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# Class Note for MATH 831 with Professor Kachi at KU

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Math 831 ABSTRACT ALGEBRA II PROGRESS CHECK 7 II Draft 648pm February 28 Sat 2009 Instructor Yasuyuki Kachi Line 82255 CHAPTER 2 RINGS AND RING AXIOMS 0 CATEGORIES 0 Now we have familiarized ourselves with the complex number field C along with the notions of abstract elds and subfields In the present lecture we introduce a new algebraic system called a ring Like a field a ring is an algebraic system which carries both additive and multiplicative structures In particular we introduce a commutative ring as a ring whose multiplicative structure is commutative As in the diagram in page 5 a field is indeed a special example of a commutative ring In other words the notion of a commutative ring is a generalization of the notion of a field 0 We slightly shift our gears Today we are going to see lots of arrows7 Mathematicians working in any branch of mathematics set their eyes on objects7 In case of linear algebra the objects are k vector spaces where k is a fixed field See Appendix to Chapter 2 for some rudiments of the notion of arbitrary vector spaces defined over an arbitrary field However in linear algebra what is really important that comes with the notion of k vector spaces is the notion of k linear mappings V a W 101 o The same remark that applies to linear algebra also applies to other branches of mathematics For example in topology the objects are topological spaces However in topology what is really important that comes with the notion of topological spaces is the notion of continuous mappings X 7 Y In the ring theory the theory we are now ready to explore the theory we set our eyes on the objects are 7 clearly 7 rings What is really important that comes with the notion of rings is the notion of the so called 7 an homomorphisms R 7 S The definition of a ring homomorphism is formal simple and it does not require you to really know anything to understand it at the formal level However it is perhaps too formal though very natural from the definition of rings and you may not see how it has anything to do with our broad ultimate goal the study of algebraic equations The truth is the notion of ring homomorphisms is an indispensable tool we rely on when we try to achieve that goal Speaking of an algebraic equation here is a slight shift of paradigm for the second time 7 the first time was when we allowed complex number coef cients in algebraic equations Needless to recall an algebraic equation is of form Q where f is a polynomial with C coef cients that is HZ adzd 04d712 d71 0412 040 with ad 0414 a1 040 E C and also 6 C Remember the content of the fundamental theorem of algebra Yes it states that C as a field is algebraically closed But what was really the meaning of this in a plain language Yes any algebraic equation with 2 unknown with C coef cients has a root in C The statemen C is an algebraically closed field77 is a paraphrase of that fact wrapped up in some fancy7 terminology I would like to hereby offer a second paraphrase of the fundamental theorem of algebra This one is formulated in a plain language meaning general topology Historically this was not Clear at all 102 Fundamental theorem of algebra paraphrased II Let d E N with d 21 Let ad 0414 a1 040 E C Let fz adzd oszZd l 0412 a0 Regard Hz as amapping f C a C Then f is surjective o In today7s lecture we cover one extremely important underlying concept which governs the scenes behind77 the commutative ring theory The notion of ring homomorphisms7 and the notion of polynomial mappings are the ip Sides of the same coin 77 In today7s lecture we take a certain path to see what this really means In other words we use the notion of polynomial mappings as a motivating device to introduce the notion of commutative rings and ring homomorphisms 0 We jump start with the following which is what mathematicians usually refer to as category theory What I do here is not really anything to be called a theory7 Below is about recognizing one common avor7 which almost universally exists in mathematics regardless of sub disciplines lwant you to agree with the fact that such a common avor exists and also I want you to be comfortable with it In mathematics we study many different subjects and deal with many different objects but there is one common truth about armws7 which almost universally applies to various different mathematical contexts and settings Below we use one odd sounding word that is morphism7 A morphism simply means an arrow We could have used the word arrow7 to mean an arrow but the mathematicians7 community somehow felt they had to coin some alternative word just to make it sound more formal 103 0 Linear algebra 7 The category of kvector spaces In linear algebra you work with the category of vector spaces or more precisely the category of vector spaces de ned over a field k 77 There is a shorter form of a vector space de ned over k which is a k vector space The precise definition of a k vector space is found in Appendix to Chapter 2 Hence we may also call it the category of k vector spaces It is a system consisting of a an object 7 which is a k vector space V and 7 b a morphism V E W which is a k linear mapping 0 This category has the following feature7 Feature i Composition V 4gt W and W L U are both k linear mapping gt V Lf U is a k linear mapping ii The identity V L V is a k linear mapping 0 Set theory 7 The category of sets In set theory you work with the category of sets The category of sets is a system consisting of a an object 7 X which is a set and 7 b a morphism X E Y which is a mapping 104 o This category has the following feature7 Feature i Composition X 4gt Y and Y L Z are both mapping gt X A Z is a mapping ii The identity X L X is a mapping 0 General Topology 7 The category of topological spaces In general topology you work with the category of topological spaces The category of topological spaces is a system consisting of a an object 7 X which is a topological space and 7 b a morphism X E Y which is a continuous mapping 0 This category has the following feature7 Feature i Composition X 4gt Y and Y L Z are both continuous mapping gt X A Z is a continuous mapping ii The identity X L X is a continuous mapping 0 Next we talk about one category which has a bearing on our business the category 77 of groups lts object is a group We are going to look at this one closely 105 1 THE CATEGORY OF GROUPS 0 We recall the de nition of a group the group axioms De nition A set G is called a w if G has a distinguishable element 1 the identity element and two operations one called the inversion Which takes at E G into x 1 E G and the other called multiplcation Which takes at y E G into any 6 G such that the following G1 through G3 hold G1 ltyzgt ltxygtz G2 110 x 1 and G3 3amp4 l x lac 0 Most importantly in the above we do not assume the commutativity of the multiplication AbG any yx A group satisfying the additional axiom AbG above is called an Abelian group A group Which is not an Abelian group is called a non Abelian group 0 We Will see dozens of examples of groups both Abelian and non Abelian shortly But first we also make precise the morphism part77 of the category of groups that is the notion of a group homomorphism De nition Let G1 and G2 be groups A mapping 15 G1 G2 is called a group homomorphism if my w my holds in G2 for arbitrary at y 6 G1 106 0 Group theory 7 The category of groups In group theory you work with the category of groups The category of groups is a system consisting of a an object 7 which is a group G and b a morphism 7 G1 G2 which is a group homomorphism o This category has the following feature7 Feature i Composition G1 L G2 and G2 L G3 are both group homomorphism gt G1 ampgt G3 is a group homomorphism ii The identity G L G is a group homomorphism De nition Let G be a group and H its non empty subset H is said to be a subgroup of G if the following 8G1 and 8G2 hold 8G1 any 6 H whenever an E H and y E H 8G2 x 1 E H whenever an E H l 1 Let b G1 6 G2 be a group homomorphism Verify that Ker x G1 xl is a subgroup of G1 2 Let b G1 a G2 be a group homomorphism Verify that mas EG2 G1 is a subgroup of G2 Math 830 covers the notion of groups There the notion of normal subgroups7 and the fact that Ker qt is a normal subgroup of G17 are covered 107 ll Let G be a group and H its subgroup Verify that the natural injective mapping b H gt G is a group homomorphism De nition A group isomorphism is a group homomorphism b G1 a G2 which is bijective Ill Verify that for a group isomorphism b G1 a G2 its inverse mapping 1V1 G2 a G1 is a group isomorphism 0 When a group isomorphism b G1 a G2 exists for two groups G1 and G2 then we say that G1 and G2 are isomrphic to each other Write G1 2 G2 The notion of two groups being isomorphic is a re exive symmetric and transitive relation i G 2 G always ii G1 2 G2 implies G2 2 G1 and iii G1 2 G2 and G2 2 G3 imply G1 2 G3 0 Finite groups A little digression A group G which has finitely many elements is called a finite group The number of elements of G is called the order of G For a given natural number m it makes sense to classify finite groups77 having order m up to an isomorphism Or the L 77 same to say it makes sense to classify the isomorphism types of finite groups having order m The task is much harder than what the simplicity of the definition of groups may suggest save that there are some trivial cases such as when m is a prime number then there is only one isomorphism type or when m is the square of a prime number then the group is Abelian The Classi cation of nite Abelian groups is elementary See Fact 5 one page before the de nition of special linear groups 108 However note that for a given natural number m it is not easy to decompose m into primes when m is large Can you decide whether each of m 3250433 232 and m 1 is a prime number just by speculation Classification of finite groups up to an isomorphism was an extremely challenging problem complete in some perfectly reasonable sense only a few decades ago by experts of finite group theory This is not a topic we address in this course Below is the table of the number of groups haVing order m up to an isomorphism for 1 g m S 32 m1gt 1 mfg gt 2 m2 gt 1 m10 gt 2 m3 gt 1 m11gt 1 m4 gt 2 m12 gt 5 m5 gt 1 m13 gt 1 m6 gt 2 m14 gt 2 7717 gt 1 m15 gt 1 m8 gt 5 m16 gt 14 m17 gt 1 m25 gt 2 m18 gt 5 m26 gt 2 m19 gt 1 m27 gt 5 m20 gt 5 m28 gt 4 m21gt 2 m29 gt 1 m22 gt 2 m30 gt 4 m23 gt 1 m31gt 1 m24 gt 15 m32 gt 51 gtilt Table 1 The number of isomorphism types of groups with order m 1g m g 32 More precisely Classi cation of nite groups with no nontrivial normal subgroups nite simple groups 1 R Shafarevich Basic Notions of Algebra Springer 1997 p 152 109 0 From the table you speculate that regardless of m the number of isomorphism types is always at least 1 This is true Indeed for an arbitrary m the subset of C consisting of the m th roots of unity is a group of order m Such a group is denoted pm See Example below 0 From the table you speculate that when m is prime then the number of isomorphism types is always 1 This is true The converse is not true Look at m 15 0 From the table notice that the number of isomorphism types for m 32 is disproportionately large as compared to the corresponding numbers for m less than 32 Later we will construct one group of order 32 which exhibits some interesting kind of symmetry 0 Let us return to a general discussion of finite or infinite groups Below is an observation that a field naturally yields two Abelian groups De nition Let F be any field Define Fquot F0 aEF 10 An alternative notation for Fquot is GL1 or sometimes Gm Thus F GL1F Fact 1 Fquot is an Abelian group with respect to multiplication IV Prove Fact 1 above Example 3 GL1C a E C Oz 31 0 is an Abelian group with respect to multiplication Example R GL1R a E R 17 0 is an Abelian group with respect to multiplication The proof is elementary This is a possible topic in Math 830 Some literatures use the notation FX 110 Example Qquot GL1Q a E Q 17 0 is an Abelian group with respect to multiplication 0 Below are variations Example RaER agt0 is an Abelian group with respect to multiplication This is a subgroup of Rquot Example U1ae c all is an Abelian group with respect to multiplication This is a subgroup of Cquot 0 Below are related examples Example exp lt27rrxTlgt E C TEQ is an infinite Abelian group with respect to multiplication This is a subgroup of U1 I do not know the common notation for this one Some write QZ however this is an additive group7 notation If you accept this additive notation then it means you will also accept RZ for U1 If you are only looking at isomorphism types then such a manner is acceptable But the price you pay is this way you put yourself in a position to always identify R and the additive group R say which ordltltxgt is frequently not preferable See Fact 2 below A candidate will be U1 Example Let m be a natural number