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# ALGEBRA MATH371

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Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman October 26 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 The Degree of a Field Extension Degree of Field Extension De nition 0101 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 multiplication 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 F Extensions of degree 2 are called quadratic7 of degree are called cubic7 ect The term degree comes from the case when K F oz 2 for an algebraic oz over F and so 1 oz Oin l form a basis for the vector space Where n is the degree of the irreducible polynomial ln this case we also call the degree the degree of oz over F l Degree of Field Extension vs Dimension of V Theorem 0102 fa is algebraic ouerF then FltOtgt F is the degree of the irreducible polynomial of oz Proof Immediate D l Adjoining 21 Square Root l Theorem 0103 Suppose F does not have char acteristic 2 Then any extension F C K of degree 2 can be obtained by adjoining a square root K F6 where 62 D E F Conversely if6 is an element of 3 an extenston ofF and if 62 E F but 6 9 F then F6 ts a quadratte extension Proof We rst show that every quadratic extension is obtained by adjoining a root of a quadratic polynomial E To do this7 we choose any element oz of K which is not in F Then 1 oz is a linearly independent set over F Since K has dimension 2 as a vector space over F Lox is a basis for K over F and K FM lt follows that a2 is a linear combination of 1 oz say 2 oz boz C7 with 90 E F Then oz is aroot of ns 2be Since 2 y O in F 7 we can use the quadratic formula oz bmgt to solve the equation 62 3 C O This is proved by direct calculation There are two choices for the square root7 one of which gives our cho sen root oz Let 6 denote that choice 6 V192 4ae 4 2a I Then 6 E K and it also generates K over F Its square is the discriminant 2 4616 which is in F The last proposition is clear D Product of Degrees Theorem 0104 Let F C K C L be elds Then L F L K K These are called towers of eld extensions Proof Let B 11 yn be a bsis for L as a K vector space and let C 1 ajm be a basis for K as an F vector space So L K n and K F m We will show that the set of mm products P 2417 is a basis of L as an F vector space7 and this will prove the proposition The same reasoning will work if B or C is in nite Let oz be an element of L Since B is a basis for L 5 over K we can write or 51111 nyn with 5239 E K7 in a unique way Since C is a basis for K over F each 6239 1126131 39 39 39 am m for some azj E Thus oz Emaijajiyj This shows that P spans L as an F vector space However we also know that the 57 is uniquely determined by Q7 and since B is a basis for K over F the elements azj are uniquely determined by 57 So they are uniquely determined by oz This shows that P is linearly indepen dent Hence P is a basis for L as an F vector space and we are done B One case to observe is the case where F C K and or E K Then we have F C FOz C K And Fa is an intermediary eld 6 Corollary 0105 Let K be an extension of F of nite degree n and let oz E K Then oz is algebraic over F and it s degree divides n Proof Immediate MAYBE GIVE AS HOMEWORK D Corollary 0106 Every irreducible polynomial in RM has degree 1 or 2 Proof D l Algebraic Sub eld l Theorem 0107 Let K be an extension of F The elements ofK which are algebraic over F form a sub eld of K Proof Let oz 6 be algebraic elements of K We must ShOW Ot 57 Oi 7 Oi7 ofl are all algebraic also 7 Note that since oz is algebraic that Fa F lt 00 Moreover 6 is algebraic over F and hence also algebraic over F oz Hence F oz which is generated by 6 over Fa is a nite extension ofFa le Fa Fa lt 00 and hence we know that Foz F Foz FaFa F lt 00 Hence every element of Fa is algebraic over F and all the elements we needed to check were algebraic are in this eld E Algebraic Extension De nition 0108 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 Transitivity of Algebraic Extensions 8 Theorem 0109 Let F C K C L be elds IfL ls algebralc over K and K ls algebralc over F then L ls algebralc over F Proof We need to show that every element or E L is algebraic over F We are given that F is algebraic over K and hence sonie equation of the form 1 ornan10l a10lan0 holds With a0 an1 E K Therefore or is algebraic over the eld Fa0an1 generated by a0 an1 over F Note that each coef cient a being in K is alge braic over F We consider the chain of elds FCFa0CFa0a1CCFa0an1CFa0 obtained by adjoining the elements a0 a714 or in suc cession For each a 1 is algebraic over F a0 oz be cause it is algebraic over F Also or is algebraic over F a0 owl So each extension in the chain is nite 9 We therefore know that the degree of Flta0 111 oz over F is nite and hence oz is algebraic over F D 02 Constructions with Ruler and Compass ln ancient Greece one of the favorite elds of study was constructions with a ruler and strait edge The idea was that you started with two points one unit apart Then given any two points you could draw a circle centered at one and going through the other and you could draw a line between to points The questions then arose7 what shapes could you make One of the most famous open questions from ancient Greece was7 given a constructed angle7 is it possible to trisect the angle ie construct an angle 13 its size Using what we know about elds we will now show that 10 this is impossible Speci cally we have the following de nitions l Ruler and Compass Constructions De nition 02010 a Two points in the plane are given to start with These points are considered con structed b lf two points are constructed we may draw a line through them or a circle with center at one and pass ing through the other Such lines and circles are con sidered constructed c The points and intersection of lines and circles which have been constructed are considered to be constructed All such constructed objects are said to be constructed with a strait edge and compass l 1 To A Line Through a Point l Theorem 02011 Given a constructed line I and a 11 potnt p we can construct a ltne perpendicular to l and passtng through p Proof Construction 1342 on Page 501 of the book D l Parallel Line Through A Point l Theorem 02012 Gtuen a constructed ltne l and a potnt p not on l we can construct a ltne parallel to l and passtng through p Proof Construction 1343 on Page 502 of the book D l Translation of Distance l Theorem 02013 Gtuen two potnts p q and a con structed ltne l unth a potnt r on l we can nd a potnt s on l such that dtstance between rs ts the same as the dtstance between p and 1 Proof Construction 1344 on Page 502 of the book D l Constructible Real Numbers l 12 De nition 02014 We say that a real number a s constructible if a is the distance between two con structible points Theorem 02015 A porntp 11 is constructr ble if and only rfo and b or constructz39ble numbers Proof Given a point p we can construct its coordinates by dropping perpendiculars to the axes Conversely if a b are constructible numbers we can construct p by marking off a b on the axes and constructing perpendiculars D Constructible Sub eld of R Theorem 02016 The constructr ble numbers for a sub eld of R Proof We will show that if a b E R are constructible nunibers then a 97 ab7 a b if a gt b and cf1 if a y O are constructible nurnbers 13 Addition and subtraction are dong by marking lengths on a line and using Construction 1344 on page 502 For inverses and multiplication see Proof of Proposition 1346 p 503 D l Constructible Sub eld Is Closed Under Squar Theorem 02017 If a is constructihle then so is xlal Proof Proposition1347 Page 503 D l Enlarging Constructible Sub elds l Theorem 02018 Suppose four points are given whose coordinates are in a snh eld F of R Let A B be lines or circles drawn using the given points Then 14 the points of intersection of A and B have coordinates in F or in a eld of the form Fr where r E F Proof The line through a0 0 a1 1 has linear equa tion 01 00W 50gt lt51 50gtlt 00gt The circle with center at 10190 and passing through 11191 has quadratic equation 55 00gt2 y bogt2 a1 00gt2 ltb1 bogt2 The intersection of two lines can be found by solving two linear equations with coe icients in F and hence is in F To nd the intersection of a line and a circle we use the linear equation to eliminate one variable from the equa tion of the circle leaving a quadratic equation in one un known with coe icients in F Hence the solutions to the equation are in F VD where D is the discriminant of 15 the equation and if D lt O the line and the circle don7t intersect Now consider the intersection of two circles say 55 a1gt2 1 b1gt2 7 and a a22 y 52gt T where 1239 9239 n E F ln general the solutions to a pair of quadratic equations is an equation of degree 4 However in this case we see that the difference between the two equations is linear and hence can be used to eliminate one of the variables in one of the equations as in the case of a circle and a line D l Chain Of Constructible Sub elds l Theorem 02019 Let a1am be construetz ble real numbers There is a chain of sub elds Q F0 C F1CF2CCFnKsuchthat a K is a sub eld ofR hja1amEK e For eaehi O n l the eld Fl is obtained form E by adjoining a square rot of a positiue number r2 E E which is not a square in E ConuerselyletQF0 CF1CF2 C CFnK he a chain of suh elds ofR satisfying Then euery element ofK is constructible Proof Introduce coordinates so that the initial points have coordinates in If we construct a number a this is done in a series of drawing lines and circles and taking their intersections But each single intersection produces adds at most one more level to the tower by previous results Similarly if we have such a tower all elements must be constructible Also by previous results D 17 l Degree of Constructible Real Numbers l Corollary 02020 fa is a eonstruetz ble real num ber then it is algebraic and its degree on is a power of 2 Proof In the Chain of elds in the previous theorem EH 2 and hence if K Fa we see that K must be a power of 2 D Now lets turn to the trisection of the angle We have to be a little careful in how we de ne this as it is the case that some angles can be trisecteol l Constructibility Angle l De nition 02021 We say an angle 6 is constructible if the length 6056 is constructible Trisecting The Angle NOW in general to solve the trisection problem we want to take an angle 6 18 and7 once we are given 6056 to start wiht7 we want to come up with a method for showing 6 is constructible from it So in order to show that the trisection problem has no solution it su ices to nd a single constructible angle 6 such that 6 is not constructible The angle we we will choose is 6 60 and so we will show that it is impossible to construct a 200 angle We will do this by showing that 005200 is an algebraic num ber of degree 3 The addition formulas for sin and cos can be used to prove 00536 400556 36056 Setting oz 200 we get 1 43 3 2 or or or 0 so3 3a 1 Lemma 02022 The polynomial gs 83 6 1 239s irreducible over Q Proof lt suf ces to check for factors ob where a b E Z But then we must have a divides 8 and b divides 1 ie is lzl lt isn7t hard to see that none of these work D This then tells us that in or is of degree 3 and hence cant be constructed A similar technique can be used to show the following l Constructing A Regular p gon l Corollary 02023 Let p be a prime integer If the 20 regular p gon can be constructed by ruler and compass then p 2 1 for some tnteger r Proof Let 6 be the angle 27rp Let C coslt6gtt5tn6 Then C is a root ofp71 1072 1 O which is irreducible If the p you is constructible then so are 52716 6056 and hence C lies in a real eld extension of degree 2 over Call this eld K and consider K This extension has degree 2 over K and hence K Q 2 But C 6056 t5tnlt6gt and hence is in K So the degree of C rnust divide 2M1 Hence p 12 forsonier D Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman November 16 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 Cubic Equations 011 Story Now we are going to consider the case of the cubic How ever before we do it is interesting to consider a little bit of the history behind the solution 0 ln the 1500s one way mathematicians proved their skills was by challenging each other to solve problems 0 The rst person known to have solved cubic equa tions algebraically was del Ferro but he told nobody o On his death bead del Ferro told his student Fior who wasnt as good a mathematician 0 At the time there were several types of cubics because negative numbers weren7t used 0 Fior had only been shown how to solve 3 a3 b o Fior began to boast that he was able to solve cu bics and a challenge between him and Tartaglia was arranged in 1535 0 At the time Tartaglia had been gured out how to solve a different type of cubic7 those of the form 33 1362 b o For the contest between Tartaglia and Fior7 each man was to submit thirty questions for the other to solve o Fior submitted 30 of the same type Which lead Tartaglia to gure out a solution Tartaglia offered several different types which Fior 0 At the beginning Tartaglia couldn7t solve Fiors prob lenis7 but then in the early hours of 13 February 1535 3 he discovered the general method to solve 3 a3 b Tartaglia was then able to solve all thirty of Fiorls problems in less than two hours 0 Fior had made little headway with Tartaglia7s ques tions so Tartaglia was declared the winner but Tartaglia didnt take the prize as the honor of winning was enough 0 At this point Cardan enters the story He was a pub lic lecturer of mathematics at the Piatti Foundation in Milan 0 He heard about the contest and when it was clear that Tartaglia had discovered a general method he set out to try and nd it for himself but was unable o A few years later7 in 15397 Cardan contacted Tartaglia7 requesting that the method could be included in a book he was publishing that year 4 Tartaglia declined stating his intention to publish his formula in a book of his own that he was going to write at a later date Cardan accepted this and then asked to be shown the method7 promising to keep it secret Tartaglia7 however7 refused An upset Cardan then wrote to Tartaglia challenging him to a debate but7 at the same time7 hinting that he had been discussing Tartaglia7s brilliance with the governor of Milan who was one of Cardanls powerful patrons When Tartaglia got this letter he radically changed his mind7 realizing that association with the in uen tial Milanese governor could be very rewarding and could maybe get him out of the modest