An Elementary Proof of a Theorem of Jacobson

By KENNETH ROGERS (Honolulu)

1. Introduction

In [9], JACOBSOI~ proved the follo~_ug as a corollary to his theorem on algebraic algebras:

Theorem (JACOBSON). Let R be a ring in which for every element r there is an integer n(r) > 1, depending on r, such that r = ~ ( ~ . Then R is commutative.

Subsequently there have been many papers writ ten on this subject, by HuA [8], KAPLXNSKY [10], HERSTEIN [3, 4, 5, 6], N A g A y A ~ [11, 12] and others: of particular interest are the generalisations due to HERSTEIN and NAgaY~A. Most of HERSTEIN'S work is also found in his book [7]. In papers [5, 6] he has reduced to a more elementary form the proof of JACOBSON'S theorem: in [5] he needs ZORN'S Lemma, and the proof owes much to A R ~ [1], as HERSTEIN points out, while in [6] only division rings are considered. We shall give a serf-contained proof of JACOBSON'S theorem, using only basic ideas in algebra: in particular, we give a new proof of WEDDERBURN'S theorem, which is a special case of the above.

Theorem (WEDDERBURN). Every ~nite ~iv'~sion ~ng i8 a f~e~.

2. Preliminaries

We first develop the ideas needed to prove WEDDERBURN'S theorem, which includes a result needed later also. Hence Lemma 1 is stated in general terms rather than just for division rings: in this sense it generalises a result of HERSTEIN [6, 7].

We need the following elementary facts: a finite division ring R is of prime characteristic p, has p~ elements, where n is the dimension of R over the prime field Z~, and the elements o f / ~ all satisfy x ~ ~ x. In particular, a finite field contains at most one subfield of q elements, since t ~ - t has at most q roots in a field. We also need the fact tha t every non-zero subring of a finite division ring is a division ring, since a finite associative system with cancellation laws is a group.

15"

224 Kenneth Rogers

Lemma 1. I f R is a non-commutative ring o/ characteristic p which contains a finite field F, and i / x lies in F but not in the center o / R , then there exists a non-zero element w in R such that

W X ~ - X~ W~

for some x' in F, with x' ~ x.

Proo f i Let ~ : r - - > x r ~ r x , so ~ 0 and ~--- -L~--R~. Since L| = R~L~, we have

hence

where p~ ---- # (~v). For any element z of F, L, commutes with L, and Rz, hence with ~, and so the factorisation

t~" - - t = 1 -I (t - - ~) zeav

in Fit] yields

0 = ~ ' - - ~ = 1-[ ( ~ - - L . ) . z e F

Since ~ ~ O, there is a shortest zero subproduct containing ~, say

0 = (~ (~ - - L~I) �9 �9 �9 (~ - - L~),

0 # ~ (~ - - L . I ) �9 �9 �9 (~ - - L.~_~) .

Thus there is an element r with w ~ (r) ~(~--L,1) �9 . . ( ~ - - L~_I) :4= 0 but w ( 6 - - L ~ ) = O. Thus we have w in R* with x w - - w x ~ zw for some z in F*, and so w x ~ ( x - - z ) w -= x'w, with x' in F , x' =4= x.

P r o o f o f WEDDERBURN'S t h e o r e m . Suppose there exist non- commutative finite division rings, and let R be one with the least number of elements. Then R has characteristic p, and every proper subring is a smaller finite division ring than R and so is commu- tative. We refer to them as subfields, therefore. One such is the center: E = { c l c e R , c r = r e , V r e R } .

