RADICALS OF A RING
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yA' 35ir'ector of tne^evarhepartmen. t of MaSHe^ETcc ~~7 Tie an "of The Gra3lia"5e"~ScaooT ± Crawford, Phyllis J., Radicals of a Ring, Master of Science (Mathematics), May, 1971, pp. 48, bibliography, 7* titles. The problem with which this investigation is concerned is that of determining the properties of three radicals de- fined on an arbitrary ring and determining when these radicals coincide. The three radicals discussed are the nil radical, the Jacobson radical, and the Brown-McCoy radical. ,:.n an arbitrary ring R, the prime radical Rvery element of £>(R) is nilpotent, that is, If a •; p(f<) then there Is some positive integer n so that aP « o, Thus *>YR) ir= a ri.1J 3deal since every element of 0(R) is nilpotent. If A Ife an Ideal in R such that there is some positive integer K so that a' ~ o for all a e A} then A Is a nilpotent Ideal and A £ ^R)« A. nil radical of R, */>(R) is defined as the sum of all nilpotent ideals in R, Thus '//(R) Is a nil ideal oi. eit contains every nilpotent id.eal of the ring, Then an upper nil radical of R, l{ is defined as the sum of all nil ideals in K so that every nil radical of R is contained in l(. The -tower nil radical of R, is defined as the Intersection ol. 8j.j rij.i i'aaj cais of R so that every nil radical of R con- conf; trucuion of ^(R)., the conclusion reached 1« that for right or left ideals, then st = ^(R) = ??(R) = V. so that there is exactly one nil radical of R. The Jacobson radical of a ring R, j?(R)> is defined as the intersection of the annihilators of the left irreducible modules of R and coincides with the intersection of the annihilators of the right irreducible modules of R. Thus j?(R) coincides with the intersection of all maximal regular left-ideals in R. If P is a left ideal in R such that there is an a e R so that x - xa e P for all x e R, then P is a regular left ideal. Similarly, if x e R and there is some y e R such that x+y+xy=o (x + y + yx = o) then x is right (left) quasi-regular. Thus $(R) consists of those elements of R that are both right and left quasi-regular. Since every nilpotent element is quasi-regular, then K c j?(R) so that £ c ??(R) c K c £(R). if A is an ideal in R and A is considered as a subring of R, then J?(A) = In a ring R, an element a e R is G-regular if a e G(a) = {ar - r + 2(xj_ayi - x1y±) [r,xi,yi e R}. The Brown-McCoy radical of R, /ft(R), is defined as the set of all b e R so that the ideal (b) generated by b is G-regular. Thus $(R) c ^(R). if A is an ideal in R and A is considered as a subring of R, then A) = 57?(R) nA. If R is a ring with the descending chain condition for right or left ideals, then s£ = K = #(R) ~ ^(R)- Therefore, in a ring with the descending chain condition for right or left ideals, the nil radical is the only radical of the ring. RADICALS OF A RING- THESIS Presented to the Graduate Council of the North Texas State University in Partial Fulfillment of the Requirements For the Degree of MASTER OF SCIENCE By Phyllis Crawford, B. S, Denton, Texas May, 1971 INTRODUCTION . In an arbitrary ring R, a particular ideal of R can "be defined so that such an ideal is called a radical of R. Since different radicals can be defined on a ring R, the conditions on R under which they coincide will be investi- gated, along with the properties of three of these radicals. Some basic concepts of rings will be used in the following theorems. The sum of two right (left or two-sided) n ideals A and B in a ring R is A + B = {Z) (a. + b. ) | a. e A, b. € B i=l ill- 1 and n ranges over all positive integers}. If {A^} is an in- finite collection of rlght(left or two-sided) ideals in R, T,A. is the ideal whose elements^ consist of finite summations of elements of the A.. The product of two right (left or two- n sided) ideals A.B in R is AB = {S a.b.la. e A, b. e B for 1=1 111 i all positive integers n}. If r e R and A is a right (left or two-sided) ideal in R, then Ar = {ar|a e A} and rA = {ra|a € A}. In general a two-sided ideal in R will be referred to as an Ideal in R. If R is a ring in which the descending chain condition for right (left) Ideals holds, then R Is said to "be right (left) Artinian. If R is a ring in which the ascending chain condition for right(left) Ideals holds, then R is said to be right (left) Noetherian. iii TABLE OF CONTENTS Chapter Page I. NIL RADICALS 1 II. JACOBSON RADICAL 22 III. BROWN-McCOY RADICAL 39 BIBLIOGRAPHY 47 IV CHAPTER I NIL. RADICALS Definition 1.1: An ideal P In a ring R is a prime ideal if for ideals A,B in R, ABjSTP implies A £ P or B«™P. Theorem 1.1: If P is an ideal in a ring R, the follow- ing are equivalent: (I) P is a prime ideal* (ii) If a,b e R such that aRb cP^ then a e P or b e P; (ill) If (a), ("b) are principal ideals in R such that (a)(b)CP, then a € P or b € Pj (iv) If U,V are right ideals in R such that W<=P, then » UcP or V^P; (v) If U,V are left Ideals in R such that W<^P, then U c.P or Vc P. Proof: Assume that P is a prime ideal in a ring R, then let a,b e R such that aRbcp. Then aRb = {arb|r e R}, such that RaRbR £P so that (RaR) (RbR) CP, Then since P is a prime ideal, RaR ^P or RbR«=P. For (a) the principal ideal In R generated by a, (a)^RaR, And for (b) the principal ideal in R generated by b, (b)3c RbR. If RaRc.P, then (a)5cP and since P is prime, (a) CP and thus a e P. If RbR cp, then (b) CP and then (b)cp so that b e P. Then since RaRCP or RbRCp, it follows that a e P or b e P. Therefore, (i) im- plies (ii). 