Popular in Course
Popular in Mathematics (M)
This 17 page Class Notes was uploaded by Claudine Friesen on Monday September 28, 2015. The Class Notes belongs to MATH371 at University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/215397/math371-university-of-pennsylvania in Mathematics (M) at University of Pennsylvania.
Reviews for ALGEBRA
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/28/15
1 RINGS 1 1 Rings Theorem 11 Substitution Principle Let p R a R be a ring homomor phism a Given an element 04 E R there is a unique homomorphism ltlgt RM a R which agrees with the map p on constant polynomials and sends p a a b Given elements 041 04 E R there is a unique ring homomorphism ltlgt Rp1 an such that ltlgtlR b and 041 Lemma 12 For every ring B there is a unique ring homomorphism Z a R Lemma 13 IfR is a ring and a E R then ra r E R a is an ideal Theorem 14 A ring R is a eld if and only if it has emactly two ideals Corollary 15 Let F be a eld and R a non zero ring Then every homo morphism ltp F a R is injective Lemma 16 Every ideal in Z is principle Theorem 17 Let gp be a monic polynomial in RM and let 04 be an element ofR such that ga 0 Then z 7 oz divides Theorem 18 IfF is a eld then every ideal ofF p is principle Corollary 19 Let F be a eld and let fg E which are both non zero Then there is a unique monic dp E called the greatest common divisor offg such that a d generates the ideal f7 g of generated by fg b d divides f andg c Ifh is any divisor off andg then h divides d d There are pq E such that d pf qg Theorem 110 Let I be an ideal of a ring R a There is a unique ring structure on the set RI such that the canonical map 7139 R a RI sending a a a I is a homomorphism 1 RINGS 2 b The kernel 0f7T is I Theorem 111 Mapping Property of Quotient Rings Let f R a R be a ring homomorphism with kernel I and let J be an ideal which is contained in I Denote RJ by F a There is a unique homomorphism F a R such that fir f b First Isomorphism Theorem If J I then maps E isomorphically to the image off Theorem 112 Corresgondence Theorem Let E RJ and let 7T denote the canonical map R a R a There is a bijective correspondence between the set ofideals ofR which contain J and the set of all ideals ofR given by I a 7TI andT a W 1T b Third Isomorphism Theorem IfI C R corresponds to T C D then RI and RI are isomorphic rings De nition 113 Let R be a ring extension of R and let oz 6 R WE de ne RM EnOil rl E R Theorem 114 Let R C R and let oz 6 R Then there is a unique map p RM a R such that p is the identity on R and takes z 04 Further FM ima De nition 115 C RM Theorem 116 Let R be a ring and let fp be a monic polynomial of positive degree with coe cients in B Let RM be the ring obtained by ad joining an element satisfying fa 0 The elements ofROi are in bijective correspondence with vectors r07 77quot 1 E R via a map M where M707 39 quot 7amp4 To T104 r2042 MAO 1 Theorem 117 Let R be a ring and let ab E R such that ab 0 Further let c 6 RM be such that wac 1 where w R a BM Then if RM 31 0 we must have wb 0 De nition 118 We say b E R is a Zero Divisor if there is a non zero a E R such that ab 0 1 RINGS 3 11 Integral Domains De nition 119 A ring R is an Integral Domain if it has no zero divisors le 07 landifab0thena00rb0 Theorem 120 IfR is an integral domain then it satis es the eaneelation law Vabea 31 0 ab ac a b 0 Theorem 121 IfR is an integral domain then so is Theorem 122 Let R be an integral domain with nitely many elements is a eld Theorem 123 Let R be an integral domain Then there epists an embed ding RaF into a eldF De nition 124 Let R be an integral domain A fraction in R will be a pair ab where ab E R and b 31 0 Two fractions allbL7 lgb2 are called equivalent allbL lgb2 if 1le agbl There is a ring structure on F ab equivalence classes of fractions such that F is a eld It is called the Field of Fractions of R Theorem 125 Let R be an integral domain with eld of fractions F and let p R a K be an injectiye homomorphism from B into a eld K Then lb ltPaltr7b 1 De nition 126 An ideal M is Maximal if M 31 R but M is not contained in any other ideals other than M7 R Theorem 127 IfR is an integral domain then M is a mapimal ideal if and only if RM is a eld Theorem 128 The maximal ideals of the ring Z of integers are the principle ideals generated by the prime integers Theorem 129 The mapimal ideals of the polynomial ring CM are the prin ciple ideals generated by the linear polynomials z 7 a 1 RINGS 4 Theorem 130 Hilbert7s Nullstellensatz The mamimal ideals of the poly nomial ring Cz1 an are in a bijective correspondence with points of com pleccn dimensional space a lta1 7angt a Ma 