teacherls job he then held 5 Tartaglia wrote back trying to get an invitation to meet the Milanese governor and Cardan invited him to come to the court When he came to llilan7 after much persuasion he agreed to tell Cardan his method if Cardan swore never to reveal it and to only ever write it down in code so that on his death bed no one could discover it from his papers Cardan agreed and Tartaglia divulged his formula in the form of a poem7 to help protect the secret7 should the paper fall into the wrong hands Tartaglia then left for home7 but by the time he got back he felt he had made a mistake revealing his secret when Cardan wrote to him to try and continue their friendship Tartaglia rebuffed him and ridiculed the books Cardan had published since Tartaglia had left Milan 0 Based on Tartaglia7s formula7 Cardan and Ferrari7 his assistant7 made remarkable progress nding proofs of all cases of the cubic in solving the quartic equation Tartaglia though made no move to publish his for mula despite the fact that7 it had become well known that such a method existed Tartaglia probably Wished to keep his formula in reserve for any upcoming de bates Cardan and Ferrari then travelled to Bologna in 1543 and learnt from della Nave that it had been del Ferro7 not Tartaglia7 Who had been the rst to solve the cubic equation When he learned this Cardan felt that although he had svvorn not to reveal Tartaglia7s method he felt that nothing prevented him from publishing del Ferrols formula 0 ln 1545 Cardan published a book which contained solutions to both the cubic and quartic equations and all of the additional work he had completed on Tartaglia7s formula 0 ln the book Del Ferro and Tartaglia are credited with their discoveries7 as is Ferrari7 and the story written down in the text 0 Tartaglia however was furious when he discovered that Cardan had disregarded his oath and his intense dislike of Cardan turned into a pathological hatred 012 Solution We will being by showing the ad hoc construction of Tartaglia First lets start with a cubic fltgt 3 1232 1136 a0 and lets set O Computation is simpler if a2 0 so lets substitute 1 1 a2 3 We then see that it su ioes to consider ag 3paq where p q are in our eld F Next we want to make the substitution 1 u v and we see u v U3 03gt 3W pgtltu v q Now v 0 if we have 31w p O and u v3 q O Ano l7 as we have two variables we might hope to nd so lutions to these two equations We solve the rst equation to get 1 pSu and sub stituting into the second we get 33u6 p3 33u3q 0 By some miracle this is a quadratic equation in ug Set ting y u3 we see 3312 33qy p3 0 So and 013 Galois Theory Now lets consider the Galois theory of an irreducible cubic polynomial We can assume that x 3pq and let K be the splitting eld of f over F and let a1 a2 a3 be the roots We then know that fltgt 3 p q 6 a1 a2 a3 expanding this equation we see that m d2 d3 0 a1on mm alozg p 061042043 q The rst of these show that a3 is in the eld Fltoz1 a2 So we have a Chain F C Foz1 C Foz1oz2 K 11 Lets call L Fltoz1 We then see that we break into two cases Either L K or L lt K Lets consider 3 2 The three roots of this polynomial are on 7 a2 CV2 a3 C2 where is the real square root and C eQMZ As E R we see that if we let F Q we are in the second case That a2 Z L Fltoz1 However7 if we let F then we see that am E FOz1 and we are in the rst case Now lets try and factor f in L We know that 1 lt36 a1gthltgt lt36 OM36 MW a3 12 where Maj m in L we know a1 E L and as it is a root7 f factors in L Now if Maj factors in L then we see that 062 Org E L and so L K So L lt K if and only if Maj is irreducible over L And in this case the degree La L 2 And siniilarly7 because f was assumed to be irreducible we have L F 3 HencewehaveKF3ifLKandKF6if LltK Now as CKF rnust perniute the roots 04 a2 a3 we nd that CKF g 83 And what is more CKF K F l by main theorem of Galois theory because K is a splitting eld 13 So ifL lt Kwe have K F 6 83 and so CKF 6 lfL Kthen K F 3 3so GKFZ3 Intermediary Fields Now lets determinethe intermediate elds in the case K F 6 in the case K F 3 there are none 83 has 3 conjugate subgroups of order 2 And these corre spond to the natural intermediary elds Fa1 Fa2 Fa3 They are isomorphic but not equal However7 we still have one more subgroup of 837 and that is Z This subgroup must correspond to a eld F C L C K Further7 as CKL Z3 we have K L 3 and therefore L F 2 So L is a quadratic extension of F and hence is gen erated by a square root Further as there is only one subgroup of 83 of order 3 we see that this is essentially the only element of F with a square root in K Lets call this square root 6 Further we know that L is the xed eld of 2 So an even permutation of the roots xes 6 while an odd perniutation Changes its sign We therefore see that 6 ltOZ1 OZQgtltOZQ OZ3gtltOZ3 01 Further 6 is not xed by every element of G K F and so 6 Z F But 62 is xed by every element of CKF andso626F Discriminant 15 De nition 0131 For any cubic polynomial SC d1a d2a d3 D ltOt1 a22d2 a32d3 Ot1gt2 is called the Discriminant lt is the number which is zero if and only if f has a multiple root ln this way it is analogous to the discriminant for a quadratic polynomial lt can be shown that the discriminant of f 3 pa q is D 4p3 27g2 Theorem 0132 The discriminant of an irreducible cubic polynomial gs E is a square in F if and only if the degree of the splitting eld is 5 Proof lf D is not a square in F then 6 Z F and hence F6 F 2 Since 6 E K7 K F is divisible by 2 and hence is 6 16 On the other hand if 6 E F then every element of the Galois group xes 6 Ano l7 since odd permutations of the roots Change the sign of 6 they are not in G and so G83HenoeKF3 D 02 TODO 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman October 17 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 Integral Domains and action Fields 011 Theorems Now what we are going to do is to consider ways in which a ring R can be embedded in a eld F Now we saw previously that if a E R is a zero divisor then we cant adjoin an inverse without killing some elements And so we cant embed a ring with zero divisors into a eld This turns out to be the only obstacle though Integral Domain De nition 0111 A ring R is an integral doniain if it has no zero divisors le O y 1 and if ab 0 then aOorbO l Cancellation Law l 2 Theorem 0112 IfR is an integral domain then it satis es the cancellation law Va 9 ca 75 O ab ac gt b c ProofWehaveab acab cOsob c0as ay O D Integral Domain of Polynomials Theorem 0113 Let R be an integral domain Then RM is an integral domain Proof D Finite Integral Domain is a Field Theorem 0114 An integral domain with nitely many elements is a eld Proof D l Field of Fractions l 3 Theorem 0115 Let R be an integral donidin Then there exists an embedding gb R gt F into d d eld F Proof The way we are going to show this is to mimic how the rational numbers are created from the integers l De nition of Field of Fractions l De nition 0116 A Fraction Will be a pair 6119 Where a b E R and b y 0 Two fractions Llbl7 lgb2 are called equivalent albl N lgb2 if Clle agbl First we want to check that N is an equivalence relations Re exivity and Symmetry are obvious so all that is left is transitivity Suppose albl N lgb2 and lgb2 N agbg We then have Clle 02191 and 02193 Q3192 HEDGE 01192133 4 agblbg agbgbl by the assumptions But because we are in an integral domain we can cancel the b1 and we get 12193 13192 and hence N is an equivalence relation The Field of Fractions of R is de ned to be the set of equivalence classes of fractions of R We de ne ad be bd We have to check that the axioms of a eld are satis abcd acbd 11 cd ed and that replacing ab cd by equivalent elements doesn7t change the answer7 but this is easy and we wont do it here D As an example consider the the case of the ring K where K is an integral domain Then in this case the eld of fractions is Kltgt f g are polynomials with g y O and is the equ 5 Extending a map onto eld of fractions Theorem 0117 Let R he an integral domain with eld offrdetions F and let go R gt K be any injectioe homomorphism from B into a eld K Then the rule Mab 6092509 de nes the unique extension ofgo to a homomorphism from F gt R Proof First we need to check that this extension is well de ned Since the denominator in ab is not allowed to be zero and since 90 is injective we have 900 y O for all ab 71 Hence god is invertible in K and hence goltagtgoltbgt is an element of K Next we need to check that equivalent fractions have the same image But albl N agbg H Clle agbl and hence 90a190b2 lt02gt90ltb1gt and Malbl 90ltalgt90ltblgt71 90lta2gtsl7ltb2gt71 agbg as required Hence I is a homomorphism And showing that I is unique is easy D 02 Maximal Ideals 021 De nitions Now lets look at surjective maps from a ring R to a eld F If 90 R gt F is such a map then we know by the rst isomorphism theorem that F Rkergp So in particular we can recover F go from R kerltgp So we must look at ideals M such that R M is a eld By the Correspondence Theorem 43 the ideals of R M correspond to the ideals of R which contain M So R M is a eld if R has exactly two ideals containing M le M R Maximal Ideal De nition 0211 An ideal M is maximal if M y R but M is not contained in any other ideals than M R Quotient by Maximal Ideal Lemma 0212 An ldedl M ls mammal if and only if RM ls a eld And the zero ldedl ls mammal if and only lfR ls a eld Proof This is immediate from the de nition of maximal ideal D 022 Examples l Maximal Ideals in Integers l Theorem 0221 The maximal ideals of the rtng Z of lntegers are the prlnolple ldedls generated by the prlme integers 8 Proof If a is an ideal and a g b an ideal7 then a divides b D l Maximal Ideals in Polynomial rings over C l Theorem 0222 The marimal ideals of the poly nomial ring CM are the principle ideals generated by the linear polynomials c a The ideal generated by c a is the kernel of the substitution homomorphism sa CM gt C sending r w fa So there is a bi jectioe correspondence between marimal ideals in CM and C Proof First notice that if M is a maximal ideal then we know that M is a principle ideal generated by a monic polynomial f of least degree Because CM is a PlD But since every complex polynomial has a root7 f is divisible by some cc a But then f is in the principle ideal cc a 9 and hence M g cc a so in fact M cc a as M is maximal Next we show that the kernel of the substitution homo morphism sa is generated by cc a To say that a polynomial f is in the kernel of sa is to say that a is a root of g or that St a divides 9 Thus cc a generates sa Since the image of sa is a eld this shows that cc a is a maximal ideal D l Hilbert s Nullstellensatz l Theorem 0223 Hilbertls Nullstellensatz The max imdl ideals of the polynomial ring C1cn are in a hijectioe correspondence with points of complex n diniensiondl space A point 6 o1angt in CC corresponds to the kernel of the substitution map sa C1n gt C which sends x w n The 10 Kernel Ma of this map is the ideal generated by the linear polynomials 1 o1n on Proof Let a E C and let Ma be the kernel of the sub stitution niap 5a Since 5a is surjective and C is a eld Ma is a maximal ideal Next let us verify that Ma is gen erated by the linear polynomials as asserted To do so we expand in powers of 1 a1 ojn on writing fltgt fltagtZ Cii azZ C jlti a gtltj ajgt i m The existence of such an expansion can be gotten by let ting a a u and substituting it into and then re placing every n with 13 a Now notice that every ternion the right except a is divisible by at least 1 of 36 azr So if f is in the kernel of sa7 ie a 0 then is in 11 the ideal which is generated by 1 a1 331 an Hence aZ39 l g generate Ma Next we have to prove that every maximal ideal is of the form aZ39 l g for some point a E C To do so lets let M be a maximal ideal and let K denote the eld 1331 chlM We consider the restrictions of the canonical map 7r 1331 ch gt K to the subring 7m Cm gt K Lemma 0224 The kernel of 7m ls either 0 or else ll ls a maximal ideal Proof Assume the kernel isnlt O and O f E kerlt7r Now since K isnt the zero ring kerlt7r isnt the Whole ring So f is non constant and hence divisible by a linear polynomial Say f 36 my 12 Then 7Tlt a 7rg O in K Since K is a eld either 7Tlt aZ39 O or 7rltggt 0 So either 33 azr E kerltmgt or 9362 E kerlt7r So by induc tion on the degree of f ke m contains a linear polyno mial D We are now going to show that kerlt7r isnt the zero ideal and hence M contains a linear polynomial of the form 36 61 Because i was arbitrary this will then show that M is contained in the kernel of a substitution map and hence must be equal to the kernel as it is a maximal idea Suppose ke m 0 Then 7m maps Chm isomorphi cally to its image Which is a subring of K Hence 7m can be extended to the eld of fractions of Chm Hence K contains a sub eld isomorphic to C367 the eld of ratio nal functions Now the monomials Xi 211 form a basis for 1331 as a vector space of C So in particular 1331 331 has a countable basis as a vector space over C And7 as K is a quotient of 1331 ch there is a countable family which spans K namely the residue of the X2 However we will show that there are uncountably many linearly independent elements of lt will follow that C33 cant be isomorphic to a sub eld of K We need the assumption that C is uncountable and then the following two lemmas 14 Lemma 0225 The uncountabe many rationalfune Mons 6 oz 1oz E C are linearly independent Proof A rational function fg de nes an actual function by evaluation at points of the complex plane Where g y O The rational function 33 004 has a pole at on which means that it takes on arbitrarily large values near or It is also bounded near any other point Now consider the following linear combination 77 2 Cf ZIZ OZZ39 z391 where m an are distinct complex numbers and say 61 y 0 Then the rst term is unbounded near or While the other functions are all bounded near it Hence the linear combination doesnlt de ne the zero function E 15 Lemma 0226 Let V be a vector space wblcb ls spanned by a countable famlly vb vn7 of vec tors Then every set L of llnearly lndependent vectors tn V ls nlte or covntably ln nlte Proof Let L be a linearly independent subset of V and let Vn be the span of the rst n vectors and let Ln L H V So Ln is a linearly independent subset of a nite dimensional vector space and hence Ln is nite More over7 L U Ln n And7 the union of countably many nite sets is nite or countable D D 03 TODO 0 Come up with A BUNCH of examples more than I can use so that l don7t run out of time 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman