.60 (Division Algorithm in \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) ). Let \( \preccurlyeq \) be a monomial order on \( {\mathbb{N}}^{n} \), and let \( k\left\lbrack X\right\rbrack = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . If \( f\left( X\right) \in k\left\lbrack X\right\rbr... | Proof. Once a monomial order is chosen, so that leading terms are defined, the algorithm is a straightforward generalization of the division algorithm in one variable. First, reduce \( {\;\operatorname{mod}\;{g}_{1}} \) as many times as possible, then reduce \( {\;\operatorname{mod}\;{g}_{2}} \) , then reduce \( {\;\op... | Yes |
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