m 2 1 pm 046C 0quot1 is a finite Abelian group of order m with respect to multiplication This is a subgroup of U1 Again the additive notation for this one will be ZmZ lll Example Let p be a prime number ppoo 046C apzl forsome EN is an in nite Abelian group with respect to multiplication This is a subgroup of U1 0 The last one in the above list of examples is interesting in its own right Let Mp gtA HomAbgp I va CK X2 up a 3 X isagroup homomorphism A ppm forms an Abelian group with respect to the following multiplication For two of its elements namely for two group homomorphisms X1 Mpoo 6 CCquot X2 2 upoo E Cquot we de ne m z a lt9 mm mm mm aeupw It is straightforward to see that X1X2 is a group homomorphism thus it is an A as A element of Hpoogt Moreover with respect to this multiplication the set Hpoogt indeed forms an Abelian group What is interesting is the fact that this group A ppm is not isomorphic to the original group up A V 1 Verify that ppm indeed forms an Abelian group Identify the A multiplicative identity element 1 in the group Mm A 2 Prove the above statement ppm ppm 112 A 0 We continue with the example of up and lt up Even though the following may at first look artificial let us try the following In the above construction of A A Mpwgt out of Mpoo we replace Mpoo with Mpoogt Thus consider AA A ppm HomAbgp lt Mpoogt 3 A b ppm 6 3 b isagroup homomorphism AA lt up forms an Abelian group with respect to the usual77 multiplication For two of its elements namely for two group homomorphisms 451 We a 0 m We a 0 we de ne lt 1 2gt wear a w lt 1 2gtltgtltgt lt 1ltxgtgtlt 2ltxgtgt Xe M Now for an arbitrary Oz 6 Mpoo we may naturally de ne an element pa of as AA Mpoogt namely a homomorphism A pa 1 1 C a A as follows Let X E llpoo be arbitrary Regard it as a homomorphism X2 upoo 3 Now define AA Thus we have defined the mapping p which takes 04 E ypoo into pa 6 ppm p Mp 1 113 V1 Prove that p is a group homomorphism Also prove Ker p 1 0 Characters of a nite Abelian group The above example is actually the so called pro finite version of the following notion that is applicable primarily to nite Abelian groups We now move on to the case of nite Abelian groups From a strictly logical point of view this should have come first prior to the discussion of upco First let me introduce one new terminology and also apply one old notion not too old it was new two pages ago to a new situation These are practically nothing new except the name De nition Let G be an arbitrary finite Abelian group A group homomorphism from G to 3 is called a character of G We denote the set of characters of G as HomAbgp G 3 or simply GA Conceptually this is nothing new We may introduce the multiplication structure on this set GA in exactly the same manner as we did for ppm A in the previous page For two of the elements of GA namely for two group homomorphisms X1GgtCC XgGgtCC we define X1X2gtZG a 6 X1X2a X1agtltxgagt lta Ggt It is straightforward to see that with respect to this multiplication GA forms an as Abelian group We have seen this three times in a row Still conceptually there is nothing new 114 0 Case G um Below for simplicity we concentrate on the group G Mm where m E N is arbitrary As we will see shortly in Fact 5 below pm is not too far away from a A general nite Abelian group Now we can prove that GA Mm as a group is isomorphic to G pm A Vll Prove that the group um is isomorphic to pm Let us remind ourselves that the same is not true in general for an infinite Abelian group G as we have previously observed for G Mpoo A 0 Now let X 6 pm be arbitrary Here is an elementary but important fact Formula m when Xlgt 2 Ma OLEI m 0 when X7 1 Proof Let us recall that X 1 means that X is the character 1 Mm 3 that takes any element in pm into 1 E 3 Thus the above is obvious when X 1 because the index set um in the summation symbol is a set consisting of m elements Now when X 31 1 then X takes Cm 6 pm into some complex number which does not equal 1 Since X Cmgtm XltltCmmgt Xl 1 it follows that X Cm is an m th root of unity thus X Cm Cmy for some natural number j with 0 lt j lt m Thus the above sum becomes x1 xltCmgt XltltCm2gt XCmm 1 1 any Cm2j Gwen Cmm71j Let d denote the greatest common divisor of m and j Let md n and jd k Thus guy C7k and n and k are co prime 115 Then the above sum becomes 1 Cnk was ak Cnn71k 1 Cnk Cams Cn3k Cnn71k l Cnni1k where the number of is exactly d and the quantities inside those are identical To write the quantity inside one more time 1 Cnk was Cnfk Cnn71k Note that n and k are co prime Hence by re ordering terms this last quantity is expressed as 1 C cl2 of Hoquot1 which clearly equals 0 Hence the proof of Formula above is complete D 0 Formula above holds for an arbitrary character X 6 pm So let us choose two arbitrary characters X1 X2 6 pm and we may substitute X with X1X2 1 in the sum in Formula 2 W W 0 61quotquot 0 when X1 m m when X1 X2gta Since X2 04 is a root of unity and in particular it belongs to U1 it follows 71 7 X2 04 gt X2 04 Accordingly we have arrived at the following paraphrase of Formula above 116 Orthogonality formula of characters The original version A Let X1 X2 6 pm be two arbitrary characters of the group um Then 1 when X1 X2gt O EMm 0 when X1 31 X2 0 Now there is a dual version of the above Orthogonality Formula Orthogonality formula 7 II The dual version Let a1 042 6 pm be two arbitrary elements Then 1 when 041 a2 X HmA 0 when 041 31 042 A V111 For G um recall that GA pm is isomorphic to G um Use this fact to prove the following 1 De ne the group homomorphism p G 6 GAA for G um analogously AA to the group homomorphism p up 6 ppm which we de ned earlier Prove that p is a group isomorphism 2 Use this isomorphism p and the original version of Orthogonality Formula above to deduce the dual version of the formula above 0 The above two formulas are valid for an arbitrary finite Abelian group Such a generalization is straightforward essentially by virtue of the fact that an arbitrary finite Abelian group is a direct product of mm s see Fact 5 below 117 0 Meanwhile how to create the equivalent counterparts of the above formulas for nite non Abelian groups is not at all obvious It is not self evident to generalize the notion of characters to non Abelian groups in the first place If G is a non Abelian group then there is no natural group structure de ned on the set of group homomorphisms from G to 0 There is a way to rebuild everything from the scratch so the new framework is applicable to finite non Abelian groups and at the same time it reproduces our formulas as a special case applied to finite Abelian groups Such a framework is already at the doorstep77 of representation theory So Called irreducible representations of G turns out to be where the notion of Characters of G resides in 118 IX This exercise problem requires you to have an access to a computer mathematics software Maple Mathematica etc However any standard calculator routinely used in undergraduate calculus courses Will also do 1 Use a computer to calculate each of a C3 7 C322 b c5 7 52 i 53 94 2 c c7 72 i 93 94 i 95 i 96 2 d C11 7 11 of 411 of 7 C106 7 C11 9 C13 C132 C133 C134 C135 C136 C137 2 C138 C139 C1310 C1311 C1312 a f C17 C172 C173 C174 C175 C176 C177 C178 C179 C1710 C1711 C1712 C1713 C1714 C1715 C1716 r g C19 C192 C193 C194 C195 C196 C197 C198 C199 C1910 C1911 C1912 C1913 C1914 C1915 C1916 11917 C1918 119 The answers are supposed to be integers Then calculate what is inside the bracket in each line The exact value of what is inside each bracket is what I really want you to find If I tell you to just calculate what is inside the bracket then your computer calculator will likely give you a decimal answer and you will not realize that its square is an integer That is why I tell you to find their squares first The reason I tell you to re do calculation for the content of the bracket is to determine the sign of that quantity Now read the pattern of the signs in the above formation and define a function XpltjgtZ 17253747H397p71 E 17 71 for each prime number p with p 2 3 and make a conjecture about the value of the sum p 1 XPO by 1 g H which conforms to what you have got for p 3 5 7 ll l3 17 19 The correct forrnula may depend on the parity of p 7 l 2 In making a conjecture use the following information The sum of those terms that come with positive signs inside the bracket in each of a through g is rewritten as an 3 am 5 94 ltcgt c7 94 79 ltdgt C11 c104 of 411 1025 cu 13 car as 413 41325 c1936 n 17 M4 mg 417 41725 41736 417 417 ltggt lt19 ltlt19gt4 W 419 41925 41936 419 419 419 120 The correct formula for the sum is found by Gauss and it is called the Gauss sum Xp is called the Legendre character or the GaussiLegendre character 2 One instant application for p 5 Write explicitly 5 as in a fraction whose numerator and denominator involve integers and square root signs only In other words write explicitly each of 2 W and sin 2 7r 5 5 COS in a fraction whose numerator and denominator involve integers and square root signs only This result and the half angle formula will give us a way to write 5 gm 7r 7r or equivalently each of cos W and sin W for each m E N 3 Let m be a natural number with m 2 3 Consider the subset of C 2 mil a0 a1 Cm a2 amil a07a17a27 quot397am71 E Q Prove that QCm is a subfield of C Note that by definition an arbitrary element of QCm is written as 2 mil a0 a1 Cm a2 amil using some rational numbers 10 a1 a2 am1 but not uniquely 4 ln 3 assume m p is a prime number still assume 1 2 3 Based on the result of 1 find one integer d with d 31 1 such that i ldl is either 1 a prime number or the product of finitely many distinct prime numbers and ii W 6 QCp Recall we we and do for p 3 121 X In Planet X they use a language whose words consist of two letters only namely a and b Their words satisfy the following rule i Any word is of finite length and consists of two letters a and b some words consist only of a some words consist only of 1 ii a word can start with either the letter a or the letter b iii a word can end with either the letter a or the letter b iv you see at most six consecutive a s in any word and v you see at most two consecutive b s in any word An arbitrary sequence made of a and b satisfying the above rules through v is listed as a word7 in their dictionary In addition vi there is one special word consisting of 0 number of letters which they denote 17 which is in their dictionary Agree that their vocabulary is infinite indeed for an arbitrary m E N there exists a word consisting of exactly m letters Now in addition just like we do it sometimes up set upset counter example counterexample they know they can put two words together and write them in juxtaposition and it may yield another word7 sometimes but not always lndeed sometimes the resulting sequence of letters7 as it is violates either rule iv or rule v above So it cannot be called a word So they came up with the following deletion rule7 vii Delete seven consecutive a s or three consecutive b s and continue such a deleting procedure until you get a word that fulfills the above rules You will not see many problems with real life metaphor like this one in this course Problems with real life metaphor in mathematics are often referred to as word problems I have to say three things about it a Poorly formulated word problems7 are misleading and have no de nite translation to genuinely mathematical problems b Good or bad word problems represent one stereotype of college level mathematics c The name word problem7 sounds primitive Based on these I have a stigma that this exercise problem be called a word problem 122 For example ltaabbgt bum aabbbbab aabab aaa bb batma aaa bbbaaaa aaa aaaa 1 With this deletion rule vii they know they can always put any two words together and it yields a word that satisfies through v above They call this procedure the multiplication of two words 1 Prove that the set of words7 forms an in nite non Abelian group G with respect to the multiplication7 with 17 as the multiplicative identity element 2 We human mathematicians analyze their language and realize that it has to do with the exotic group structure7 of the set M21 More precisely we write each of their words more ef ciently as in z z z albm1a2bm2 aTme where each of 1 and mi are integers such that 0lt 1 6 16136 i2rgt lltmi 2 1r71gt 0 mT 2 To this word we experimentally assign the following complex number lt m mT zl2m1 l22m12m2 ls2m12m2 m2 H A s lt 7 7 that is expltm1mmrgt exp 27 V 71 T m m mv fi 12212 439 12 123 An easier way to understand this assignment is to describe it verbally as follows Suppose the word is called w A1 Each b appearing in the word w has the contribution 3 which is a complex number A2 Each 1 appearing in the word w has the following contribution Count the number of b s farther left to that a and then consecutively square 7 that number of times as a complex number If there is no b s farther left to that a then the contribution of that a is 7 A3 Multiply contributions from all the