1 01212n7 an Theorem 131 If ab E Z have no factor in common other than i1 then there are c7 d such that ac bd 1 Theorem 132 Let p be a prime integer and let ab be integers Then ifp divides ab p divides a orp divides b Theorem 133 Fundamental Theorem of Arithmetic Every integer a 31 0 can be written as a product acp1pk where c is i1 and each p is prime And further up to the ordering this product is unique Theorem 134 Let F be a eld a If two polynomials fg E have no common non constant factors then there are polynomials r7 5 E such that rf sg 1 b If an irreducible polynomialp E divides a product fg thenp divides one of the factors c Every nonzero polynomial f E can be written as a product Cpl 39Pn where c E and the p are monic irreducible polynomials andn 2 0 This factorization is unique ecccept for the ordering of terms Theorem 135 Let F be a eld and let fx be a polynomial of degree n with coe cients in F Then f has at most n roots in F De nition 136 Let R be an integral domain If a7 b E R we say a divides b if 3r 6 Rar b We say that a is a proper divisor of b if b qa for some q E R and nei ther q E R and neither q nor a is a unit 1 RINGS 5 We say a non zero element a E R is irreducible if it is not a unit and if it has no proper divisor We say that a7a E R are associates if a divides a and a divides a It is easy to show that if a7a are associates then a ua for some unit u E R Theorem 137 Let R be an integral domain u is a unit gt 1 aa are associates gt a a a divides b gt a D b a is a proper divisor ofb gt 1 Q a 2 b Theorem 138 Let R be an integral domain Then the following are equiv alent a For every a E R a 31 0 ifa is not a unit then a b1 bn where each b is irreducible b B does not contain an in nite increasing chain of principle ideals 11 Q 12 Q as Q De nition 139 We say that an integral domain R is a Unique Factorization Domain UFD if i Existence of factors is true for R ii If a E R and a p1 10 and a ql qm where phqj are irreducible Then in n and after reordering phql are associates for each i De nition 140 Let R be an integral domian p E R is prime ifp 31 0 and Va7 b E R ifp divides ab then p divides a or p divides b Theorem 141 Let R be an integral domain such that eccistence offactor ization holds Then R is a UFD if and only if ever irreducible element is prime 2 FIELDS 6 Theorem 142 Let R be a UFD and let a p1 1 b ql qm be prime factorizations Then a divides b if and only ifm 2 n and after reordering phqi are associates for alli S n Theorem 143 Let R be a UFD and ab E R such that at least one ofa 31 0 or b 31 0 holds Then there CCElStS a greatest common divisor d of ab such that i d divides a and b ii if an element e divides a and b then e divides d De nition 144 Let R be an integral domain We say R is a principle ideal domain PlD if every ideal is principle Theorem 145 In an integral domain all prime elements are irreducible In a PID all irreducible elements are prime Theorem 146 Every PID is a UFD Lemma 147 Let R be any ring Then the union of an increasing chain of ideals is an ideal Theorem 148 Let R be a PID a Let 0 7amp1 E R Then Rp is a eld if and only ifp is irreducible b The mamimal ideals ofR are those generated by irreducible elements 2 Fields De nition 21 Let F be a eld And let oz 6 K such that K 2 F We say that 04 is algebraic over F if there is a polynomial over F which is satis ed by a le if 1 anan1a a1aa0 for some a07an1 E F We say that 04 is transcendental over F if it is not algebraic over F 2 FIELDS 7 Lemma 22 Let F C K be elds with 04 E K Then if w FM n K it W a we have 04 is transcendental if and only ifltp is injectiye Or more speci cally if the kernel ofltp is 0 De nition 23 Let F C K be elds with 04 E K Further let w FM n K it W a We then know that herltpa is principle as is a principle ideal domain So in particular it is generated by a single element faz E But because K is a eld we must have faz is irreducible because otherwise K would have a zero divisor Hence fax is the only irreducible polynomial in faz because every element of the ideal is a multiple of fa and we call fa the lrreducible Polynomial for 04 over F De nition 24 Let FltOtgt be the smallest eld containing both 04 and F Similarly let F041 704 be the smallest eld containing 041 704 and F Lemma 25 Recall that Fla is the ring ZaneL an E F and is the smallest ring containing both F and 04 We then have FltOtgt is isomorphic to the eld offractions of Fla In particular we have that if 04 is transcendental then a Fla is an isomorphism and hence FOt is isomorphic to the eld of rational func tions Theorem 26 a Suppose that 04 is algebraic over F and let fd be its irreducible