(1) Every subfield o / R is sell-conjugate in R. For let F be a maximal subfield of R: since E and F generate a commutative subring of R con- ta]ning F, they generate F only, hence F contain~ E. E itseff is not maxima], since every extension E[r] is commutative, so F D E. For x e F - - E , there is w e R* with w x = x'w, for some x' e F , x'~= x. Hence w x w - X e F , w C F , so x e u r - l F w n F . I f the field w - l F w con- rained an element outside F, these two fields would generate R: but then x commutes with both these fields, so x would lie in E, contrary

An Elementary Proof of a Theorem of Jacobson 225

to assumption. Hence w-IFw ---- F, for a t least one e lement w outside F. Hence the normaliser NR(F) ~ ( r l r ~ R, r F - ~ Fr} is a subring of R proper ly containing F , hence NR (F) ~- R. I f K is an y subfield of R, it lies in a max ima l subfield F , and so for all r e R* we have r -1Kr Cr -1Fr ~- F. Thus K and r - l K r are subfields o f f wi th the same n u m b er of elements,

hence K = r-l K r.

(2) Every commutator r - l s - l rs lies in the center of R. Suppose r l s - l r s ~= 1: t h e n rs =~ sr, so the fields E[r] and E[s] toge ther genera te R, b y the min imal i ty of R. B u t t hen E [r] n E [8] = E, as a n y common e lement commutes wi th all o f R. Since r- ls- lr �9 s e E[s] , and r -1 �9 s - i t s e E[r], we have r- ls- lrs in E , as claimed.

(3) R is actually commutative. Since R is non-commuta t ive , there are e lements x and y wi th x y ~= y x, hence x y = c y x, where o e E. Also f rom (2), we have x (x ~ y) ---- 0' (x ~- y ) x for some c' e E. B y subtrac-

t ion we obta in

�9 ~ = d x ~ + ( r - - c) y x .

Since x ~= 0, we deduce t h a t

( 1 - - c ' ) x - ~ ( d - - o ) y ,

which does imply t h a t x and y COmmUte, unless the central e lements 1 - - d and 0 ' - - c are bo th zero, in which case 1----d----0. Bu t this means t h a t x y - ~ y x, giving the final cont radic t ion which shows t h a t no such non-commuta t ive division ring exists.

3. Approach to Jacobson's theorem

B y "J/~COBSOl~ condi t ion" on a ring R we mean the condi t ion assumed in JACOBSON's theorem. To show R commuta t ive is to show eve ry sub- ring (x , y) is, where (x, y) denotes the subring genera ted b y x and y: it consists of all finite sums of "words" in x and y.

Lemma 2. A finitely generated ring R with Jac~bson's condition has squarefree characteristic P l ' " P ~ , for some primes p~ and some 8, and is the direct sum of subrings Rs o/ characteristio p~.

P r o o f . I f r " - - - - r and s ~ - s , t h en r - - - - r k and s----s ~ for k ---- ( n - 1 ) ( m - 1) -t- 1, so for any finite n u m b e r of e lements x in R we m a y use the same value for n(x). Fo r r in R and an y integer k > 1, there exists n > 1 such t h a t r = r n and k r = (kr)" =- k~r ~, and hence ( k " - - k ) r = k"r" - - Icr = O. Thus, if R = ( z l , . . . , x~), there are

integers ks > 1 wi th ks xs -~ 0 for i ---- 1 , . . . , m. Since integers commute

226 Kenneth Rogers

wi th elements of R, i t follows t h a t (k 1 �9 �9 �9 kin)R = {0}, so let the charac- terist ic of R be p ~ l . . , p{,. F o r r e /~ , there exists n > 1, such t h a t

P l P ~ ' " P s ' r = (P l " " " P s ' r ) "

= (ivy- �9 �9 ps" r) i+/v~"-1~

= ( i P l . �9 �9 p s ) l + / V ( " - l } �9 r l + N ( , * - l )

= 0 ,

since we m a y choose N so t h a t 1 + N ( n - 1) exceeds all the ]~. Now,

the integers P~ = ( P l " " "Ps)/P~ are re la t ively pr ime, hence there are integers ui such t h a t u 1P1 + �9 �9 �9 + us Ps = 1. Hence r = r 1 + �9 �9 �9 + rs, where r~ = u~P~r, and so