letters in the word w in the usual fashion in the complex number field C The resulting complex number is the number we assign to w Let us denote that resulting complex number w For example to w ababbaaaba we assign w c7csc72cscs72392392c723923927222472392 1 1 10 which is simplified to 73 that is 91 recall 7 T Now this way we have defined a set theoretic mapping b G 6 C Prove that the image of G under the mapping b equals M21 3 Define a unique group structure on the set M21 in such a way that a 1 E C serves as its multiplicative identity and moreover b b is a group homomorphism Note that the usual77 group structure of 21 as a subgroup of Cquot will not do Verify that such an exotic group structure of 121 is non Abelian Thus you have constructed a unique non Abelian group of order 21 see Table 1 4 What kind of words7 in G are in the kernel of b 124 De nition Let F be any field We may ignore the multiplicative structure of F and regard the additive structure of F as the group operation of F Indeed the group axioms G1 through G3 are all satis ed if you modify the way of writing namely if you modify the writing at y to any modify 0 to l and also modify 7 x to ab l This way F is a group F also satisfies the axiom AbG under the same modification so F is indeed an Abelian group We call this group the additive group of F Sometimes we refer to it as the underlying additive group structure of F We usually denote the additive group of F just by F Sometimes the notation Ga is used 0 Usually for a field F the multiplicative group of F and the additive group of F do not resemble each other However as we have touched in Chapter 1 l page 23 we have the following curious fact for F R Fact 2 There exists a group isomorphism b R the additive group 6 R lndeed b exp 1Y1 log Alternatively there exists a group homomorphism b R the additive group a R which is injective and lm b R In particular R the additive group 2 R 0 Below we introduce one series of groups which is extremely fundamental l said modify only for the sake of verifying G1 through G3 and AbG When we talk about the additive group of F we preserve the additive way of writing7 y 0 and em 125 De nition The general linear group For n E N n 2 l and for a field F let GL7 denote the set of n gtlt n matrices Whose entries are in F and Whose determinant does not equal 0 0411 Olin A Otij E F 7 0 04711 39 39 39 awn 0 Here is an equivalent definition of GL7F Define GL7F as the set of invertible n gtlt n matrices Whose entries are in F 0411 0417 GL7F A 041 E F and there exists 04711 awn 511 517 B a ij 6 Fa nl nn such that AB BA I For example GL2F Z GL7 forms a group GL7 is called the general linear group of degree 11 over The formula det AB lt det Agtlt det B Which look abcd F adbc7 0 familiar from linear algebra is valid over any field F and it has the following paraphrase The determinant mapping b det GL7F a F GL1F is a well de ned mapping and it is a group homomorphism 77 126 XI Let F be a field and L its subfield F 2 L Then clearly GL7F Q GL7L Verify that GL7L is asubgroup of GL7F In particular GLAR is a subgroup of GL7 Xll Verify that ab R either a7 0 or b7 0 is a subgroup of GL2R Verify that G is isomorphic to 3 GL1 To repeat GL1C is isomorphic to a subgroup of GL2 0 Even though the exponential mapping is non existent for arbitrary fields we still have the following Fact 3 Let F be an arbitrary field Then a aep is a subgroup of GL2F Prove that G and the additive group of F are isomorphic to each other In particular G is an Abelian group Xlll l Prove Fact 3 above a aEZ Prove that G is a subgroup of GL2 G is Abelian 2 Let 3 Let 1 G 0 abc Z 0 GHQ b c l Prove that G is a subgroup of GL3 ls G Abelian Probably the isomorphism type of this G is the most basic group in the world 127 XIV Verify that the exponential mapping b exp C the additive group 7 3 is a well defined group homomorphism Prove that b surjective Prove that b is not injective o The role of general linear groups and its various subgroups de ned below in mathematics is enormous When you see the definition of GL7 for the first time you may sense that this is just one example or one series of examples of a group The truth is it is way more than just an example We offer the following Fact 4 as one evidence one of many evidences indeed Fact 4 Cayley s theorem Let G be an arbitrary finite group Then G is isomorphic to a subgroup of GL7 R for some n In particular G is isomorphic to a subgroup of GL7 C for some n 0 There is a discipline of mathematics called representation theory The word representation7 refers to representations of groups primarily groups in a certain broad class 7 Lie groups 7 and of certain algebraic systems which are inseparably linked with the notion of Lie groups 7 Lie algebras 7 even though it must be noted that representation of finite groups is another big part of representation theory Mathematicians7 community broadly accepts the specific use of this word representation7 that is it means a realization of a given group G 7 either a Lie group or a finite group or some other type of a group 7 as a subgroup of GL7 F The theory of Lie groups which is sometimes but not always understood as synonymous with representation theory is a big subject by itself We will not head to this direction Let me simply point out that GL7 R and GL7 C are two representative examples of Lie groups Also some of the examples we provide below are Lie groups lndeed let F be an arbitrary eld Then an arbitrary nite group G is isomorphic to a subgroup of GLn F for some n 128 0 Fact 4 Cayley s theorem paraphrased Any nite group is representable as a subgroup of a general linear group over R and hence over C o The proof of Cayley7s theorem is elementary and it is certainly appropriate to cover in this course However it could also be treated in Math 830 Just for this reason we will not worry about it for now Also elementary representative examples of finite groups represented as subgroups of GL7 R usually fit into Math 830 contexts For this reason we will not list such examples that would otherwise be natural to be placed here Instead below we offer one exercise problem concerning a representation of one specific group of order 32 in GL4 XV We will construct one out of 51 up to an isomorphism group of order 32 as follows 1 First let in lm lm im Prove that this is a non Abelian subgroup of GL2 R of order 8 2 Second let GD DA B AB D Prove that G is a non Abelian subgroup of GL4 R of order 32 Note that wehave A B 7A 7Bgt 3 Detect two distinct subgroups of G that are isomorphic to D Though this is a subjective statement Cayley s theorem is one theorem whose proof is elemen tary elementary enough that it belongs to our standard graduate algebra curriculum yet what it states is striking 129 Q 43M iltl l l3 l Prove that this is a subgroup of G of order 8 Prove that this is not isomorphic to D 5 Let b 660 be de ned as b A 8 B B 8 A Prove that b is a group homomorphism such that b o b is the identity on G In particular b is an isomorphism Let QL Q Prove id 2 3 6 Prove that there is no subgroup of G other than G itself that contains both Q and Qi You may use the fact that the order of a subgroup of a group G divides the order of G Lagrange7s theorem Thus it suf ces to prove that any subgroup H of G that contains the union Q U QL has at least 17 elements According to the result of 5 the union Q U QL already accounts for 14 Thus your job is to offer three elements of G Which are outside of the union Q U Qi and Which have to belong to H 130 XVI This exercise concerns a representation of nite Abelian groups For notational convenience we write 041 0 0 0 0 a2 0 0 Diag Oak 0 0 0 3 0 k1 i 0 0 0 Gaul Let 1 2 7 E N be natural numbers all greater than 1 not necessarily all distinct Prove that the set Glt 17 27 agn l Diagltozkgtk1 046ka foreach 13k N forms a finite Abelian subgroup of GL7 Note that a general element of G 17 27 7 is DiagltCkjkgt1 e GLltC jke 0 1 2 k71 foreach igk l 111 0 V V n 0 02quot 0 Diag ltltCZkgtJkgtk1 0 0 5 Agree that for n 1 the group G 1 is exactly M1 131 Fact 5 l The isomorphism types of C l 2 n as n and 1 2 7 all vary exhaust the set of isomorphism types of nite Abelian groups 2 More strongly the isomorphism types of G p1j1 p2 Pnj as n varies inside the set of natural numbers 191 p2 1 all vary inside the set of prime numbers and jl jg jn all vary inside the set of natural numbers exhaust the set of isomorphism types of finite Abelian groups 0 The isomorphism type of G 1 2 n is usually denoted Ml X M XX Mn The second part 2 of Fact 5 above is easily verified once you agree with the first part 1 of Fact 5 above and the isomorphism Mp1j1p2j2pnjn 2 ulpljl gtlt Mp2j2 gtlt gtlt Hpnjn whenever 191 p2 1 are distinct primes and jl jg jn E N For example M10800 3 M16 X M25 X M27 Fact 5 above recovers Cayley7s theorem for finite Abelian groups The representation of a finite Abelian group using diagonal matrices as above is called a diagonal representation However please do not be misled that diagonal representations are the only kind of representations for a finite Abelian group For example um has the following representation 27rj 27rj cos 7 sin m m j 012m71 27Tj 27Tj sin cos m m in GL2R The proof is elementary although we will not cover it here This is a topic in Math 830 132 XVII Let A and B be the following m gtlt m square matrices 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 A a 0 0 0 0 1 11 0 0 0 01 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 B 0 0 0 0 1 171 0 0 0 01 1 Prove mil m detA 71gt and detB 71gt 2 Prove Ajy il for j12 mil Bjy il for j12 mil Am I and B 7 3 Prove that for j E Z V V T V T A 147 and B 7ltBm gt 4 Prove that the subset I A A2 A3 A4 Am l g GLWGR is a subgroup of GLm R Which is isomorphic to Mm 133 5 Concretely write down the matrix 1 m 1 m I 71 HA 71 Bl using the binomial theorem Does it look familiar 6 Let m 3 thus Concretely write down the linear combination PI IA TAZ Also concretely write down p1 11 rA2gt p1 11 rA2gtT p1 11 r142 191 TA 1142 Does the latter look familiar o I emphasize the fact that Cayley7s theorem holds for an arbitrary finite group Abelian or non Abelian 1n realizing the isomorphsm type of a finite non Abelian group G as a subgroup of GL7 C be aware that that subgroup will have to include non diagonal matrices 134 XVIII I Let f C2 6 C be the mapping de ned as fltll23w3 Here we prefer the column notation We could have written fz Let all ew l ill l lil flil Prove that G is a non Abelian subgroup of GL2 C of order 18 2 Do the same as part I for the following f C2 a C fltizigtz4w4 Define G exactly as in Describe G Find the order of G Note The group G as in problem XVIII above arisen from a function f is called the subgroup of invariants of GL2 C defined by f In the above problem in each of l and 2 G was a finite group On the contrary if you do the same for fz w 22 1112 or 22 1112 then you will obtain an infinite group as G More broadly suppose f is a polynomial function in 11 variables It is natural to ask Under what condition on f the subgroup ofinvariants G Q GL7 C defined by f is finite It may or may not be obvious but the question actually belongs to commutative ring theory According to Table I there are ve isomorphism types of groups of order 18 Two are Abelian three are nonAbelian This is one of the three nonAbelian groups of order 18 One ultimate result in this direction is known as MatsumuraiMonsky s theorem 135 0 Next we introduce the so called special linear groups De nition The special linear group For n E N n 2 l and for a field F de ne 0411 Olin SLnF A g g e GLnF detA 1 04711 39 39 39 awn Once again by Virtue of the formula det AB det Agtlt det B it follows that SL7 forms a subgroup of GL7 SL7 is called the special linear group of degree 11 over A slick way to define SL7 is SLnF Ker det GL7F a Fquot o For example b SL2F d 0 Characterization of SL2 abcd F adbcl a b Let F be any field Recall that for A C d where a b cd F its transpose is defined as T i a c A 7 b d In other words you just exchange the two entries off the main diagonal Fact 5 Characterization of SL2 Let Mf 01 SL2F A Z abcd F AMATM 136 XIX Prove Fact 5 above XX In this problem F R 1a Prove that the set of matrices cos isin sinei cos 66R forms a subgroup of SL2 Prove that this subgroup is isomorphic to U1 1b Let t it t 7 it cosht sinht ltt Rgt Establish cosh t1 cosh 152 sinh t1 sinh t2 cosh 151 152 ltcosht1gt ltsinht2gt ltsinht1gt ltcosht2gt sinh 151 152 ltcoshtgt2 7 ltsinhtgt2 1 Use these to prove that the set of matrices cosht sinht sinht cosht t6 R forms a subgroup of SL2 and 1c Prove that the set of matrices forms a subgroup of SL2 137 1 1 71 7 7 0 cos 6 7 sin 6 1 1 7 0 7 sin 6 cos 6 using 6 7 E R With 7 2 0 2a Express the matrix J in a form 62quot 1 1 27r 7 210 Express the matrix 6 1 in a form 2 1 e 7 1 62739 7 1 62quot 1 t 1 7 0 cosht sinht 627 7 1 7 27r i 1 6 1 1 0 7 sinht cosht 62739 7 1 using If 7 E R With 7 2 0 b d of SL2R Prove that there 3 Consider an arbitrary element 2 exist 6 u 7 E R With 7 gt 0 such that a b 7 u