October 3 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 Rings Today we are going to introduce a new structure which you haven7t seen before They are called Rings and they are a generalization of the integers 011 De nitions l De nition of Ring l De nition 0111 We say ltR gtltO1gt is a Ring if oRisasetWithO1ER ogtltRgtltR gtR o R ltR O is an abelian group VyzERgtltygtltzgtltyxz VscERagtlt11gtlta Distributive Laws Vmyz Rcgtlt yzgtltygtlt z Vcyz Ryz xay xaszgtltx We say a ring is Commutative if Va 9 E R De nition of Subring De nition 0112 Let ltR gtltO1gt be a ring We say that S g R is a Subring of ltR gtltO1gt if 01 E S VyESyEX ESAmeES 012 Examples Integers 3 One of the most important examples of a commutative ring is the ring of integers Z Some properties of the ring of integers which are inter esting are 0 Z is commutative 0 Z has no subrings This is because if S g Z is a subring then it contains 0 1 and hence contains 1 1 1 n times for all 71 And similarly contains 1 1 and hence contains all the integers 1 Gaussian Integers Gaussian lntegers De nition 0121 We say that abi 611962 is the ring of Gaussian Integers The Gaussian integers are the points of a square lat tice in the complex plane DRAW THE COMPLEX PLANE WITH THE POINTS OF THE GAUSSIAN IN TEGERS Subring of C generated by oz Similar to the ring of Gaussian integers7 given any com plex number or E C we can form Za EZ39SnCZZ39OZZ WM E Z and n E 0 1 2 Notice that Za is the smallest subring of C containing oz We call Za the subring generated by oz ASK IF THEY WANT ME TO PROVE THA l AlgebraicTranscendental Number De nition 0122 We say that oz E C is algebraic if Elaoa1an E Zaoa1a1aga2anan 0 We say that oz is transcendental if it is not algebraic Some examples of algebraic numbers are 239 V 17 While some examples of transcendental numbers are 6 7r lf oz is transcendental then there is a bijection Ma W WC From the ring generated by oz to the ring of polynomials With coe icients in Z l Polynomial Rings l Ring of Polynomials over a Ring ln general we have that if R is a ring then RM Ez39gna z Li G R is also a ring Zero Ring De nition 0123 We de ne ltO 00gt to be the Zero Ring We then have that Theorem 0124 IfR ts a ring such that O 1 then R ts the zero ring Proof First note that Oh O for any element of a ring This is because 0a 0a O Oa Oh so subtracting 0a from both sides we see 0a 0 Now assume 0 1 in aringR Thenfor allaER7a1aOaO D Fields 7 Every eld is also a ring ln fact elds are special types of rings because every element other than 0 has a multiplica tive inverse There is a special name for such elements l De nition of Unit l De nition 0125 Let ltR gtlt01gt be a ring We say that u E R is a unit of R if Elu 1 E Ru gtlt u 1u 1 gtlt u1 So a eld is a commutative ring Where every element other than 0 is a unit Matrixes Real Matrixes Our last example of a ring are the real n X n matrixes This is an important example of a ring because it is an example of a non commutative ring 02 Homomorphisms and Ideals 021 De nitions l Ring Homomorphism l 8 De nition 0211 Let R X O 1 PH X 0 1 be rings A Homomorphism from R to R is a map gb R gt R such that 9M0 0397 9251 139 WWI agt39 bgt7 WWI agtgtlt39 ltbgt for all a b E R We say that gb is an isomorphism if it is a bijective ho momorphism lf there is an isomorphism between R R then we say R and R are isomorphic 022 Theorems The most important ring homomorphisms are those which are obtained by evaluating a polynomial at a value For example7 if a E R then we have lt25 2 WC W W is a homomorphism gb RM gt R This leads to the following theorem l Substitution Principle l Theorem 0221 Substitution Principle Let go R gt R be a ring homomorphism a Given an element oz E R there is a unique ho momorphism I RM gt R which agrees with the map 90 on constant polynomials and sends a w oz h More generally given elements a1 ozn E R there is a unique homomorphism I Rpm ajn gt R from the polynomial ring in n over R to R which agrees with go on constant polynomials and which sends ZIZ WOZ Proof First note that the proof of b is obtained by sim ply repeatedly applying a Note that if 13 exists then we must for each polynomial that En Scampi because we have determined Where I send coe icients of the polynomial as well as Where it sends 13 and it is a ring homomorphism Hence7 if 13 exists it must be unique To prove its existence lets take the above formulas as the de nition for I and show that it is in fact a homo morphism from RM to R 0 Note that 131 901 1 o lt is obviously true that I commutes With addition 0 Let f Eha g Zjbjscj Then 009 ltE 7ja bj jgt st m s lb aw ltE Spla gtaillzj9 D 11 Corollary 0222 Leta 31 Zt n y y1ym denote sets of variables Then there ts a unique mor phtsm so I Rhquot 14 gt thtllyl which ts the tdenttty on B U azLJ y Proof Note that B is a subring of BM and BM is a subring of BMW so B is a subring of BM Let go B gt BMW be the inclusion map The substitu tion principle tells us that there is a unique extension of go to a map from I Bt y gt BMW sending 6 to 6 and y to y While agreeing With 90 on B However7 we also know that there is an inclusion niap w BM gt B y And so by the substitution principle there must be a unique extension lJ BMW gt BLT y sending y to y and agreeing With w on 12 But then lKlD and IMF are both the identity on R and on 13 y so they must be the identity on their respective domains And hence inverse isomorphisms D We also have a relationship between polynomials as ele ments of RM and polynomials as functions via the above substitution maps This can be seen by the following the orem Theorem 0223 Let R denote the ring of contin uous real valued functions on R The map 902R1 Ztn gtR sending a polynomial to its associated polynomial func tion is an injectipe homomorphism Proof Immediate D Another set of important ring homomorphisms comes from the following l Integers are an Initial Object l 13 Theorem 0224 For every ring B there is exactly one homomorphism go Z gt R It is the map which takes 9001 131R1R n times ifri gt O and go ri gori Proof First we will show uniqueness Suppose there are two homomorphisms 9mm gt R Then we must have 901 13 Now lets assunie go and w agree on all numbers between 337 n g 13 g n Then 9001 1 90ltngt901gt ltngt lt1gt WW 1 And so by induction go psi on Z Now lets consider the map de ned by 14 This is the only possible homomorphism from Z gt B So all that is left is to show that it is an actual homomor phism Addition First notice that 901 1 901 901 Now assume that if m n 7 and both are positive then 90m n 90m Now let 5 t 7quot 1 Then we have 90ltStgt 90ltT1gt 90ltTgt90lt1gt 90lt8ltt 1gtgt90lt1gt 90lt8gt90ltt 1gt And so by induction we have 90 commutes with addition Multiplication First notice that 901 gtllt 7 901 gtllt 907quot 907quot Now assume that 90m gtllt n 90m gtllt We then have 90ltltm1gtmgt 90ltmnmgt 90ltmgtwltngtsoltmgt lt90ltmgt90lt1gtgtw 15 And so by induction and the previous result concerning addition we have 90 commutes With multiplication and hence is a homomorphism D 03 Ideals and Kernels 031 De nitions l Kernel of a Homomorphism l De nition 0311 Let gb R gt R be a ring homomorphism We the de ne the Kernel of gb to be TERngrO 1 De nition of Ideal l De nition 0312 An ldeal of a ring R is a set I such that o I O is a subgroup of R 0 olfaEIandTERthenTaEI Lemma 0313 I g R is an ideal if and only if I y Q and Va E I M E RgtE T a E I Proof Immediate D Lemma 0314 fa E R then the set Ta 7 E R is an ideal This ideal is denoted a or Ra or all It is called a Principle Ideal Proof Immediate D Lemma 0315 Let go R gt R be a homomor phism Then herltgogt is an ideal MAYBE Give examples of kernels in the polynomial ling7 or the ring l Zero Unit and Proper Ideals l De nition 0316 Let R be a ring Then the set 0 is an ideal of R called the zero ideal and denoted R is also an ideal called the unit ideal and denoted lt is the only idea which contains a unit An ideal is called a proper ideal if it is not 0 or Theorem 0317 a Let F be a eld Then the only ldedls ofF are 0 and 1 b If a ring R has emetly two ldedls then it ls a eld Proof a HOMEWORK The two facts we need to prove about R to show that it is a eld are 1 R is not the zero ring 2 Every non zero element of R has an inverse To show 1 observe that if R is the zero ring then it has only one element As such the zero ring has only one ideal and so our ring isnt the zero ring This means that O y 1 and so 0 y 1 and hence these must be the only two ideals of our ring Let a E R be a non zero elernent Then a y 0 be cause a E a Therefore a This implies that 1 E a and hence 1 ra for some 7 E R Hence a has an inverse and we are done D Corollary 0318 Let F be a eld and R a non zero rmg Then every homomorphism go F gt R is mjeetz Ue Proof We know that either kerltgpgt O or kerltgpgt 1 But we cant have ke gp 1 because then go is the constant function and in particular we have 900 901 and so R must be the zero ring Now 90m 901 Then 90a b O and so a b E kerltgpgt O and hence a b and go is injective D Theorem 0319 Every ideal m the ring Z is prin ciple Proof We know that every subgroup of the additive group Z is of the form HZ by previous theorenis from last semester Hence every ideal must be of this form and hence principle D Characteristic of a Ring 20 De nition 03110 The Characteristic of a ring R is the generator of the kernel of the unique homomorphism go Z gt R Notice that for instance the characteristic of Zp is p l Remainder Theorem for Polynomials Theorem 03111 Let R be a ring and let g he polynomials in Assume that the leading coe l cients off is a unit in R Then there are polynomials q r E RM such that gltgt fltgtqtrgt rltgt such that the degree ofr is less than the degree of f or r 0 Proof HOMEWORK Proposition 10319 D l Division by Roots of a Polynomial l 21 Corollary 03112 Let 936 be a nionie polynomial in RM and let o be an element off such that 9a 0 Then 6 oz divides g E RM Proof We know that 933 Oi 6 where 6 E R But then 9a 6 O D l Ideals of Polynomial Ring over a Field are Pri Theorem 03113 Let F be a eld Every ideal in the ring is principle Proof Let I be an ideal of F Since 0 is principle we can assume that I y Next lets Choose an non zero polynomial f E F of minimal degree Now we want to show that I First observe that we must have g I Now let let 936 E I We then have by the remainder theo 22 rern that gc faqa rc Where ra has degree less than that of f because otherwise rltccgt O and so gltgt E But then we have 936 E I and has degree less than that of gtlt Hence D l Greatest Common Divisor of Polynomials ove Corollary 03114 Let F be a eld and let g E which are both non zero Then there is a unique monic da E called the greatest common divisor off and 9 such that a d generates the ideal f g of generated by f and g h d divides f and g c Ifh is any divisor off and 9 then h divides d d There are polynomials p q E such that d pfqg Proof Let d be such that f g 04 TODO o Flush out the outline of math 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman November 2 1 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 Quintic Equations The motivation for Galois work was the problem of the insolubility of the quintic in terms of radials Able had shown a few years before that the general quintic had no solution but no one had been able to come up with a spe ci c polynomial with rational coef cients which couldnt be solve The rst thing we have to do though is to de ne what it means for a number to be expressible as by radicals l De nition By Radicals l De nition 0101 Let oz E C We say that oz is ex pressible by radicals over F if there is a tower of sub elds ofC FF0CF1Cquot39CFT where i oz E Fr ii For every j 1TFj Fj71lt jgtWhere E Fj1 Notice the similarity with the de nition of those elements which can be constructed with a ruler and a strait edge Except there the only radicals which were allowed were square roots Notice that the nth roots of unity Cm 62 are al lowed in expressions by radicals And7 if n T5 then 15 m and so at the cost of adding more steps we can assume all roots are pth roots for some prinie integers p Also note that there is a great deal of ambiguity in any expression by roots as there are 71 different 7b for every non zero b However as the notation is in and of itself very cumbersome we won7t be using the 7b often and so we won7t bother trying to make it more precise l One Root Solvable By Radicals All Are l Theorem 0102 Let gs be an irreducible poly nomial over a eld F If one root off in K can be expressed by radical so can any other Proof Suppose that one root oz can be expressed by rad icals7 say using the tower F F0 C C FT Choose a eld L which contains FT and which is a splitting eld of some polynomial of the form over F Then L is also the splitting eld of fg over F oz Let 0 be a root of f in another eld K and let L be a splitting eld 4 of fg over Fo Then we can extend the isomorphism FltOtgt gt Fo to an isomorphism go L gt L The tower of elds F 90F0 C 90F1 C C 90FT shows that o is expressible by radicals D l Galois Groups of Prime Roots of Unity are C Theorem 0103 Let p be a prime integer and let Cp emp For my sub eld F ofC the Galois group of FCp over F is a cyclic group Proof Let G be the Galois group of over F We de ne a map 1 G gt IF as follows Let 0 E G be an automorphism lt will carry C to another root of the polynomial mp4 a 17 say to Ci The exponent i is determined as an integer modulo p because C has mul tiplioitive order p Let 110 i 5 Now let 739 E G be such that 117 j7 ie Cl We then have we altcjgt aw lt27 so 110739 110 v739 Further vid 1 as C1 So7 since 1 commutes with multiplication and is not the zero map we have 1 G gt F3 However it is also injec tive as C generates K over F D l Galois De nition By Radicals l Theorem 0104 Let oz he a complex number which can he expressed by radicals over F Then a tower of elds F F0 C C FT K can he found which witness this and in addition iii For each j Fj is a Galois extension of F74 and the Galois group CFjFj1 is a cyclic group 6 Proof Suppose we have a tower F F0 C C FT in which FT 1751 r As we have remarked we may assume that 5 E F74 for some prime pj Let Cp 2mpi be a pjth root of 1 We can then form a new chain of elds adjoining Cpl Cm 61 r in that order We then know that each of these extensions is Galois the rst by the previous theorem and the later by what we proved about Kummer extensions D Lets consider the Galois group of a product of polynomi als over F Let K be a splitting eld of fg Then K contains a splitting eld of K of f and F of 9 So we have the following