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Unformatted text preview: Abstract Algebra I Notes
UIUC MATH 500, F'08 Jingjin Yu 1 Groups 08/25/08  08/27/08 Definition 1.1 Semigroups, monoids, groups, rings and commutative rings. For a map G G the following properties: 1) 2) 3) 4) 5) Closure, Association, Identity, Inverse, Commutative. G is called : G, consider A set G with a composition law G G a a a a semigroup if it satisfies 12, monoid if it satisfies 13, group if it satisfies 14, abelian group if it satisfies 15. Definition 1.2 A subgroup H of group G is a subset of G that is also a group with the same map of G. GG Example 1.3 0: semigroup; 0: monoid; : group. H s.t. f xy f xf y . If f is injective, Definition 1.4 A group homomorphism is a function f : G surjective, and bijective, then we call them monomorphism , epimorphism , isomorphism . , is an injective homomorphism; but there does not exist a nonzero homomorExample 1.5 , , . Suppose not then f 1 0 and nf 1n f n 1n f 1 f 1n 0. But no phism , matter what f 1 is, it cannot be infinity and there are some n such that 0 f 1n f 1n 1. This is not possible since f 1n .
Definition 1.6 A permutation of a set X is a bijection X 1......n a1 . . . . . . an 1 X: Definition 1.7 The symmetric group on a set X is the collection of all permutations on X with notation SymmX . If X 1, . . . , n, we write Sn for SymmX . Sn 1 by extending any permutation f with f n 1 n 1. Remark. We have injection : Sn We may define S 1, 2, . . ., then S SymmX since the permutation n 1 Sn . If we let X 2, 1, 4, 3, 6, 5 . . . SymmX , but S . Theorem 1.8 Let f : G H be a homomorphism. (1) f is a monomorphism if and only if ker f (2) f is an isomorphism if and only if there eixsts a homomorphism f 1 s.t. f f 1 Proof. (2) ( ) f is invertible, we just need to verify that it is a homomorphism: f 1 ab Clear. (1) ( ) Clear. ( ) Suppose f x f y , then f xy 1 f 1 e. 1H , f 1 f 1G . 1H , by assumption xy 1 1 x y. f 1 af 1 b. ( ) 08/29/08 Definition 1.9 let a1 , . . . , ar be distinct elements of X 1, 2, . . . , n, and let Y X a1 , . . . , ar . If f Sn fixes every element in Y and f ai ai 1 for i 1, r 1 and f ar a1 , then f is called an rcycle and we write f a1 , . . . , ar . A 2cycle is called a transposition. Definition 1.10 Two permutations , are disjoint if (1) k (2) k k k k k k, and k. Proposition 1.11 Every permutation Sn is a composite (product) of disjoint cycles. Proof. By induction on the number of elements that moved by . (case m (case m 0) 0) a1 , . . . , al1. We claim that al ai ai1 , which imply that al1 1 is the identity. Choose a1 with a1 a1 and let a2 a1 , . . . ai 1 ai until we have al a1 . Suppose not and al ai , 1 i l, then al al1 and ai1 , contradicting that al is the first repetition. We may then get a cycle that is disjoint from the rest of the permutation and apply induction hypothesis. Definition 1.12 G is a group and a, g Example 1.13 G Sn , let G, then gag1 is a conjugate of a. a1 , . . . , an Sn, then for Sn, 1 a1 , . . . , an .
2 If a if a ai , then 1a ai , hence it is fixed by and 1 a 1a a. On the other hand, ai , then 1 a ai ai 1 . Therefore 1 1 , . . . , n ).
G, a gag1 , g is an automorphism. gag1 gbg1 . Obviously g is surjective; it is injective since Definition 1.14 An automorphism is an isomorphism between a group G and itself. Proposition 1.15 Define g : G Proof. gag1 a homomorphism since gabg1 a b. gag1 gbg1 Definition 1.16 InnG g : g G is the set of inner automorphism. Definition 1.17 A subgroup K of G is normal if gkg1 K for all g G, k K. Lemma 1.18 K is normal if and only if gK Kg for all g G. Theorem 1.19 K G then aKbK abK for all a, b G. gK : g G forms a group that is the quotient group of G over Definition 1.20 The collection GK
K. Theorem 1.21 (First isomorphism theorem) Let f : G G ker f and ker f imf H be homomorphism of groups, then G, imf H. G, N Theorem 1.22 (Second isomorphism theorem) Let K N K K and N K N K N K . Theorem 1.23 (Third isomorphism theorem) Let K, N G, then KN is a subgroup of G, G, K GK N K GN N then N K is normal in GK and . 09/03/08 Lemma 1.24 Every Sn is a product of transpositions. Proof. We get the disjoint cycles and then from them it is easy to get the transpositions. Definition 1.25 If G is a group and X is a set, then a (left) group action of G on X is a binary function X denoted g, x g x which satisfies associativity and e x x for all x X. GX 3 Definition 1.26 Orbit of x X is Ox gx : g G. Stablizer Gx g G : x
X gx. Proposition 1.27 If G acts on a set X, then i Oxi , in which different Oxi 's are pairwise disjoint. Remark. Every transposition is a conjugate of another one. Conjugation take a rcycle to another rcycle. Conjugation takes subgroups to subgroups. 09/05/08  09/08/2008 Remark. Category: see notes. Low importance for now. 09/10/08 Definition 1.28 For G X R, G is finitely generated if there exists a presentation such that X is finite. G is finitely presented if more over R is finite. G be a subgroup then H acts on G on the right. The orbit of g Lemma 1.29 (1) Let H coset gH gh : h H ; G is a disjoint union of left cosets gH . Lemma 1.30 (2)Let now X gkH. GH G is the left
X, g, kH kH : k G, then G acts on X on the left by G X
G a subgroup, then gH H, k Lemma 1.31 (3)Let G be a finite group and H gH H for all g G. Definition 1.32 For a subgroup H denoted G : H . g1 k is a bijection and G, the number of left cosets of H is called the index of H in G, Theorem 1.33 (Lagrange) For a finite group G and a subgroup H, G : H GH . Proof. From (3) we know cosets are of size H and they are disjoint by (1). Therefore, the number of them is G H . The proof of (1), (2), and (3) are trivial. Definition 1.34 For a subgroup H of group G, NG H is the normalizer of H in G. That is, NG H is the largest subgroup of G such that H NG H . Corollary 1.35 The number of conjugates gHg1 of H is G : NG H . 4 Proof. We first show that if G acts on a set X, then Ox G : Gx . Gx is a subgroup of G, by Lagrange, GGx G : Gx . We wish to establish a bijection between GGx gGx and Ox (not a homomorphism). gx as the mapping. It is well defined since gGx hGx g hf x for some We may define : gGx f Gx ; but then f x x hence gx hx. is an injection since if gx hx, then h1 gx x h1 g Gx h1 gGx Gx gGx hGx . is obviously surjective. Therefore it is a bijection. With above, and the fact that G acts on H with conjugation with stablizer is NG H . Here our set is X gHg1 and we let x H. 09/12/08  09/15/2008 Theorem 1.36 (Cauchy) If G has a prime p as a factor, then G has an element of order p. Sketch of Proof. If G is abelian with trivial center, then the abelian case of Cauchy applies. For G p, then it is obvious. If not, let G mp with nontrivial center, we prove via induction over G . If G Gx , induction gives that and for any x G, check the centralizer Gx . This is a subgroup of G and if p Gx has an element of order p. If this is not the case for all Gx , then p does not divide i G : Gxi and by G C G i G : Gxi , the center C G contain p as a factor. The induction hypothesis again applies. Lemma 1.37 G p2 , p prime, then G is abelian. 1 i G : Gxi but then p2 1 mp. If C G p2 Proof. G has nontrivial center since otherwise G we are done since its G and abelian. If C G p (cyclic and abelian, by definition of center), C G is normal in G, GC G is of order p and is cyclic; let aC G be a generator of the quotient group. Now an1 cn1 , g1 an2 cn2 for some for any g1 , g2 G, g1 aC Gn1 , g2 aC Gn2 for some n1 , n2 . Then g1 1 2 n1 n2 n2 n1 n2 n2 n1 n1 an1 n2 c1 c2 an2 c2 a c1 g2 g1 . G is then abelian. c1 , c2 C G. Then g1 g2 an1 c1 a c2 Definition 1.38 A Sylow psubgroup of a finite group G is a maximal psubgroup P . Lemma 1.39 (5.33) Let P be a Sylow psubgroup of a finite group G, then (1) Every conjugate of P is again a Sylow psubgroup. (2) NG P P is prime to P . (3) If a G has order some powers of p and if aP a1 P , then a P . g1 Qg implies P Proof. gP g1 is not Sylow (not maximal), then Q (1) Suppose Q Contradiction. P but P Q. (2) Suppose not, then NG P P contains p as a factor and by Cauchy, NG P P contains an element aP of order p. aP is then a subgroup of NG P P of order p. The elements of P, aP, a2 P, . . . , ap1 P is a subgroup of NG P . Therefore, P is not Sylow. 5 (3) aP a1 P a NG P ; but by (2) a cannot be of order divisible by p. Theorem 1.40 (5.34) Let G be a finite group of order pe1 . . . pet , and let P be a Sylow psubgroup of G for t 1 some prime p pj . (1) Every Sylow psubgroup is conjugate to P . (2) If there are rj Sylow pj subgroups, then rj is a divisor of G pj j and rj
e 1 mod pj . Proof. (2) Let X P1 P, P2 , . . . , Prj be the Sylow psubgroups conjugate to P . Let Q be any Sylow psubgroup Q : QPi . of G, let it act on X by conjugation. The orbit of an element of X, Pi , satisfies OPi P , if a P makes If we let Q P , the only orbit of size 1 is OP ; this is true because for any Pi aPi a1 Pi , then by 5.33(3) a Pi . Therefore there must be some a P s.t. aPi a1 Pi , otherwise X 1 pm 1 mod p. P Pi . This gives us that rj (1) If Q X is a Sylow psubgroup, then above would suggest that X 0 mod p, contradiction. Corollary 1.41 A finite group G has a unique Sylow psubgroup P for some prime p iff P G. For g G, if Q gP g1 P , then Q is another Sylow psubgroup, contradicting that P is Proof. unique. Therefore gP g1 P for all g G, P G. Since for all g G, gP g1 P , P has no conjugate and is unique.
