32 The law of quadratic reciprocity “What'sp?” “T'm pushing196poos for quadratic reciprocity
32 The law of quadratic reciprocity “What’s up?” “I’m pushing 196 proofs for quadratic reciprocity
Every finite division ring is a field Chapter 6 Rings are important structures in modern algebra.If a ring R has a mul- tiplicative unit element 1 and every nonzero element has a multiplicative inverse,then R is called a division ring.So,all that is missing in R from being a field is the commutativity of multiplication.The best-known exam- ple of a noncommutative division ring is the ring of quaternions discovered by Hamilton.But,as the chapter title says,every such division ring must of necessity be infinite.If R is finite,then the axioms force the multiplication to be commutative. This result which is now a classic has caught the imagination of many math- ematicians,because,as Herstein writes:"It is so unexpectedly interrelating two seemingly unrelated things,the number of elements in a certain alge- braic system and the multiplication of that system." Theorem.Every finite division ring is commutative. Ernst Witt This beautiful theorem which is usually attributed to MacLagan Wedder- burn has been proved by many people using a variety of different ideas. Wedderburn himself gave three proofs in 1905,and another proof was given by Leonard E.Dickson in the same year.More proofs were later given by Emil Artin,Hans Zassenhaus,Nicolas Bourbaki,and many others.One proof stands out for its simplicity and elegance.It was found by Ernst Witt in 1931 and combines two elementary ideas towards a glorious finish. Proof.Our first ingredient comes from a blend of linear algebra and basic group theory.For an arbitrary element s E R,let Cs be the set Ix E R:zs sx}of elements which commute with s;Cs is called the centralizer of s.Clearly,Cs contains 0 and 1 and is a sub-division ring of R.The center Z is the set of elements which commute with all elements of R.thus Z=RC.In particular,all elements of Z commute.0 and 1 are in Z,and so Z is a finite field.Let us set Z=q. We can regard R and Cs as vector spaces over the field Z and deduce that R=g",where n is the dimension of the vector space R over Z,and similarly Cs g"for suitable integers ns >1. Now let us assume that R is not a field.This means that for some s ER the centralizer Cs is not all of R,or,what is the same,ns<n. On the set R*:=R\10}we consider the relation r'rr=x-rx for some x E R". M.Aigner,G.M.Ziegler,Proofs from THE BOOK,DOI 10.1007/978-3-662-44205-0_6, @Springer-Verlag Berlin Heidelberg 2014
Every finite division ring is a field Chapter 6 Ernst Witt Rings are important structures in modern algebra. If a ring R has a multiplicative unit element 1 and every nonzero element has a multiplicative inverse, then R is called a division ring. So, all that is missing in R from being a field is the commutativity of multiplication. The best-known example of a noncommutative division ring is the ring of quaternions discovered by Hamilton. But, as the chapter title says, every such division ring must of necessity be infinite. If R is finite, then the axioms force the multiplication to be commutative. This result which is now a classic has caught the imagination of many mathematicians, because, as Herstein writes: “It is so unexpectedly interrelating two seemingly unrelated things, the number of elements in a certain algebraic system and the multiplication of that system.” Theorem. Every finite division ring is commutative. This beautiful theorem which is usually attributed to MacLagan Wedderburn has been proved by many people using a variety of different ideas. Wedderburn himself gave three proofs in 1905, and another proof was given by Leonard E. Dickson in the same year. More proofs were later given by Emil Artin, Hans Zassenhaus, Nicolas Bourbaki, and many others. One proof stands out for its