diagram Kl l Splitting Fields of fIgJ l Theorem 0105 With the above notation let G GKF and H GF F and Q GK F i G and H are quotients on ii Q is isomorphic to a subgroup of the product G X H Proof The rst assertion follows from the fact that K and F are intermediate elds which are Galois extensions of F Let us denote the canonical homomorphism Q gt GQ gt H by subscripts o w of and o w 09 Then of describes the way that o operates on the roots of f and 09 describes the way it operates on the roots of 9 We then get a map Q gt G X H by a W ofog lf of 09 are both the identity then a operates trivially on the roots of fg and hence a 1 This shows that Q gt G X H is injective and so it is isomorphic to a subgroup of G X H D l De nition Simple Group l De nition 0106 Recall that a group is called Simple if it is not the trivial group and if it contains no proper normal subgroups l Simple and Abelian Groups and Solvability b Theorem 0107 Let f be a polyuomlal ouerF whose Galols group G ls a slmple hon Abellah group Let F be a Galols errtenslou ofF wlth Abellah Galols group Let K be a spllttlhg eld off ouer F Then the Ga lols group of GltK F ls lsomorphle to G Proof We rst reduce ourselves to the case that F F l is a prime number To do this we suppose that the lemma has been proved in that case and we choose a cyclic quo tient group H of CF F of prime order Such a quo 9 tient exists because CF F is abelian This quotient determines an intermediate eld F1 C F which is a Ga lois extension of F and such that CFlF H Let K1 be the splitting eld of f over F1 Then since F1 F is a prime CKlFl G So we may replace F by F1 and K by K1 Induction on F F will complete the proof So we may assume F F p is prime and that H GltF F is a cyclic group of order p The splitting eld K will contain a splitting eld of f over F 7 call it K We are then in the situation of the previous theo rem So the Galois group Q of K over F is a subgroup of G X H and it maps surjectively to G lt follows that G divides Q and that Q divides G gtlt H pG lf G Q then counting degrees shows K K ln this case K contains the Galois extension F and hence H is 10 a quotient of G Since G is a nonabelian simple group this is impossible The only remaining possibility is that Q G X H Applying the main theory to the chain of eld F C F C K we conclude that GK F G as required D This proposition is key because it tells us that if the Ga lois group of f is a simple non Abelian group then we will not make any progress towards solving for its roots if we replace F by an abelian extension l S5 Not Solvable By Radicals l Theorem 0108 The roots of a quintie polynomial gs whose Galois group is S5 or A5 can not be ex pressed by radicals over F Proof Let K be a splitting eld of lf G S5 then the discriminant of f is not a square in F ln this case we replace F by F 6 where 6 is a square root of the 11 discriminant in K The Galois group of GltKF6 is A5 lt is obviously enough to show that the roots cant be expressed by radicals over F 6 and so this reduces the case of S5 to that of A5 Suppose that the Galois group 0 f is A5 but that some root oz of f is expressible by radicals over F Say that or E FT where F F0 C C FT and each extension in the chain is Galois with cyclic Galois group Now since the Galois group of f over F is a simple group ie A5 we see by the induction and the previous theorem that the Galois group of f over E is A5 to for all 239 However7 since or E FT we see that f is not irreducible over FT So in particular the Galois group of f over FT 12 will not act transitively on the ve roots of f in the split ting eld Hence the Galois group cant be the alternating group A5 gtlt D Now we will come up with a speci c quintic polyno niial whose Galois group is S5 The fact that G acts transitively on the roots a1oz2oz3oz4oz5 and that 5 is prime greatly limits the possible Galois extensions l S5 and transpose l Lemma 0109 IfG contains a transpose then G S5 Proof By a transpose we mean a permutation which in terchanges two indices This isnt hard to show and we will leave it for you to try at home D l Condition for Not Solvable By Radicals l Corollary 01010 Suppose our trreduetble quthtte has roots a1a2a3a4a5 and K ts the spltttthg eld If Fa1 Otg Otg lt K then CKF S5 Proof For let F Fa1 Otg Otg The only nontrivial permutation xing a1 Otg Otg is the transposition 45 If F y K this permutation must be in CKF Thus CK F contains a transposition and is S5 D l 3 Real Roots Implies Quinitc Is Not Solvable Corollary 01011 Let fltgt be an trreduetble quth tte polyuomtdl ouer Q wtth erdetly 5 real roots Then the Gdlots group ts the symmetrte group and hence the roots can t be expressed by rddtedls Proof Call the real roots a1 Otg Otg ThenF Qlta1 Otg Ot3gt Q R But since a4 a5 aren7t real we see that F lt K Where K is the splitting eld of f over Hence CKF S5 and f has no roots which can be expressed by radicals D Example Example lt2 42 4 5 1633 has exactly 3 real roots And we can add a small constant to it Without changing the number of real roots So 365 1633 2 has exactly 3 roots and is irreducible 02 What To Do Next 0 Show that V V o Latices and Boolean Algebras 0 Valuation Rings and Completions 0 Modules 0 More Advanced Galois Theory ln nite Galois Theory o Algebras 03 TODO 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman September 19 2006 1 01 TALK SLOWLY AND WRITE NEATLYH 02 Introduce Bilinear Forms 021 De nitions Now that we have dealt with the structure theorem for nitely generated abelian groups it is time to go back to vector spaces an study another element of theIn7 bilinear forrns Bilinear forms are meant to be a generalization of the dot product on R So before we continue recall 1 Dot Product De nition 1 De nition 0211 Let XY E R then we de ne X Y XtY 1311 2y2 nyn The important features of the dot product are Bilinearity X1 X2gt39YgtltX139YgtltX239Ygt ltXltMY2gtgtltXgtltXgt cXYcXYXCY Symmetry X Y Y X Positivity X y O gt X X gt 0 Notice that bilinearity says that if we x one element of the dot product then Y R gt R is a linear transformation And it is this property which we will focus on rst l Bilinear Form De nition l De nition 0212 Let V be a vector space over F We de ne a bilinear form to be a function f V X V gt F such that Vv1v2w E Vfv1v2w fv1w fltv2w Vvw1w2 E Vfvw1 7112 fvw1 fvw2 Vuw 6 V6 E FgtfCvwgt CfWUJ ucwl We will often use the notation 11111 for u w 3 Symmetric Bilinear Form De nition De nition 0213 We say a bilinear form lt gt is Symmetric if VUWXMW ltw70gt SkewSymmetric Bilinear Form De nition De nition 0214 We say a bilinear form lt gt is Skew Symmetric if Vvltv v 0 022 Theorems Explanation of SkewSymmetric De nition Lemma 0221 Let V be a vector space over a eld F of characteristic 75 2 Let gt be a btltnear form on V Then gt 13 skew symmetric if and only if V11111 E Vlt11111gt lt11111gt Proof 3 Well we then know that Olt1111111111gt lt1111gtlt111111gtlt11111gtlt11111gt lt11111gtlt11111gt So we have 0 11111 11111 and hence 11111 ltwgtvgt 3 Well we then know that 1111 lt1111gt and so 2lt1111gt 0 So either 2 O or 1111 0 But we are assuming that the characteristic isnlt 2 and so we must have 1111 0 It is for this last part that we need the characteristic isn7t 2 In the case of characteristic 2 we see that the condition that 11111 lt11111gt is the same as saying that ltgt is symmetric because a a for all a But7 as 5 we will see there are in general skew symmetric matrixes over elds of characteristic 2 which are not symmetric Hence the above de nition is the right one D 03 Matrix Representation of Bilinear Forms 031 Theorems The most common examples of bilinear forms are those which act on the space F n of column vectors as follows Let A be an n X 71 matrix Then X Y XtAY noticethat this isa1gtlt1 matrix l WOI k out Why on the b0 The rst thing we need to check is that this is in fact a bilinear form l Matrix Form is Bilinear l 6 Lemma 0311 Let V on n dtmenstonal vector space over F and let XY E V be represented as column vectors relative to some basis Further let A be an n X n matrtr tn F Then X Y XtAY ts a btltn ear form Proof We need to Check the following lt X1X2 Y gt X1X2tAY XHXQAY XfAYHQ ltXY1Y2gt XtAY1Y2XtAY1XtAi2 ltXY1gtlt ltCX Y cXtAY XtAeY eltX Y X CYgt And so in fact lt gt is a linear transform D 7 Now7 given a nite dimension vector space we want to show that any given bilinear form is of the above form l Bilinear forms have matrixes l Lemma 0312 Let gt be a bilinear form on V a nite dimensional vector space and lets let B b1 bn be a basis for V Then there is a matrir A such that X Y XtAY where X Y are consid ered column vectors relative to the basis B Proof We want to show that there is a matrix A such that W in ii E 71 Q Fgtlt1gt 39 quot gtngttAlty1gt 39 quot 7yngt ltEi nibia Emmy Well we know that any matrix we come up with will cor respond to a bilinear form So in particular if we can come up with a matrix which agrees with our bilinear form on the basis elements then the bilinear form associ ated to the matrix must be the one we want Speci cally what we need is A am where am 11 11739 Then by bilinearity X 1tAY X Y for all vec tors X Y D De nition 0313 We say that A 1 17gt is the Matrix Associated to the Bilinear forni lt gt relative to the basis 91 bn 04 Change of Base One of the most important questions regarding these Ina trixes is what happens when we change bases This leads us to the following theorem 041 Theorem l Change of Base l Theorem 0411 Let A be the matrix associated to a hilihear form gt with respect to a basis Then the 9 matrtres whteh represent the same form wtth respect to dt ereht hasts are those of the form QAQt for some Q E GLnF Proof Let P be the element of GLnF which represents the linear transformation which changes the base So we have X PX Y PY and ltXYgt ltXYgt as X X and Y Y are just different representation of the same vectors We then know that X Y XtAY P 1XtAP 1Y XtP 1tAP 1Y But we also know that X Y ltX Ygt XtAY for A the matrix representing the bilinear for relative to 10 the new basis Hence letting Q P711t we must have A vot n 05 Dot Product l Dot product matrix Now lets consider What happens to the dot product if we change basis Recall that X Y X 1tY And so we have that the matrix associated to the stan dard dot product is just the identity matrix Orthogonal De nition 0502 Recall that a matrix is said to be orthogonal if PtP I or P 1 Pt Lemma 0503 If you change base relattoe to an orthogonal change of base then the dot product ts pre served 11 Proof Let P be the orthogonal change of base Well I is the matrix associated with the identity so by previous theorems this means that the matrix associated with the dot product under the new basis is P 1tIP 1 PM PPt I So changing the basis by an orthogonal matrix preserves the dot product D Similarly we have Lemma 0504 The matrtres which represent the dot product are those of the form PPt for P E GEAR Proof By previous theorem D Now while this is nice7 it doesnt tell us a whole lot as we dont know what matrixes of the form PP1t should look like So we will have to use other properties of the dot product to help us out Recall the three conditions on the dot product which were important First off was Bilinearity But this isnt a help ful as we know that X 1tAY is bilinear for every A The next is symmetry This is in fact useful l De nition Symmetric Matrix l De nition 0505 We say a matrix is symmetric if A At Lemma 0506 A bilinear form is symmetric if and only if the matrir associated to it is symmetric Proof Symmetry is equivalent to ltXYgt XtAY YtAX Y X but we have YtAX YlAXl XtAtY because the transpose of a 1 X 1 matrix is itself Hence being a symmetric bilinear for is equivalent to VX YXtAY XtAtY and this is equivalent to A At D The third condition is that X X gt 0 if X 7 O Posi tivity l Positive De nite l De nition 0507 We call a bilinear form ltgt on V Positive De nite if V11 E Vv 75 Oltvvgt gt O 06 Orthogonal 061 De nitions l Orthonormal basis l De nition 0611 Given a bilinear form ltgt on a vector space V we say that two vectors 11711 E V are orthogonal o J in if v w 0 Let B v1 m be a basis for V We then say that B is an orthonormal basis if V1 150 07gt 0 Vixm vi 1 062 Theorems Lemma 0621 IfB is an orthonormal basis for V with respect to gt then the matrir associated to gt relative to B is the identity Proof Immediate D Now we are going to show that for any positive de nite bilinear form an orthonormal basis exists l Orthonormal basis always exist for symmetric 15 Theorem 0622 Let gt be a positive de nite sym metric bilinear form on a nite dimensional vector space V Then there is an orthonormal basis for V Proof The method we are going to use is called the Gram Schmidt procedure We are going to start with a basis B v1 vn Step 1 The rst step will be to normalize U1 Now we know that 01110111 02111 v1 But7 because we know that that lt gt is positive o le nite7 111111 gt O and so we know 111111 is a real number and hence if we let 1111 V ltU1gtv1gtvl then we see that wl w1gt 1 Step 2a Now we want to look for a linear combination of 027111 which is orthogonal to wl The value is w v2 v2w1gtw1 because ltww1gt 112101 v2w1gtltw1w1gt 0 Step 2b Then normalize w and call that vector LU2 Further w1w2v3 vn is a basis for V Step k a Suppose we have de ned orthonormal vectors wl wk1 and that wl wk1vk vn is a basis Then we want to look for a linear combination of 11k wl wk1 which is orthogonal to 711239 for all i lt k The value is w vk vkw1gtw1 vkwk1gtwk1 because mm vkw gt ltvkw1gtltw1w gt ltvkwk71gtltwklw gt But whim 0 iii y j and wjwjgt 1 so we have 111102 ohm vkw gtltw w gt 0 Step kb Then normalize w and call that vector wk Further W E Spamwl wkvk1 vn and soltw1wkvk1 is a basis Hence after iterating this process 71 times we see that wl wn is an orthonormal basis for V D We then have the following theorem 18 Theorem 0623 The followlng are equloalent for a real n X n matrlr I A represents the dot product 2 There ls an lnoertlble matrlr P E GLnGR such that A PtP 5 A ls symmetrlc and posltloe de nlte Proof We have already shown that 1 and 2 are equiv alent Further7 the fact that 1 gt 3 is by virtue of the fact that the dot product satis es positivity and symmetry So all that is left is to show that 3 gt Well the rst thing to notice is that if A is positive de nite then so is the form X Y X 1tAY So in particular there is an orthonormal basis B With respect to lt Now we know then that with respect to the basis B the matrix associated to lt gt is I because B is orthonormal But at the same time we know that if P is the matrix associated