Theorem 1.42 (Sylow, 5.36) If G is a finite group of order pe m in which p is prime and p Sylow psubgroup of G has order pe . m, then every Proof. G : P G : NG P NG P : P . We know the number of conjugates of P in G is G : NG P and p G : NG P . P is the unique Sylow psubgroup in NG P by previous corollary, so p NG P : P . Therefore p G : P , and the rest follows. 09/17/08 Theorem 1.43 G mpe , p m, then there is a Sylow psubgroup of G with order pe . pe1 pe2 . . . for distinct primes s.t. there is a single Sylow Theorem 1.44 Let G be a group of order G 1 2 psubgroup for p p1 , p2 , . . . , pk . Then G P1 P2 . . .. Remark. See notes for proofs. 6 Definition 1.45 A normal series of a group G is a sequence of subgroups such that G G0 G1 G2 . . . Gn G0 G1 , G1 G2 , . . . , Gn1 Gn . A group is called solvable if it has a normal series with all factor groups having prime order. Definition 1.46 A composition series is a normal series with simple factor groups. The composition factors are the nontrivial factors. Proposition 1.47 Every finite group has a composition series Proof. Proof via induction. Let G be a smallest such group; take H be its maximal normal subgroup such that GH is simple (existence by the remark below). Then H has a composition series and G as well. Contradiction. Remark. Correspondence theorem is useful G1 H 09/19/08  09/22/08 Definition 1.48 Refinement of a normal series Gi is a sequence Nj that contains the original sequence as a subsequence. Lemma 1.49 (Zassenhaus Lemma, Butterfly Lemma) Given four subgroups A A , B B of a group G, then AA B AA B , B B A B B A and there is an isomorphism AA AA B B B B B B A . A 1, The factor groups of this series are the quotient groups G0 H G1 G0 with H normal in G0 . Theorem 1.50 (Schreier Refinement Theorem) Any two normal series for a group G have equivalent refinements. Theorem 1.51 (JordanHolder) Any two composition series of a group G are equivalent. Remark. The above theorems can be proved in that sequence. Lemma 1.52 Let P be a sylow psubgroup of G, and let N NG p, then NG N N. N and xP x1 P N since P N . P is normal in N , Proof. Let x NG N , then xN x1 hence it is the only Sylow pgroup in N (since all Sylow pgroups of G are conjugates); therefore P P and x N G P N . xP x1 P Lemma 1.53 In a group with n G e, the only subgroup H with NG H 7 H is H G. Proof. We show that if H G, then H NG H . Since H G and G 0 G 1 G . . . n G e, i s.t. H i and i 1 H. Let a iH, then for any g G, g, a gag1 aa i 1 H. If g h H, we have hah1 a1 h for some h H. Then ah1 a1 h1 h H aHa1 H a NG H . But a H, so H NG H . Lemma 1.54 A group if nilpotent if and only if the derived series stablize at the identity. TFAE: (1) The group G has a descending central series with n G (2) The group G is a direct product of its psubgroups ( Proof. e from some n. Every Sylow psubgroup is normal in G). 1 2: For any Sylow psubgroup P of G, let N NGP . By Lemma 1.52, NGN N . By Lemma 1.53, NG N N N G. So NG P G, implying that every Sylow psubgroup is normal in G. Let the Sylow psubgroups be P1 , . . . , Pn . We have Pi Pj 1 for i j since any nonzero element of Pi must have order greater than 1 and divides pi , therefore Pi , Pj cannot share any nonzero elements. Pj , ai aj ai 1 a ai aj ai 1 aj 1 a aj 1 Pj . Take any elements ai Pi , aj Pj , since ai Pj ai 1 j j ai aj ai 1 aj 1 ai a Pi . This implies ai aj ai 1 aj 1 1 ai aj aj ai . Thus Similarly, aj ai 1 aj 1 a i i any element of G can be expressed uniquely in the form a1 a2 . . . an , meaning that G is a direct product of its psubgroups. 2 1: We only need to show that each Pi is nilpotent, the direct product properties then that G is nilpotent as well. So we only need to show that any pgroup is nilpotent, which is proved in the next three results. Lemma 1.55 If G is a nilpotent group, then Z G 1.
G, i , G Proof. If G is nilpotent, then from the lower central series, there exists i gi g ggi . Since i 1, Z G 1. element gi i , g G, gi ggi 1 g1 1 Lemma 1.56 If GZ G is nilpotent, then G is nilpotent. 1. For any GZ G Z G. Let the lower central series of G be i and that of GZ G be Proof. G , . . . , , . . . , . We show inductively that i Z GZ G . We have i 1 Z GZ G i , GZ GZ G. 1 n i i An element of i , GZ GZ G has the form gi ggi 1 g1 Z G
hence i 1 Z GZ G gi Z GgZ Ggi 1 Z Gg1 Z G i . This gives us that n i, GZ GZ G iZ GZ G, GZ G i , GZ G Z G and n 1 1. Corollary 1.57 Every pgroup is nilpotent. 8 Proof. Let P acts on itself via conjugation, it is easy to see that P has nontrivial center and this is true for all P Z P (also pgroup). We can apply the previous lemma inductively to get that P is nilpotent. Lemma 1.58 A group is solvable if and only if the lower central series stablizes at the identity, or TFAE: (1) The group G has a derived series with Gn e for some n. (2) The composition factors of G are cyclic of prime order. Proof. F1 G0 G1 , then F1 is abelian; it then has a normal series with prime factors since all subgroups of an abelian group are normal. Applying correspondence theorem then gives us the refinement from G0 to G1 we need. 1 2: Derived series has abelian factor groups, the refinement then gives what we need. Let 2 1: Assume that G has a normal series G G0 G1 . . . Gn e with prime (cyclic and abelian) factor groups. We need to show that Gi Gi . This is straightforward by JordanHolder: the refinement of G0 G1 . . . is the same as G G0 G1 . . . Gn e; since there are no normal subgroups between G and G1 , we have the conclusion. Proposition 1.59 A nilpotent group G is solvable. Proof. G G0 G 1 G 2 G Nilpotent G1 . . ., we have Gi1 , Gi1 Gi ... n G e for some n. For derived series, i 1 G G, i G. So Gn1 n G e. Lemma 1.60 G solvable then its subgroups are solvable. Proof. H G, then the derived series of H is contained in the derived series of G and must reach e. Lemma 1.61 G is solvable, N G, then GN is solvable.
G1 ... N ... 1. N 1 refines to the normal series of G: G0 Proof. N G, then G 0 N 1 N G . . . N N gives us the normal series of GN . G Lemma 1.62 G a group, N N N hence Gk Proof. k has a normal series that completes the whole normal sereis of G. normal, G G, N, GN both solvable, then G is solvable. Show that Gi N N GN i , this gives that for some k, Gk N N N . Since N Theorem 1.63 The direct product of nilpotent groups is again nilpotent. Proof. This is straightfoward if we go to lower central series. 9 2 Rings 09/24/08 Remark. For the material covered here, the role of prime and maximal ideals are very important; for the part involving PID/UFD, irreducible element generates prime ideal is key to many proofs. The noetherian properties are introduced at last; with Zorn's lemma, some additional results can be proved. Definition 2.1 Semigroups, monoids, groups, rings and commutative rings. G, consider the following properties: For a map G G 1) Closure, 2) Association, 3) Identity, 4) Inverse, 5) Commutative. A set G with a composition law G G a semigroup if it satisfies 12, a monoid if it satisfies 13, a group if it satisfies 14, a abelian group if it satisfies 15. G is called : A ring R is a set together with two maps: : R R R (addition), and R (multiplication), . : RR such that the set R is an abelian group under addition (satisfying properties 15) and a monoid under multiplication (satisfying properties 13). Also, the distributive law holds: ab c ab ac, a, b, c R R is an commutative ring if the multiplication is commutative. Example 2.2 Example of rings. , , , are rings. x is the ring of polynomials in the variable x, with coefficients in 2 is not a ring since there is no multiplicative identity. n is a commutative ring. 2 6 0, 2, 4 6 د , 6 and 2 6 are rings. 1R in 2 6 : 0, 1, 2, 3, 4, 5 4 0 2 4 0 0 0 0 2 0 4 2 4 0 2 4 Definition 2.3 . A subset S 1) a b S, a, b S, 2) ab S, a, b S, and 3) 1S 1R . R of a ring R is called a subring if . If S is a subring, then S is itself a ring. 10 Example 2.4 Subring. is a subring of . S 2 6 is a ring, but it is not a subring of R 6 because 4 1S 1R 1. Definition 2.5 An integral domain is a commutative ring R with : 1) 1R 0R , and a 0 or b 0. 2) a, b R, ab 0 Remark. The second criterion above is equivalent to for a, b, c R, c 0, then ac bc the cancellation law holds. Property 2) in the above definition is equivalent to saying that for a, b, c R, ac bc, c 0 a a b. That is, b. Definition 2.6 a, b R, we say b divides a in R if there exists q R such that a qb (if R is commutative, a qb bq). A divisor of 1R is called a unit; a divisor of 0R is called a zero divisor if q 0. With above definition, if b is a unit, then qb Example 2.7 Let R 2 , a, n 1, a R is a unit a, n 1. a R is a zero Definition 2.8 A field is a commutative ring with 1 Proposition 2.9 Every field is an integral domain. Proof. Let F be a field and a, b Therefore, b 1b qab q0 0. 0 such that every nonzero element is a unit. 1R for some q R. F s.t. ab 0. Suppose that a 0, then a is a unit in F with qa 1F . Proposition 2.10 Every finite integral domain is a field. R by f : x ax. The Proof. Let R be the domain and a R, a 0. Define a homomorphism f : R map is well defined since every element x is mapped to ax (not one to many). The map is an injection since ax y 0; R is a domain so a 0 xy 0 x y. Since the map is from finite R to R ax ay and injective, it is also surjective. Since the map is a bijection, there exist x R, ax 1R . Hence a is a unit. Example 2.11 Integral domains that are not fields: , x. Remark. [the following till next lecture not covered in class] Every subring of a domain is then itself a domain since the cancellation law carries over to the subring. Since fields are domains, it follows that every subring of a field is a domain. The converse is also true: every domain is a subring of a field. We have: Theorem 2.12 If R is a domain, then there is a field F containing R as a subring. Moreover, F can be chosen s.t. for each f F , there are a, b R with b 0 and f ab1 (b1 is the multiplicative inverse of b). Remark. In section 3.4 of text, we have the following results: 11 Theorem 2.13 Assume that k is a field and that f x, gx kx with f x polynomials q x, r x kx with gx q xf x r x and either r x 0 or degr degf . Corollary 2.14 Assume that R is a field and that f x, gx Then there exist q x, r x kx with gx and either r x 0 or degr degf . f x q xx u f u. 0. Then there are unique Rx with f x 0 is a monic polynomial. q xf x r x Lemma 2.15 Let f x kx, where k is a field, and let u k. Then there is q x kx with Proposition 2.16 If f x kx, where k is a field, then a is a root of f x in k if and only if x a divides f x in kx. Theorem 2.17 Let k be a field and let f x kx. If f x has degree n, then f x has at most n roots in k. Remark. In Corollary 2.14, q x, r x are not stated as unique, therefore above is not true for polynomials in Rx for arbitrary commutative ring R. Corollary 2.18 Every nth root of unity in e2ikn where k 0, 1, . . . , n 1. is equal to cos 2k i sin 2k , n n Corollary 2.19 Let k be any field, perhaps finite. If f x, gx f a ga for n 1 elements a k, then f x gx. Proof. Let hx kx, if degf degg n, and if f x gx, hx cannot have more than n roots in k. E 0 of finite field E is cyclic. Lemma 2.20 (3.30)The multiplicative group E Proof. The proof is by showing that E has at most one cyclic group of order equal to each unique divisor E of . For a divisor of E , say d, suppose there are two subgroups S, T of order d. Then the elements of d, and xd 1 has too many roots in S T are all roots of xd 1 in the multiplicative group. But S T (by theorem 3.25). Therefore, there is at most one subgroup of order as each divisor of E . Then by E theorem 2.86 in the text, E is cyclic. Theorem 2.21 If k is a field, and f x, gx kx, then their gcd, dx, is a linear combination of f x and gx in kx; that is there are sx, tx kx s.t. dx Monic gcdf x, gx is unique. 12 sxf x txgx. Lemma 2.22 Let k be a field, let px, f x irreducible, then dx 1 if px f x; dx kx, and let dx p, f be there gcd. If px is a monic px if px f x. Theorem 2.23 [Euclid's Lemma] Let k be a field and let f x, gx kx. If px is irreducible in kx, and px f xgx, then px f x or px gx. This works similarly if px f1 xf2 x . . . fm x. 1 sp tf g spg tf g. p f g, then p g. Proof. Assume that px f x, then gcdp, f 1
Definition 2.24 Let k be a field. f x, gx kx are relative primes if gcdf, g 09/26/08 Definition 2.25 : R (1) 1R (2) a b (3) ab 1S , a b, and ab. S is a homomorphism of rings if 1. We have that ker Definition 2.26 I (1) 0 I, (2) (3) r R : r 0 and im r : r R. R is a left Rideal if a, b I, a b I, and a I, r R, ra I. Example 2.27 ker R is a left Rideal.