R = R1 + "'" + R . ,

where R~ = P ~ R , which is a subring of R of character is t ic p~. F o r e lements r~ and r~ of Rt and Rj , wi th i ~= ], we have

r~rj P~r~ P~r~ P ~ P j " ' O,

and hence the sum ~ R, is direct , because

] ~ r ~ = 0 ~ r~ = - - 2 ~ rtr~ = 0 ~ r, = O,

b y JACOBSO~I'S condition. This proves the L e m m a : the reader m a y wish to compare this wi th the same resul t for the special case where r" = r for all r in R, as in AYOUR & AYOUB [2].

L e m m a 3. Let R be a r ing wi th Jacobson 's condit ion, x ~ R and x" = x. T h e n the iden t i t y r = x"-Xr + (r - - x " - l r ) expresses 12 as a s u m o /or tho - gonal subringa

R = I~ + A~,,

where I , ---- x " - l R and A , = {r - - x " - l r ] r e R}. A lso , I , contains x and

has iden t i t y element x ~-1, whi le x A ~ = A , x = {0}. The s u m Ix + A~ is direct.

P r o o f . Note t h a t e ---- x "-1 has e a = e and hence commutes wi th all o f R: for it is well known t h a t

( e r e - - e r ) ~ = c r e t e + e r e r - - e r e r - - e r e r e = O,

and hence e r e - - e r = 0, b y JXCOBSO~'s condition. Similarly, e r e - - r e = 0

hence er = re. Therefore b o t h I~ and A~ are subrings of R : t h e y are closed under addi t ion, and, for mult ipl icat ion, we have :

e r . e 8 = e 2 r 8 ~ - . e . r S ,

An Elementary Proof of a Theorem of Jacobsen 227

and

( r - - e r ) ( s - - e s ) = r s - - r e , - - e r s + e r e s = r s - - e r s - - e r s + era

= r 8 ~ e . r s .

These subrings are orthogonal, because

e r ( s - - e s ) = e r s - - e r e s = e r s - - e 2 r s = e r s - - e r s = O ,

and similarly the other way. Since x = x" = e x, x lies in I| and since e . e r = e n r = e r , e----1 on I| We know that x A x = A x x = { O } , as x �9 I~. Note also that for x 4= 0 we have I~ ~= {0}, whereas A, = {0} ff x ~-1 = 1R. Finally, the sum is direct because

r �9 I~ n A~ :~ r = x '~- lr = x n - n x r = O.

In showing that x and y commute, we use the ring (x), which is commu- tative, finite and 'JACOBSen'. Hence we prove:

Lemma 4. A commutative ring R, not {0}, which satisfies Jacobson's caradition r = r ~ (~) and is finitely generated, is a direct sum o/ f ini te fields.

P r o o f . By commutat ivi ty and x ~ = x for each generator, we know that R has only finitely many "words". Since, by Lemma 2, R has finite characteristic, it is finite and is a direct sum of finite, commutative, 'JACOBSOI~' rings of characteristic p~. Since it suffices to show each of these is a direct sum of finite fields, we may assume that R has prime characteristic 10 and use induction on ~ ( R ) . I f ~ ( R ) = p, then R - - - - { k r i k - - 0 . . . . , p - 1} and is therefore a division ring, for ff 0 = k r . k ' r -= k k ' r ~, then 10 divides kk' and one of kr, k 'r is 0, because r Z ~ 0, by JACOBSO~I's condition. Hence R = GF[p]. I f ~: (R) > 10, con- sider R ~ Ix + A=, for each x �9 R*. The induction hypothesis applies to each of I~ and A~, unless one of them is {0}: but I~ ~= {0}, which leaves the case where for all x ~= 0 we have A| and therefore R----I~, and so x ~-1 = 1R for all x in R*. In that case, all non-zero elements are invertible, so R is a finite field. This completes the induction.