to the change of base from B to the standard basis of R then A PtA P PtP and so A satis es D 07 Geometry Associated To A Positive Form 071 De nitions Suppose we have a bilinear form ltgt on a real vector space Then7 it is possible to de ne the length of a vector as follows l Euclidian space de nition l De nition 0711 Let U E V and let ltgt be a positive de nite bilinear form Then we can de ne lvl ltltvavgtgt We often call a real vector space with a length a Euclidian space 072 Theorems lt isn7t hard to see the following in equalities In equalities Lemma 0721 Let gt be a positive de nite bilineai form on a real vector space V Then we have 39ehwanz Inequality ltvwgt v w iiangle Inequality lv wl v w Proof Immediate MAYBE MAKE A HOMEWORK D Now given a subspace of W we have that 21 Lemma 0722 Let V be a vector space and a bilinear form on V Then if W g V is a subspace then gt restricts to a bilinear form on lt Further if gt is positive de nite or symmetric on V then gt is positive de nite or symmetric on W Proof Immediate D Now this will then allow us to de ne a relative angle be tween two vectors Suppose we have V is a vectors space with a positive de nite bilinear form lt gt and let U in E V Then we can restrict ltgt to Spanvw Now if 11711 are linearly dependent we say that the angle is 0 Otherwise we can de ne the angel between the two vec tors is determined by 11711 v w 0056 FIGURE OUT WHY 08 TODO 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman September 26 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 Hermitian Forms Now that we have de ned what a Euclidean space is we want to consider the analog for complex vector spaces Speci cally we want to look at what the dot product should be if our space is a complex vector space instead of a real one Well if we have 331 ngt E C then its length as an element of R2 is a b agbg 193 1cnc n Where E is the complex conjugation of as l Standard Hermitian Product l De nition 0101 This suggests that the right gener 2 alization for dot product on complex vector spaces is the Standard Herrnitian product Lemma 0102 Now this product has the following nice properties 0 X Y agrees with the dot product on the reals o VX y OltXXgt is a positive real Le second variable ltXcYgt cltXYgt and ltXY1 Y2 ltXY1gt ltXgtY2gt u the rst variable cX Y EltX Y and X1X2Ygt X1Ygt ltX27Ygt mitiah Symmetry X Y Y X Proof Immediate D Now we can de ne the generalization of a bilinear form to complex vector spaces l Hermitian Form l 3 De nition 0103 We de ne a Hermitian Form on a complex vector space V to be an function ltgt V X V gt V such that ltgt satis es 0 Linearity in the second variable 0 Conjugate linearity in the rst variable 0 Hermitian symmetry Just as in the real case we can de ne l Matrix Associated t0 De nition 0104 Let ltgt be a Hermitian form on a complex nite dimensional vector space V Further let B 111 vn be a basis for V We then de ne the matrix A associated to the form re as A aw Where aij 15117 Just as in the case of bilinear forms we have 4 Lemma 0105 Let gt be a Hermitian form on a complex nite dimensional vector space V Further let B 111 vn be a basis forl and let vw E V be such that v BX and w BY then 11711 YtAY Proof This is exactly the same as in the case of bilinear forms D There is one difference though between the Herrnitian form and the bilinear form case While in a bilinear form any matrix gave rise to a bilinear forni7 in the case of a Herrnitian form we know that an WW ltUjavigt W which leads us to the following de nition l Adjoint Of a matri De nition 0106 Let A be a complex niatrix then we de ne the M of A to be Aquot Kt Lemma 0107 We then have the following proper ttes 0f adjotnts AB AB AB BA 1471 Ailyk A A Proof Immediate Ask the class if they want to see any of these D De nition 0108 A matrix is self adjoth or Hermitian if A A 6 Theorem 0109 Let gt be a Hermitian form on a complex nite dimensional vector space V Let A be the matrir associated to gt relative to a basis Then A is a Hermitian Matrix and X Y X AY Further ifA is Hermitian then XAY is a Hermitian form Proof The proof is essentially identical to the one for real symmetric bilinear forms The only thing which is left to check is Hermitian symmetry and I Will leave it to you to check that D Lemma 01010 The real hermitian matrires are the real symmetric matrires Proof Immediate D 02 Change of Base Change of Base for Hermitian Form Theorem 02011 Let A be the matrtr associated to a Hermtttah form wtth respect to a hasts B Then the matrtres whteh represent the same form wtth respect to dt ereht hasts are those of the form A QAQquot for some trwertthle matrtr Q E GLnC Proof Let P E GLnC be the matrix from B to B Let X Y represent 11711 with respect to B and X Y represent 11711 with respect to B Further let A be the matrix associated with lt gt relative to the basis B Then we have PX X PYY and XAY 11111 X A Y PXA PY XPA PY and so we have A PA P or A QAQ Where Q 1371 D l Unitary Matrices l De nition 02012 For Hermitian forms7 the analog of orthogonal matrixes are called Unitary Matrixes A matrix is unitary if PP I or 13 131 For example i 001 1N 1 Z39 is unitary Lemma 02013 The unitary matrixes for a group Un PPP I 9 Proof Immediate D Change of base and Standard Hermitian Prod Corollary 02014 A change of base preserves the standard hermtttan product te XY X Y if and only if the change of base matrta ts nnttary Proof Immediate D 03 Carry over from bilinear forms Orthogonal Positive De nite de nitions De nition 03015 Let ltgt be a Hermitian form We can say that two vectors 11711 are orthogonal if and only if v w O 10 Similarly we can de ne to be positive de nite if and only if 1111 is a positive real number if v y O l Hermitian Space l De nition 03016 We say a complex vector space V with a positive de nite herniitian form is a Hermitian Space l Carry over from bilinear forms l Theorem 03017 We then have Sylvesters theo rem carries over to the ease of hermiticm forms A Hermiticm form has a orthonormal basis if and only if it is positive de nitie If W g V and gt restricted to W is non degenerate then v W e WL 11 Proof This proof is identical to the one for real vector spaces D 04 Spectral Theorem Notice that now we have two seemingly different inter pretation of an n X 71 matrix Such a matrix can be associated to o A bilinear form ltgt V X V gt F o A linear transformation T V gt V Now we are going to combine these two uses of ma trixes 041 Theorems Recall from last semester l Linear Operator change of base l Theorem 0411 Let T V a V be a linear operator on V arid let M be the matricc associated to it with respect to a basis B 117 7on Now let P E GLnF be the matricc associated with a change of base from B 12 to B w17wn Then the matrir associated to T with respect tot he basis B is M PMP l Proof Let i Sali Ebiwi So in particular we know that To expressed with basis B M i expressed with basis B P o Mi expressed with basis B P o MP 1i expressed with basis B Hermitian Operators Unitary Operators Theorem 0412 Let T V a V be a linear operator on a hermitian space V Further let M be the matrip associated to T with respect to an orthonormal basis Then a The matricc M is hermitian if and only if lto7Twgt ltTowgt for all raw 6 V In this case we say T is a Hermitian Operator b The matrip M is unitary if and only if ltowgt ltTo7Twgt for allow E V In this case we say T is a Unitary Operator Proof Let XY be the coordinate vectors so that i BXw BY and hence lto7wgt XiY and Ti BMX Part a We have lto7Twgt XMY and ltTo7wgt MXY XMY In particular we therefore have V01 E V lto7Twgt ltTo7wgt if and only if M M or M is herrnitian Part b Similarly we have ltTo7Twgt MXMY XMMY So7 in particular we therefore have V01 E VltToTwgt lto7wgt if and only if MM I7 or M is unitary 1 l Spectral Theorem Theorem 0413 Spectral Theorem a Let T be a hermitian operator on a hermitian vector space V Then there is an orthonormal basis for V consisting of eigenvectors of T b Matrir form Let M be a hermitian matrim There is a unitary matria P such that PMP is a real diagonal matrim Proof WE WILL PROVE THIS NEXT TIME D 05 TODO 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman October 19 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 Examples of Fields We are now ready to begin the study of elds which will eventually lead us to Galois theory Back when l took this class this was by far the coolest part of it ln the study of elds we will often consider pairs of eld F g K However unlike the case of groups where we often start with a group G and look at subgroups7 in the case of elds we will be more interested in starting with a eld F and looking as different extensions of F to elds K l Extension Field l De nition 0101 Let F be a eld If F C K then we say K is a extension eld of F The three most important classes of elds we will en counter are the following 2 1 Number Fields A number eld K is any sub eld of C As any sub eld of C contains 1 it must also con tain The number elds most commonly studied are those were every element is algebraic 2 Finite Fields A eld having only nitely many elements If K is nite then the kernel of the unique homo morphism Z gt K is a prime ideal since Z is in nite So in particular K contains a sub eld isomorphic to Zp 7p for some prime p And hence we can consider K as an extension of 731 3 Function Fields Certain extensions of the eld Speci cally7 suppose we have an irreducible polyno mial in f E CLT y which is not a polynomial in 1 alone Since it is irreducible in CM y we know that it is irreducible in Cy and hence7 by the gener alized Guassls Lemma we have that we have that f is also irreducible in F C33 the fraction eld of ln particular this means that the idea E F y is maximal and hence is a eld 02 Algebraic and Transcendental Elements Similarly to the case of the complex numbers we de ne AlgebraicTranscendental Extensions De nition 0202 Let F be a eld And let or E K 4 such that K 2 F We say that oz is algebraic over F if there is a polynomial over F which is satis ed by oz le if oz an1oz 1alozoz0 for some do 111 E F We say that oz is transcendental over F if it is not al gebraic over F We can think of the two possibilities in terms of the eval uation homomorphism Speci cally Lemma 0203 Let F C K be elds wlth oz E K Then if 90a Fltvl gt K 1 W fltagt we have oz ls transcendental if and only if 90 ls injec tloe 07 more speci cally if the kernel ofgo ls O Proo f lmmediate D Similarly we have l Irreducible Polynomial l De nition 0204 Let F C K be elds with or E K Further let waiFl Sl gtKfgtwflt04gt We then know that kerltgpa is principle as F is a prin ciple ideal domain So in particular it is generated by a single element E F But because K is a eld we must have is irre ducible because otherwise K would have a zero divisor Hence is the only irreducible polynomial in because every element of the ideal is a multiple of and we call fay the lrreducible Polynomial for oz over F lt is important to note that the base eld we are working over is crucial when determining the irreducible polyno 6 mial of an element Or for that matter even if a polyno mial is irreducible For example Let oz Then the irreducible polynomial for oz over Q is 4 1 However7 over the irreducible polyno mial is 2 1 And What is more7 over 4 1 is not irreducible as 34 1 62 t2 FW De nition 0205 Let Fltozgt be the smallest eld con taining both oz and F Similarly let Fa1 an be the smallest eld containing a1 an and F Connection between FM and 1704 l Lemma 0206 Recall that FM ts the ring Eanan an E F 7 and ls the smallest rlng contalnlng hoth F and oz We then have Fa ls lsomorphlc to the eld offractlons of F oz In partlcnlar we have that lf oz ls transcendental then gt Fa ls an lsomorphlsm and hence Fa ls lsomorphlc to the eld of ratlonal functlons Proof Immediate D Notice that this means that if oz and 6 are both tran scendental over F then F oz F For example this means that 3 e which is not at all obvious at rst glance However7 if oz and 6 are algebraic the case is very different l Isomorphism 0f Equivalent Extensions Theorem 0207 a Suppose that oz ls algebralc over 8 F and let gs be its irreducible polynomial over F The map gt F0i is an isomorphism and F0i is a eld Thus F0i Fm b More generally let a1 an be algebraic elements of a eld extension K ofF Then Fa1 an Fa1 ozn Proof Let go be the map which takes to u Since generates herltgogt7 we know that is isomor phic to the image of go which is F Since f is irreducible we know that is a maximal ideal and hence is a eld Since F oz is the smallest eld containing F oz we must have Fa The second part follows from the rst and is let as an exercise D l Field Extensions as Vector Spaces l 9 Theorem 0208 Let oz be an algebraic ouer F and let gs be its irreducible polynomial Suppose fltgt n71 has degree n Then 1 oz oz is a basis for F0i as a vector space over F Proof This is a special case of the same theorem for rings which we have already seen D The following is one of the most important results con cerning eld extensions l Isomorphism 0f Equivalent Extensions Theorem 0209 Let oz E K and 6 E L be alge braic elements of two extensions of F There is an isomorphism of elds a F0i gt F6 which is the identity on F and which sends oz w 6 if and only i the irreducible polynomials for oz and 6 over F are equal Proof Assume f is the irreducible polynomial for or and 6 over F We therefore get two isomorphisms so I FlSUlf gt Flal and w I FlSUlf gt Fl l The map a W4 is the required isomorphism sending or to 6 and preserving F Conversely7 if such an isomorphism a which sends or to 6 and is the identity on F exists7 then ag 0 if and only if f O and so or 6 have the same irreducible polynomial D l Isomorphism of Field Extensions De nition 02010 Let K K be eld extensions of F A11 isomorphism go K gt K which restricts to the identity on F is called an isomorphism of eld exte of an F isomorphism l Field Extension Automorphisms and Roots of Theorem 02011 Let go K gt K be an isomor phism of eld extensions ofF and let gs be a poly nomial with eoe leients in F Let oz be a root off in K and let 0 90m be its image in K Then 0 is also a root off Proof Say ann39 39 39aiao The Mail ai and 900i oz Since 90 is a homomorphism7 we can expand as follows 0 900 90ltflt0zgtgt Mama W1 00 12 wltangt9004gt 39 3990ltai900zgt90ao ana 39 a10ao Hence 0 is a root of D 03 TODO o Flush out the outline of math 0 Come up with A BUNCH of examples more than I can use so that I don7t run out of time 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman November 14 2006 1 l TALK SLOWLY