cos 7 sine r gt 0 c d 0 r 1 sine cos 4 Express the matrix 21 5 in a form 1 V W a 7 u cos 7 sine 0 r 1 sine cos using 6 u 7 E R With 7 2 0 138 5 For A it verify A2 1521 Moreover let 1 1 1 m 1 7 2 3 2m f29c 2x32x 2mI 2m 1 7 k 7 Z k 35 k0 Verify f2mA 2m 1 k Z M A k0 t2 153 1527 t2 t4 t2m72 1 t 1 gt 112x13x1 12m1 8 13x1511 277171 7 2 153 1527 0 lit 7 u 2x 3x 12m1 1 6 Write 2 WA to mean m1 Ilmf2mA that is k0 oo 1 2m 1 27M 11m 27M 1 1 k0 k m oo k0 k Verify 0 1 k sinht EVA ltcoshtgt1 k0 6t 8 6t 7 6 t 2t 0 6 6t 7 Bit Here When t 0 the quantity T should read as 1 139 In short the matrix 1 t s k arises as E 7 J y 7 k 0 k 0 t 2 34 7 Express the matrlx 0 12 3 73 i 3th k0 in a form 8 For B If verify A2 762 Moreover let 7 i2 1 3 4m f4mlt71232 4mI 4m 1 k k at k0 Verify f4mB 4m 1 k Z M B k0 62 64 64m 63 65 64m71 1 2 4 7 4m 76 3 7 5 477171 63 65 g4m71 62 64 64m 6 3 5 7 quot 477171 1 2 4 7 39 4m 140 oo 1 9 Write 7B to mean iim f4 B that is M k0 m oo 0 4m 1 k 1 k 2TB 2102 160 k0 Verify 0 1 k sin B lt cos I lt 6 gt B 7 cos 6 7 sin 7 sin 6 cos 6 39 In short the matrix cos 7 sin sine1 cos 6 E R 00 k 1 0 7amp7 arisesas 0 k0 10 The results of 3 6 and 9 may suggest that an arbitrary matrix 2 2 e SL2R may arise as 1331 215 1113 36W Decide Whether this is true or false If your answer is false then give a concrete matrix 2 in SL2 R Which cannot be expressed as f 1 t s 7 6 k k 6 7t k0 141 XXI Let aim whim WM Mamwt Maxim Mwl ow il These six matrices are all regarded as 4 gtlt 4 matrices in F Note that if F R then these all belong to the group D 8 D77 of order 32 inside GL4 R which we have constructed earlier in problem XIV l Prove M17 M27 M37 M47 M57 M6 6 2 Verify that the above six matrices and their negatives exhaust the list of 4 gtlt 4 matrices M in F satisfying the following conditions i M is skew symmetric namely MT 7M ii each row of M has one and only one non zero entry and it is either l or 71 and iii each column of M has one and only one non zero entry and it is either 1 or 71 Agree that if a matrix M satisfies ii and iii then it follows det M 74 0 3 Verify that the above six matrices form an F basis of the vector space of skew symmetric 4 gtlt 4 matrices in F 142 4 Let M be any of M1 M2 M3 M4 M5 or M6 Let 01 12 13 14 i 7 b1 b2 b3 b4 T 7 i A 7 01 C2 03 C4 1139 bi Ci E F i M d1 d2 d3 d4 Prove that Sp4 FM is a subgroup of SL4F Do it just for M M1 Use Formula on page 12 in Chapter 1 Prove that Sp4 FM does not equal SL4 Thus Fact 5 above does not generalize to yield a characterization of SL4 You may regard the containment Sp4 Q SL4 as a weak analog77 of Fact 5 The name for Sp4 FM is the symplectic group With respect to It follows that the isomorphism type of Sp4 does not depend on the choice of M M The isomorphism type of Sp4 FM is denoted as Sp4 The notation Sp4 never signifies a particular subgroup of SL4 V for each 5 Let F R For simplicity denote Sp4R for Sp4R J M j 1 2 3 4 5 6 Let H aIbM5cM2dM3 abcd R a2b202d2 1 and Hia1bM6cM1dM4 abcd R a2b202d2 1 Prove that H and HL are both subgroups of SL4R Prove H Q Sp4lR1 Sp4R4 Sp4R6 and Hi g Sp4R2 m Sp4R3 m Sp4R5 143 6 True or false question Does the equality hold in each of the two containments in 5 6a H Sp4R1 Sp4R4 Sp4R6 6b Hi Sp4R2 m swam m Sp4R5 7 0 Part 6 of problem XXI is interesting First 6a and 6b are equivalent by symmetry So only worry about 6a The question 6a is the same as solving the following system consisting of eighteen 18 quadratic algebraic equations in the l6 dimensional space 16 n R 7 alaaZaa3a4b17b27b35b47015027035047d17d27d37d4gt 1i bi Ci E R Z a1 a2 a3 a4 7 Cl 02 03 C4 7 b1 52 bs b4 i 1 d1 d2 as d4 7 1 11 12 13 14 7 b1 b2 b3 b4 7 C1 02 C3 C4 7 0 d1 12 13 d4 7 0 a1 a2 as a4 7 b1 b2 bs b4 7 d1 d2 as d4 7 0 01 C2 Cs 04 i 0 a1 a4 a2 a3 7 01 C4 02 CS 7 b1 54 52 bs i 0 d1 d4 d2 as i 0 a1 a4 a2 as 7 b1 b4 b2 bs 01 C4 C2 Cs 7 0 d1 d4 d2 as i 0 a1 a4 a2 as 7 b1 b4 b2 bs d1 d4 d2 as i 1 01 C4 C2 Cs 7 1 11 13 7 12 14 0 C1 C3 7 02 C4 0 b1 b3 b2 b4 d1 d3 d2 d4 a1 as 7 a2 a4 1 b1 bs 7 b2 b4 71 01 Cs 02 C4 d1 as d2 d4 11 13 7 12 14 0 b1 b3 7 b2 b4 0 d1 d3 d2 d4 C1 C3 02 C4 144 It is straightforward to see that for any a b c d E R satisfying a2b202d2 1 a1 a2 a3 a4 a 7b 70 7d b1 b2 b3 b4 7 b a d 0 01 02 03 C4 7 c 7d a b d1 d2 d3 d4 d C 7b a is a root of the above system of eighteen quadratic equations This is actually the content of H Q Sp4ltRgt1 SP4R4 SP4R6 This tells us that the solution set of the above system of eighteen quadratic equations is at least three dimensional Now it is dif cult to see if the above solutions exhaust the solution set This is because none of the above eighteen equations is linear Try whichever way suits you to solve the above system You will realize that it is too much to handle So the lesson is that there is a huge difference between linear and quadratic when it comes to solving systems of equations This is what I partially meant by the paragraph at the introduction of Chapter 1 The nature of systems of algebraic equations is complicated Having said 0 Mathematicians are seldom content when a complicated mathematical object is left as it is In many different mathematical contexts mathematicians find the idea of linearization useful They like to do linearization whenever applicable to reduce the level of dif culty of the problems they are tackling In our case there a linearized version of the the above system of equations Such a system does not tell you right way the solution set of the original quadratic system of equations but it instantly tells us that the original quadratic system has a solution set which is of dimension exactly three Although this is not one of the major topics we cover in this course the name of the theory that enables us to do such a trick is the theory of Lie algebra I brie y mentioned this name earlier when l remarked about the role of the general linear groups in mathematics 145 o It is very natural to seek a characterization of the group H and Hi as the intersection of some symplectic groups which re ects the solution to a system of linear as opposed to quadratic equations There such a characterization The key concept here is the unitary group The unitary group is a larger size analog of U1 namely a matrix version of complex numbers having absolute value 1 To offer a little sneak preview H and H i are mutually isomorphic as groups and their isomorphism type is called the the special unitary group SU2 an alternative notation for this group is U1 Now inside SL4 C not inside SL4 R there is a big subgroup called the special unitary group SU4 As we will later see H is characterized as the unitary part of the intersection of the three complexified symplectic groups Sp4CCj j l 4 6 inside SL4 If you replace j l 4 6 with j 2 3 5 then you will obtain Hi But this is not the end of the story You may now intersect any three of the six unitary symplectic groups and obtain twenty subgroups which are indeed all distinct as subsets of SL4 C all of which are isomorphic to SU2 only two among which are real meaning that they are inside SL4 R that are H and Hi The other eighteen subgroups exhibit an interesting kind of symmetry which can be seen from the viewpoint of the the multiplication law of the standard set of generators of the so called Clifford algebra This is closely related to group automorphisms of the order 32 group D 8 D we have constructed earlier in problem XlV 0 Now you have seen the matrices below at least twice a 7b 7c 7d a 7c 7b 7d b a d 7c c a 7d b c 7d a b and b d a 7c d c 7b a d 7b c a The first time was in problem H in Chapter 1 2 You have also seen the matrix a2 b2 7 02 7 d2 2bc ad 2bd 7 ac 2bc 7 ad a2 7 b2 c2 7 d2 2cd ab 2bd ac 2cd 7 ab a2 7 b2 7 02 d2 146 once This was in problem X in Chapter 1 l Now it is time to streamline these Remember the set of matrices a 7b b a J a b E R is in one to one correspondence with the set of complex numbers a x 71 b In that regard those 2 gtlt 2 matrices falling into this type is special7 In view of the evolution of number systems which we discussed at the introduction in Chapter 1 it is utterly natural to inquire what kind of 3 gtlt 3 and 4 gtlt 4 matrices would possibly best substitute the role of 2 71b 7 Here is my answer The 3 gtlt 3 matrix a2 b2 7 02 7 d2 2bc ad 2bd 7 ac 2bc 7 ad a2 7 b2 02 7 d2 2cd ab i 2bd ac 2cd 7 ab a2 7 b2 7 02 d2l and the two 4 gtlt 4 matrices a 7b 7c 7d a 7c 7b 7d b a d 70 7 and L l ldi bil d 7b c a are naturally the correct generalization of 2 71b The above 2 gtlt 2 3 gtlt 3 n 9 l amp w and 4 gtlt 4 matrices are considered to be all in the same family they exhibit the same characteristic or they share the same DNA type Here I am not saying that the set of the above 3 gtlt 3 matrices forms a number system which is larger than C 147 o The skew eld axioms The skew eld of quaternion numbers H The obstacle is that the entries of the above 3 gtlt 3 matrix are quadratic in a b c and d So let us instead look at one of the two 4 gtlt 4 matrices above We need to choose one out of the two so let me randomly pick a 7b 70 7d b a d 70 a 7d a b J 39 d c 7b lhad to toss a coin for this As a b c and d vary inside R these matrices form an algebraic system which satisfies all the field axioms except one the commutativity of multiplication To be precise the set a 7b 70 7d b a d 70 a 7da b l abcd R d c 7b a endowed with the usual matrix addition and matrix multiplication satisfies the following set of axioms called the axiom of skew fields SkewFl a Q Q a SkewF2 a 5 lta gt v AM WM SkewF3 lta gtv av m alt vgt a m SkewF4 0aa laala SkewFB 04 704 0 04071 07104 1 the second part of SkewFB assumes a 74 0 The algebraic system H is called the skew eld of quaternion numbers or alternatively the divison ring of quaternions or the quaternion algebra The reason for the naming quaternion will be explained shortly see also footnote next page 148 0 Agree that if the commutativity law for multiplication oz n is added to the above SkewFl through SkewF5 then the whole set of axioms will precisely be the field axioms However for H the commutativity law for multiplication fails Thus H is not a field This will become transparent once we see the following basic multiplication laws77 of H which actually offers an easier way to understand the multiplicative strucutre of H Denote 1000 07100 1i0100l i1 0 0 ol 0010 0001 l0001l l00710l 00710 00071 7i00071l k7i0010l l 1000 07100 l0100l l1000l In our old notation l is I i is M5 j is M2 and k is M3 Then inside H the following hold ji7ijk kj7jki ik7kij i2 j2 k2 Now these rules completely dictate the multiplicative structure of H lndeed an arbitrary element of H is written uniquely as a1bicjdk ltabcd Rgt W Hamilton 180571865 discovered H by means of offering these multiplication laws his version is ij ijz39 k jk ikj 239 ki iik j but is equivalent to ours His construction of H was symbolical meaning he did it without use of matrices Indeed it was before matrix algebra became in common use Here is a quote of David F Anderson an algebraist who is knowledgeable of Hamilton s career On the evening of October 16 1843 a man and his wife were strolling along the Royal Canal in Dublin He had a flash ofgenius took out his penknife and carved the equations 2392 j2 k2 ijk 71 on a stone on the Brougham Bridge The man was Sir William Rowan Hamilton and these equations de ne what is now called the division ring of quaternions a number system which behaves like the real and complex numbers but multiplication is not commutative 149 So to recover the full multiplication laW on H all you have to do is form 11 bi cj dk 11 m cj wk and then distribute Here keep in mind that the multiplications of L j k are non commutative so When you distribute you must do it as in matrix algebra way Treat a b c and d as scalars multiplied to matrices To perform a1bicjdk a1bic jdk lt gt gt aa 11 lb10 ac lj ad 1k w 1 by m by m W m caj1 cb N 00 jj cd da k1 db m dc kj dd kk aa1abz lCj adk w z bb 71bc kbdj caj cbk cc71cdiz dak dbijdcz dd71 aalabzacj Fadls ebb1bai bdj bck icel fadz caj cbk ddl dci dbj dak aaibbicciddgtl abba7cddcgti acbdca7dbgtj ad7bccb da gtk 150 To highlight the result lta1bicjdkgt a1bicjdkgt aa7bb7cc7ddgt1 abba7cd dcgti acbdca7dbgtj ad7bccb dagtk As for the reciprocal as long as a b c d E R satis es a2 b2 02 d2 7E 0 namely as long as at least one of a b c d is non zero the reciprocal of 11 bi cj dk existsin H andit is 11 bi cj dkyl l i 4 Wlta1 bl w dkgt 0 Most importantly agree that the above are an exact paraphrase of the following Fact 6 l The two matrices a 7 b 7 c 7 d a 7 b 7 c 7 d b a d 7 c b a d 7 c c 7 d a b and c 7d a b d c 7 b a d c 7 b a multiply to yield a matrix a 7b 7c 7d b an d 7 CH CH 7 d a b d C 7 b a where 151 2 has an inverse Where bH d aa 7 bb 7 cc 7 dd ab ba 7 cd dd 10 bd ca 7 db ad 7 bd CH dd Provided a2 b2 02 d2 31 0 the matrix b d a 7 b 7 c 7 d b a d 7 c c 7 d a b d c 7b a 7 b 7 c 7 d b a d 70 c 7 d a b d c b a a 7b a2b2c2d2 a2b2c2d239 152 o The multiplicative group H Just like we have de ned the multiplicative group Fquot for a field F we can de ne the multiplicative group H for