simplicity and elegance. It was found by Ernst Witt in 1931 and combines two elementary ideas towards a glorious finish. Proof. Our first ingredient comes from a blend of linear algebra and basic group theory. For an arbitrary element s ∈ R, let Cs be the set {x ∈ R : xs = sx} of elements which commute with s; Cs is called the centralizer of s. Clearly, Cs contains 0 and 1 and is a sub-division ring of R. The center Z is the set of elements which commute with all elements of R, thus Z = s∈R Cs. In particular, all elements of Z commute, 0 and 1 are in Z, and so Z is a finite field. Let us set |Z| = q. We can regard R and Cs as vector spaces over the field Z and deduce that |R| = qn, where n is the dimension of the vector space R over Z, and similarly |Cs| = qns for suitable integers ns ≥ 1. Now let us assume that R is not a field. This means that for some s ∈ R the centralizer Cs is not all of R, or, what is the same, ns < n. On the set R∗ := R\{0} we consider the relation r ∼ r :⇐⇒ r = x−1rx for some x ∈ R∗. M. Aigner, G.M. Ziegler, Proofs from THE BOOK, DOI 10.1007/978-3-662-44205-0_6, © Springer-Verlag Berlin Heidelberg 2014
34 Every finite division ring is a field It is easy to check that is an equivalence relation.Let Ag={x-1sx:x∈R*} be the equivalence class containing s.We note that As=1 precisely when s is in the center 2.So by our assumption,there are classes As with |As|≥2.Consider now for s∈R*the map fs:x-→x-1 sx from R onto As.For x,y ER*we find x-1sz=y1sy (yz-1)s=s(yz-1) →yr-1∈Cg→y∈Cgx, for C:=Cs\[0),where C=fz:2C:}has size C.Hence any element r-sx is the image of precisely C*=g".-1 elements in R* under the map fs.and we deduce R*=As C.In particular,we note that R-g”-1 IC:I 1=A.l is an integer for all s. We know that the equivalence classes partition R*.We now group the central elements Z*together and denote by A1,...,At the equivalence classes containing more than one element.By our assumption we know t 1.Since R"=+=Axl.we have proved the so-called class formula g”-1=g-1+g”-1 Eiq"-1' (1) where we have 1<N for all k. With(1)we have left abstract algebra and are back to the natural numbers. Next we claim that gTk-1 g"-1 implies nk|n.Indeed,write n ank+r with 0<r<nk,then g"-1 gank+r-1 implies gk-11(gmk+r-1)-(gk-1)=gk(ga-1)mk+r-1), and thus g"-1 g(a-1)nx+r-1,since o"and g"-1 are relatively prime.Continuing in this way we find g"k-1g"-1 with 0<r<nk, which is only possible for r=0,that is,nk|n.In summary,we note nk n for all k. (2) Now comes the second ingredient:the complex numbers C.Consider the polynomial x"-1.Its roots in C are called the n-th roots of unity.Since An =1,all these roots A have 1 and lie therefore on the unit circle of the complex plane.In fact,they are precisely the numberske cos(2kπ/n)+isin(2kπ/m,0≤k≤n-1(see the box on the next page).Some of the roots A satisfy Ad 1 for d<n;for example,the rootλ=-1 satisfiesλ2=1.For a root入，let d be the smallest positive exponent with Ad =1,that is,d is the order of A in the group of the roots of unity.Then d n,by Lagrange's theorem("the order of every element of
34 Every finite division ring is a field It is easy to check that ∼ is an equivalence relation. Let As := {x−1sx : x ∈ R∗} be the equivalence class containing s. We note that | A s | = 1 precisely when s is in the center Z. So by our assumption, there are classes A s with | A s| ≥ 2. Consider now for s ∈ R ∗ the map fs : x −→ x − 1sx from R ∗ onto A s. For x, y ∈ R ∗ we find x − 1sx = y − 1sy ⇐⇒ (yx − 1 ) s = s (yx − 1 ) ⇐⇒ yx − 1 ∈ C ∗s ⇐⇒ y ∈ C ∗s x, for C ∗s : = C s\{ 0 }, where C ∗s x = {zx : z ∈ C ∗s } has size | C ∗s |. Hence any element x − 1sx is the image of precisely | C ∗s | = q n s − 1 elements in R ∗ under the map fs, and we deduce | R ∗ | = | A s| | C ∗s |. In particular, we note that | R ∗ | | C ∗s | = q n − 1 q n s − 1 = | A s | is an integer for all s . We know that the equivalence classes