AND WRITE NEATLY l 01 Proof of the Main Theorem Let f be a monic polynomial of degree n with coef cients in a eld F We recall that a splitting eld of E is a eld of the form K Foz1ozn such that lt33 a1 lt33 an in We now want to show that any two splitting elds of a given poly nomial f are isomorphic This follows from the fact that a eld extension of the form F oz is determined by the irreducible polynomial for oz over F and from some bookkeeping The bookkeeping is notationally a little confusing but not hard l Extending Isomorphism l De nition 0101 Let go F gt F be an isomor phism Then 90 extends to an isomorphism F gt FM by Z and w 261736 Where Tn won We denote by the image of under this map Notice that as 90 is an isomorphism7 f is irreducible if and only if is irreducible Lemma 0102 Let gs E be an irreducible polynomial Let o is a root of f in an extension eld K ofF dnd leta be a root of ce in an extension eld F of 7 Then there is a unique isomorphism 9011Flt0zgt gtFltgt which restricts to go on a the sub eld F and which sends oz to a Proof We know that F 0i is isomorphic to the quotient F f and similarly 75 is isomorphic to FM 3 The rings F and FM are isomorphic as we saw7 and since f and 7 correspond under this isomorphism7 so do the ideals and Hence the rings F and FM are isomorphic7 and combining these isomor phisms yields the isomorphism 901 D l Isomorphism of Splitting Fields l Theorem 0103 Let go F gt F be an isomorphism of elds Let fltgt E be nonconstant and let be the corresponding polynomial in Let K and K be the splitting elds for gs and Then there is an isomorphism 1b K gt K which restricts to go on F Proof lf factors into linear factors over F 7 then also factors into linear factors ln the case K F andKF7sogo1b 4 Now assume that f does not split completely Choose an irreducible factor 936 of f of degree gt 1 The corre sponding polynomial mot will make an irreducible factor of Let oz be a root of g in K and write F1 FltOtgt Make a similar choice for a and let E 7a in F Then by the previous lemma we can extend go to an iso morphism 901 F1 gt E which sends oz w 5 Being a splitting eld for f over F 7 K is also a splitting eld of f over the larger eld F1 and similarly F is a splitting eld for 7 over Fl Therefore we may replace F F 90 by 17171 90 and proceed by induction D Corollary 0104 Any two splitting elds of fltgt E are isomorphic Proof Set F F and go id in the previous theorem D l Splitting Fields are Galois l 5 Theorem 0105 Let K be the spllttlng eld of a polynomlal gs E Then K ls a Galols exten slon of F That ls K F l Number of Isomorphisms l Lemma 0106 Wlth the notatlon of the preolons lemma the number of lsomorphlsms w K gt K extendlng go ls equal to K F Proof We proceed as in the proof of the previous theo rem Choose an irreducible factor 9a of f and one of the roots or of 9a in K Let F1 FltOtgt Any iso morphism w K gt K extending 90 Will send F1 to some sub eld E of K This eld K Will have the form 75 Where a Moo is a root of mot in K Conversely to extend go to 1b we may start by Choosing any root 5 of mot in K We then extend go to a map 901 3 F1 gt 71 FW by setting 901m a We use induction on K Since K F1 lt K F the induction hypothesis tells us that for this particular choice of 901 there are K F1 extensions of 901 to an iso morphism w K gt K On the other hand7 g has distinct roots in K because 9 y are irreducible So the number of choices for a is the degree of g Which is F 1 F So there are F1 F choices for the isornorphisni 901 This gives us a total of K F1F1 F K F extensions of go to w K gt K D Galois Group of Polynomial De nition 0107 Since any two splitting elds K of E are isornorphic7 the Galois group CKF depends7 up to isomorphism7 only on f lt is often re ferred to as the Galois group of the polynomial over F l Collection of Equivalences l Corollary 0108 Let KF he a nite eld exten sion the following are equiualent i K is a Galois extension of F ii K is the splitting eld of an irreducible polynomial g 6 FM ii K is the splitting eld of a polynomial x E iii F is the xed eld for the action of the Galois group GKF on K iii F is the xed eld for an action of a nite group of automorphisms of K l Main Theorem of Galois Theory l Theorem 0109 Main Theorem Let K be a Galois extension of a eld F and let G GKF be its Gdlois group the function H w KH is d hijectiue nidp 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 if H GKL then KzLH dndLFGH Proof Let K F be a Galois extension We have to show that the maps L w GKL and H w KH are inverse functions between sets of intermediate elds and the set of subgroups of G CKF To do so we verify the compositions of the two maps in either direc tion are the identity Let L be an intermediate eld The corresponding sub group of G is H CKL By ole nition7 H aots triv ially on L7 so L g KH On the other hanol7 K is a Galois extension of L by previous results Hence K L H However also by previous results we have H K K H and so L K H ln the other olireotion7 suppose that we start With a sub group H C G and let L KH Then H C CKL But K KH K L GKL Therefore H GKL We therefore have that these two maps are inverses Fur ther sinoe K is a Galois extension of L K H 7 K L HandLFGH D There are a few properties of the Main Theoreni which 10 are worth discussing First notice that the correspon dence between elds and groups is order reversing That is if LL are two elds and H CKL and H CKL then L C L if and only ifH C H Images Of Intern Theorem 01010 Let KF be a Galois extension and let L be an intermediate eld Let H CKL be the corresponding subgroup ofG CKF Then a Let a be an element of G The subgroup of G which corresponds to the conjugate sub eld 0L is the conjugate subgroup oHo l In other words GltK0Lgt oHo l b L is a Galois extension ofF if and only ifH is a normal subgroup of G When this is so then CLF is isomorphic to the quotient group GH SEE DIAGRAM ON PAGE 559 Proof Part a Let 0L L lf 739 is an element of H CKL then 070 1 is in H CKL To check this we mush show that 07071 xes every element 0 E L By the de nition of 0L 0 0a for some or E L Then 070 1o 070z 0a 0 as required Hence H I D UHJ 1 and by counting elements we have H aHa 1 Part b Now suppose that H is normal Then H aHa 1 for all 0 E G Hence CKL GltKO39Lgt this implies that L 0L for all 0 Thus every F automorphism of K car ries L to itself and hence de nes an F automorphism of L by restriction The restriction de nes a homomorphism 7rG gtGLF i2 lts kernel is the set of 0 E G which induce the identity on L7 which is H Therefore GH is isomorphic to CLF Counting degrees and orders we have L 1 Fl lGH S GLFgt Hence L is a Galois extension and that GH CLF Conversely suppose that L F is Galois Then L is a splitting eld of some polynomial 936 E F So L wa k where the 5239 are the roots ofg in K An F autoniorphisn1 a of K perrnutes these roots and there fore carries L to itself L al By Part a7 H UHJ 1 and thus H is normal D 02 Kummer Extensions Now we will consider extensions which contain pth roots of unity where p is a prime For now we will assume that all elds are sub elds of C 13 De nition 02011 Let F g C be a sub eld of C which contains a primitive pth root of unity Cp e2mp l Extensions Generated by Single Root of xp o Lemma 02012 If oz ts a root of gs 33p a then ozCpozC ozC 1oz are the roots of So the splttttng eld of 33p a ts generated by a stngle root K FM Proof This is easy to see as ugly a for all m And further as C17 y C for all m y m these are all the roots So it su ices to show that Q E Foz osz 0ng4 But this is true because Cp oszVoz D l Splitting Field of xp a l Theorem 02013 Let F g C and let F contaln a pth root of untty Fmther let a E F be an element whteh ts not apth power tn F Then the splttttng eld 14 of og 33p a has degree p over F and its Galois group is a cyclic group of order p Proof Let K be a splitting eld of f and let or be one of its roots in K Assume that or is not in F Then there is an automorphism o of K F which does not x or Since the roots of f are CZon O p 1 0a Cma for some m y 0 We now compute powers of o Remembering that o is an automorphism and that 0C C because C E F we nd that 02oz 0Cma C ooz C2ma Similarly7 o ltozgt Cm for all 239 Since C is a pth root of unity the smallest positive power of a which xes or is op Hence the order of o in the Galois group is at least p On the other hand7 or generates K over F and or is a root of the polynomial cop a of degree p7 so K F g p This shos at the same time that K F p7 that top a is 15 irreducible over F 7 and that CK F is cyclic of order p D l Galois Extensions of Degree p l Theorem 02014 Let F be a snb eld 0fC which contains a pth root of nntty Cp and let KF be a G0 lots extension of degree p Then K ts obtained by adjoining a pth root to F Proof The Galois group G has prinie order p K F so it is a cyclic group Any element 07 not the identity7 will generate it Let us View K as an F vector space Then a is a linear operator on K For7 since a is an F autoniorphisn17 aoz aoza and 0ca acaa odor for all C E F and oz 6 E K Since G is a cyclic group of order p 0p 1 An eigenvalue for this operator must satisfy the relation p 1 which means that is a power 16 of C By hypothesis7 these eigenvalues are in the eld F Moreover7 there is at least one eigenvalue different from 1 This is a fact about any linear operator T such that some power of T is the identity7 because such a linear operator can be diagonalized lts eigenvalues are the en tries of the diagonal matrix A Which represents it lf T is not the identity7 as is the case here7 then A y I7 so some diagonal entry is different from 1 We choose an eigenvector oz With an eigenvalue Cl y 1 Then 0a Ga and hence 0o aozp group Clpozp of So a xes of Since a generates G the ele rnent ozp is in the xed eld KG Which is F We have therefore found an element or E K Whose pth power is in F Since 0a y oz the element oz is not in F But7 since K F l is prinie7 oz generates K D Kummer Extensions De nition 02015 Extensions of the above type are called Kummer Extensions 03 TODO 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman October 10 2006 1 l TALK SLOWLY AND WRITE NEATLY 01 Quotient Rings and Relations 011 Theorems De nition 0111 Let ltR X be a ring and let I g R be an ideal De ne RI cosets of R as a group under Theorem 0112 Let I he an ideal of a ring R 0 There is a unique ring structure on the set RI such that the canonical map 7r R gt RI sending a w a I is a homomorphism h The kernel of7r is I Proof Part a The rst thing to note is that R is an abelian group and I is a normal subgroup of R and so given co set Ea1andhblwe can de ne the coset 6Erszr6alseb1 just as we did in the case of groups So all that is left is to de ne 65 The problem is that ingeneral PrsrEaIsEbI is not an ideal But7 it is contained in a single ideal abI To see this note that if we let 7 a u and 5 b 1 then r5aubvabavbuuvEab1 Further7 this is the unique coset which P is a subset of because the cosets partition R Hence we can let ab ab I And the rest of the proof of part a is the same as the case for groups Part b So obviously I g herlt7r Now suppose a E herlt7r We thenhaveIOIIandsoaEI D Theorem 0113 Mapping Property of Quotient Rings Let f R gt R be a ring homomorphism with kernel I and let J be an ideal which is contained in I Denote R J by R a There is a unique homomorphism 7 E gt R such that 77F f f x V FRJ RR RI 1 First Isomorphism Theorem If J I then 7 maps E isomorphically to the image of f Proof TODO Theorem 1042 MAYBE GIVE AS HOMEWORK D 4 Theorem 0114 Correspondence Theorem Let RJ and let 7r denote the canonlcal map R gt R a There ls a Injecter correspondence between the set of lcleals of R whlch contaln I and the set of all ldeals of R gloen I w 7a and 7 w WU h Thlrd Isomorphlsm Theorem If I C R corre sponds to I C R then RI and EI are lsomor phlc rlngs Proof Part a We need to Check i If I is an ideal of R which contains J then 7rI is an ideal of E ii If I is an ideal of E then 7r 1I is an ideal of R iii 7r 17rI I and 7rlt7r 1I I Now we know the image of a subgroup is a subgroup so to show that 7rI is an ideal we just need to show that 7rI is closed under multiplication by elements of E Let T E E and let E E I We can write 7Wquot T and Mas E for some 6 G Liquot G R But then W 7rra and m E I because 13 is Notice that we havent used the assumption that I g J yet however the fact that 7r is surjective is essential lteni ii Let go E gt EI Now since 90 is surjective so is go 0 7r Further kerltgo 0 7r r E R 7rr E kerltgpgt I 7r 1I Hence 7r 1I is an ideal Part b The rst isornorphisni theoreni also shows us that R 7F1 I is isomorphic to EI Iteni iii First note that for any map of sets 7r 17rI D I and 7rlt7r 1I C I And fuithei7 for any surjective niap 7T7 7r7r 1I I To see W 17TI C I we need that I D J Let a E W 17TI Then Maj E 7rI and so there is an element y E I such that My Since 7r is ahoniomorphisni we have 7r y O and hence 13 y E J ke Since y E I and J g I this means that a E I and we are done D 012 Examples One intuitive way to think about what is going on when we take the map 7r R gt RI is that we are setting all 7 the elements of I equal to O This is witnessed by the fact that I Kerlt7r One easy example is consider the ring 2 This ring has characteristic p and can be gotten from the ring Z by setting every element of p or equivalently every multi ple of p to 0 Another example is consider the map Ca RM gt R which is evaluation of 13 at oz Then Ezrazccl E Kerlt a ltgt Ca2 a l O ltgt Ezrazrozl O ltgt 04Z a z 02 Adjoining an element Another procedure we will now study for creating rings is to adjoin an element to the ring De nition 0201 Let R be a ring such that R g R 8 where R is a ring Then we say that R is a ring extension of R De nition 0202 Let R be a ring extension of R and let oz E R R We then de ne RM Signno n E R We say that RM is generated by R and oz Now the by the substitution principle we have Theorem 0203 Let R g R and let oz E R Then there ts a unique map 90 RM gt R such that go ts the identity on R and takes 6 w oz Further RM Proof Immediate from the Substitution Theorem D So7 we have that RM is a universal solution to the prob lem of adjoining a new element 021 Examples Construction of Complex Numbers The example which we want to keep in mind when con sidering adjoining new elements is case of the complex numbers over the reals Speci cally we nd that C So we know that there is a map 90 RM gt C which is constant on R and takes 13 w 239 But we also know that go is surjective and so C Rker90 But pc Ejltnrjj E ke gp if and only if Ejltnrjz37 O and