H even though H is not a field but it is only a skew field First as we have observed any non zero element of H has the reciprocal in H with the reciprocation rule a page ago Taking this into account we agree that the multiplication rule also a page ago of two elements of H completely dictates the group structure of the set Hquot H 0 We call Hquot the multiplicative group of H It is clear from the skew field axioms that any skew field has the multiplicative group structure defined on the complement of 0 in that skew field However note that unlike the case of Fquot where F is a field the group Hquot is non Abelian To repeat the definition of Hquot Ha1bicjdk H abcd R a2b2c2d27 0 The group H has an important subgroup defined as the subset of H consisting of elements 11 bi cj dk such that a2 b2 02 d2 equals 1 U1Ha1bicjdk H abcd R a2b202d21 a 7b 70 7d 39 39 39 4 b a d 70 Under the identi cation of a1 bi Cj dk With 0 7d a b a d c 7b this latter group U1 is exactly H in problem XXI In particular U1 is representable as a subgroup of SL4 153 0 Recall that at the starting point of our discussion we made a random choice between two matrices namely I chose a 7b 70 7d b a d 70 A T c 7d a b d c 7b a over a 70 7b 7d L 7 c a 7d b A T b d a 70 d 7b c 1 Here the attachment of 17 does not mean that the one with 17 is a subordinate of the one without The roles of A and AL are completely symmetric there is no intrinsic distinction between the two Indeed the above theory that worked for the matrices of A type goes word by word for the matrices of AL type The set of matrices of AL type is identified with the skew field H In short the set of 4 gtlt 4 matrices whose entries are in R carries two replicas of H inside Each replica has the group U1llll and those are exactly H and HL in problem XXI o The 3 gtlt 3 analog Now I finally came to the stage of explaining the link between the 3 gtlt 3 matrices oftype a2 b2 7 02 7 d2 2bc ad 2bd 7 ac 2bc 7 ad a2 7 b2 02 7 d2 2cd ab 2bd ac 2cd 7 ab 12 7 b2 7 02 d2 and the skew field H I made the statement The above type of a 3 gtlt 3 matrix shares the same DNA type77 as the two types of 4 gtlt 4 matrices we just discussed above Below is my justification 154 o The multiplication of two of the 3 gtlt 3 matrices of the above type is exactly dictated by the rule the same rule that dictated the multiplication a 7b 7c 7d a 7b 70 7d b a d 7c b a d 70 c 7d a b c 7d a b d c 7b a d c 7b a More precisely consider a2 b2 7 02 7 d2 2bc ad 2bd 7 ac 2bc 7 ad a2 7 b2 02 7 d2 2cd ab 2bd ac 2cd 7 ab a2 7 b2 7 02 d2 and a2 b2 7 02 7 d2 2bc ad 2bd 7 ac 2bc 7 ad a2 7 b2 02 7 d2 2cd ab 2bd ac 2cd 7 ab a2 7 b2 7 02 12 Multiply out these two matrices the second one from the right of the first one Then it Will come out an 7 b2 7 CH2 7 d2 2bncn 7 andn 2bndn 7 anon 2bncn 7 and an 7 b2 7 CH2 7 d2 2Cndn 7 anbn 2bndn 7 anon 2Cndn 7 anbn an 7 b2 7 CH2 7 d2 Where a b c and d are exactly determined by the same formula 155 d dd 7 bb 7 cc 7 dd b db bd 7 cd dd 0 10 bd ca 7 db d dd 7 bd CH dd To offer the same statement in a little more memorable way Fact 7 For a b c d E R4 de ne d2 b2 7 02 7 d2 2bc dd 2bd 7 dc Altabcdgt 250 7 ad a2 7 b2 c2 7 d2 2cd db i 2bd dc 2cd 7 db 12 7 b2 7 02 d2l Then for a b c d and d U 0 d E R4 we have Alta7b767d Aa7b7C7d Aa7b7C7d where d dd 7 bb 7 cc 7 dd b db bd 7 cd dd 0 10 bd ca 7 db d dd 7 bd CH dd 0 Proof of Fact 7 is nothing but calculation The job is actually a little tiresome However there is a modestly ef cient way to carry it out Problem XXII below 156 XXII Prove Fact 7 A separate work sheet will be provided XXIII I In the skew field H solve the quadratic equation 2210 2 In an arbitrary skew field write a general form of a linear equation with unknown 2 Note that terms 0425 and 0 23 cannot be combined7 o The orthogonal groups The unitary groups We have observed that there is a parallelism between the multiplications of 3 gtlt 3 matrices of a certain type and the multiplications of 4 gtlt 4 matrices of a certain type This observation as it is is nothing more than just some nifty matrix algebra You may wonder if there is a group theory interpretation of the above observation There is actually a clear cut one line re formulation of Fact 7 relying on the notion of group homomorphisms The key to obtain such a re formulation is the notion of orthogonal groups De nition The orthogonal group over R For n E N n 2 I define all aln 0 A g g e CLAIR AAT I anl 39 39 39 awn In the above definition AAT I can be substituted by AT A I 0 To understand this de nition let us work with the case n 2 first Let A 2 b Then we can honestly write out the condition AAT I 2 Z 2 SI 5 3 157 That is That is a2b2 acbd acbd 02d2 To conclude the condition AAT I reads a2b2l acbd0 02d2l Note a b c d E R Thus from the first and the third equations it follows that there exist 739 E R and 6 E R such that acosT b7sin r csin6 d7cos6 The negative signs for b and d do not have to be there ljust want it to be there Then the equation in the middle reads ltcosTgt sine ltsin r ltcos6gt 0 By addition formula Chapter 1 2 this becomes sin 6 739 0 Hence a 739 e 0 in i27r i37r i47r or the same to say 158 67Tm7r for some m E Z Thus lt 7 cos 6 sine or COST SHIT lt cos 6 7 sin 6 In other words either i a b 7 cos T 7 sin T or c d 7 sin T cos T lt gt a b 7 cos 6 sin 6 H c d sin 6 7 cos 6 To conclude Fact Description of 02 02 cesT 7s1nT T R sinT cosT cos sine U sin 7cos6 gER 0 We can paraphrase the statement as follows Fact Alternative description of 02 0 Characterization of On Let f R 6 R be the mapping de ned as 961 1177 Again we are using the column notation Let all aln G g g e GLMR anl awn all aln 1 1 anl awn 71 71 Prove that G On 160 0 Let us return to SL2R We W111 construct the inverse mapping of 0 7amp7 cos isin eXp396 0 66R sm6 cos 66R 1 exp A Z 7 Al k0 The target of the mapping exp is a subgroup of SL2 R called 502 exp is a group homomorphism from the additive group to 502 gtkgtkgtk Iwilladdmore gtkgtkgtk 161 2 THE CATEGORY A1 OF AFFINE LINES 0 As announced we create a new category Below we choose to work over the complex number field C However much of the theory goes without a change even if you opt to work with an arbitrary algebraically closed field F instead of C De nition of the category A1 We de ne the new category A1 as follows a An object7 of the category A1 is C We use the notation A1 for C whenever it is regarded as an object of A1 We call A1 an af ne line b A morphism7 A1 E A1 is a polynomial mapping 0 This category has the following familiar Feature7 i Composition A1 L A1 A1 L A1 polynomial mappings gt A1 A A1 a polynomial mapping ii ldentity A1 i A1 a polynomial mapping 0 The above Feature7 may be trivial7 However the following problem is worthy to consider Suppose that a mapping f C 6 C satisfies the property that f o f is a polynomial mapping Then is it true that f is a polynomial mapping T 1 Offer an example of a mapping f C a C such that f o f is a polynomial mapping whereas f is not a polynomial mapping 2 Offer an example of a mapping f R a R such that f o f is a polynomial mapping whereas f is not a polynomial mapping 3 Offer an example of a mapping f R R such that f o f is a polynomial mapping with Q coef cients whereas f is not a polynomial mapping and moreover fQ Z QR 162 Theorem 1 Let f A1 a A1 be a morphism in A1 Then there exists another morphism g A1 E A1 in A1 such that gofldA1 and fogldA1 if and only if there exist a E 3 and 6 C such that fz a2 Q 0 Any morphism f satisfying one of the tWO equivalent conditions in the statement of Theorem 1 above is said to be invertible ll Prove Theorem 1 Theorem 2 Assume that tWO morphisms szl Al and ngl Al in A1 are both invertible Then the following hold a Their composite g o f is also invertible b By assumption f and g are written as fz az and 92 yz6 using some 04 y 6 E C Then 9 o f is expressed as lt90fgtlt2gt A2 M Where A and M are the complex numbers uniquely decided by the identity A M v 5 a B 0 1 0 1 0 1 in GL2 In short the composition laW of invertible polynomial mappings and 1 the multiplication laW of matrices of type 3 J are entirely parallel 163 Ill Verify that the set of matrices a C C is a subgroup of GL2 0 In View of problem Ill we offer the following paraphrase of Theorem 2 Corollary Let Aut A1 f A1 E A1 f invertible This set forms a group with respect to the composition The group is isomorphic to 04 0 1 0 In this context here are two questions one may naturally ask a e 0 5 6 C Q GL2CCgt Question 1 Can we create a category which contains our A1 as a subcategory and in which we can establish a group isomorphism between the group of invertible7 morphisms in that enlarged category and the general linear group GL2 CC Question 2 Does there exist an algorithm which resembles matrix multiplication that dictates the compositions of non invertible f in the category A1 0 I would like to brie y discuss the higher dimensional analog We will eventually look at theories in dimensions greater than 1 Pointing out the following at this stage is worthwhile because it suggests the non triViality of our subject 164 Remark By analogy we may consider the category A2 An object in A2 is 2 which I redenote as A2 A morphism in A2 is a pair of two variable polynomial mappings lt fz w 92 w Then the notion of invertibility in the category A2 makes perfect sense Now by analogy you may somewhat naively suspect that lt fz w 92 w gt 58le l3 l lil lil where 04 y 6 A and M are constants in C such that 3 E GL2C of form would be the only invertible morphism in the category A2 This is false Here is a counterexample f z w z 92 w w z o lnvertible morphisms in the category A2 are completely classified and they are called De Jonquierre transforms IV Verify that Hz w 92 111 as in is invertible in A2 by concretely writing out the inverse Also find the determinant of the Jacobian matrix as a function on 2 and w 165 Z w 22 w gzw T 27 lt22wgt2 Regard this as a morphism in A2 Can we write this morphism as a nite times composition of morphisms of type fz w 2 92 w w where is some polynomial which does not involve w but only 2 and type fz w 04 2 A 92 w y 6 w M where 04 y 6 A and M are constants in C such that 3 E GL2C 7 Conclude that the morphism in is invertible in A2 2 Let fz w 22 w gtkgtk 92 w z 7 23 7 222w 7 1112 The same question as in Rely on the Jacobian matrix Conclude that the morphism in is not invertible in A2 0 Now the same question for A3 or even for A with n 2 4 makes sense That is describe a general invertible morphism in A3 or in A n 2 4 This is an open problem Have I already convinced you that the notion of polynomials looks innocent but it is non trivial 0 To address Questions 1 and 2 it is crucial to look at the set of polynomials l must add to the best of my knowledge 166 3 THE CATEGORY OF COMMUTATIVE RINGS De nition A set B is called a commutative ring if B has tWO distinguishable elements 0 1 0 31 1 and three operations one called negation Which takes a E B into 7a E B one called addition Which takes 04 6 B into a B E B and one called multiplication Which takes 04 6 B into 045 E B such that the following CR1 through CR5 hold CR1 a a a a CR2 a lt vgt lta gt alt vgt a gt CR3 alt v a m CR4 0aa 1aa CR5 altiagt0 o A eld is a commutative ring If B 0 forms a group With respect to multiplication then B is a field In this sense a field is an example of a commutative ring On the other hand not all commutative rings are fields See problem o The adjective commutative7 in the name commutative ring7 is a re ection of the fact that 045 Ba the commutativity of multiplication is a part of the axioms the second part of CR1 1 Verify that the set of integers Z is a commutative ring Z is not a field This can be the de nition of a eld namely a eld can be de ned as a commutative ring B such that B 0 forms a group With respect to multiplication 167 ll Let B be an arbitrary commutative ring De ne Bquot aEB thereexists 56B suchthat a l Verify 0 Z B Verify that Bquot forms a group Verify that B is a field if and only if 13 B0 0 Below is the utmost important fact Fact Let 2 denote the set of polynomials in z z adzd adL12d 1 0412 040 0411504171 mahao 6 C d E N Then C forms a commutative ring with respect to the usual addition and multiplication of polynomials We call C the polynomial ring over C o The category of commutative rings Here we introduce the category of commutative rings The category of commutative rings and the category A1 are inseparably linked as we will see shortly We have already defined its objects We now need to define its morphisms De nition Let A and B be commutative rings A mapping b A a B is called a ring homomorphism if Rm Ma 5 Na w and Rm New Ma 445 both hold in B for arbitrary 04 6 A and moreover RH3 451 1 168 o Commutative ring theory 7 The category of commutative rings In commutative ring theory you work with the category of commutative rings The category of commutative rings is a system consisting of a an object 7 B which is a commutative ring and b a morphism 7 A E B which is a ring homomorphism o This category