partition R ∗. We now group the central elements Z ∗ together and denote by A 1,...,A t the equivalence classes containing more than one element. By our assumption we know t ≥ 1. Since |R∗| = |Z∗| + tk=1 |Ak|, we have proved the so-called class formula q n − 1 = q − 1 + t k=1 q n − 1 q n k − 1 , (1) where we have 1 < q n − 1 q n k − 1 ∈ N for all k . With (1) we have left abstract algebra and are back to the natural numbers. Next we claim that q n k − 1 | q n − 1 implies n k | n. Indeed, write n = an k + r with 0 ≤ r<n k, then q n k − 1 | qan k + r − 1 implies q n k − 1 | ( qan k + r − 1) − ( q n k − 1) = q n k ( q ( a −1) n k + r − 1) , and thus q n k − 1 | q ( a −1) n k + r − 1, since q n k and q n k − 1 are relatively prime. Continuing in this way we find q n k − 1 | q r − 1 with 0 ≤ r<n k , which is only possible for r = 0, that is, n k | n. In summary, we note n k | n for all k. (2) Now comes the second ingredient: the complex numbers C. Consider the polynomial x n − 1. Its roots in C are called the n -th roots of unity. Since λ n = 1, all these roots λ have | λ | = 1 and lie therefore on the unit circle of the complex plane. In fact, they are precisely the numbers λ k = e 2kπi n = cos(2kπ/n) + isin(2kπ/n ) , 0 ≤ k ≤ n − 1 (see the box on the next page). Some of the roots λ satisfy λ d = 1 for d<n; for example, the root λ = − 1 satisfies λ 2 = 1. For a root λ, let d be the smallest positive exponent with λ d = 1, that is, d is the order of λ in the group of the roots of unity. Then d | n, by Lagrange’s theorem (“the order of every element of
Every finite division ring is a field 35 a group divides the order of the group"-see the box in Chapter 1).Note that there are roots of order n,such as=e 2=reip Roots of unity Any complex number z=+iy may be written in the"polar"form y=rsin o z ret r(coso+isin), wherer==2+yis the distance of z to the origin,and is the angle measured from the positive z-axis.The n-th roots of unity are therefore of the form g=e=cos(2k/n)+isin(2kn/n), 0≤k≤n-1, since for allk AR e2kri cos(2kn)+isin(2km)=1. We obtain these roots geometrically by inscribing a regular n-gon into the unit circle.Note that=C forallk,wheree Thus the n-th roots of unity form a cyclic group2,....=1} of order n. The roots of unity for n =6 Now we group all roots of order d together and set pa(x）= Π(e-). λof order d Note that the definition of oa(z)is independent of n.Since every root has some order d,we conclude that x”-1=Πa(: (3) dn Here is the crucial observation:The coefficients of the polynomials on(r) are integers (that is,n()EZIx]for all n),where in addition the constant coefficient is either 1 or-1. Let us carefully verify this claim.For n =1 we have 1 as the only root, and so (x)=z-1.Now we proceed by induction,where we assume d(x)EZ[]for all d<n,and that the constant coefficient of oa(x)is 1 or-1.By(3), z"-1=p(x)on(x) (4) iena-盒n，以a国=艺o=1wn=-1 k=0 Since-1=poao,we see ao {1,-1).Suppose we already know that ao,a1,...,ak-1EZ.Computing the coefficient of x on both sides of (4)
Every finite division ring is a field 35 a group divides the order of the group” — see the box in Chapter 1). Note that there are roots of order n, such as λ1 = e 2πi n . Roots of unity Any complex number z = x + iy may be written in the “polar” form z = reiϕ = r(cos ϕ + isinϕ), where r = |z| = x2 + y2 is the distance of z to the origin, and ϕ is the angle measured from the positive x-axis. The n-th roots of unity are therefore of the form λk = e 2kπi n = cos(2kπ/n) + isin(2kπ/n), 0 ≤ k ≤ n − 1, since for all k λn k = e2kπi = cos(2kπ) + isin(2kπ)=1. We obtain these roots geometrically by inscribing a regular n-gon into the unit circle. Note that λk = ζk for all k, where ζ = e 2πi n . Thus the n-th roots of unity form a cyclic group {ζ,ζ2,...,ζn−1, ζn = 1} of order n. x = r cosϕ ϕ z = reiϕ ⎫ ⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎭ y = r sin ϕ r = |z| ⎧ ⎪⎪⎪ ⎪⎪⎪ ⎪⎪⎪ ⎨ ⎪⎪⎪ ⎪⎪⎪ ⎪⎪⎪ ⎩ !" # −1 λ2 λ1 = ζ 1 The roots of unity for n = 6 Now we group all roots of order d together and set φd(x) := λ of order d (x − λ). Note that the definition of φd(x) is independent of n. Since every root has some order d, we conclude that xn − 1 = d | n φd(x). (3) Here is the crucial observation: The coefficients of the polynomials φn(x) are integers (that is, φn(x) ∈ Z[x] for all n), where in addition the constant coefficient is either 1 or −1. Let us carefully verify this claim. For n = 1 we have 1 as the only root, and so φ1(x) = x − 1. Now we proceed by induction, where we assume φd(x) ∈ Z[x] for all d<n, and that the constant coefficient of φd(x) is 1 or −1. By (3), xn − 1 = p(x) φn(x) (4) where p(x) = j=0 pjxj , φn(x) = n − k=0 akxk, with p0 = 1 or p0 = −1. Since −1 = p0a0, we see a0 ∈ {1, −1}. Suppose we already know that a0, a1,...,ak−1 ∈ Z. Computing the coefficient of xk on both sides of (4)
36 Every finite division ring is a field we find k ∑pyk-j=∑P,ake-+oak∈Z j= j=1 By assumption,all ao,...,ak-1(and all pj)are in Z.Thus poak and hence ak must also be integers,since po is 1 or-1. We are ready for the coup de grace.Let nkn be one of the numbers appearing in (1).Then x”-1=Πpa(a)=(ak-1)pn(x)Π oa(z). d n d|n,dnk,d≠n We conclude that in Z we have the divisibility relations on(g)gm-1 and 41g-1 gnk-1 (5) Since(5)holds for all k,we deduce from the class formula (1) on(q)1q-1, but this cannot be.Why?We know on(x)=II(x-A)where A runs through all roots of rn-1 of order n.Let Aa+ib be one of those roots. Byn>l(because of R≠Z)we have入≠l,which implies that the real part a is smaller than 1.Now 2 a2+b2 =1,and hence 1g-2=g-a-2=(g-a)2+b2 =g2-2ag+a2+b2=g2-2ag+1 >q2-2g+1 (because of a 1) l9-4>la-1 =(q-1)2, and so >-1 holds for all roots of order n.This implies lm(g=Π9->g-1, which means that on(g)cannot be a divisor of q-1,contradiction and end of proof. ▣ References [1]L.E.DICKSON:On finite algebras,Nachrichten der Akad.Wissenschaften Gottingen Math.-Phys.Klasse (1905),1-36;Collected Mathematical Papers Vol.III,Chelsea Publ.Comp,The Bronx,NY 1975,539-574. [2]J.H.M.WEDDERBURN:A theorem on finite algebras,Trans.Amer.Math. S0c.6(1905),349-352. [3]E.WITT:Uber die Kommutativitiit endlicher Schiefkorper,Abh.Math.Sem. Univ.Hamburg 8(1931).413
36 Every finite division ring is a field we find k j=0 pjak−j = k j=1 pjak−j + p0ak ∈ Z. By assumption, all a0,...,ak−1 (and all pj ) are in Z. Thus p0ak and hence ak must also be integers, since p0 is 1 or −1. We are ready for the coup de grâce. Let nk | n be one of the numbers appearing in (1). Then xn − 1 = d | n φd(x)=(xnk − 1)φn(x) d | n, dnk, d=n φd(x). We conclude that in Z we have the divisibility relations φn(q) | qn − 1 and φn(q) $ $ qn − 1 qnk − 1 . (5) Since (5) holds for all k, we deduce from the class formula (1) φn(q)| q − 1, but this cannot be. Why? We know φn(x) = (x − λ) where λ runs through all roots of xn −1 of order n. Let λ% = a+ib be one of those roots. By n > 1 (because of R = Z) we have λ% = 1, which implies that the real part a is smaller than 1. Now |λ%| 2 = a2 + b2 = 1, and hence |q − λ%| 2 = |q − a − ib| 2 = (q − a) 2 + b2 = q2 − 2aq + a2 + b2 = q2 − 2aq + 1 > q2 − 2q + 1 (because of a < 1) = (q − 1)2, and so |q − λ%| > q − 1 holds for all roots of order n. This implies 1 q μ |q − μ| > |q − 1| |φn(q)| = λ |q − λ| > q − 1, which means that φn(q) cannot be a divisor of q − 1, contradiction and end of proof. References [1] L. E. DICKSON: On finite algebras, Nachrichten der Akad. Wissenschaften Göttingen Math.-Phys. Klasse (1905), 1-36; Collected Mathematical Papers Vol. III, Chelsea Publ. Comp, The Bronx, NY 1975, 539-574. [2] J. H. M. WEDDERBURN: A theorem on finite algebras, Trans. Amer. Math. Soc. 6 (1905), 349-352. [3] E. WITT: Über die Kommutativität endlicher Schiefkörper, Abh. Math. Sem. Univ. Hamburg 8 (1931), 413.