S. That is, kernel of a ring homomorphism R S is an Rideal in in The only ideals of a field f is 0 and f ; but f may contain nonzero proper subrings. For example, . Theorem 2.28 (Correspondence Theorem of Rings) The RIideals, J I , are in bijection with Rideals J : I J R. Proof. [trivial, to be filled in, for now, see book (page 320, proposition 6.1) for details.] R, and ab I a I or b I. Definition 2.29 An ideal I Example 2.30 Prime ideals. (1) 5 (2) 6 is a prime ideal. R is a prime ideal if I is not a prime ideal; 6 6 , 6 2 3, but 2, 3 6 . 13 Proposition 2.31 I Proof. R is a prime ideal RI is a domain. " " For ab RI, if a I b I ab I 0 I, then ab assume that a I; we then have a I I. " " For ab I, since 0 I a I or b I. ab I I. I is prime hence a I or b I. We may I or b I I; therefore, a I b I and RI is a domain, a I Definition 2.32 An ideal I J R. R is a maximal ideal if I R and if for all Rideal J, I J R I J or Proposition 2.33 Every maximal ideal is a prime ideal. Proof. Suppose I R is a maximal ideal in R. Suppose I is not prime; then ab I, a I, b I. Then I, a is a larger ideal in R; since J is maximal, I, a R. Then 1R is generated by elements y I and a such that 1R y ra, r R. But then b yb rab. Both y and rab belong to I, so b I, contradiction. Proposition 2.34 An Rideal I is maximal Proof. R, which gives 1R ra y for some r " " For a RI, I is maximal hence I, a ra y I 1 I in RI. This gives r I a I 1 I, making a I a unit. RI is a field. R, y I. Then " " RI is a field, then the only ideals in RI are RI and 0. By correspondence theorem for rings, I is a maximal ideal in R. Example 2.35 x, y y x is a domain but not a field (since for example, x 1 has no inverse in it), y is a prime ideal but not a maximal ideal. Note: 0 y x, y x, y. 09/29/08 In , if p ab, then p a or p b. This property generalizes. P I P or J P. Now ab Lemma 2.36 Let P R be a prime ideal, then IJ P , if I, J are not in P , then exist a Proof. For IJ a P or b P , contradiction. P I, b P J. a b
: a IJ P , P prime hence R be ideals, then I J is an Rideal. I J Let I, J a1 b1 a2 b2 . . . ar br : ai I, bi J, r is an Rideal. 14 I, b J is an Rideal. IJ Definition 2.37 a, b R are called associates if exists a unit u R, s.t. a Lemma 2.38 R is a domain. a, b R are associates if and only if a Proof. " " a, b are associates, then a ub for some u R. Then a ub. b. b. Similarly b a; hence a b.
ub uva. R is a domain, we may cancel a " " a b then a ub, b va for some u, v R. Hence a and get uv 1. Therefore, u, v are units in R. Definition 2.39 An element r R is irreducible if r is not a zero or a unit and if r or b is a unit. We then have r a or r b. Lemma 2.40 If p with p 0 is a prime ideal, then p is irreducible. ab, then a is a unit p R is prime ideal hence p is not a unit; otherwise p R. Let p ab. We get Proof. p a, p b. Since ab p, we may assume that a p; then a rp for some r R. We then have a p. Hence p a and p, a are associates. This gives that p is irreducible.
Definition 2.41 R ia principal ideal domain (PID) if R is a domain and every ideal in R is generated by a single element in R. , are PIDs; x, y is not a PID. In fact, every field is a PID since the only ideals in a Example 2.42 field is the zero ideal and the whole field. Remark. For any PID R and a, b R, we can prove that they have a gcd and there are r, s R s.t. p x or p y. This works ra sb. Furthermore, for any irreducible element p R and xy R, p xy for any PID, hence for any field. We formally prove it in the following theorem. Theorem 2.43 (3.57) (1) Every , r, s R. R has a gcd, , which is a linear combination of and in the form: ab for some ab R, then p r s for some a or p b. (2) Furthermore, if p is an irreducible element of R and p Proof. (1) If one or both of , are zero, then the conclusion is trivially true; assume they are not zeros. Let , , the ideal generated by and . R is a PID, then I for some R. We claim that I gcd, . Obviously , and is a linear combination of the form r s for some r, s R, for some R, since , . On the other hand, for any common divisor of , , . Therefore r s r s and . similarly (2) p ab, p is irreducible, then either p a or p b, therefore p
15 a or p b. Proposition 2.44 In a PID R, p is a prime ideal if and only if p is irreducible. Proof. " " By Lemma 2.40. " " We show that if p is irreducible then p is maximal in R, hence prime. Suppose not, then exists ideal J such that p I J R. R is a PID, therefore J q for some q R. We have q rq for some r R. p is irreducible, therefore either r or q must be a unit. If r is a unit, then I p q J; if q is a unit, then J q R. Both contradict the assumption that I J R. Definition 2.45 A domain R is a unique factorization domain (UFD) if (1) Every r R is either 0, or a unit, or a product of irreducible elements. (2) Every nonzero element has a unique factorization into irreducible elements. That is, if r up1 p2 . . . pm vq1 q2 . . . qn with u, v units and pi , qi irreducible, then m n and there exists a bijection : p1 , p2 , . . . , pn q1 , q2, . . . , qn such that pi and qj pi are associates. Proposition 2.46 (6.17) Let R be a ring in which ever nonzero r R is a product of irreducible elements, then R is UFD if and only if p is a prime ideal for every irreducible element p. Proof. rp for some r R. p is irreducible, " " Suppose R is a UFD, p R irreducible. If ab p, then ab therefore it must be an associate of an irreducible factor of either a or b. Therefore, a p or b p. p is prime. " " Let x R have two factorizations up1 p2 . . . pm vq1 q2 . . . qn . We prove via induction on maxm, n. p1 is an irreducible element in R hence p1 is prime by assumption. vq1 q2 . . . qn x p1 , therefore some qi p1 . Assume qi rp1 ; qi is also irreducible, therefore p1 , qi are associates. By induction hypothesis, we have that the two factorizations are equivalent and unique. 10/01/08 Proposition 2.47 (6.18) (1) If R is a commutative ring and I1 is an ideal. I2 . . . In . . . is an ascending chain of ideals, then J I2 . . . In . . .. n 1 Ii (2) If R is a PID, then there is no infinite strictly ascending chain of ideals I1 16 (3) Let R be a PID, then every nonzero r Proof. (1) Straightforward verification. 1. 0 J, 2. 3. R is a product of irreducible. (2) R, a In for some n, therefore ra In J. Given any chain Ij , R is a PID, J x for x In for some n. Then J
it cannot be infinite. a, b J, a Im , b In for some m, n, therefore a b Imaxm,n a J, r J, and In and the chain stops there; (3) Suppose r R is not irreducible, then there exists r1 s1 such that neither r1 or s1 is a unit. At least r1 (if one of r1 and s1 is not irreducible; suppose r1 is not irreducible. Then r r1 and r r r1 , then s1 must be a unit; but s1 is not a unit). Repeating this we get an infinite chain r1 r2 r3 . . ., which contradicts (2). Therefore, r is a product irreducible elements. Theorem 2.48 (6.19) R is PID R is UFD. Proof. The proof follows directly from above previous two results. Definition 2.49 A domain R is a Euclidean domain if there exists a degree function d : R0 with the following properties: (1) da dab for all a, b R0, and 0, q, r (2) For all a, b R, b Example 2.50 , R s.t. a qb r and either r 0 or dr db. i, x are Euclidean domains; x, y, 19 are not. Theorem 2.51 Every Euclidean domain is a PID. Proof. Let I be an Rideal, choose b I s.t. db is minimal; we claim that I b. To see this, let a I, R Euclidean then a qb r with r 0 or dr db. Suppose r 0, since a, b I, we have r I and by the assumption that db is minimal, dr db. Therefore, r must be 0 and a b. Definition 2.52 Let R be a UFD. A polynomial f x Rx is primitive if its coefficients are relatively prime, that is, if f x an xn . . . a1 x a0 , then the only common divisors of a0 , a1 , . . . , an are units. Remark. In a UFD, a common divisor is well defined, and we can talk about a greatest common divisor of a and b. d is a gcd for a and b if d a and d b implies d d. In a UFD, if a upe1 . . . per and b vpf1 . . . pfr . r r 1 1 maxei ,fi r is a gcd for a and b. Then d i 1 pi Proposition 2.53 For a UFD R, if px RxR is irreducible, then px is primitive. 17 Proof. Suppose px is not primitive, then q R s.t. px qgx, degg 0, where q is not a unit. Since px is irreducible, px and q are associates; this makes gx a unit. Contradiction. Lemma 2.54 (Gauss's Lemma, 6.23 in the text) Let R be a UFD, if f x, gx then f xgx is primitive.
Rx are primitive, Proof. If p is a prime ideal in R, then the ring homomorphism : R Rp, a a p induces Rpx s.t. the coefficients of ai of f x R are mapped to ai . If a ring homomorphism : Rx hx h0 h1 x . . . hr xr is not primitive, then exists irreducible p R s.t. p hi for all i. Therefore, hi hi p 0 p. hx 0 Rpx since all the coefficients are 0. Suppose now that f xgx is not prmitive, then f g f g 0 Rpx. Rpx is a domain (this can be proved by showing that R is a domain then Rx is a domain), therefore f 0 or g 0. That is, either f x or gx are not primitive. Contradiction. 10/03/2008 Definition 2.55 For f x Rx, let cf be a gcd of the coefficients, and let f x f x kx is primitive. cf is called the content of f . Lemma 2.56 (6.24) cf f x. The factorization is unique, if f x (1) For f x Rx, let f x primitive, then cf and r are associates and f and g are associates. bf x, b R, b rg x with r cf f x so that R, g (3) Let f x Rx, g x Rx, g primitive. If g Proof. (2) For f x, gx Rx, cf g and cf cg are associates and f gզ and f g are associates. 0, then g f. f u vf . u vf (1) rg cf , with r c v such that u, v are relative primes. Then ug unit. Similarly, v is a unit. Therefore, c, r are associates and g , f are associates. u u is a (3) We may write bcf f x bf x hxg x chh xg x. h xg x and f x are associates, g x f x g x f x. therefore exists unit r, rh xg x f x (2) f g cf gf gզ cf cgf g , since f , g are primitive, so does f g by Gauss's Lemma. Therefore, by (1) we have that cf g and cf cg are associates and f gզ and f g are associates. Theorem 2.57 If R is a UFD, then Rx is a UFD. Proof. (1) Every nonzero element f x this by induction on degf . Rx that is not a unit is a product of irreducible elements.