has the following feature7 Feature i Composition B1 L B2 and B2 L B3 are both ring homomorphism gt B1 ampgt B3 is a ring homomorphism ii The identity B A is a ring homomorphism 0 There is a notion of so called isomorphisms7 for commutative rings which is parallel to the notion of isomorphisms7 for groups De nition A ring isomorphism is a ring homomorphism b A a B which is bijective Ill Verify that for a ring isomorphism b A 6 B its inverse mapping 1V1 B a A is a ring isomorphism 0 When a ring isomorphism b A a B exists for two commutative rings A and B then we say that A and B are isomrphic to each other Write A 2 B The notion of two commutative rings being isomorphic is a re exive symmetric and transitive relation i B 2 B always ii B1 2 B2 implies B2 2 B1 and iii B1 2 B2 and B2 2 B3 imply B1 2 B3 169 o Kernels Images Let b A 6 B be a ring homomorphism We de ne its kernel and its was follows Ker a A 1504 0 lm a B aeA o The following de nition is of primary importance De nition Ideals Let B be a commutative ring A non empty subset I Q B is called an ideal of B if ldel 04 B E I holds Whenever Oz 6 I B E I lde2 045 E I holds Whenever Oz 6 B B E I Two trivial examples of ideals are 0 called the zero ideal and B itself Later we Will use alternative notations for these two ideals of B as follows i 0 for 0 and ii 1 for B At this stage let us not worry about the meaning of these notations Fact Let b A a B be a ring homomorphism Then Ker b is an ideal of A Moreover Ker b 31 A IV 1 Mimic the definition of the underlying additive group structure of a field to define the underlying additive group structure of a commutative ring 170 2 Let B be a commutative ring and I its ideal Verify that the following holds 704 E I whenever Oz 6 I In particular I is a subgroup of the additive group B For this matter you need to first prove the following general fact valid inside any commutative ring B lt71gta 0 whenever aEB 3 Prove Fact above De nition Subrings Let B be a commutative ring A non empty subset B0 Q B is called a subring of B if SR1 04 B 6 B0 holds whenever Oz 6 B0 6 B0 SR2 704 6 B0 holds whenever Oz 6 B0 SR3 045 6 B0 holds whenever Oz 6 B0 6 B0 SR4 1 6 B0 0 Note in particular that B0 is a subgroup of the additive group B o A trivial example of a subring of B is B itself Note that the set 0 is never a subring of B Fact Let b A a B be a ring homomorphism Then 1m b is a subring of B 0 Now we have seen two definitions One ideals and two subrings These two notions look alike However they are fundamentally different Indeed we have 171 V Let B be a commutative ring Verify that the only subset of B which is an ideal of B and at the same time a subring of B is B itself Fact Characterization of a eld 1 Let B be a commutative ring Then B is a field if and only if the only ideals of B are the two trivial ones 0 and B 2 Let b B1 6 B2 be a ring homomorphism Suppose B1 is a eld Prove that b is injective In other words prove Ker VI Prove Fact above 0 Meanwhile a field F can have non trivial subrings lndeed any subfield of F is a subring of F Example B Z This is a commutative ring Examples of ideals of B other than the WM ones are 22a B a B 33a B a B 44a B a B 55a B aEB and so on More generally let m be an integer m 2 0 regarded as an element of B Z Let mma B a B This is the only possible form of an ideal of B Z We will eventually see proof of this fact but not now Meanwhile the only subring of B Z is B itself 172 Example B This is a commutative ring Examples of ideals of B other than the trivial ones are zz B EB lt271gtlt271gt B B 22gt22feB 5 EB ltzswgtltzswgtswiwa and so on More generally let 7 be any polynomial having 2 as variable and having CC coe icients regarded as an element of B Let nn B EB It turns out that this is the only possible form of an ideal of B Again we will eventually see proof of this fact but not now Meanwhile the following are examples of subrings C a0 constant polynomial 040 E C the coef cient field 22 OddZZd 0411122 1 2 04122 040 04d04d71quot395041040 E C d E N a or more generally zm adzmd ad712md71 01le 00 Otd adil a1a0 e C d E N 173 Also any subring of the coef cient field C is a subring of Vll In the commutative ring B Cz consider the subset B0 consisting of elements of form adzd adnzd l 222 040 with ad 04114 a2 040 E C namely elements d dil 2 adz 1712 0422 0412 are with 041 0 Prove that B0 is a subring of B We usually denote the subring B0 by C 22 23 The rationale for this notation is that any polynomial n22 23 where n 71u v is a two variable polynomial having C coef cients which belongs to B0 is written as Vlll In the commutative ring B Cz consider the subset B1 consisting of elements of form 712 22 22 23 where n 71u v is a two variable polynomial having C coe icients Prove that B1 is a subring of B We usually denote the subring B1 by Cz 22 22 23 IX In the commutative ring B Cz consider the subset B2 consisting of polynomials having R coe icients namely adzd ad71zd 1 112 a0 where ad 1111 a1 a0 6 R Prove that B2 is a subring of B We usually denote the subring B2 by R We could have de ned R independently of C 2 Then we could have declaired that the former is isomorphic to a subring of the latter Rh 2 Rz Q Czi 174 o Quotient of a commutative ring by an ideal In commutative ring theory we do not consider the quotient of a commutative ring B by its subring The quotient makes sense as a mere additive group it does not inherit a commutative ring structure from B Rather we consider the quotient of a commutative ring B by its ideal I Thus we consider BI This one does carry a commutative ring structure De nition Let B be a commutative ring Let I be its ideal Consider the equivalence relation on B a E B ltgt a 7 6 I Then the set of equivalence classes BI MI aEB carries a natural structure of a commutative ring Indeed the addition and the multiplication are a1 1a 1 041 5 a 139 ALso 1 serves as 17 in the commutative ring BI I 0 Most importantly the natural surjective mapping 7r B a B I that takes a E B into 04 E BI becomes a ring homomorphism Theorem Isomorphism theorem Let b A a B be aring homomorphism Then b induces aring isomorphism AKer a Im Or the same to say b induces an injective ring homomorphism AKer a B whose image is the same as the image of the original homomorphism Im 175 Example Let Rz denotes the subring of 2 consisting of those polynomials having R coef cients Define b R a C as follows d d J b 2 1sz 2a V71 1 ER j0 j0 To describe b verbally b takes a polynomial with real number coef cients and with variable 2 into the complex number arisen as a result of substituting z with x 71 77 In particular 151 l and 152 xil Then b is a surjective mapping Indeed the images of polynomials of form 12 b a b E R under b exhaust C Moreover it is straightforward to see that b is a ring homomorphism Clearly its kernel is Ker 22 1 Thus by the isomorphism theorem b induces the following isomorphism of commutative rings Rz22 1 2 C Notational remark Here I make one notational remark As we proceed further we deal with polynomials having a larger number of variables Remember that in Chapter 1 we set a convention to use the letters at and y for real variables and the letter 2 and occasionally w for a complex variable In this Chapter up to this point this convention was carried over However we already experience some dif culty to keep up with this In the above 2 as opposed to x was used for the variable of the polynomial ring This was essentially because it was to our advantage to regard the set of polynomials with real coef cients as a subring of the ring of polynomials with complex coef cients and 2 was used for the variable of the latter 176 Now I would like to relax the above convention It is going to become more and more cumbersome to consciously avoid the letters at and y So from now on even when we deal with polynomials with complex coef cients we frequently use at and y for their variables Sometimes we need more letters for extra variables so we use at y z w and occasionally 5 u or 1 too For 11 variables we use 1 7 or yi yn and so on 0 Having said the above isomorphism is rewritten as Rxac2 1 2 C 0 In Chapter 1 we have de ned the complex number field C in at least three different ways plain way Declaration 7 l and Declaration 7 ll Here is the fourth way Declaration 7 III We hereby de ne the complex number eld C as the commutative ring R x2 l C R x2 1 This is indeed a eld By abuse of notation we denote the class of a constant polynomial a where a E R in the eld C by 1 Keeping this notation intact the subset of C consisting of a where a E R forms a subring of C which is isomorphic to R We denote the class of x in C by xTl It follows that an arbitrary element of C is written uniquely as a xle with a b E R Inside C the following holds 71 77 o If we adopt Declaration 7 III77 above as the definition of C then the commutativity of the multiplication of C is self evident lndeed C is defined as the quotient of a commutative ring by an ideal which is a priori a commutative ring This is in good contrast with Declaration 7 I H77 in Chapter 1 where we had to prove the commutativity of multiplication of C 177 0 We need the notion of the polynomial ring with two variables De nition Let C am y be the set of polynomials with two variables x and y lts elements are of form Otij 041739 E i0j0 Note that there are only finitely many terms in the sum This set C am y forms a commutative ring with respect to the usual77 addition and multiplication of polynomials We call it the polynomial ring in two variables over C We now have to call C the polynomial ring in one variable over C o The polynomial ring C am y contains C and C y as subrings o It is self evident as to how to generalize this to the three or more variables De nition Let C 1 gun be the set of polynomials with 11 variables 1 2 at Its elements are of form 7 r2 r1 2 Z Z 1 1 22 M E C 250 2392 0 2391 0 This set C 1 gun forms a commutative ring We call it the polynomial ring in 11 variables over C This definition includes Ch and C y as two special cases namely 11 l and n 2 respectively X Establish the following isomorphisms l Ct2 t3 2 Cac y 203 7 242 2 Ctt2 t2t3 2 Cac ylt37y7y2gt 178 0 Now we explore the connection between the category of commutative rings and the category of A1 Below is an extremely curious fact which offers the first evidence that the two categories are related Observation The set G or Ch Ch 15 ring isomorphism blc ldc forms a group with respect to compositions This group is isomorphic to 3 f This is intriguing This suggests that there might be some hidden governing a e o n e c g GL2C principle that we have not seen yet We are eager to understand it For this our first task is to carefully look at the condition imposed on b in the above set G Below is seemingly a superficial definition On the contrary it is actually vital for us to accept the definition to recognize the importance of the concept underlying the definition and to substantially rely on it in our subsequent discussion 0 The category of Balgebras Where B is a commutative ring In the above we defined the set of ring isomorphisms from C to C itself such that gt ltzgtC Ido This forms a group G and we call it the automorphism group of C over C Let us call its element an automorphism of C over C or alternatively a C algebra automorphism Now what is really an equivalent translation of the condition which is more appealing to us Yes the condition is nothing but the requirement that the image of the single element x in Ch under b completely determines77 the mapping b 179 o Substitution principle A little more precisely suppose b takes the element x into a polynomial n E Now let 5 ad d badil iilquot 0422041040 be an arbitrary polynomial in Then the image of the same polynomial 5 under b is the polynomial substitution Otde Oldil dil 042712 0417 040 This follows directly from the axiom of ring homomorphisms RHl through RH3 We would call such a mechanism the substitution principle As you may wonder we could have instead considered the entire set of ring isomorphisms from C to C itself Such a set forms a group We would call it the absolute automorphism group of C This is much bigger a group than G the automorphism group of C over C The reason is as follows The set consisting of ring isomorphisms from C to C itself forms a group It is called the automorphism group of the field C Any element in the automorphism group of C naturally extends to an absolute automorphism of C Let us be concrete Let us take what is probably the most representative example of an automorphism of C the complex conjugate automorphism UzC C defined as 039 a a This and Idc are actually the only R automorphisms of C Now to an arbitrary polynomial awed adiwd l 042962 04196 are in Cx we may assign the following new polynomial 071961 T1436 cm cm V0 180 Such an assignment de nes an automorphism of the polynomial ring x which we may also denote as 039 by a slight abuse of notation U C a This automorphism 039 is outside of the group G which as we saw is isomorphic to the