so 1333 E ke go H 62 1 divides pltgt Hence er 3321 and we see that C R2 1 l Construction of In nitesimals l Similar to the constriction of the complex numbers we can just the previous results to add in nitesimals to our ring De nition 0211 Let R be a ring 6 E R is called an ln nitesinial or nilpotent element if 62 0 Theorem 0212 If R is a ring and R is a ring extension of R such that there is an e E R R which is nilpotent then RH Proof This is immediate from the Substitution Principle D 022 Theorems This leads to the following theorem Theorem 0221 Let R be a ring and let fltgt he a monie polynomial of positive degree n with eoe l 11 cients in B Let RM denote the ring obtained by ad joining an element satisfying a O The elements of RM are in hijectioe correspondence with vectors r0 rn1 E R via d map 77 Where 77ro 3171 r0 Tia r2oz2 73171047171 Proof Theorem D It is more di cult to analyze the structure of rings ob tained by adjoining an element satisfying a non nionic polynomial Now we Will consider a special case of ad joining an element satisfying a non nionic polynomial7 that of adjoining an inverse Let R be a ring and let a E R a has an inverse oz if and only if ooz 1 So7 in the ring Rooc 1 c is an inverse to a Let us call ring RM a 1 RM Now while RM doesnt have a basis in the sense of the previous theorem we do know that every element b E R is of the form I Signal However7 this does up a new questions lf we adjoin an element 13 with some relation to our ring B when is R a subring of RM I Now we know that there is a canonical niap w R gt RM gt RCI And it is Clear that the kerlt gt R H I lt then follows from the previous theorem that if I where pltgt is a rnonio polynomial then kerlt gt O and so w is injective However7 this is not always the case For example sup pose we wanted to adjoin an element oz which was the inverse of 0 Then in this case we must have RM O the O ring7 because that is the only ring in which 0 has an inverse To be precise this gets us Theorem 0222 Let R be a ring and let ab E R such that ab 0 Further let 0 E RM be such that wac 1 where w R gt RM Then if RM 75 O we must have Mb 0 Proof We have RlSCllaf 1 g RlCl gt gt Rlal But I ba 1 E ate 1 the ideal and hence 1W 0 D 14 De nition 0223 We say an element 9 of a ring R is a Zero Divisor if there is a non zero element a such that w0 03 TODO o Flush out the outline of math 0 Come up with A BUNCH of examples more than I can use so that I dont run out of time 0 Go through Langjs book on the same topics Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman September 14 2006 1 01 TALK SLOWLY AND WRITE NEATLYH 02 Abelian Group 021 De nitions 1 De nition of Abelian Group 1 De nition 0211 We say agroup A 0 gt is Abelian if Vab E Aabba 1 Spanning Notation 1 De nition 0212 Let Gbe agroup and let 91 92 E G We then de ne lt91 92 subgroup of G gener ated 91 92gt 39 39 39 Torsion Subgroup 1 De nition 0213 Let G 0 be an abelian group and let a E G We say a is a torsion element if it has nite order The collection of all torsion elements of an Abelian group A is called the Torsion Subgroup of A A group A is said to be of exponent m if every element of A has period dividing m Lemma 0214 IfA 239s abelian then the torsion sub group ofA is a subgroup ofA Proof Notice that if no 0 then ncf1 O and if no O and mb 0 then nmlta b O D De nition 0215 We say an abelian group A is free if c A Spanltu1 urgt o EZ39STCZZ39UZ O gt oi O for all 2 g r We then say A has rank r 022 Theorems l Subgroups of Free Groups are Free l Theorem 0221 Dedekind Let F be afree abellau group of rank r and let G be a subgroup of F Then G ls a free abellau group of rank 5 g r Further F has a set of generators uzr urgt such that G ls generated by U1 a11u1 algllg alrur U2 012U2 agrur uS assuS asrur for some azj such that all an whleh are posltlue Proof Let uzr urgt be a basis for F for notational convenience we will say that 13 E STClt lu for all a E Now Choose some non zero element b 2290 lu of By reordering the indexes and possibly taking inverses of some of the basis elements we can assume that 01 1 gt 0 Now lets look at all the values of 01 1 for b E G and choose a minimum positive value which we know must exist because 01 1 gt 0 Call the smallest such integer all and pick some element 1 such that 001 1 all We therefore have that if 13 Z STCltiU E G then an divides Clt IZ 1 This is because Clt IZ 1 p 111 a where a lt all And so if all didn7t divide Clt IZ 1 then pv1 a would have a coordinate in ul with coe icient less than all 5 So in particular7 for all a E G there is a q such that 6 qvl CZC 2u2 Crur lf 7 1 we are done Assume 7 gt 1 but the theorem is true for all 7 1 The theorem breaks into two cases Case 1 G g Spamm un Then we are done by induction Case 2 G g Spamug un Then let F1 Spanltu2 1 and G1 G H F1 This theorem then applies to F1 G1 as G1 g F1 and so we have that G1 Spanltv2 ms with 5 g r and further U2 CLQQUQ 123113 agrur U3 133113 agrur Us 55115 asrur with agg ass all positive Hence we know that G Spanltv1 mg by the in ductive hypothesis So all that is left is to check that 111 1 are inde pendent lf not then there are some EZ39STdZ39UZ39 0 where dZ39EZ But we know that if 7 1 this can7t happen And hence by induction we that ltU5 vrgt are independent So this means that there must be some linear combination of 11 which adds to O and what is more this must have the coeffecient of U1 non zero d1 U1 d2U2 39 39 39 dn Un But7 putting any such linear combination in terms of the m we see that d1 O as 111 is the only element with ul component and the us are linearly independent gtlt Hence the v1 ms are a linear independent basis for F as a free abelian group D l Theorem about basis l Theorem 0222 Let F ltu1 urgt and let 1 b1u1 brur with gedltb1 br 1 Then there exists v2 vr E F such that F 11112 1 Proof Set 5 b1 b2 lbrl lf 5 1 then the result is trivial as v lu for some 239 Now lets assume this theorem is true for all 0 g 7 lt 5 8 As 5 gt 1 we know that at least 2 of the bs are non zero With out loss of generality we can assume b1 2 b2 gt 0 although we may have to Change the signs of some of the basis elements Now lets let wl u1w2 u1 u2wj uj ifj Z 3 We then Clearly have F ltw1wrgt and also that U ltb1 b2w1 b2w2 brwr Further7 gedltb1 b2 b2 br 1 and lbl b2b2lbrl lt8 And so the result follows by induction D l Relation of Basis of Free Group to Subgroup l Theorem 0223 Let F be a nitely generated free abeltau group of rank r and let G be a subgroup ofF 0f 9 rank s with O lt s g r Then there exists v1 vr E F such that F v1vgt G h1v1hrvrgt where h1 h are all positive and satisfy th divides hi1 Proof Let ul u be a set of generators for F Take 6 E G and write 5 11 can We then de ne 6ltulimiurgtltgt gcdc1 ay Lemma 0224 6ltulimiurgtltgt is independent of the choice of generators Proof We know by the de nition of 13 that 6 6lt 1quotquot Tgt ay1u1 yrur for some integers r 10 Now let wl wr generate F Then we can represent each uzr in terms of wl wr Hence a 6lt 1quotquot Tgtltgtltyi w1 yiwa and so 6ltU17W7UTgt divides 6ltw17quot397wrgt And so by symmetry we must have 6ltU17W7UTgt 6ltw17W7wTgt D We will then simply write 633 for 6ltulvmvurgtltgg Now take any nonzero yl E G such that 6y1 is min imal Set hl 6y1 We then have yl h1z1u1 zrur where gcdltz1 Zr 1 So in particular by the previ ous lemma we know that if we set 111 21111 Zrur then there exist v3 1 such that 11111311quot F 7 7 11 If 7 1 then we are done because we have 5 17 F 111 G lth1U1gt So lets do induction on 7quot Suppose 7 gt 1 Let F1 1 v and let G1 F1 H G Then G1 is a subgroup of F1 There are two cases we must consider Case 1 RankltG1 O In this case we know that G1 O and so G hm Case 2 RankltG1 gt O In this case we know by the inductive hypothesis that there are 112 vrgt such that F1 ltU2 lrgt G1lth2 U2 bran and h divides hHl But we therefore know that F v1v2vrgt Now lets consider a y E G Well then we know we can write ya1v1agv arv But we know a1 qh1m for some m lt hl and further 6y qyl g m Hence m 0 because we choose yl to be such that 6y was minimal So hl divides a1 gtllt 7 So we have y qh1v1 by try Hence G1 lth1U17 h2v2 a hrvrgt and all that is left to show is that hl divides hg 13 let hg ahl b where 0 g b lt hl Then if we let Ot h1 U1 hg Ug E G We have nglty0gt nglth1h2gt gedlth1 9 Hence by the niininiality of hl we have b 0 So by induction we are done D 03 The Structure Theorem for Finitely Gen erated Abelian Groups 031 Theorem From the previous theorem we have Existence of Decomposition By Increasing Div Theorem 0311 Euery nitely generated Abeltau group A can be expressed as the dtreet sum of eyelte groups AEJZWQBZMQBHQBZhj where m dtutdes h 1 14 Proof Let A be a nitely generated Abelian group such that A ltu1ungt Now consider the map 90 Z gt A Where 9062 m and 6239 is the ith basis element This map is obviously surjective and so we know that A 5 Z kergo But ke go is a subgroup of Z and hence there is a basis fl for Z such that kerlt90gt lth1f17 39 39 39 a hrfrgt and h divides hHl Hence A g ltltf1gtlth1f1gtgt 69 9 ltfrgtlthrfrgtgt 9 9 ltfngtgt hence A ZmZh1Zhj and we are done D But what is more7 is that this decomposition is unique Speci cally we have Uniqueness of Decomposition By Increasing D Theorem 0312 Let A be d hitely generated Ahelidh group Let AZT Ze1 Zhn where a divides e2 and AZ5Zd1Zdm where d divides dHl Then 5 r m ii and a d for all i But before we show this we need another theoreni Existence of Decomposition By Sylow Subgror 16 Theorem 0313 Let G be a nite abelian group of order p flpg2 where the pfs are distinct primes Then G P1 Q9 P2 Q9 Where B are the subgroups of elements of G with order a power ofpzr Proof Let a E G be an element of order p f1 where p1 and f1 are relatively prime We can then nd u u such that uf1 up 1 We then have 13 uflsc upihp but we know that uflst has order p and upihp has order f1 So we know we can break a 1 331 where 1 has order p and 61 has order f1 And so by iterating this process we can break a 51 2 where b has order at P2 So7 if we can show that this decomposition is unique we are done Supposewe have b1b2 a bgt1kbWh 1quot 9239 b has order pf We then have that there are 61 C2 0 Where each 6239 has order a power of pi But this means that C C1C 1C 1 ai1 And so p1plf11p 1 c 0 But this is only true if 0239 0 Hence CZ39 bi b O and so bi b for all 139 And so we are done D Decomposition of Cyclic Group Into Sylow Su Corollary 0314 Let 6 pi 43 Then Z gt g ZPi1gt 39 ZP Proof We know WHRm where B 2 E Ze orderz is a power of p2 ln particular if 13 generates Ze then we have 55 91 QB QB Qn where each Q has order a power of p2 but every y E Ze ma and so in particular this means that every 2 E Ze of order pzr mzsc So in particular npgimzqzu But as 7117 is relatively prinie to pg there is an in a teger 5 such that 5 gtllt np E 1 mod So z npgimzq mzqzr because or der of In particular this means that B is generated by a sin gle element and hence B 3 Z and we have Zlt6gt g Zltp1gt Zi lm D Now we can return to the proof of the uniqueness of the decomposition Proof Uniqueness of Decomposition By Increa Proof First to see that r 5 notice that A ZS e9 T 2 e9 T Where T is the torsion group of A So if 7n A gt 2 then tmlt7rr Zr Aher br AT But similarly if 71 A gt ZS then tmlt7rs Z5 Aherq s AT So we must have ZS 2 Z and hence r 5 Now lets just look at the torsion group T First we need a lemma Lemma 0315 Let G be any group Suppose as y E G and my ya and the order OfZC ts relatively prime to the order of y Then lt93 y gt WA and ts cyclic of orderrordery Proof Let ordera n and ordery m First observe that my Q 3311 7 However7 because But then we also have ym a m n are coprinie there is a or such that am E 1m0dn And so ya1m 3 And we can get y in an identical 1118111181quot and SO WE see lt93 ygt ltygt Now we know the order of lt33 y 3 mm as every element isoftheforrnzcaybwithO a n 10gbgm l Suppose we have 36yl 1 Then we know that 01 th 36 and so 71 divides mt But as m n are coprinie this means that n divides t We similarly get that n divides t and so t is divisible by mn Hence the order of my is at least mn But we have already shown that it is at most mm and hence the order must be exactly mn D For clarity lets let 6239 p1ialll pwiawl We there for know that Mi 3 ach or all i g k by our assump tion on the 6239 But we also know that UT Z39SMQBijZleilajlilll So in particular we have T j wlt mZltpjltigtajlilgtgt Hence we have that if H is any pj Sylow subgroup of T H 5 ngpjltigtajlgt Further if we have a decomposition of the form then this implies that it must have come uniquely from the decomposition T Zel Q9 Zem by the nature of the es So7 in order to prove the theorem7 all that is left is to prove that the decomposition of the pj Sylow groups is unique Decomposition of Abelian p Groups Theorem 0316 Let A be an abelian group of order pa where p is prime Suppose Altuigt ltu1 gt and Altvigt ltvlgt where 0rderu 0rdervj gj fiZf2Zquot39kagt1 and 91292Z 39Zglgt1 Then kl and f g for alli Proof The rst thing to notice is that f1f2quot39fka9192quot 91 Now we will prove the theorem by induction Base Case If a 1 this result is trivial as f1 91 a Inductive Case Assume this is true for a lt or and now assume a OZ Let AP as E A p36 0 So we have Ap PfHUi QB 39 39 39 Pfkiluw Ap ltp91 1v1gt a9 9 1411 Hence AP ZpW and sol k Now lets consider AP p33 a E A Now if y is such that gay gt 1 but 97H 1 and H is such that fix gt 1 but f 1 1 then we have AP pin QB Q9 pm AP ltPv1gt m ltvagt But then by the inductive hypothesis we have R y andfi 1g 1foralli c And so we are done With this theorem D And this also proves the main theorem D 04 Modules 041 De nition gt ONLY IF TIME 05 TODO 0 Speci cally gure out what to show other than the structure theorem 0 Come up with A BUNCH of examples more than I can use so that I dont run out of time Lecture Notes Math 371 Algebra Fall 2006 by Nathanael Leedom Ackerman October 17 2006 1 1 TALK SLOWLY AND WRITE NEATLY 1 01 Factorization 011 Factorization of Integers and Polynomials Now we are going to discuss some of the nice properties of the integers with regards to factorization One of the most important ones is that for any a b with a y 0 there are q r such that Properties of the Integers o b aq r o 0 g r lt a From this fact we get several important results Including Theorem 0111 If a b have no factor tn common other than 1 then there are c at such that acbd 1 and 2 Theorem 0112 Let p be d prinie integer and let a b be integers Then ifp divides ob we must have p divides a orp divides b Which will get us i Fundamental Theorem of Arithmetic