18 We prove (2) R, R is UFD. degf 0. f x cf f x. cf R, if f x is an irreducible element then we are done. x gh such that g, h are not units. g, h R, otherwise f x being Suppose not, then f primitive implies that g or h is a unit. Therefore, degg, deg h degf . We may then apply induction hypothesis on g , h to conclude that every f x is a product of irreducibles. Every irreducible px Rx generates a prime ideal. In other words, we want to show that for every p f or p g. We prove this for two cases based on degp. px Rx, p f g p cf g p cf cg. p, cf , cg degp 0. f x cf f x, gx cgg x. p f g R and R is a UFD, p is prime, cf cg p, then cf p or cg p. Therefore, p cf or p cg. 0. Let I p, f Rxpx Rxf x Rx, let mx I be of minimal degree, degp apply quotient/remainder over Qx, in which Q FracR, to f and m, f x mxq x r x, mxq x r x Rx, b R, s.t. q x, r x Qx. Clearing denominators gives bf x 0 or deg r x deg mx. bf x I, mx I r x I. Therefore r x 0 since r x mx is of minimal degree in I. We have that bf x mxq x cmm xq x. It follows that m f x by Lemma 2.56(3). Similarly m x px. px is irreducible, m x is a unit or an associate of px. Two cases: m sxpx r xf x. mgx sxpxgx r xf xgx, i (m is a unit). m p, f px cmm gx. px is primitive (since it is px f xgx hence px mgx irreducible, px cpp implies that cp is a unit), px gx by Lemma 2.56(3). px f x. ii (m is an associate of px). m x f x
0. Then f degf Corollary 2.58 For k a field, kx1 , x2 , . . . , xn is a UFD. Proof. By induction on n. k is a field; the only ideas are 0 and k, which are both generated by a single element. k is then a PID then UFD. We have kx is also a UFD as the base case, and we may use kx1 , x2 , . . . , xn1 as induction hypothesis to get that kx1 , x2 , . . . , xn is a UFD. Corollary 2.59 (Gauss) Let R be a UFD, let Q FracR be the field of fraction of R. For f x Rx, if f x GxH x is a factorization of f x in Qx, then f x has a factorization f x gxhx Rx such that deg g deg G, deg h deg H. Proof. By Lemma 2.56(1), we may factor Gx qG x, H x q H x s.t. G , H Rx are primitive. Then G H is also primitive in Rx. f x cf f x Rx, this forces qq R. Therefore, qq G Rx; a factorization of f x Rx is given by f x qq G xH x. Rest follows. 10/06/08 19 Definition 2.60 A commutative ring R satisfies the ascending chain condition (ACC) if every ascending chain of ideals I1 I2 I3 . . . In . . . stabilizes. Definition 2.61 An Rideal I is finitely generated or f.g. if there exist a1 , a2 , . . . , an I a1 , a2 , . . . , an . Proposition 2.62 Let R be a ring, the following are equal: (1) R satisfies ACC, (2) R satisfies the maximum condition: every nonempty family F of Rideals has a maximal element, and (3) Every Rideal is f.g.. Proof. R such that 2). R satisfies ACC, let F be a family of ideals in R. For I1 F, since I1 is not maximal, (1) (1 I2 F, I1 I2 . Similarly, I3 F s.t. I2 I3 . Repeating this we get an infinite ascending chain I1 I2 . . . In . . ., contradiction. 3). Let I R, Let F f.g.Rideals I . F has a maximal element M , M I, and M is (2) (2 f.g.; if M I, then exist a I, a M , then M, a is a larger ideal than M , contradiction. Therefore, M I and I is f.g..
Ii , J is an ideal in R and J is f.g. Then (3) Every Rideal is f.g., for any ascending chain Ii , let J there are only finitely different Ii 's in the chain and the chain must stabilize. Definition 2.63 A commutative ring R with any of the above properties is called noetherian. Corollary 2.64 Let R be a noetherian ring. Then every ideal I is contained in some maximal Rideal of M . Proof. Let F be the family of proper ideals containing I, then F has a maximal element M that contains I. If M is not maximal ideal in R then exists J, M J R. J F , contradiction. Corollary 2.65 If R is noetherian and J is an Rideal, then RJ is noetherian. Proof. Let I J be an RJideal with J I R, the generators for I certainly generates I J. Theorem 2.66 (6.42 in the text, Hilbert Basis Theorem) R is commutative noetherian ring, then Rx is also a commutative noetherian ring. Proof. Let I Rx be an ideal, assume that I is not f.g., we will obtain a contradiction. Rx of minimal degree, then f0 I (I is not f.g.). Choose f1 I f0 of smallest Choose f0 I degree and repeat this procedure, we obtain a sequence of ideals f0 f0 , f1 f0 , f1 , f2 . . . such that deg f0 degf1 . . .. For f0 , f1 , . . . , fm I, consider the ideal a0 , . . . , am R, in which ai is the leading coefficient of fi . 20 R is noetherian, then the chain formed by a0 , . . . , am stablizes. Say am 1 r0 a0 . . . rm am , consider f fm 1 m a0 , . . . , am , let am 1 rifixxdeg f m 1 deg fi , then f I f0 , . . . , fm since the rest of the terms are linear combinations of elements in f0 , . . . , fm . If we write fk ak xdk f i 0 lower terms, then am 1 xd lower terms m 0 riai xd lower termsxdeg d i am 1 xd m 0 ri ai xd lower terms i
m 1 i m 1 m 1 m 1 deg di lower terms. this gives us that deg f deg fm 1 , therefore f f0, . . . , fm , yielding a contradiction.
y onr X that is With Zorn's Lemma, 6.46, 6.48, 6.53 can be proved. Definition 2.67 X is a partially ordered set if there is a relation x Reflexive, Antisymmetric (x Transitive. Definition 2.68 A partially ordered set X is a chain if for all x, y y, y x x y), and X, either x y or y x. Axiom 2.69 (Zorn's Lemma) If X is a nonempty partially ordered set in which every chain has an upper bound then X has a maximal element. Remark. The standard way of applying Zorn's Lemma to a set A is to construct a family F with desired property, then show that any chain C in F has an upper bound M (for example, if the partial order is inclusion, the union of all sets in C is the upper bound) and that M is in F hence in C. Zorn's Lemma then says that F contains a maximal such with the desired property. Definition 2.70 A poset if wellordered if every nonempty subset S of X contains a smallest element. Theorem 2.71 The following are equivalent: (1) Zorn's Lemma. (2) The well ordering principle: Every set X has some wellordering of its elements. (3) The axiom of choice. (4) Hausdorff maximal principle. 10/08/08 Proposition 2.72 (6.46) In a commutative ring R with 1 21 0, every ideal is contained in a maximal ideal. Proof. Let F I : I I be the set of proper Rideals that contain I. The set F is partially ordered w.r.t. inclusion. F has a maximal element if every chain in F has an upper bound (Zorn's Lemma). Let C I and I is an upper bound for C. Moreover, be a chain of ideals in F, define I I I, then clearly I F I is a proper Rideal. Indeed, C is proper, for if 1 I then 1 I for some I; but the ideals in C I I are proper. Contradiction. Let M F be maximal, then M is a maximal Rideal. Suppose I M J R, then J F , contradiction. Definition 2.73 Let V be a vector space V over some field k, and let Y V be a (possibly infinite) subset. (1) Y is linearly independent if every finite subset of Y is linearly independent. (2) Y spans V is each v V is a linear combination of finitely many elements of Y . We write V when V is spanned by Y . (3) A basis of a vector space V is a linearly independent subset that spans V . Example 2.74 V kx as a vector space over k has a basis Y Y 1, x, x2 , . . .. Proposition 2.75 (6.48) Let V be a vector space over k, then V has a basis and every linearly independent subset B V is contained in a basis for V . B : B B , the family of all the linearly independent subsets of V that contain B. Proof. Let X to The family X is nonempty, since B X. We want show that every chain C X has an upper bound. Let Bj : j J be a chain of X and let B B for all j J and B is an upper C j J Bj . Then Bj bound for C if B is linearly independent. y1 , y2, . . . , ym with y1, y2, . . . , ym linearly dependent. Assume B is not linearly independent, say B 1, 2, . . . , m. We have m 1 Bji y1, y2 , . . . , ym. There must Then there are Bji 's such that yi Bji , i i j1 , j2 , . . . , jm s.t. Bj y1 , y2 , . . . , ym (since we are working in a chain with the relationship be some j being inclusion). Contradicting that Bj 's are all linearly independent. We have shown that every chain C X has an upper bound, by Zorn's Lemma, X has a maximal element M . By definition of X, M is linearly independent. M also spans V ; for if not, then there exists a V M and B M M, a. That is M plus a is a larger linearly independent subset of V that contains B, contradicting the maximality of M .
Lemma 2.76 (6.52) Let R be a commutative ring and let F be the family of all those ideals in R that is not finitely generated. If F , then F has a maximal element. Proof. We show that every chain C in F has an upper bound in F. Let I I C I, then I is an upper a1 , a2 , . . . , am , ai Ii is f.g., as before, there exits I0 C s.t. bound for C if I is not f.g.. Suppose I a1 , a2 , . . . , am I0 . But then I I0 I and I0 is f.g., contradiction. I is an upper bound. The rest follows Zorn's Lemma.