polynomial over F The map a Fla is an iso morphism and Fla is a eld Thus Fla FltOtgt b More generally let 041 704 be algebraic elements of a eld eactension K ofF Then Fa1an Fa1an Theorem 27 Let 04 be an algebraic over F and let fd be its irreducible polynomial Suppose fx has degree n Then 1047 7oL l is a basis for Fla as a vector space over F 2 FIELDS 8 Theorem 28 Let oz 6 K and E L be algebraic elements of two eptensions ofF There is an isomorphism of elds o FOt a F which is the identity on F and which sends 04 w B if and only the irreducible polynomials for 04 and over F are equal De nition 29 Let KK be eld extensions of F An isomorphism p K a K which restricts to the identity on F is called an lsomorphism of eld extensions of an F isomorphism Theorem 210 Let p K a K be an isomorphism of eld eptensions ofF and let fd be a polynomial with coe cients in F Let 04 be a root off in K and let cv ltpa be its image in K Then cv is also a root off De nition 211 Let K be a eld extension of a eld F We can always regard K as a vector space over F where addition is eld addition and mul tiplication by F is simply multiplication We say that the degree of K as an extension of F is the dimension of the vector space denoted K Extensions of degree 2 are called quadratic7 of degree are called cubic7 ect The term degree comes from the case when K FOt for an algebraic 04 over F and so 1047 7oL l form a basis for the vector space where n is the degree of the irreducible polynomial In this case we also call the degree the degree of 04 over F Theorem 212 fa is algebraic over F then Fa F is the degree of the irreducible polynomial ofa Theorem 213 Let F C K C L be elds Then L F L These are called towers of eld ecctensions Corollary 214 Let K be an edtension ofF of nite degree n and let oz 6 K Then 04 is algebraic over F and it s degree diuides n 2 FIELDS 9 Corollary 215 Every irreducible polynomial in RM has degree 1 or 2 Theorem 216 Let K be an eptension ofF The elements ofK which are algebraic over F form a sub eld of K De nition 217 We say that an extension K of F is an algebraic extensions and K is algebraic over F if every element of K is algebraic over F Theorem 218 Let F C K C L be elds IfL is algebraic ouer K and K is algebraic over F then L is algebraic over F Theorem 219 Let F be a eld Let F Then I f for some f E and F is a eld if and only if the f is irreducible Corollary 220 Let F be a eld and let f E be irreducible Then K is a eld eatension ofF and the residue ofz is a root off in K Theorem 221 Let F be a eld and let f be a monic polynomial in of positive degree Then there is a eld eptension K ofF such that f factors into linear factors ouer K Theorem 222 Let fg E and let K be a eld eptension of F a Division with remainder ofg by f gives the same answer whether car ried out in or in b f diuides g in if and only iff diuides g in c The monic greatest common divisor d of fg is the same wether com puted in or in d ff and g have a common root in K then they are not relatiuely prime in Fz Conversely if f and g are not relatiuely prime in then there epists an extension eld L in which they have a common root e ff is irreducible in and f and g have a common root in K then f diuides g in De nition 223 Let F be a eld and let f E It t anx anil nil alz ao 2 FIELDS 10 then we de ne fp nanznil n 7 1an1x 2 a1 Where we interpret n as the image of n E Z under the unique ring homo morphism Z a F It can be shown that things like the product rule hold for these formal deriva tives Lemma 224 Let F be a eld and let fp E Let oz 6 F be a root of Then 04 is a multiple root ie that x 7 002 divides fp if and only ifa is a root offp and of f p Theorem 225 Let fx 6 where F is a eld Then there epists a eld eptension K ofF in which f has a multiple root if and only iff and f are not relatively prime Theorem 226 Let f be an irreducible polynomial in Then f has no multiple roots in any eld eptension ofF unless the derivative f is the zero polynomial In particular ifF has characteristic 0 then f has no multiple root De nition 227 Let K be a eld extension of F Let 041 704 be a se quence ofelements of K We say that 041 704 are Algebraically Dependent if there is a polynomial f E Fx1 7zn such that flt0 17 7an 0 We say 041 04 are algebraically independent otherwise Lemma 228 Let F Q K 0417ozn are algebraically independent if and only if the substitution map p Fp1 an a K which takes fz1 7x to flt0 17 704 has kerltp 0 Corollary 229 If a1an are algebraically independent over F then F041 7an is isomorphic to Fx17 7x the eld of rational functions in 1 an De nition 230 An extension 0 the form F041 7an where 041 04 are