i Theorem 0113 Fundamental Theorem of Arith metic Every integer a 75 O can be written as d product a where c is i1 and each pl is prime And further up to the ordering this product is unique Proof First we need to show a prime factorization exists It su ices to consider the case when a gt 1 And7 we can assume that this is the case for all b lt a Now there are two cases Case 1 a is prime Then we are done Case 2 a bb then both bb are less than a and hence have a prime factorization Next we need to show that prinie factorizations are unique Let ipipniq1qm Where all p25 qj are prime We use the fact that if p divides ab then p divides a or p divides b This means that for each p17 pi Inust divide sonie qu Hence7 because qu is prime p1 Q for some t We can then cancel p1 from both sides and proceed by induction D l Polynomial Rings over Fields l Theorem 0114 Let F be a eld and let de note the polynomial ring tn one variable over F 4 a If two polynomlals f 9 have no common non constant factor then there are polynomlals r s E such thatrfsg1 b If an lrreduclhle polynomlal p E dlyldes a product fg then p dlyldes one of the factors c Eyery nonzero polynomlal f E can he wrltten as a product Cpl Pn where c ls a nonzero constant and the pl are monlc lrreduclhle polynomlals tn and n 2 O Thls factorlzatlon ls unlque except for the orderlng of the terms Proof PROVE B MAYBE MAYBE GIVE A7 C AS HOMEWORK ASSUMING Same as the case for integers MAYBE PROVE THIS INSTEAD OF THE INTEGER CASE D Similarly we have l Number of Roots of a Polynomial l Theorem 0115 Let F be a eld and let gs be a polynomial of degree n with ooe lolents m F Then f has at most 71 roots m F Proof An element or E F is a rot of f if and only if to oz divides So if we can then write to ozq where q is a polynomial of degree n 1 ll 6 is another root of f then ozq O and so 6 is a root of q as the product of non zero elements is non zero in a eld Hence by induction on n we see that f has at most 71 rots D 02 Unique Factorization Domains Principle Ideal Domains and Euclidean Domains Now that we have have seen that the integers and poly nornials rings over elds behave sirnilarly it is natural to 6 ask which other rings behave in a similar manner So that we have the cancellation law we Will assume that all of these rings are integral domains l Divisors Irreducible Elements Associates De nition 0206 Let R be an integral doniain lf a b E R we say that a divides b if EIT E Rm b We say that a is a proper divisor of b if b qa for some q E R and neither q nor a is a unit We say a non zero element a of R is irreducible if it is not a unit and if it has no proper divisor We say that a a E R are associates if a divides a and a divides a lt is easy to show that if a a are associates then a ua for some unit u E R 7 We can think of these ideas in terms of the principle ideals Whichtheygenerate l Connections to Ideals l Theorem 0207 a is a unit ltgt 1 a and a are associates ltgt a a a divides b ltgt a D b a is a proper divisor ofb ltgt 1 gt a gt 9 Proof Immediate D We then have the following result which compares the process of factoring With ideas l Factorization and Chains of Ideals l Theorem 0208 Let R be an integral domain Then the following are equivalent a For every non zero element a E R which is not a unit factors as a b1 bn where each bi is irreducible b B does not contain an in nite increasing chain of principle ideals a1 lt a2 lt a3 lt Proof Suppose R contains an in nite increasing sequence a1 lt a2 lt a3 lt Then an lt 1 for all n because an lt anH g Since aml lt an7 an is a proper divisor of an1 Say 111 anon Where anon are not units This provides a non terminating sequence of factorizations a1 agbg agbgbg a4b4b3b2 Conversely such a factorization gives us an increasing chain of ideals D 9 We call these conditions either the ascending chain condition for princiI or that existence of factorizations holds in R 021 Examples NonUFD To see an example of when the existence of factorization fails consider the following ring F12 331333 2 This ring is essentially F with 27k adjoined for each k E w le we adjoin the equations Z21 36 This doesnt have the ascending chain condition because we see that 5131 lt 5132 lt is an in nite ascending chain of principle ideals 022 Unique Factorization Domains l Unique Factorization Domains 10 De nition 0221 We say that an integral dornain R is a Unique Factorization Domain UFD if i Existence of factors is true for R ii lfa E Randap1pn anda qlqmwhere pbqj are all irreducible7 then m n and with a suitable ordering pzr and Q are associates for each 239 To see how this can fail consider ZW B 61 b 5 a b E Z We then have 6231 51 5 and it isnt hard to show that 2 3 1 V 5 1 x 5 are all irreducible But because l1 are the only units it is also clear that these are not associates of each other Hence ZN 5 is not a unique factorization doniain 11 The crucial property of primes which we use in UFDls is l Prime Elements l De nition 0222 Let R be an integral domain We say that p E R is prime ifp y O and for all ab E R such that p divides ab7 either p divides a or p divides b Theorem 0223 Let R be an integral domain Sap pose existence of factorization holds in R Then R is a UFD if and only if every irreducible element is prime Proof Proposition1128 MAYBE GIVE AS HOME WORK D Theorem 0224 Let R be a UFD and let a p1pn and let b ql qm be prime factorizations ofa and b Then a divides b if and only ifm Z n and under a suitable ordering of the factors qzr of b pi is an associate of qi for all i g n 12 Proof Immediate D 1 Greatest Common Divisors in UFD s l Corollary 0225 Let R be a UFD and let a b E R where we don t have both a O and b 0 Then there exists a greatest common divisor d of 11 with the following properties i d divides a and 9 ii if an element e ofR divides a and b then e divides d It follows form the de nition that any two greatest com mon divisors are associate7 but a greatest common divisor does not need to be of the form oq bp For example ZM is a UFD and 2 13 have a greatest com mon divisor of 1 but 1 is not a linear combination of 2 and 13 With integer coef cients 023 Principle Ideal Domains Principle Ideal Domains De nition 0231 We say an integral domain R is a Principle ldeal Domain PlD if every ideal is principle Theorem 0232 a In an integral domain aprime element is irreducible b In a PID an irreducible element is prime Proof Proposition 11211 MAYBE GIVE AS HOMEWORK D PID implies UFD Theorem 0233 A Principle Ideal Domain is a Unique Factorization Domain Proof Suppose that R is a PlD Then every irreducible element of R is prime So we only need to prove the ex istence of factorizations for R and uniqueness is taken care of We have shown that this is equivalent to there being no in nite ascending chain We will argue by contradiction so lets assume that 11 lt 12 lt is such a chain Lemma 0234 Let R be any ring The union of an increasing chain of ideals 1C12C is an ideal Proof Let IUIn new ifuv G then uv E In forsorne In and souv E mg n Let A new Then A is an ideal and in particular A b for some I E R as R is a PlD But then there must be some n such that b E an But then we have an C an C A I Q an So we have an an for all j contradicting that a1 lt a2 lt was an in nite ascending chain D Note however that there are unique factorization domains which are not principle ideal doniains like PID 7S and l Theorem 0235 a Letp he a nonzero element of a principle ideal domain R Then Rp is a eld if and only ifp is irreducible h The maximal ideals are the principle ideals gen erated by irreducible elements Proof Since an ideal M is maximal if and only if R M is a eld then the two parts are equivalent So we will prove the second part We Will prove part a a 2 b if and only if a di vides b The only divisors of an irreducible element p are the unites and the associates of p Therefore the only principle ideals which contain p are p and Since every ideal of R is principle this shows that an irreducible element generates a maximal ideal Conversely let I aq Where neither a nor q are a unit Then we have b lt a lt 1 and so 9 is not maxi mal D 024 Euclidean Domains The next step we want to take is to generalize the division with remainder procedure Now in order to do this we need a sense of size in a ring Recall that for the following rings we have the following sizes absolute value if R Z degree of a polynomial if R F absolute value2 R So in particular we have a the following de nition Euclidian DO De nition 0241 Let R be an integral domain We say that R is a Euclidean Domain if there is a size func tion azR O gtO12 Such that if a b E R and a y 0 then there are p q E R with b aq r and either 7 O or 07quot lt 0a As examples we have Z F as are Euclidean Domains Theorem 0242 A Euclidean Domain is a princi ple ideal domain and hence also a unique factorization domain Proof Proposition 11220 MAYBE GIVE AS HOMEWORK OR SAY lS llVllVlEDlATE D 03 Gauss s Lemma We know that Q is a eld and hence we also know that is a unique factorization domain So in particular7 given any polynomial Ma E Zltgt we know that it can be factored in But we can then ask the question7 Can this factorization be done inside Z The answer is YES and that is What we Will now prove l Primitive Polynomials l De nition 0303 A polynomial Ziazrasi E ZM is called primitive if gedlta1 an l and n is positive l Primitive Decomposition of Polynomials l Theorem 0304 Euery non zero polynomial fltgt E tt can be written uniquely as the product fltgt Cfoltgt where c is a rational number called the content of and f0a E ZM and is primitiue Further the polynomial f has integer eoe leients if and only if c is an integer Proof First What we do is factor out all the denominators to get a polynomial f1 E ZM such that 1f1a in 20 for some n E Z Then factor out the greatest common divisor of the coef cients of f1 to get a polynomial f0 where c gtllt f0 This proves existence To show uniqueness lets assume we have cf0 dgo where f0 9033 are primitive We then want to show that c d and f0c 9056 By cross multiplication it suf ces to consider the case when 0 d are integers Let a2 9 be the coef cients of f0go respec tively We therefore have that cazr db for all 239 But we also have that gcda gcdb 1 Hence gcdca c and gcddb d and we have c ld and f0 lzgo But because the leading coef cient of f0 and of go are positive this means that c d and f0 go D l Gauss s Lemma l Theorem 0305 Gaussls Lemma A product ofprim itioe polynomials in ZM is primitive Proof Let f g E ZM be primitive and let h fg We know that the leading coef cient of h is positive as the leading coef cient of g and f are We will show that no prime p divides all the coef cients of This will show that the gcd of the coef cients of h is 1 and hence that h is primitive Consider the homomorphism 90p ZM gt fpljc If we can show that gopUL y 0 then we are done But we know that mpg y O and 90109 y 0 because f g are primitive and hence the gcd of their coef cients is 1 But7 7p is an integral domain and we have gopUL mpg 90109 and 22 so goph y O D Theorem 0306 a Let fg be polynomials in tc and let f0 go be the associated primitive polyno mials in ZM Iff diyides g in tc then f0 diyides go in ZM b Let f be a primitive polynomial in ZM and let 9 be any polynomial with integer coe lcients Sup pose that f diyides g in QM say 9 fq with q E Then q E ZM and hence f diyides g in Zi il c Let fg be polynomials in If they have a common nonconstant factor in QM then they have a common nonconstant factor in ZM too Proof To prove a Clear denominators so that f g be come primitive and then a is a consequence of 23 ln order to prove b we apply the previous theorem to get q cqo where c E Q and go is primitive We then have fqo is primitive and g ch0 shows that fqo go the primitive polynomial associated to g and g 090 But as g E ZM this means that c E Z and hence q E To prove c suppose that f 9 have a common factor h We may assume that h is primitive as we are working in and if h divides f 9 then so does ch But then by part b we have that h divides f g in ZM D Corollary 0307 If a nonconstani polynomial f is irreducible in ZM then it is irreducible in cc Proof lmmediate D l Irreducibility in mg l Theorem 0308 Let f be an integer polynomial with positiue leading coe lcieni Then f is irreducible in ZM if and only if either a f is a prime integer or h f is a primitive polynomial which is irreducible in Ql il Proof Suppose f is irreducible We ay write f efo Where f0 is primitive Since f is irreducible this cant be a proper factorization so either f0 1 or e 1 D Theorem 0309 Every irreducible element of ZM is a prime element Proof Let f be irreducible and suppose f divides gh where gh E Case 1 f p a prime integer Let g ego and h ho Where go ho are primitive We have that goho is also primitive So in particular there is some coef cient a Which is not divisible by p But we have p divides gh 25 so p divides cda And in particular this means that p divides c or p divides d Hence p divides g or p divides h M f is a primitive polynomial which is irreducible in We then have f is a prime element of and so f divides g or f divides f in Hence f divides g or f divides h in D Theorem 03010 The polynomial ring ZM is a unique factorization domain Euery nonzero polyno mial f E ZM which is not i1 can be written as the product ipl pmq1c qna where the pi are prime integers and the bias are ir reducible primitiue polynomials This expression is unique up to rearrangement of factors 26 Proof Immediate D Now if we let R be any UFD and let F be the eld of frac tions of R then the results above can be copied we just have to allow some ambiguity to deal with units Speci callywe have l Generalization t0 Arbitrary UFD S l Theorem 03011 Let R be a unique factorization domain with eld of fractions F a Let g be polynomials in and let f0go be the associated primitive polynomials in If f divides g in then f0 divides go in BM b Let f be a primitive polynomial in RM and let 9 be any polynomial in Suppose that f divides g in FM say 9 fq with q E Then 1 E RM and hence f divides g in c Let g be polynomials in If they have a 27 common noneonstant factor in F then they have a common noneonstant factor in HM also a If a noneonstant polynomial f is irreducible in BM then it is irreducible in also e RM is a unique factorization domain Proof This follows the previous proofs concerning and D Corollary 03012 The polynomial ring Z1 an and Fa 1a n where F is a eld are unique fac torization domains 04 TODO 0 Go through Langjs book on the same topics

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