automorphism group of A1 in the category A1 If we count this type of automorphisms namely those automorphisms of C that are not dictated by the aforementioned substitution principle then we will not get the correct parallelism between the object A1 of the category A1 and the polynomial ring C as an object of the category of commutative rings Since our focus is such a parallelism it is in our interest to consider only those automorphisms of C that follows the substitution principle In other words we only look at those automorphisms b C a C satisfying blc ldc in what follows 0 Let us be self critical on the category we are currently working on that is the category of commutative rings As long as we stay in the category of commutative rings we will not be able to exclude those automorphisms that do not follow the substitution principle if your standpoint is that all the key definitions pertaining to homomorphisms such as the automorphism groups should be extracted formally or categorically from arrows7 Actually this is the standpoint we take This leads us to the following concept Let us return to the general set up o The category of A algebras De nition Let A be a xed commutative ring We define the category of A algebras as a system consisting of a an object 7 B which is a commutative ring together with a ring homomorphism 153 Z A 6 B called the structure homomorphism of B as an A algebra and 181 b a morphism 7 11 Bl Bg Which is a ring homomorphism compatible With77 the structure homomorphisms bl of Bl and 2 of Bg namely a ring homomorphism 11 such that the composition of A nial 31 and B1 ab B2 coincides With A Bg In short 11 0 Z531 4532 We call a morphism in the category of A algebras an A homomorphism o This category has the following feature7 Feature i Composition Bl L 32 and 32 L BS are both A homomorphism gt Bl ampgt BS is an A homomorphism ii The identity B A is an A homomorphism De nition An A isomorphism is an A homomorphism 11 Bl 6 32 Which is bijective 0 Every commutative ring carries a Z algebra structure Now we claim What is in the title Every commutative ring carries a unique Z algebra structure Moreover every ring homomorphism is a Z homomorphism 77 XI Make this statement precise Prove the statement 182 0 Let us return to the discussion of automorphisms of First of all Ch as a commutative ring carries a C algebra structure This is because C is a subring of Hence the natural injective mapping C 6 Ch serves as the structure homomorphism of Ch as a C algebra Now we de ne AutoAlg C gt as the set of C algebra isomorphisms from Ch to Ch itself Clearly this is a group with respect to composition We call it the C automorphism group of Ch Now our Observation above is much more neatly phrased Observation The E automorphism group of the C algebra Ch is isomorphic to the following subgroup of GL2 C 3 f a e o 5 e c g GL2C x11 1 Let A H 6 M a i y E C A E C Q GL3C 0 l OQQ Prove that H forms a subgroup of GL3C This H is called the group of af ne linear transforms on A2 2 Let G AutCAlgltCac 24 2 Cx y E Cx y 15 ring isomorphism blc ldc be the group of Eautomorphisms of the C algebra C y ldentify one subgroup of the group G which is isomorphic to H Does that subgroup equal the entire group G 7 If the answer is no then give one concrete element of G which is in the complement of that subgroup 183 0 Next for a reason which is undisclosed for the time being we will look at the following analog namely the set of C algebra homomorphisms not from Ch to Ch itself but this time from Ch to C Such a set is not a group any more So it is just a set Category wise we once again rely on the category of C algebras which serves as our source category From the categorical point of View there is one unlikely thing that you might not have expected to see here For the first time the category of sets is actually in use In fact the category of sets will serve as our target category So it will complement the role of the category of C algebras De nition Define HomOAlg C at C as the set of C algebra homomorphisms from Ch to C Here we regard C as a C algebra with the identity homomorphism over C as its structure homomorphism 0 To repeat the above HomOAlg C at C b Ch E C 15 ring homomorphism blc ldc 0 Key Fact Substitution principle II b E HomOAlg C at C is uniquely determined by Mac 6 C Namely for 451 m e HomCAlg CM cc 451 m if and only if 1 M35 Proof Indeed since b is a ring homomorphism over C 041 n i0 184 0 An important consequence 7 The notion of regular functions Recall that A1 stands for C as a set An important consequence of Key fact above is as follows There is a natural bijective mapping ev HomOAlg ltC C E A1 which is given by 152 C a C gt gt o A little psychology The above consequence makes us feel reasonable enough to view elements of C as regular functions on A1 We may call C as the ring of regular functions on A1 0 We push this direction a little further I hinted earlier that C and A1 are the ip sides of the same coin They reside in two different categories whose natures are completely different 7 one in the category of CC algebras and one in the category of polynomial mappings A1 However the two are inseparably linked as though they are mirror images of each other The discussion below will make it precise The key word is Duality o The word duality One feature of abstract algebra is the existence of the notion of duality The notion of duality plays an important role in many different contexts in abstract algebra Strangely enough there is no clear cut mathematical de nition of the word duality7 We have seen a kind of duality in the group theory characters of nite Abelian groups Now we observe a slightly different type of a duality in the commutative algebra context 185 Duality x lt gt A1 Let b C y a C be a C algebra homomorphism Here we could have denoted the both rings as C In the following argument we need to carefully distinguish the source ring of b and the target ring of 15 So for convenience we assigned two different letters Then there is a naturally induced morphism in the category of A1 43V A1 6 A1 This morphism is the polynomial mapping My which is an element of Che regarded as a polynomial mapping with variable x Here again we will be better off if we distinguish the two A1 s namely the source A1 and the target A1 Otherwise it will be confusing As we did in the previous page we regard Ch as the ring of regular functions of the source A1 and regard Cy as the ring of regular functions of the target A1 Not the other way round It is a good idea that we attach the subscripts x and y to the two A1 to clarify this So the above W is written as 43V A a A The subscript x in A is understood as the name of the coordinate So A is A1 with x as its coordinate Similarly A is A1 with y as its coordinate Now better still psychologically we may reverse the arrow and write it like V A E A To repeat for a given morphism in the category of C algebras b Cy Cx we assigned a morphism in the category of A1 45V A E A 186 Let me repeat it one more time as y E ac V A E Ai 0 Now conversely suppose we are given a morphism in the category of A1 1 A E Ai Then we may de ne the following morphism in the category of CC algebras 11 y a x which takes y into 11ac regarded as an element of Here remember the substitution principle According to that principle assigning Mac suf ces to determine uniquely a CC algebra homomorphism from y to 0 In sum we have defined the two assignments both denoted by 0 V b y E C gt VZ A E A310 and VCy ltcx lt szlt Ai o The following is the precise statement of our duality o Duality gtW b and 1W 11 In particular VVV v and wVVV wv 0 You may feel like what we are doing here is a general nonsense My response is this is not nonsense This has a substance However to see that this really has a substance we must go one step further abstraction 187 De nition Bvalued points Let B be an arbitrary C algebra De ne the set of B valued points of A1 A as the set of morphisms from C to B in the category of Ealgebras A1B HomCAlgltCCIEBgt If we introduce the polynomial ring over B write it Bac then this set equals the set of morphisms from B to B in the category of ialgebras A1B HomBAlg B x B Xlll Let B be an arbitrary commutative ring Define Bx Bac y and more generally B 1 Observe that these all carry natural B algebra structures Observe that for A B 1 71 A is B isomorphic to Bac1 De nition Af ne spectrum We can view A1 as an assignment that takes an arbitrary CC algebra into the set A1 When we regard A1 as such an assignment we specifically denote A1 Spec 0 Below is the key notion in category theory 0 Functoriality B1 L B2 CC algebra homomorphism gt 1 NW 1 4dgt A B1 A B2 a mapping Later we de ne the notion of tensor products in the category of commutative rings The fact that the two sets HomCAlgCm B and HomgAlg B m B are equal is equivalent to m B C 188 0 Feature lt the axioms of functors gt i Composition Given Bl L Bg 32 L BS in the category of CC algebras A1 2 o A1 1 and A1 2 o bl coincide as mappings A1Bl a A1Bg ii Identity 1 B L MB A MB A1B induces this coincides With ldA1B 0 We say A1 is a covariant functor from the category of C algebras 77 to the category of sets To write it symbolically A1 0419 Set Selfcriticism Does such a generalization have any Virtue at all 189 Example Suppose we are given a diagram A1 it A a A 111 Where 111 and 1amp2 are given by 111y y2 1205 1353 bac2 cm d Where a b c and d E C are constants such that a 31 0 To duplicate the same information in less formal but more intuitively appealing way 242 z a3b2cacd o This arises from the dual diagram in the category of C algebras C 96 it CM lt CM 1 Question Does there exist a fourth Al on Which all the functions on A and on A become regular In other words does there exist a fourth A1 that complements the above diagram and yields the following commutative square Af EA i it A a A 111 Here I gave up using Greek letters for the coef cients of polynomials 190 Answer There exists such A1 if and only if D 4ac3 27am2 MW 7 1902 718ade 0 o This D arises as the determinant 3 2 b c D 2b 2b272ac b073ad c be 7 3ad 02 7 2bd ii Substitute a 1 and b 0 Then D becomes 403 27d2 o D is called the discriminant This is the cubic analog of b2 7 4ac 77 Today we learn how the above formula for D is derived and how it fits into a broader context symmetric polynomials and the more general notion of resultants in the Quiz 0 Proof of this requires some ingredients which are the major topics 1 plan to cover in the subsequent part of the semester Two of such ingredients are i The Jacobian criterion and ii Luroth7s theorem These can be treated from at least two mutually complementary perspectives One the theory of Riemann surfaces and two the theory of algebraic function fields 0 So D 0 is a special case As for the general case we need a remedy Luckily we already have the category of CC algebras handy The category of CC algebras offers a perfect tailor made answer to our problem 191 o Remedy There exists some commutative CC algebra A such that A lt C always Accordingly we obtain back the dual diagram Spec A gt Spec C l l9 Spec y gt Spec 2 7 f 0 As an A we can no longer choose a polynomial ring unless D 0 o A little closer look at the above square diagram The smallest A that fits into the above diagram is A C 20 ylty2 7 1163 bx2 cac This is regardless of whether D 0 or D 31 0 Now if D 0 then there exists an injective CC algebra homomorphism t i C 20 ylty2 7 1163 be 0x Example Assume a b c d lt1 1 0 0 Then D 0 There exists an injective CC algebra homomorphism t L ad212 7153 The image is the following C subalgebra of t CC 252 t3 7 t 192 Example Assume a b c d lt1 3 3 1 Then D 0 There exists an injective CC algebra homomorphism H 2 3 t Cxyy l The image is the following C subalgebra of t Ct2 t3 0 The whole point I am making is if D 31 0 then we cannot construct such an injective CC homomorphism from the CC algebra A Cylty27ax3bx2cacdgt to apolynomial ring C This is actually the content of Luroth7s theorem Now regardless it is important to acknowledge that we still have the CC algebra A Thanks to A we have found the right substitute of the fourth A that is Spec A 0 So what really is Spec A Answer Spec A is a covariant functor which takes an arbitrary CC algebra B for example C itself into the following set Spec A B y E BXB y2iltax3b2 cxdgt 0 called the set of B Valued points of Spec A In particular SpecAC y E Cxc y27ltax3bx2 cxdgt 0 193 What you see here is an algebraic equation The latter set is exactly the set of roots of an algebraic equation a polynomial in x and y equals 077 inside A2 that is C2 the usual any coordinate system Our first impression is that it suf ces to look at the set of C Valued points of Spec A so considering the set of B Valued points of SpecA for an arbitrary CC algebra B is super uous The reason we do what seems to be a genereal nonsense77 is this way SpecA becomes an object of the category which properly generalizes Al the category of polynomial mappings o The category of af ne CC models An af ne CC model is