Theorem 2.77 (I. S. Cohen, 6.53 in the text) A commutative ring R is noetherian if and only if every prime ideal in R is finitely generated. 22 Proof. We need to show the "if" part only. Let F be the family of ideals that is not f.g., we obtain a contradiction from F . By previous lemma, there exists maximal M F that is not f.g., we show that M is a prime ideal. Suppose not, then there exist a, b M, ab M . Then M Ra M, M Rb M and M Ra, M Rb are f.g., by the maximality of M . But M RaM Rb M is then f.g., contradiction. 23 3 Fields 10/13/08 Definition 3.1 A field is a commutative ring in which 1 of a field K is a subring k of K that is also a field. 0 and every nonzero element is a unit. A subfield Definition 3.2 For a given field K with a subfield k, we call K an extension field of field k and K k a field extension or an extension of fields. An extension field K of a field k is a finite extension of k if K is a finitedimensional vector space over k. The dimension of k, denoted by K : k, is called the degree of K k. Remark. This is saying that for any field extension K k, we may treat elements of K as vectors and elements of k as scalars. Example 3.3 Dimension of field extensions. (1) (2) . is of infinite dimension over . i : i a bi : a, b , i2 1. i is of finite dimension over . Theorem 3.4 If R is a commutative ring, and I is an Rideal, then RI can be made a ring with multiplicative operation a I b I ab I . Proof. Sketch: The underlying abelian groups of R and I give RI as a quotient group under addition; we only need to verify that RI is a monoid under the given multiplication, which is straightforward. RI is called the quotient ring of R modulo I. Theorem 3.5 For any homomorphism f : R A of rings, Rkerf imf . Proof. To prove we need to show that we have a homomorphism and it is bijective. If we forget about imf is an isomorphism of groups. It then suffices to verify that the map is the multiplication, Rkerf compatible with multiplication. K, the image is a subring of K and therefore Remark. In any homomorphism between a ring and a field R K, it is must be a domain (straightforward to verify). If we have a homomorphism between two fields F more special since the kernel is an ideal in F but there are only two ideals, 0, F in F . Definition 3.6 For a field k, let k0 be the intersection of all the subfields of k, then k0 is itself a subfield, called the prime field of k. Example 3.7 k it must contain all , k0 . Any subfield of R must contain 1; then it must contain since very nonzero element is a unit in a field. or k0
p with addition; then Proposition 3.8 For a field k, the prime field k0 is either k0 number. 24 p where p is a prime Proof. Consider the ring homomorphism f: k, n1k . is PID. We have two cases: which is the mapping f :n then ker f and ker f m for some m . The later is true because k is an injective mapping. Since k is a field, imf is a domain case m 0: We have that f : m (cancellation law certainly holds for any subring of a field; therefore any subring of a field must be a domain). m is then a prime ideal in . Let prime p m, we then have p p is isomorphic a subfield P 0, 1k , 21k , . . . , p 11k (21k is 2 multiplies 1k ) of k. Now any subfield of k must contain 1k and therefore all of P . P is then the prime field of k. case m 0: f is injective, therefore, k contains all the integers. For k to be a field, it then contains the fraction ring of , which gives us . We then have k0 . If k0 , we say k is a field of characteristic 0 ; if k0 p , k is a field of characteristic p. The later case is sometimes also called positive characteristics, finite characteristic, or nonzero characteristic. Proposition 3.9 If k is finite field, then k pn for a prime p and n 0. Proof. Assume p k and q k for distinct primes p, q, then by Cauchy's for abelian group, there are elements a, b in k with order p and q, respectively (a, b is of some prime order so they are not the 0 element). ap ap1k . Simiarly, bq1k 0k . Since k is a field, then k is an integral domain, a is of order p so 0k 0 and a 0 p1k 0; p1k q1k 0. But p, q 1 r, s, s.t. sp tq 1. But therefore, ap1k 1k sp1k tq1k 0k , contradiction. Proposition 3.10 Let k be a field and let I irreducible in kx. Proof. gxhx with degg degp and " " Suppose kxI is a field. If px is not irreducible, then px px gx degp degg. degh degp. We must have that gx I I, otherwise gx I Same holds for hx. Now gx I hx I gxhx I px I 0 I. But kxI is a field then an integral domain, in which the product of nonzero elements is nonzero. Contradiction. " " Suppose px is irreducible, we show that kxI is a field by verifying that the definition is satisfied. 0. I px is a proper ideal and 1 (1) 1 irreducible. Therefore, 1 I 0 I. px kx. Then kxI is a field if and only if px is I for if not, then px must be a unit and then not (2) Every nonzero element is a unit. For the above choice of f x, px f x, otherwise f x px and f x I I. Since px is irreducible, f x px by definition. Therefore p, f are relatively 1. Now 1 I sxf x txpx I prime and there are polynomials s, t with sf tp sxf x I sx I f x I , and we have found the inverse of f x I. 25 Proof. [Second proof] We may use Proposition 2.34. That is, RI is a field if and only if I is maximal. " " We have that RI is a field then px is maximal and then prime. We want to show that px is irreducible. For any gx, hx s.t. px gxhx, obviously px gx, px hx. px is prime, then gxhx px hence gx px or hx px. We then have gx px or hx px. Therefore one of g, h is a unit, and px is irreducible. " " We want to show that px irreducible then px is maximal. For any f x px, f x q xpx r x, where degr x degpx. Therefore px, r x 1 and px, f x R. Since f x is arbitrary, px is maximal. , kx Example 3.11 k xpx i by f x I
10/15/08 x, f i. choose px x2 1, px is irreducible. Let I px, then px. Let K Proposition 3.12 (3.117) Let k be a field and let px kxI, and write x I. (a) K is a field with a subfield k (b) is a root of px. kx be irreducible polynomial of degree d; I
k. a I : a k s.t. k (d) px is the unique monic irreducible polynomial in kx s.t. is a root of px. (e) K is a vector space over k of dim Proof. K s.t. for a k, a a I. The (a) K is a field by previous proposition. Define the map : k map is injective by Cor 3.53 in the textbook(we first verify that defines a homomorphism of rings, then we have that ker is an ideal in k. But k is a field, the only ideals in k is zero ideal or k itself. 1 I 0 I, not everything maps to 0 I, therefore the kernel is not k; hence it is 0 Since 1 and we have an injection.); the kernel is then trivial. Therefore, by first isomorphism theorem of rings, k . k k0 k . is an isomorphism from k
d i i 0 ai x . (c) If is a root of gx kx then px gx. K : k d with basis 1, , 2 , . . . , d1 . (b) We may write px p Therefore, is a root of px. px I a0 a1 x I . . . ad x I d a0 a1 x I a2 x2 I . . . ad xd I a0 a1 x . . . ad xd I px I I. We evaluate p as 26 (c) If px gx, px irreducible gcdpx, gx 1 and exist sx, tx s.t. 1 sp gt. Since is the root of both gx and px, s p t g 0. Contradiction. Therefore, px gx. Alternatively (more clearly), follow proof in (b), 0 I g gx I. gx px, hence px gx. (d) By (c), if there is another monic irreducible gx with as a root, then px gx and gx px. Thus gx cpx for some constant c. Since both px and gx are monic, the leading coefficient are both 1, which implies c 1. Thus px gx and px is unique. (e) We first show that every f x in kx is a linear combination of i 's. Every elements of K kxI has the form f x I. We may write f x q xpx r x with degr x deg px d. Then f x I r x I r0 r1 x r2 x2 . . . rd1 xd1 I r0 r1 x I r2 x2 I . . . rd1 xd1 I r0 r1 x I r2 x I 2 . . . rd1 x I d1 r0 r1 r2 2 . . . rd1 d1 . We have shown that every f x can be represented as a linear combination of i with i d. We now show that the basis is linearly independent. Suppose this is not true; then there exist c0 , c1 , . . . , cd1 1 that are not all zeros such that d0 ci i 0. Let gx c0 c1 x c2 x2 . . . cd1 xd1 , then i d1 i gx I, therefore gx I and px gx. But deggx d degpx; 0 I i 0 ci I therefore gx 0 and every ci must be 0. Contradiction. Definition 3.13 Let K k be an extension of fields. K is algebraic over k if there is some nonzero polynomial f x kx having as a root; otherwise, is transcendental over k, An extension K k is algebraic if every K is algebraic over k. Example 3.14 Algebraic and transcendental elements and algebraic field. (1) (2) ; since it is a root of f x is transcendental over (proof is not trivial).
5 3 , is a root of f x 3 is algebraic over x2 3. (3) K Frackx, then K k if a field extension. x K k, is transcendental over k. Example 3.15 x4 16x2 4. x 5 3x 5 3x 5 3x 5 3 Definition 3.16 If K k is an extension and K, then k is the intersecion of all those subfields of k that contain k and ; we call k the subfield of K obtained by adjoining to k. Example 3.17 3 3 is the smallest subfield of that contains 3. Proposition 3.18 If K k is a finite field extension, then K k is an algebraic extension. 27 Proof. For a finite K k, K is a vector space of some dimension n over k. Then for any K, the list of n 1 vectors 1, , 2 , . . . , n must be linearly dependent. Then there are a0 , a1 , . . . an , not all 0, such that ai i 0. Then is a root of f x ai xi . Remark. This suggest that for K, px irr, k cannot have degree higher than K : k. This is true because 1, , 2 , . . . , n are dependent; then we can obtain a f x as in above proof with as a root. f x n; but px f x by the definition of px. Hence degp n. Example 3.19 2 : 2 with possible basis 1, is algebraic over s.t. deg irr, 2. 2. Any element of the form a b 2 2 Theorem 3.20 (3.120) Let K k be an extension of fields and let K be algebraic over k. (a) There exists a unique monic irreducible polynomial px I px, then kxI k given by f x I f . (b) If is another root of px, then k Proof. (a) For algebraic K, consider the homomorphism of rings : kx f x K f kԽ . kx having as a root. Moreoever, if The kernel of is a principle ideal (any ideal I kx is a principle one, otherwise px, q x I, p q 1 p, q R, I cannot be proper); we may let and q p, then there are some s, t s.t. sp tq ker hx for some hx kx. Then kxhx im K. im is a subring of K and thus a domain; this gives that hx is irreducible, and we can get px monic irreducible and px hx. Finally, im k, since im is a subfield of K (since kxhx is a field since hx is irreducible) containing k (if we let f x c k) and (if we let f x x); it is the smallest since every such subfield k. Uniqueness follows Proposition must contain all the polynomials of . We then get kxI 3.12(d). (b) The two isomorphisms, kxI k and kxI kԽ induces a third: k kԽ . Definition 3.21 For given K k, K algebraic over k, let irr, k kx be the unique monic irreducible polynomial in kx that has as a root. irr, k is called the minimal polynomial of over k. Example 3.22 irr 5 3, x4 16x2 4. Theorem 3.23 Let k be a field and let f x kx be a nonzero polynomial. Then there exists a field K containing k as a subfield and with f x a product of linear factors in K x. Proof. We prove via induction on the degf . 1). If degf 1, f x is already in linear factor form; K 28 k then suffices. (degf (degf 1). If degf 1, write f x pxgx in which px is irreducible. By Proposition 3.12(a), F kxpx is a field containing k and a root of px (a root in F is z x I, I px). Hence, in F x, we have px x z hx and f x x z hxgx. Induction hypothesis then gives us a field K containing F in which hxgx, and hence f x, is a product of linear factors in K x. 10/17/08 Theorem 3.24 (3.121) Let k E K be fields s.t. K E and E k are both finite. Then K k is finite and K : k K : E E : k. Proof. We first show that every element of K can be expressed as linear combination with a basis of size K : E E : k over k, then we show that the basis is linearly independent. (1) Let the basis of K : E be 1 , 2 , . . . , m and that of E : k be 1 , 2 , . . . , n . Then we may write any m n element x K as x i 0 ci i . For each ci , we may write it using basis over k as ci j 0 dij j . i j spans K over k. Then x i,j dij i j and X (2) To see that X is linearly independent, we see that for any linear combination of all basis elements in X to be 0, since i 's are linearly independent, dij j must all be zero. This in turn requires that all dij 's are zero. Definition 3.25 Let k K be fields, let f x kx. Then f x splits over K if f x ax z1 x z2 . . . x zn , where z1 , z2 , . . . , zn are in K and a k is nonzero. For a given field k and given polynomial f x kx, E k is a splitting field of f x over k if f x splits over E but not over any proper subfield of E. Proof. By Kronecher's theorem, there is a field extension K k such hat f x splits in K x; say, f x ax 1 x 2 . . . x n . Then the subfield E k1 , 2 , . . . , n is a splitting field of f x over k. Proposition 3.27 (3.126) Let p be a prime, and let k be a field. If f x xp c kx, then either f x is irreducible in kx or c has a pth root, say, , in k. Therefore, if k contains the pth roots of unity, then k is a splitting field of f x. Proof. Suppose that f x is not irreducible over k, then f x 2 , . . . , p1 be the set of pth roots of unity. We have 1, , f x f x gxhx, 1 d deg g p. Let Corollary 3.26 (3.124) Let k be a field, and let f x kx. Then a splitting field of f x over k exists. x x . . . x p1 gxhx
29 Let b be the constant term of gx be b, Ԩbps ctp Ԩbs ct p . Therefore, c has a pth root in k as Ԩbs ct . If k, E k is a splitting field of f x.
c csd cdp Theorem 3.28 (Galois, 3.127 in the text) Let p be a prime and n pn . Proof. Let q over K. Let E pn , k p , g x K be the subset 0. Then there exists a field of size p is prime and d We get b d , in which is again a pth root of unity (p 1). Then Ԩbp d p dp cd . p, therefore p, d are relative primes; gcdd, p 1 and we have 1 sd tp with s, t integers.