algebraically independent is called a Pure Transcendental extension De nition 231 A transcendence basis for a eld K of F is a set of elements 041 704 such that K is algebraic over F041 7an 2 FIELDS 11 Theorem 232 Let 041 704m and 617 6 be elements in a eld ecc tension K ofF which are algebraically independent IfK is algebraic ouer F l7 6 then m S n and 041 70 can be completed to a transcen dence basis for K by adding n 7 in many of the 61 Corollary 233 Any two transcendence basis for a eld eactension F Q K have the same number of elements 21 Finite Fields De nition 234 We say that q p7 is the order of a eld K When dealing with nite elds p will always be a prime and q will be the order of the eld we are talking about Fields with q p7 elements are often denoted 1 Theorem 235 Letp be a prime and let q p7 be a power ofp with r 2 1 Let K be a eld with order q a There epists a eld of order g b Any two elds of order q are isomorphic c Let K be a eld of order q The multiplicatiue group Kgtlt of nonzero elements ofK is a cyclic group of order q 7 1 d The elements ofK are roots of the polynomial 7 x This polynomial has distinct roots and it factors into linear factors in K e Euery irreducible polynomial of degree r in fpm is a factor of 7 x The irreducible factors of 7 z in fpm are precisely the irreducible polynomials in fpm whose degree diuides r f A eld K of order q contains a sub eld of order q pk if and only if h diuides r Corollary 236 Let K be a nite eld Then there is an element a E K such that for all b E Kb 31 0 there is an n E w such that a b De nition 237 A generator for the cyclic group 7 is called an primitive element modulo p 2 FIELDS 12 Theorem 238 Letp be a prime and let q p7 a The polynomial 7 x has no multiple root in any eld L of character istic p b Let L be a eld of characteristicp and let K be the set of roots oqu7x in L Then K is a sub eld of L Theorem 239 Let L be a eld of characteristic p and let q p7 Then in the polynomial ring Lxy we have x y 7 xq yq Lemma 240 Let h be an integer dividing r say r ks and let q pl 1 pk Then 7 z divides 7 d De nition 241 A eld F is algebraically closed if every polynomial fd E has a root in F Theorem 242 Fundamental Theorem of Algebra Every nonconstant poly nomial with complecc coe cients has a complecc root De nition 243 Let F be a eld F is an algebraic closure of F if 0 F is algebraically closed 0 F is algebraic over F Corollary 244 Let F be a sub eld ofC Then the subsetF ofC consisting of all numbers algebraic over F is an algebraic closure ofF Theorem 245 Every eld F has an algebraic closure and if K1K2 are algebraic closures ofF there is an isomorphism p K1 7 K2 which is the identity map on F Corollary 246 LetF be an algebraic closure of F anc et K be any alge braic eactension ofF Then there is a subecctension K C F which is isomor phic to K Lemma 247 Let fd be a complecc polynomial Then takes on a minimum value at some point x0 6 C 2 FIELDS 13 22 Field Extensions De nition 248 If K is a eld extension of F we say De nition 249 An F automorphism of K is an automorphism of K which is the identity on F De nition 250 The group of all F automorphisms of K is called the Galois Group of the eld extension Theorem 251 For any nite extension KF the order oflGKFl divides the degree K F of the eld extension De nition 252 A nite eld extension KF is called a Galois Extension if lGKFl KiFl De nition 253 Let G be a group of automorphisms of K The set of elements xed by every element of G is called the xed eld of G KGa Kltpaaforallltp G Corollary 254 Let KF be a Galois extension with Galois group G The xed eld ofG is F De nition 255 Let fx 6 be a nonconstant monic polynomial A splitting eld for fx over F is an extension K of F such that i fx factors into linear factors in K fx x 7 041 7 04 with 04139 E K ii K is generated by the roots of fx K F041 7an Theorem 256 IfK is a splitting eld of a polynomial fx over F then K is a Galois extension ofF Conversely every Galois extension is a splitting eld of some polynomial fx 6 Corollary 257 Every nite extension is contained in a Galois extension Corollary 258 Let KF be a Galois extension and let L be an intermediate eld F C L C K Then KL is a Galois extension too 2 FIELDS 14 Theorem 259 a Let K be an extension of a eld F let fd be a polynomial with coe cients in F and let 0 be an F automorphism of K fa is a root offz in K then 0a is also a root b Let K be a eld eatension generated over F by elements 041 04 and let 0 be an F automorphism of K Ifo cces each of the generators 04 then 0 is the identity automorphism c Let K be a splitting eld of a polynomial fd over F The Galois group operates