the functor SpecA for a CC algebra A We de ne the category of af ne C models as a system consisting of a an object 7 Spec B where B is a C algebra and b a CC morphism 7 11 Spec B1 Spec B2 which isa natural transform arising from a CC algebra homomorphism B2 L B1 namely an assignment that takes a CC algebra A into a mapping SpecB1A Spec B2A that takes M B1 6 A into M 045 B2 6 A o This category has the following feature7 Feature i Composition Spec B1 L Spec B2 Spec B2 L Spec B3 both CC morphism 50 gt Spec B1 gt Spec B3 is a C morphism 1d SpecB ii The identity SpecB Spec B which arises from the identity CC algebra homomorphism B A B is a C morphism 194 0 There is SpecA which is not de ned as a functor the original notion of spectrum Later you will hear the name a ine CC schemes along with the notation Spec A without the underline on Spec7 This is an alternative equally legitimate way to make the set of roots so far inside A2 of an algebraic equation into an object of the category which properly generalizes A1 These two notions 7 the notion of a ine models MA and the notion of a ine schemes SpecA 7 are equivalent The avor of the latter is less functorial even though we may create a functor out of it lnsead mastering the latter requires us to be proficient on mechanics of ideals naturally as the theory is based on two concepts 7 maximal ideals7 and prime ideals7 here a maximal ideal of A is a prime ideal not vice cersa In Chapter 3 we will start with these notions right off the bat We will devote the entire Chapter 3 to basics on commutative ring theory or of ideal theory the language needed to gain an organized perspective of systems of algebraic equations ldeal theory itself may be actually boring or dry However in mathematics sometimes it is true that we have to go through a boring process of definitions after definitions paired with proving one lemma after another that are essentially routine drills Sometimes we have to be patient but it pays off Let me put it this way Remember that in linear algebra we needed to develop a general theory and introduce some key concepts linear independence ranks etc so as to understand the substance of the notions of systems of linear equations The avor of the ideal theory is not at all similar to the avor of linear algebra after all linear algebra suits our undergraduate curriculum Still morally g of the many purposes of the ideal theory is facilitate our understanding of the substance of the notion of systems of algebraic equations o Ultimately we will have to abandon the idea that for a given system of algebraic equations physically manipulating the equations always works We have already seen an example that supports this statement in 2 problem XXI 195 0 Alexander Grothendieck How can Spec A be the starting point of algebraic geometry A little digression Just to give you some idea of how these two Spec s are related recall that MARC is the set of C homomorphisms b A E C The kernel of b turns out to be a so called maximal ideal7 m of A meaning that m is an ideal of A such that there is no ideal of A other than 111 and A that contains m In view of this one could have looked at the set of maximal ideals of A and could have defined it as Spec A Actually if you say this then you are on the right track The actual de nition of SpecA the one without the underline is as we will see shortly the set of prime ideals of A whose de nition will also be given shortly A Grothendieck who was also a visiting scholar here at KU in 195475 defined the notion of Spec A and more generally the notion of schemes His idea is based on a formidable development of algebraic geometry by his predecessors by the 1950s I will not attempt to list the names whose work have in uenced Grothendieck because no matter how I do it it will be an incomplete list whereas Grothendieck7s theory of schemes is more or less totally self contained So in theory Spec A can be the starting point of learning algebraic geometry He and his colleague J Dieudonne have written up an introductory account of the theory of schemes In the preface they warned the readers that any prior knowledge of algebraic geometry might be potentially harmful probably as an attempt to emphasize the self contained nature of their account in an impressionable way In any case the impact of the work of Grothendieck whose early motivation seemed to be an overwriting of the pre eXisting algebraic geometry was enormous Indeed the scheme theory was a smash success save unavoidable criticism by some of his contemporaries that his theory appeared a general nonsense His new framework offered tools to solve many dif cult problems again I will not attempt to list the names of conjectures that someone solved using Grothendieck7s ideas and language To follow this spirit I could have skipped Chapter 1 and Chapter 2 except the de nition of the category of commutative rings 1 did not choose that course The book Undergraduate Algebraic Geometry by M Reid has a chapter detaling the in u ence of Grothendieck s work As an example Hironaka used Grothendieck s idea to solve the problem on resolution of singularities 196 0 Just for a little while until the end of this chapter let us stick with our notion of af ne models We are going to make the following definition on isomorphisms of af ne models If you are asked what would be the right way to make such a definition then probably you will offer the same thing as mine which is as follows 0 Right rst de nition of isomorphisms for CC models Two af ne C models Spec A1 and Spec A2 are said to be CC isomorphic whenever A1 and A2 are isomorphic as C algebras In particular there exists a morphism 11 1W Spec11 a Spec12 in the category of af ne C models which is induced from a CC algebra isomorphism 152 A1 lt A2 We call 11 W a CC isomorphism in the category of CC models 0 However there is a potential twist in this definition I could have defined the notion of isomorphisms as follows 0 Potential alternative second de nition of isomorphisms for CC models A morphism in the category of CC models 11 Spec11 a Spec12 is said to be CC isomorphic if for an arbitrary CC algebra B the induced mapping 11 Spec113 Spec123 is bijective77 197 0 Some of you might have had a second thought Wait the latter de nition seems an equally sound de nition of the notion of isomorphisms of C models Your rationale Will be as follows CC models are functors having the category of sets as their target category and bijective mappings are isomorphisms in the category of sets77 This is a sound argument My vote is let us adopt the first definition that is Spec11 Spec12 ltgt A1 2 A2 77 C The subtlety of the issue is sometimes a CC algebra homomorphism Which is not a CC isomorphism can induce a morphism of af ne C models Which yields bijective mappings at many B s Let us go back to the previous example of an injective CC algebra homomorphism M t lt C 20 ylty2 7 ac Us For simplicity call the CC algebra C 20 ylty2 7 ac Us as B Recall that the image of B under M is the following C subalgebra of t C 152 153 The CC algebra homomorphism M induces a CC morphism 1 MV Spec t a Spec B As it turns out for an arbitrary CC algebra F Which is a the induced mapping 1F is bijective However the original CC algebra homomorphism M is not a C isomorphism of C algebras Therefore I cannot call 1 a CC isomorphism of C models Indeed there is a regular function 157 on MGM This is not a regular function on m B lndeed 157 is not in the image under M of the ring B This is categorically paraphrased as follows Evaluate 1 at the CC algebra B Then 1B is not a bijective mapping because the identity ldB is not in the image under 1B of w t I actually have more to say about this example I Will return to this example after I have defined the notion of Laurent polynomials 198 o Isomorphisms of af ne Models and gluing We have seen that there are af ne C models that look close but aren7t isomorphic Now we look at its opposite phenomenon There are af ne C models that do not look alike in terms of their descriptions While they are indeed isomorphic This is actually common Example The following two af ne C models are isomorphic mun yy a a and A1 mun Indeed there is a CC isomorphism of CC algebras Clo yly 2 a CM XIV 1 Construct one CC isomorphism m yy a a a n 2 More generally let fltacgt 6 C n be arbitrary Construct one CC isomorphism Clo yly 1 a CM De nition The C algebra of Laurent polynomials Let d Cac x71 Z ajacj ajECdEN j7d Thus a general element of C 20 x l is a71f1 a0 04196 ad1xd 1 adxd 199 This set forms a C algebra with respect to the usual7 addition and multiplication Note that C 20 x l contains C and C 4 as C subalgebras Note also that those two C subalgebras are C isomorphic C 2C C 4 Example The following two af ne C models are isomorphic MC y2y27 l and MC 71 Indeed there is a C isomorphism of C algebras Cac yac2 y2 71 7 C 1 XV l Establisha C isomorphism C 20 y x2 y2 71 7 Cac x l or a C isomorphism with the reversed arrow 2 Decide whether the two R algebras Rac yac2 y2 71 and Rac x l are R isomorphic to each other Fact Consider the two natural injective C algebra homomorphisms 1 1 71 1 Cx 7 Cac x lt Cac These induce morphisms Spec C A Spec C 20 x l L Spec C 4 which are both injective at an arbitrary C algebra B 200 XVI Let us denote ac for ac l Also refresh sorne notations in the diagram Spec C A Spec C 20 x l L Spec C 4 in the previous page as 00 A i SpecC L A1 with the agreement ac ac l Let F be an arbitrary C algebra which is a field In other words F is a field which contains C as a subfield 1 Make precise of the following two statements and ii i The image of the set r1 F mop 3H under the mapping 11F as a subset of A is A ac 0 ii The image of the same set T1 Spec Cx under the mapping wF as a subset of A is A 10 0 2 Describe the composite mapping 1 AF a 0 amp T1F M AF o 71 where means the inverse mapping of 11F which is regarded as a bijective Inapping whose target is A ac 0 3 What happens if you substitute F with an arbitrary C algebra B which is not a field Explain in each of the following cases B C t and B C 15 fl 201 o Birational equivalence I explain one fundamental concept in algebraic geometry 7 birational equivalence Let us have a look at the same example one more time p C lt B C2 3 I changed the letter for the variable Let us add7 ac at to inside the bracket of the both rings M Cac x lt CC2 3 Now the induced 1 MV SpecC 6 SpecCCx2 3 is bijective at any CC algebra B regardless of whether B is a field To see this look at the CC algebra level namely the CC algebra homomorphism M Cac x lt CC2 3 x This is actually a C isomorphism Indeed the both CC algebras are equal to C 20 x 1 and M is the identity We say that the original 1 Spec C lt Spec Cx2 3 is a birational morphism 0 Suppose that for two af ne C models Spec A1 and Spec A2 there exists a model Spec A3 along with two birational morphisms Spec A1 lt Spec A3 a Spec A2 Then we say that the two models Spec A1 and Spec A2 are birationally equivalent o Rational curves 7 rst de nition An af ne CC model which is birationally equivalent to A1 is called a rational curve Morally ninety nine percent of curves are not rational 202 0 Complex holomorphic mappings versus morphisms of CC models Before we proceed we naturally unavoidably ask the following question Sure we have created the category of af ne C models and it has morphisms that are basically arrow reversed version of ring homomorphisms or rather CC algebra homomorphisms to be meticulous We know that this naturally grew out of the notion of polynomial mappings On the other hand we also know that polynomial mappings are complex holomorphic mappings and not vice versa The question is how far is our category from its complex holomorphic counterpart 7 Below we construct one concrete plain example that illustrates this In it a certain collection of af ne C models is identified with a collection of objects in another category having complex holomorphic mappings as opposed to polynomial mappings as morphisms and the collection of those identified objects which are originally C models is immersed as a set of rational points7 inside a continuously parametrized7 collection of objects here I used a metaphorical expression image how the set Q is immersed in the set C Example Let Mi xi 6 3 l Assume that M are all distinct Let 1 Vi lt expZugt E CCgtltCC uEC Moreover let 203 Fact The following conditions are equivalent i VltzwgtECXc zw0 forsome 5 E Cz w ii xZEQ for ilm XVII Let m E N with m 2 2 In the above suppose Mi Guy and i Find 5 E Cz w such that V lt2wgt CXc ltzwgt0 o If we are to use the language of the theory of Riemann surfaces the two conditions in Fact above are equivalent to the following condition The complement of nitely many points in V is isomorphic in a certain appropriate category with complex holomorphic mappings as morphisms to the complement of nitely many points in a compact Riemann surface In short V is bimeromorphic to a compact Riemann surface 77 204

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