csd tp xq x kx. By Kronecker's theorem, there exists K k s.t. gx splits E pn . To see this, we first show that gx has no multiple roots. For a we claim that E is a field of size q multiple root , x gx and x g x; but g x qxq1 1 1 (we get 1 by mod p since q pn . we work with p ). Therefore, all the roots are distinct, and E We are then left to verify that E is a subfield of K. We have: 1 E since 1 is a root of xq x. For a, b E, abq aq bq ab (since aq a aq ). We thus have ab E. K : g 0, 0 a bq For b E, a ը aq . . . bq aq bq a, q q k q k 0 mod p. p k ; therefore k a b Inverse. a E, aq1 1, then aq2 a 1 a1 a b E. All cross terms aq2 . q k ak bqk are zeros because therefore, E is a field of size pn . 10/20/08 Corollary 3.29 (3.128)For given p and n n. 0, there exists an irreducible polynomial gx px of degree Proof. Let E p be a field extension with q pn elements, then E E 0 is a group of size q 1. Moreover, this group is cyclic by Lemma 2.20. Then E for some of multiplicative order q 1. E p since on one hand p E; on the other p contains and then every nonzero elements of E. Let E by px irr, p , degree d p xf x : p (Proposition 3.12(e)). But p xpx p Theorem 3.20 in the text. Therefore d E : p n; f x is an irreducible polynomial of degree n. Lemma 3.30 (3.130) Let f x kx, where k is a field, and let E be a splitting field of f x over k. Let k be an isomorphism of fields, let : kx k x be the isomorphism :k gx a0 a1 x . . . an xn g x 30 a0 a1 x . . . an xn , that is, is an extension of to kx. Let E be a splitting field of f x over k . Then there is an isomorphism : E E extending . Proof. We prove via induction on the degree of d E : k. d 1. f x kx has all roots in k, then f x is product of linear polynomials, f x f x is also product of linear polynomials in k ; then E k is a splitting field of f x. d 1. Let z k be a root of f x in E, and let px irrz, k be the minimal irreducible polynomial with z as a root. degp 1 since z k. Then kxpx kz E and kz : k degp. Given k , p x px is again a minimal irreducible polynomial in k x (if one of p, p can be :k
factorized, then , being an isomorphism, would indicate that the other is as well). Let z be a root of k z by Theorem 3.20 (before applying Theorem p x, then extends to an isomorphism of kz k extends naturally to an isomorphism : kx k x, f x f x; 3.20, we note that : k then induces an isomorphism : kxpx xp x, f x I x I . Then k f kz kxpx k xp x k z ). Then E is again a splitting field of f x kz x since all roots of f x are still in E. Similarly E a splitting field of f x k z x. But now E : kz E : k, so we may apply inductive hypothesis that extends to some that is an isomorphism between E and E ; then also extends to . Theorem 3.31 Any two splitting fields E and E for f x over k are isomorphic. Proof. This is the case when k k in previous lemma. Corollary 3.32 (E. H. Moore, 3.132 in the text) Any two finite fields having exactly pn elements are isomorphic. Alternatively, a field of size q pn is unique up to isomorphism. Proof. Let E be a field of size q pn , then E E 0 is a cyclic group of order q 1 and q1 1 for all E ; q 0 for all E. Thus, E of size pn is a splitting field for gx xq x over p (since xq x p x) and therefore, E is a unique up to isomorphism. Extra: The number of irreducible polynomials of given degree is encoded by a Zeta function. The classical 1 . [see text/notes for now, may be added later] Riemann Zeta function is s n 1 ns , s 10/22/08 Definition 3.33 Let E k be a filed extension. An automorphism of E is an isomorphism : E fixes k if a a for all a k. Proposition 3.34 (4.1) Let k be subfield of a field K, let f x and let E kz0 , z1 , . . . , zn permutes the roots. xn an1 xn1 . . . a1 x a0 E. kx K be a splitting field for f over k. If is an automorphism of E fixing k, then 31 Proof. Suppose is a root of f x, from f we obtain that 0 f n an1 n1 . . . a1 a0 0, since ai ai by the assumption that fixes k. Therefore is also a root of f x. On the other hand, E is isomorphism and the number of roots are finite, is a bijection between zi and itself, since : E therefore permuting the roots. Definition 3.35 Let E k be a field extension. The Galois group of E over k, GalE k, is the set of all automorphisms of E that fixes k. Let f x kx has splitting field E over k, the Galois group of f over k is GalE k. Lemma 3.36 (4.2 in the test) Let E i 1, 2, . . . , n, then idE . Proof. We prove via induction on n. (1) (n 1). E kz1 , let E, then f z1 g z1 n an1 n1 . . . a1 a0 n an1 n1 . . . a1 a0 f , kz1 , z2 , . . . zn and let GalE k be such that zi 0. Then zi for (2) (n 1). Let K kz1 , z2 , . . . , zn1 , if fixes z1 , z2 , . . . , zn then fixes K by induction hypothesis; fixes E after we apply (n=1) case again. for f, g kz , gz1 f z1 g z1 f z1 g z1 . Theorem 3.37 (4.3) For f x kx, deg f Sn . Proof. Let X n. The Galois group GalE k is isomorphic to a subgroup of z1 , z2 , . . . , zn . Define : GalE k SymmX , by : X. X is the restriction of to X. We need to verify that gives an injective homomorphism. zi zi zi zi zi . Therefore is a homomorphism. The kernel of fixes zi X; by previous lemma, is the identity element. Since there is a single element in the kernel, the map is injective. We then have an injective homomorphism from GalE k to SymmX , but SymmX is a subgroup of Sn (since some elements of X may be multiple roots). We are done. 0, then an irreducible polynomial f x Lemma 3.38 (4.4) If k is a field of characteristic k repeated roots. 32 kz has no Proof. Let f x kx, f x 0 be irreducible. We may assume that deg f 1, which implies f x 0 since k 0 (that is, we don't need to consider the modulo stuff). f x has a repeated root if and only if gcdf, f 1. But f x irreducible, so f f , gcdf, f 1. Definition 3.39 Let E k be algebraic. An irreducible polynomial f x kx is separable if it has no repeated roots. An arbitrary polynomial f x kx is separable if its irreducible factors are separable. An element E is separable over k if irr, k is separable. E k is separable if is separable over k for every E. Frac p t, let f x xp t. Claim: f x is irreducible over k and is not Example 3.40 Let k p t t. Then f x xp t xp p x p . Therefore separable. Let be any root of f x s.t. p f x has repeated roots and is inseparable. To show that f x is irreducible, by proposition 3.27, f x is g t either irreducible in k or it contains a root in k; suppose it contains a root k. Say ht , then thtp gtp , the leading powers of t at two sides cannot match, contradiction. p htp gtp 10/24/2008 Theorem 3.41 (4.7) Let E k be a splitting field for f x kx, f x separable in E k. k and for a splitting field E for f (1) for : k . E :E (2) GalE k f , extends to exactly E : k isomorphisms E : k. Proof. Sketch of ideas: It works similarly as Lemma 3.30. Here the induction is on E : k. For induction step, we again pick a px irreducible, and it has degree d with d nonrepeating roots. For as a root of px, for each as a root of p x, there is an extension of isomorphism from k k to isomorphism k k Խ . There are exactly d of these. Then we can apply induction hypothesis over kx to get that there are E : k isomorphisms between E, E extending . We then obtain the result. Note that here we don't need to consider other roots of px since we are basically trying to construct E from one side and find corresponding isomorphisms on the other side. For this reason, using k or k are the same; but we don't need to use both to construct E. Corollary 3.42 (4.9) If moreover f x is irreducible, then n Proof. n degf degf GalE k . k : k and E : kk : k E : k.
xq Theorem 3.43 (4.12) Let k p, E pn , where E is the splitting field of g x n . Then Gal q p pn p is cyclic of order n with generator: Frob :
pn pn , u x over p for up Proof. We know that Gal pn p n by Theorem 3.41. Therefore, we only need to verify that Frob is an automorphism of pn that fixes p , and that Frob is of order n. Let q pn , and let G Gal q p . 33 Since q has characteristic p, we have Froba b a bp ap bp Froba Frobb, Frobab abp apbp FrobaFrobb, and Frob is homomorphism of fields (a field under ring homomorphism maps 1 and for any f a, b, ab 1 f af b 1). As a homomorphism from a field, Frob to a field, since 1 is injective (since kernel is an ideal in q ; a field only has itself or 0 as ideals. Since 1 is mapped to 1, the whole field is not the kernel.), q is finite, Frob is bijective and then an automorphism. n u Frob fixes p since for a p , ap a mod p (Fermat's little theorem). The order of Frob is n since up can satisfy upn p for all roots of upn u, then upn p has too many roots. for all u. If smaller n Example 3.44 have Gal 4 u 0 1 2 id(u) 0 1 2 4 2 2 is the splitting field of x4 x xx 1x2 x 1 xx 1x x 2 . We Frob(u) 0 1 2 x4 16x2 4 irr 5 3 3, x,
2ooo ooo Example 3.45 F x 5 5, 3 5 3 4 5 PP OOO2 OOO PPP 2 PPP n nnn nnn 2 n 3 5, 3 is a splitting field for f x over , 3, 5 5, 3 3, 3, 5 5, 3 3 Proposition 3.46 (4.13) Let k be a field and let E k be a splitting field for f x kx s.t. f x has no repeated roots. Then f x is irreducible if and only if GalE k acts transitively on the roots of f x. Proof. . Let f x be irreducible and let , be roots of f x. Since Gal 5, 3 5 5 5, 5, 3 3 kxf x k p7 ppp pp
NNN NNN ' k there exists : k k ; then E k / E
/k E 34 extends to an automorphism : E E s.t. fixes k and . . Let f x q1xq2 x . . . qtx for irreducible polynomial q1, . . . , qt . Let q1, q2 0. By assumpE, GalE k. s.t. . But then q1 0 and is a multiple root (it tion, there exists : E appears in both q1 , q2 ), contradiction. Alternative proof of Corollary 3.42: Proof. where X n deg f . Then G Gx X G n GalE k acts transitively on X as the set of roots of f , G. Example 3.47 Let f x x3 2 x, f x is irreducible (by Eisenstein). What is the splitting field of 3 2u, x3 2 2u 1u2 u 1. We have f x over ? x3 2 x 3 2x2 3 2 3 4, let x E 3 2,
2 3 2
3 3 hence E : 6, which is maximal for f of degree 3. In general, for deg f GalE S3 . 10/27/2008 Definition 3.48 A pure extension of type m is an extension kuk, um K k is a radical extension if there is a tower of fields k in which each Ki 1 Ki is a pure extension. K0 K1 ... Kt , n, GalE k Sn ; in our case k for some m 1. An extension Definition 3.49 Let f x kx have splitting field E k. We say f x is solvable by radicals if there is a radical extension k K0 K1 . . . Kt , with E Kt . Proposition 3.50 Ever irreducible cubic f x is solvable by radicals.
b Proof. [sketched] Apply change of variable twice, first time let x x 3 , then x equation g3 h3 3gh q x r 0 x3 bx2 cx d kx g h. This yields the . Obviously, solutions of g, h to g3 h3 r gh 1 q 3 35 will also be solutions to the original equation when we add up g and h. It turns out that three solutions are obtained this way; since a cubic has only three roots, these are all the solutions. Remark. The material that will be covered in the next several lectures will lead to theorem 4.26 and 4.53 in the text. 4.26 says that if f x is solvable by radicals then GalE k is solvable for characteristic k 0. 4.52 makes 4.26 stronger by showing the other ways around is also true (if and only if). We can thend draw conclusions that cubics and quartics are solvable by radicals since for f x kx, GalE k S3 (cubics) or S4 (quartics). Since S3 , S4 are both solvable, cubics and quartics are solvable by radicals. Theorem 3.51 (4.16) Let k B E s.t. B k is a splitting field for f x kx. E k is a splitting field for gx kx. Then GalE B GalE k and GalE kGalE B GalB k. Proof. Let B kz1 , . . . , zn , let GalE k, then permutes the roots of f by the proof used in PropoGalB k by B . We then claim that is sition 3.34. In particular, B B. Define : GalE k a homomorphism with ker GalE B , and is surjective. is a homomorphism can be obtained similarly as that in Theorem 3.37. We also have ker GalE B since the kernel is exactly these permutations in E that fixes B. It follows that GalE B is a normal subgroup of GalGk. is surjective by Lemma 3.30: if GalB k, then there is GalE k extending . The first isomorphism theorem then gives the result. Definition 3.52 A character of a group G in a field K is a group homomorphism : G multiplicative group of K with 0 removed). K (K is the Remark. Since a field K with zero removed is a multiplicative group (denote this as K ), then restricted to K (denoted as K ) is then a character in K. AutK Lemma 3.53 (E.Artin) A set of distinct characters 1 , . . . , n is independent. That is, if a1 , . . . , an are such that Ԧ a1 1g a2 2g . . . anng 0, for all g G, then a1 a2 ... an 0. Proof. Proof by descent. Assuming that a relation Ԧ exists with nonzero coefficient, we show that there exists a relation with fewer nonzero coefficients. Clearly, at least two coefficients in Ԧ are nonzero. Say a1 , a2 0. Let h G be s.t. 1 h 2 h, then a1 1 gh . . . an n gh Subtract 0 a1 1 h1 g . . . an n gn h 0. let a 1 Ԧ1 h from the second equation yields a2 2 h 1 h2 g . . . an n h 1 hn g 0, a2 2 h 1 h, . . ., we obtain a smaller set of i 's that are dependent.