Lemma 5. I / R is a finite ring in which to each element r there corresponds an integer n(r) > 1 with r = r ~(r~, then R is commutative.

P r o o f . By Lemma 2, we need only consider the case of characteristic p. Using induction on ~ (R) as above, we again find that R = GF[p] if 4 ~ ( R ) = 10, and we again succeed with induction unless A~----{0} for all x 4= 0. The remaining case is where R is a finite division ring and hence, by WV.DDV.RBUI~'S theorem, is a finite field.

228 Kenneth Rogers

4. Proof of Jaeobson's theorem

To show x y ---- y x, we need only prove JACOBSON'S theorem for <x, y>, hence Lemma 2 applies and R is a direct sum of subrings of prime characteristic. Since these commute elementwise with each other, it suf- fices to show each commutative, so we may assume char R----p. We shall assume tha t x y @= y x and derive a contradiction. Since <x> satisfies the conditions of Lemma 4, it is a direct sum of finite fields, and not all of these can commute elementwise with y, or so would x. We may therefore assume, without loss of generality that x lies in a finite sub- field of R, in particular in F ---- Z~ Ix].

Now apply Lemma 3 to <x, y> : all of Ax commutes with x, hence not all of Ix can commute with x. Going down to the ring I x for the new R, which still contains the field F, we are now in the case where x "-1 ---- 1R. But therefore 1R lies in F, so 1R ---- IF and therefore every element of F* is invertible in R. By Lemma 1, there is an element w of R* with w x = x' w, for some x' @= x, x' in F. Hence, for P (t) e Z~ [t], we have w P(x ) -= P ( z ' ) w , and in general w~P(x) = Q(x)w ~ for some Q(t)eZ,,[t]. Thus <x, w> = Z~[x, w] = set of all sums of elements of the form xJw ~, with 0 ~ ] < n(x) and 0 ~ i < n(w), hence <x, w> is finite and there- fore commutative, by Lemma 5. Thus, w x - = x w - ~ x 'w, so ( x - - x ' ) w = 0: but since every non-zero element o f F is invertible in R, it follows tha t w = 0. This contradiction completes the proof.

5. Coneluding remarks

We wish to thank Dr. E. A. BERTRAM for suggesting this problem and also Dr. L. J. WALLEN for goading us into finding a proof of WEDDER- BURN'S theorem. There may be some difficulty in using a similar approach on other generalisations of JACOBSON's theorem: Lemma 3 was a critical step in our procedure and relied upon the existence of an equation x" -~ z for each x.

References

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[2] R. C. AYOUB & C. AYOUB, On the eommutativity of rings, American Mathe- matical Monthly 71 (1964) 267--271.

[3] I. N. I:s A generalisation of a theorem of Jacobson I, Amer. J. Math. 78 (1951) 755--762.

[4] I. N. HERSTErN, A generalisation of a theorem of Jacobson HI, Amer. J. Math. 75 (1953) 105--111.

An Elementary Proof of a Theorem of Jacobson 229

[5] I. 1~. HERSTErN, An elementary proof of a theorem of Jacobson, Duke Math. J. 21 (1954) 45 48.

[6] I. N. HERS~n~, Wedderburn's theorem and a theorem of Jacobson, American Math. Monthly 68 (1961) 249--251.

[7] I. N. I:s Noncommutative rings, The Mathematical Association of America (Cams Monographs} 1968.

[8] L. K. HuA, Some properties of a s-field, Proc. Nat. Acad. Sci. USA 85 (1949} 533---537.

[9] N. JACOBSON, Structure theory for algebraic algebras of bounded degree, Ann. of Math. 46 (1945), 695--707.

[10] I. Ir~PLANSXY, A theorem on division rings, Canad. J. Math. 8 (1951) 290---292. [11] T. ~TA]ffAyAMI, On the commutativi ty of certain division rings, Canad. J.

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Eingegangen am 27. 4. 1970