faithfully on the set 041 704 Theorem 260 The Main Theorem Let K be a Galois edtension ofa eld F and let G be its Galois group The function H w KH is a bijectiue map from the set of subgroups of G to the set of intermediate elds F C L C K It s inuerse is L w GKL This correspondence has the property that ifH GKL then KL hence LF GH Theorem 261 Existence of a primitive element Let K be a nite epten sion of a eld F of characteristic 0 there is an element 7 E K such that K FW De nition 262 We call an element 7 E K such that F y K a primitive element for K over F Theorem 263 Let G be a nite group of automorphisms of a eld K and let F be its med eld Let 61 76 be the orbit of an element 61 E K under the action of G Then 6 is algebraic over F it s degree over F is r and its irreducible polynomial over F is gd x 761 x 7 6 Further note that r diuides Corollary 264 Let KF be a Galois ecctension Let gd be an irreducible polynomial in Ifg has one root in K then it factors into linear factors in 2 FIELDS 15 Theorem 265 Let G be a group of order n of automorphisms of a eld K artd let F be its cced eld Thert K F n Corollary 266 Let G be a rtite group of automorphisms of a eld K artd let F be its ped eld Thert K is a Galois eactertsiort ofF artd its Galois group is G De nition 267 Letip F a F be an isomorphism Then p extends to an isomorphism a by Z w 2w where a7 Man We denote by the image of fx under this map Lemma 268 Let fd E be art irreducible polyrtomial Let 04 is a root of fx irt art eactertsiort eld K ofF artd let E be a root of fd irt art eactertsiort eldK ofF Thert there is a urtique isomorphism p1 FOt a FE which restricts to p ort a the sub eld F artd which sertds oz to 6 Theorem 269 Let p F a F be art isomorphism of elds Let fx 6 be rtortcortstartt artd let be the corresportdirtg polyrtomial irt Let K artdK be the splittirtg elds for fx artd z Thert there is art isomorphism w K a K which restricts to p ort F Corollary 270 Arty two splittirtg elds of fx 6 are isomorphic Theorem 271 Let K be the splittirtg eld of a polyrtomial fd E Thert K is a Galois eactertsiort of F That is K F Lemma 272 Withthe rtotatiort of the previous lemma the rtumber of iso morphisms ib K a K ecctertdirtg p is equal to K F De nition 273 Since any two splitting elds K of fd E are iso morphic7 the Galois group depends7 up to isomorphism7 only on f It is often referred to as the Galois group of the polynomial over F Corollary 274 Let KF be a rtite eld ecctertsiort the followirtg are equiu alertt 2 FIELDS 16 i K is a Galois eactension ofF ii K is the splitting eld of an irreducible polynomial y 6 ii K is the splitting eld of a polynomial y 6 iii F is the cced eld for the action of the Galois group on K iii F is the cced eld for an action of a nite group of automorphisms of K Theorem 275 Main Theorem Let K be a Galois eactension of a eld F and let G be its Galois group the function H w KH is a bijectiue map form the set of subgroups of G to the set of intermediate elds F C L C K Its inuerse function is L w GKL This correspondence has the property that ifH GKL then KLH andLFGH Theorem 276 Let KF be a Galois eactension and let L be an intermediate eld Let H GKL be the corresponding subgroup ofG Then a Let U be an element of G The subgroup ofG which corresponds to the conjugate sub eld oL is the conjugate subgroup oHo l In other words GKoL oHo l b L is a Galois edtension ofF if and only ifH is a normal subgroup of G When this is so then GLF is isomorphic to the quotient group GH De nition 277 Let F Q C be a sub eld of C which contains a primitive pth root of unity p eZMZ Lemma 278 fa is a root of u x17 7a then 04904 12704 g la are the roots of So the splitting eld of x17 7 a is generated by a single root K FM 2 FIELDS 17 Theorem 279 Let F Q C and let F contain a pth root of unity Further let a E F be an element which is not a pth power in F Then the splitting eld of fd x17 7 a has degree p over F and its Galois group is a cyclic group of order p Theorem 280 Let F be a sub eld ofC which contains a pth root of unity p and let KF be a Galois eactension of degree p Then K is obtained by adjoining a pth root to F Theorem 281 Letp be a prime integer and let p eZMZ For any sub eld F ofC the Galois group ofFCp ouerF is a cyclic group Lets consider the Galois group of a product of polynomials fdg over F Let K be a splitting eld of fg Then K contains a splitting eld of K of f and F of g So we have the following diagram Theorem 282 With the above notation let G andH GF F and g GK F i G and H are quotients ofg ii 9 is isomorphic to a subgroup of the product G gtlt H
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'