K and let 1 , . . . , n be distinct field automorphism, then 1 , . . . , n Corollary 3.54 (Dedekind) Let G are independent. 36 Proof. Apply previous lemma. 10/29/2008 Lemma 3.55 (4.17) (1) Let K k1 , 2 , . . . , n be a finite extension of fields. Then there is a finite extension E k s.t. E k is a splitting field for some f x kx(Such an extension E k of smallest degree is called the normal closure of K k). Moreover, if each i is separable over k, then f x can be chosen to be a separable polynomial. (2) If K is a radical extension of k then a normal closure E k is also a radical extension. Proof. (1) Theorem 3.20 gives us pi x irri , k has i as a root for each i . Let f x p1 x . . . pn x, we have E that is a splitting field of f x containing K. If i is separable over k then from definition f x, as a product of irreducible polynomials that are separable, is itself separable. k k1 (2) We can write the radical extension K as k0 m ku1 , u2 , . . . , ut K s.t. ki 1 ki is a pure extension (ui i1 1 G Let B0 k. We construct a tower B0 k k u 1 1 u1 ku1 , 2 u1 ... ku1 , 2 u1 , . . . , r u1 B1 . GalE k k u 1 k2 ku1 , u2 ki ). We may let id, 2 , . . . , r . ... kt 1 Any j , being an automorphism then a homomorphism (that is, j ab j aj b), gives j u1 m1 j um1 . Since um1 k and j fixes k, j u1 m1 k B0 . Thus B1 B0 is a radical extension. Similarly 1 1 we may define Bi 1 Bi ui 1 , 2 ui 1 , . . . , r ui 1 , It is straightforward to see that Bi 1 Bi is a radical extension. Since E radical extension. Bt , we have that E is also a K0 K1 ... Kt K. Assume that Lemma 3.56 (4.18) Let K k be a radical extension, say k moreover Ki 1 : Ki pi is prime, and let that k contains pi th roots of unity for i 1, 2, . . . , t. If K k is a splitting field then there exits a sequence of subgroups GalK k s.t. Gi 1 Gi and Gi : Gi 1 G0 G1 ... Gt e pi . In other words, G GalK k is solvable. 37 Proof. For each i, let Gi GalK Ki . Since elements of Gi GalK Ki are automorphisms of K that fixes Ki , they certainly contain all elements that fixes Ki 1 , thus having Gi 1 GalK Ki 1 as a subgroup, therefore we have the subgroup relationship GalK k G0 G1 G2 ... Gt 1. Since K1 ku with up1 k, and k contains the p1 th roots of unity, K1 is the splitting field for f x xp1 up1 . Obviously, K is also a splitting field for f x. Theorem 3.51 then applies to say that GalK kGalK K1 GalK1 k and G1 GalK K1 is normal in G0 GalK k. Since GalK1 k K1 : k p1 by Lemma p1 is cyclic of oder p1. Repeating this for all i then yields the result. 3.55, G0 G1 Theorem 3.57 Construction of a polynomial not solvable by radicals. Let f x x be an irreducible polynomial of degree p s.t. f x has precisely two complex roots. Then the splitting field E of f x has groups GalE Sp . 10/31/2008 Items to cover today: Construct f x xp . . . x, with Galois group Sp . Compare normal extension versus splitting field extension (same). Remove condition in Lemma 3.56. Theorem 3.58 Construction of f x xp . . . x, with Galois group Sp . Let f x xp . . . x be an irreducible polynomial with precisely two complex roots. Then the splitting field E of f x has group GalE Sp . 12, 12 . . . p. Let be a root of f x, then p : and Proof. We want to construct Sp E : . Hence, GalE contains an element of order p E : . For a splitting field E , GalE p. Since GalE Sp , GalE contains an element i1 , i2 , . . . , ip . Let GalE corresponds to complex conjugation, then j1 , j2 (there exists since complex conjugation only affects the two complex roots and fixes everything else). wlog, we may assume 12 and 12 . . . p (by taking a suitable power of i1 , i2 , . . . , ip ).
Definition 3.59 An extension is normal if for every irr, k splits in E. E, algebraic over k, the polynomial px Theorem 3.60 A finite extension E k is normal if and only if E is the splitting field of some polynomial f x kx.
n Proof. ( ) Assume E k1 , . . . , n , E is the splitting field of f x i 1 irri , k . ( ) let E k be the splitting field for f x kx, let E, let gx irr, k. We have E E . Let be another root of gx. Then we have that k kԽ with isomorphism . We then have that f x kx have E as the splitting field since all of f x's roots are in E. Similarly, f x kԽ x has E Խ as 38 the splitting field. By Lemma 3.30, extends to an isomorphism between E , E Խ . But E E E Խ , therefore E E Խ . E and Remark. Lemma 3.56 has two conditions that will be removed: 1. k contains the pth roots of unity whenever p K : k, 2. K k is a splitting field. Next lemma removes the second condition. Lemma 3.61 (4.20) Let k be a field and let f x kx be solvable by radicals: There is a radical extension k K0 K1 . . . Kt with Kt containing a splitting field E of f x. If each Ki 1 Ki is a pure extension of prime type pi , and if k contains all the pi th roots of unity, then the Galois group GalE k is a quotient of a solvable group. Proof. We work with the normal closure of K k, say L. By Lemma 3.55, Lk is a splitting field extension and a radical extension. Then Lemma 3.56 gives us that GalLk is solvable. Since Lk, E k are both splitting field extensions, by Theorem 3.51, GalE k GalLkGalLE is again solvable (quotient group of a solvable group is again solvable by 4.21). 11/03/08 Remark. Summary of 4.17  4.20 Let f x kx be solvable by radicals. Then there exists a tower L K E k Assume moreover that k contains the pth roots of unity whenever p Combining the following with Lemma 3.55 (4.17), Lk is splitting field over k; Lk is radical; p normal closure ofK k radical extension overk splitting field off xoverk K : k. L : k p K : k. Lemma 3.56 (4.18) says that if E K L, then GalE k is solvable. K L, then GalE k is a quotient group of GalLk and then is Lemma 3.61 (4.20) says that if E solvable. Lemma 3.62 Let k k be a splitting field for gx Proof. Let , Galk k, say ba . a , xm 1 kx, then k k is abelian. b , where is a primitive mth root of unity, then 39 Example 3.63 Special case when k , k . m զ Gal with a mod m : a. 15 զ 8, 15 զ 5 For m 15, : not cyclic. զ 3 զ 4 2 . Abelian but Theorem 3.64 (4.26) Let f x kx be solvable by radicals and let E k be the splitting field of f x over k. Then GalE k is solvable. Proof. The idea is to construct another tower
z zz zz L L K E k zz zz zz zz K E k z zz zz by letting k k be the splitting field of xm 1 kx where m is large enough s.t. p m whenever p K : k. Then k contains all the pth roots of unity that we need to apply Lemma 3.61. We then have that GalE k is solvable. Now look at the tower k k E , since E k, k k are splitting field extensions (and k k solvable by construction), by Theorem 3.51, GalE k is normal in GalE k with quotient group isomorphic to Galk k; therefore GalE k is also solvable. Finally for the tower k E E , E k, E k are splitting field extensions, we then apply Theorem 3.51 again to get that GalE k GalE kGalE E and therefore solvable as a quotient group. Remark. Now the Fundamental Theorem of Galois Theory starts. Definition 3.65 For a field E and for a subfield H EH that is, the part of E that is fixed by H. Remark. If H GalE k AutE , then E H at least k is in the fixed field E H . k; this is true because any AutE , define the fixed field of H as a, a E : a H , GalE k fixes k; therefore k can be strict. Let E 3 2. If G GalE , then must Example 3.66 The inclusion E H 3 2. But the other two roots of f x are not real, so that fix , and so it permutes the roots of f x x 3 2. It now follows that is the identity and E G E. 3 2 Proposition 3.67 (4.28) If E is a field, then the function H then E L E H . E H is order reversing: if H L AutE , Proof. a E fixed by L must be fixed by H, but not necessarily the other way around. Definition 3.68 A rational function
g x1 ,x2 ,...,xn h x1 ,x2 ,...,xn kx is a symmetric function if it is fixed by S . n 40 Proposition 3.69 (4.30) A list 1 , . . . , n of distinct characters from E Lemma 3.70 (4.31) If G E is independent over E. Proof. 1 , . . . , r . Then the system of equations 1 , . . . , n AutE is a set of distinct automorphisms then E : E G n. r n, since E is an extension field of E G , we may assume it has a basis Suppose E : E G 1 1 x1 2 1 x2 . . . n 1 xn 1 2 x1 2 2 x2 . . . n 2 xn ... 1 r x1 2 r x2 . . . n r xn 0 0 0, has a nontrivial solution c1 , . . . , cn , since n r. Then for any E, b1 1 . . . br r then if we multiply bi with the ith row of the system and note that bi j bi for any j (since j fixes E G ), we have 1 b1 1 c1 2 b1 1 c2 . . . n b1 1 cn 0 1 b2 2 c1 2 b2 2 c2 . . . n b2 2 cn 0 ... 1 br r c1 2 br r c2 . . . n br r cn 0, summing up all the rows then gives us 1 c1 2 c2 . . . n cn which contradicts the independence of the characters 1 , . . . , n . 0 11/05/08 AutE is a finite subgroup, then E : E G Proposition 3.71 (4.32) If G Proof. G 1 G. n G . Let n; therefore we need to show E : E G From Lemma 3.70 we have E : E G id, . . . , n and let 1 , . . . , m , be independent over E G , the system 1 1 x1 1 2 x2 . . . 1 m xm 2 1 x1 2 2 x2 . . . 2 m xm ... n 1 x1 n 2 x2 . . . n m xm 0 0 0, we claim that it has no nontrivial solution. Therefore, n m and we are done. Assume that x x1 , . . . , xm is a nontrivial solution to AT x 0, we may assume that 1 i i , the first row of the system gives us Not all x1 , . . . , xm E G . Otherwise, since 1 id that x1 1 . . . xm m 0. But 1 , . . . , m are independent over E G . X has maximum number of zeros among all nontrivial solutions. 41 We may then arrange x as x x1 E G, x2 , . . . , xr 1, xr 1 0, . . . , xm 1, we have 0. 0, in which x1 , . . . , xr are not zero and the rest are zeros. Since xr j 1 x1 . . . j r Since x1 E G , some k does not fix it, k x1 x1 , applying k to above equation, k j 1 k x1 . . . k j r 0
i 1 k x1 . . . i r 0. since k j is just another element in 1 , . . . , n , we may let it be i for some i. Then the equation becomes Subtracting this from the ith equation of the linear system, we have i 1 k x1 x1 . . . i r1 k xr1 xr1 0. Since k x1 x1 0, we obtain a nontrivial solution with more zeros (each j corresponds to a unique i so we again get the system), contradiction. AutE are two finite subgroups s.t. E G Theorem 3.72 (4.33) If G, H H is injective). E words, function : H AutE , fixes E G Proof. If we can prove that for AutE , G H G fixes E G. Now for the proof, we have two directions: E ( ) Trivial inclusion. ( ) Suppose fixes E G , G. We have E G E G since all of E G are fixed by G E G , therefore E G E G . But then n other hand, Proposition 3.67 gives E G G n 1 by Proposition 3.71, contradiction. E G E : E G E H , then G H (in other fixes G, then H .
G On the E : 11/07/2008 Theorem 3.73 (4.34) Let E k be a finite extension of fields with group G equivalent (1) E k is a splitting field for some separable f x kx. (2) k EG. (3) E k is a normal extension. Proof. We have (1) (3) by Theorem 3.60, therefore we prove (1) 42 (2). GalE k. The following are ((1) (2)). E k is a splitting field, then by Theorem 3.41, G E : k. By Proposition 3.71, G . Therefore E : k G . But k G since k is fixed by all elements of G; G E : E E : E E E : k E : E GE G : k. We then have E G : k 1 then k E G. ((2) (1)). Let E, k, for all G, let 1 , . . . , n be the distinct elements of . Then let n xn an1 xn1 . . . a0 . Now any coefficient of f x is invariant under any f x i 1 x i n x xn an1 xn1 . . . a0 f x. That is, all coefficients G since f i 1 x i G then f x E G x. Since E G k by assumption, ai 's of f x are in E and are fixed by G; ai E f x kx. Then we have that E k is a splitting field for f x. f x has no repeated roots therefore any irreducible factor of f x is separable, making f x also separable. Definition 3.74 A finite extension is a Galois extension if it is normal and separable. Corollary 3.75 (4.36) If E k is Galois and B is an intermediate field, then E B is Galois. Proof. E k is a splitting field of f x kx, then E B is also a splitting field of f x B x. Theorem 3.76 (4.43) Let E k be a finite Galois extension with group G GalE k. E H is an order invesring bijection between subgroups of G and intermediate (1) The function : H GalE B . fields of E k with inverse : B (2) Let B be an intermediate field, then B k is Galois if and only if H Proof. (1) is injective by Theorem 3.72. We prove surjectiveness by showing that every intermediate field B is of form B E H for some subgroup H G. For the tower k B E, let H GalE B , E is a splitting field of some f x kx B x; then Theorem 3.73(2) gives us that B E H . (2) ( ) B k Galois then is a splitting field extension. Then GalE B is normal in GalE k. ( ) Assume EH a E : a a, H . For any G, H, a E H , since that H G, then B 1 H; a a1 1 a1 1 a1 a. Since is H G, arbitrary, a is fixed by H, so a B. This gives that for any B, px irr, k, if we denote another root of px as , then some G gives . But B, hence all roots of px is in B, making B k a normal splitting field. B k is then Galois. GalE B G. 11/10/2008 Example 3.77 Let E 3 2, , 2 1 0. Determine all subfields B E. 43 k , G GalE k S3. This is true since G is isomorphic to a subset of S3 since E is the splitting field of x3 2; E : k 3, but the largest proper subfield of S3 is A3 with 3 elements; hence E S3 . S3 1, 12, 23, 13, 123, 132 has four subgroups, so we should have four subfields B, B1 , B2 , B3 with the correspondence: E QQQ  DDD QQQQ D  QQQ  B2 mm B3 B1 BA AA m { A {{{ mmmm mm m
KKK UUUUU UUU 123 23 31 ii MMM rr iiiii MMM rrr iiii i r iii rrr rrr 1 UKKUUUU KU
S3 2, 2 12 3 2, 3 2 3 2. Choose Action of G S3 on E, let x3 2 generators and for GalE as: x 1 x 2x 3 , 1 3 2 3 2 3 2 3 2 2 3 We see that G is then isomorphic to S3 with mapping 1 1 We then have 123
ZZ 2 132 23 13 2 12 B RR ZZZZ ZZ kkkkkkk E TTTTTTTZ ZZZZZZZZZ T Z 2 T TT ZZZZZZZ k kk RRR RRR RR B1 2
3 B2 2 dd B3 dddd jj jjjdddddddddddd jj dd jddj djj ddd
3 3 2 2 11/12/2008 Example 3.78 Gal15 15 , 15 is the 15th root of unity. Find all intermediate fields. 8. We have 15 զ 1, 2, 4, 7, 8, 11, 13, 14, hence 15 : ll lll lll 5 Q QQQ mm QQ mm QQ mmm 1, 2, 4, 8SS 1, 4, 11, 14
SSS SSS S 1 RRR RRR lll RRR lll R lll 1, 11 R 1, 4 RR 1, 14 RRRllll RRRllll lRRR lRRR lll lll
k kkkk kkkk k 15 SSS
5, 3SS 15 151 k
SSSk kkk SS kkk k kkkk kkkk kk SSS SS 1, 4, 7, 13 15RR G RRR RRR R 5 3 [see notes for additional detail] Theorem 3.79 (4.46) Let E k be finite extension of fields, there exists there are finitely many fields B s.t. k B E. 44 E s.t. E k if and only if Proof. (case k is finite). E is also finite, and E k is generated by a generator for the multiplicative group E . (case k is infinite). We have two directions k1 , 2 E for 1 , 2 E, ( ). Assume that E k has finitely many intermediate fields. Let k k. Consider the extensions k k1 c2 we show that there exits E s.t. k1 , 2 k1 , 2 , c k. Since there are only finitely many intermediate k1 c2 by assumption, there exist c c s.t. k1 c2 k1 c 2 , but then k1 c2 k1 c2 , 1 c 2 k1 , 2 . ( ). Assume that E k for some E, if E k is Galois, then there are finitely many intermediate fields (by Fundamental Theorem of Galois Theory), each intermediate field corresponds to a subgroup H GalE k. If E k is not Galois, then let Lk be a normal closure for E k (such normal closure always exists when E k is a finite extension). Then Lk has finitely many intermediate fields; hence E k has finitely many intermediate fields. Remark. There is a second proof of the ( ) direction above in the note. k for some E. In particular, Theorem 3.80 (4.47) If E k is a finite separable extension then E when characteristic k 0, every finite extension E k can be written as E k for some E. k1 , 2 , . . . , n , then we can take L the splitting Proof. Let Lk be the normal closure of E k, if E n irri , k. Then Lk is finite Galois, hence has finitely many intermediate extensions; field of f x i 1 therefore, E k has finitely many subextensions and by Theorem 3.79, E k for some E. 11/14/2008 Theorem 3.81 (4.50, Hilbert's Theorem 90) Let E k be a finite Galois extension with G GalE k a E defined by N u 1 cyclic group of order n. Say G , let norm N : E G u. Then N u v if and only if u v , for some v E . Proof. Let N : E N u E G u D : E Dv E
v v the claim is then equivalent to the statement ker N imD (N and D are both homomorphisms w.r.t. multiplication) ( ) Assume u
v v imD,
N u G v
45 G v v G v G 1 ( ) Assume N u a0 For any 1, since E u, a1 , let u u, . . . , an1 u u . . . n1 u i N u 0, . . . , n 1, ai E
modulo n. Let i 1 E, ai i aiu 1 i 1 ai 1 u , for n1 i 0. i 0 ai , then u and u , provided that 1 The automorphisms 1, , . . . , n1 are independent over E. Thus, a0 a1 . . . an1 n1 n1 i exists , s.t. 0. i 0 ai Remark. Properties of norm N u. (i) If u E , then N u k since N u is fixed by G: 0 and there G, N u N u. k is a (ii) N uv N uN v (the multiplication in E is commutative since E is a field), so N : E homomorphism. (iii) If a k, then N a (iv) E : k. If G and u E , then N u N u.
an , where n p, p a prime, s.t. k contains the Remark. Additional information is given on complex and Hilbert 90 briefly in the notes. Corollary 3.82 (4.51) Let E k be a Galois extension of degree E : k pth roots of unity. Then E k is a pure extension, E kz , z p k. Proof. Let be a primitive pth root, N here , since k, ... p 1 G k, G fixes .
z z Thus (by Theorem 3.81), zp z p k. for some z
z E. But then, 1 p zp zp zp z p For z above, z k, otherwise z no intermediate fields in between. 1. Lastly, E kz since E : k p prime, there are 0. Let E k be Galois with G GalE k a Theorem 3.83 (4.53) Let k be a field with characteristic k solvable group, then E can be embedded in a radical extension of k (therefore, the Galois group of a polynomial f is solvable by radicals). f x kx, characteristic k 0 is solvable Proof. Let E k be Galois, G GalE k. a solvable group. From the composition series of G, we see that there exists H G of index G : H p. Let k kw be an extension of k that contains pth roots of unity. E E H is Galois of degree E : E H E : k. H GalE E H G is solvable. We may use induction to get that there exists a radical extension E H R1 . . . Rt s.t. E Rt . 46 Consider E H k, if k k, then k done. If not, we build a tower E H is a pure extension of degree p by Theorem 3.82 and we are
v vv vv ww www E E EH k EH k vv vv vv Let E k be a splitting field for some f x kx, then E k is a splitting field for f xxp 1. Since p is prime, E k is Galois. But then E k is Galois with group G GalE k . Let i be roots of f x, E k 1 , . . . , n E k1 , . . . , n The restriction map : GalE k GalE k is well defined group homomorphism. is injective since fixes 1 , . . . , n idE . Thus G is isomorphic to a subgroup of G. In particular, G idE can be embedded in a radical extension of k . That is, exists k T1 . . . T is solvable. As before, E s s.t. E TS (thus E Ts ). Add k k k gives us a radical extension that starts at k. 47 Remark. Some isolated topics will now be covered that is in the curriculum but not that cohorent with other materials. 11/19/2008  11/21/2008 Remark. Quartics and resolvent polynomial were discussed. In particular, methods were given as how to tell the Galois group of an irreducible quartic. See notes for detail. 12/01/2008 Remark. Algebraic closure will now be dicussed. In particular, it will be shown that closed. is algebraically Definition 3.84 A field k is algebraically closed if every nonsonstant f x kx has a root in k. An algebraic closure k of a field k is an algebraic extension kk such that k is algebraically closed. Proposition 3.85 Every polynormial f x x of odd degree has a root in Corollary 3.86 There exists no extension E of odd degree larger than 1. Proposition 3.87 Every polynomial f x .
. x of degree two has a root in Theorem 3.88 [Fundamental Theorem of Algebra] Every nonconstant f x 12/03/2008  12/05/2008 x has a root in . Remark. We will prove that every field k has a unique (up to isomorphism) algebraic closure k. Proposition 3.89 (6.54) Let K k be an extension of fields (1) For z (2) (3) (4) K, z is algebraic over k if and only if kzk is finite. kz1 , . . . , zn k is finite. For z1 , . . . , zn K, z1 , . . . , zn are algebraic over k For y, z K, K k algebraic, y z, yz, y 1 y 0 are all algebraic over k. Kalg z K : z algebraic over k is a subfield of K.
K E, E K, K k algebraic, then E k algebraic. k1
0 kn Proposition 3.90 (6.56) (1) k (2) Let k0 n k . . . kn . . . be a tower of fields such that kn 1 kn is algebraic for all n is a field and algebraic over k0 . 0. Then (3) Let K kA be obtained from k by adjoining for all is algebraic over k. A. If all 's are algebraic over k, then K 48 Lemma 3.91 (6.57) Let k be a field and let kT be the ring of polynomials in the variables t : t T . If t1 , . . . , tn T are distinct and if fi ti kti kT is a nonconstant plynomial for i 1, . . . , n, then I f1 t1 , . . . , fn tn is a proper ideal in kT . Theorem 3.92 (6.58) Given a field k, there exits an algebraic closure k of k. Lemma 3.93 (6.61) Let kk be an algebraic closure for k and let F k be an algebraic extension. Then there exists an embedding f : F k. 49 ...
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This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell University (Engineering School).
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