2 Now assume (±i) and let (a), (I)) be principal ideals in r such that (a)(b) £» P, where (a)(b) is the ideal generated by elements of the form a^b^ for e -(a) and b1 € (b). Since ( a),(b) are principal ideals then. aRb (a) (b) so that aRb P and thus a e P or b e P. Therefore, (ii) implies (iii). Assume (iii) and let U,V be right ideals in R such that UV <£ P. Suppose U^ P, then there is some u e U such that u | P. Let v e V. Since UV £ P, then RUV£, P so that UV + RUVi£ P and hence (u)(v) C UV + RUVg^ P. Then, by (iii) for (u)(v) Definition 1.2; A set M in a ring R is an ro-system if for aflb e M, there is.an x e R such that axb e M. It follows trivially that (j) is an m-system. Theorem 1,2; An ideal P in a ring "R is a prime ideal if and only if P = {x e Rjx | P} is an m-system. Proof; Let P be a prime ideal in R and let a,b e P. Then aRb ^ P since aRb g P implies a e P or b e P but a,b e P. Then aRb ^ P implies that there is some x e R so that axb e aRb and axb £ P. Then axb e P so that P is an m~system. Let P be an ideal in R so that P is an m-system. Let a,b e R such that a | P, b | P, then a3b e P. Therefore, since P is an m-system, there is an x e R such .that axb e P, then axb | P. Thus aRb P since axb e aRb but axb P. Since a,b | P implies aRb P, then if aRb CP. a e P or b e P. Then (ii) of Theorem 1 is satisfied so that P is a prime ideal. Definition l.J>l A prime ideal P in a ring R .containing an ideal A is a minimal prime ideal containing A, if for any prime ideal p' in R, A £ p" <-? P implies p' = P. Theorem 1.3: Let M be an m-system in a ring R. If an ideal A in R does not meet Mj then there is a maximal ideal with respect to M in R which contains A and does not meet M; and such an ideal is a prime ideal in R. 4 Proof; Let A be an ideal in a'ring R and let M be an m-system in R such that A n M = (j). Let F be the family of all ideals Q in R such that. A£ Q and Q n M - (I, then F 4 since A e F. F can be partially ordered by set inclusion. Let f'C F be a chain in F. Then q' = UP' where P' e f/ Q' is an ideal in R since F/ is a chain of ideals in F and A £ q' since A p' for all p' e f' Then q' (1 M = (UP/) n M = U(P'fl M) = (j). PeP' PeF' Therefore, Q' e F and P'£ Q'for all p' e F' SO that Q' is an upper bound for the totally'ordered subset FC Since f' denotes an arbitrary totally ordered subset of F, it follows ' that F is inductive. Therefore, by Zorn^s Lemma there is an ideal P e F so that P n M = (J) and P is maximal in F relative to M. Let B,C be ideals in R such that B P and C ^ P. Suppose BC 55 P? then since P is maximal with respect to M and P P + B, P^P+C then (P + B) fl M (f) and (P+C) fl M (|), Therefore, there is some m^nig e M such that m1 e (P + B) and m2 e (P + C). Then there is some P-^Pg e P and some b e B, c e C such that m1 = p1 + b and m2 = pg + C. Since M is an m-system, there is some x e R such that mnxm0 e M. 1 d. Thus m xm ]_ 2 ~ (Pi + "b) x (p2 + c) = P-^xpg + p^xc + bxp2 + bxc. Since P is an ideal in R, p1xpg e P, p^xc e P, and bxp2 e P and since BC CZ P, then bxc e P so that plxp2 + plxc + bxp2 + bxc € P* Thus rr^xrng c P, But then m^xnig t p fl M which contradicts P (1 M ~ (j). Therefore, if B J§" p and C ^ P then BC^P so that P is a prime ideal in R, Definition 1 .4: If A is an ideal in a ring R, the prime radical of A denoted by Lemma l.i: If A is an ideal in a ring R, then AS <*>(A) and Proof; Let A be an ideal in a ring R and let a e A. If M is any m-system in R such that a e M, then a e M D A so that M fl A =f (f, Therefore a e (A) so that A S\A). Let r e & (A) and let x e R. Suppose rx 0 (A), then there is some m~system M in R such that rx e M and M fl A = (j). Then M is a prime ideal by Theorem 1.2, and ASM since M (1 A = (j). Suppose 9(A) $ M, then there is some p e Theorem 1.4: If A is an Ideal in a ring R, Definitions 1.4, 1.5 and 1.6 will coincide, and the radical of A will be denoted by Proof! Let A be an ideal in a ring R and let 6**(A) be as defined in Definition 1.4. ,By Lemma 1.1, ASd^A) so that if P is any prime ideal In R so that & (A) g P, then A « P. Let P be a prime ideal In R such that ASP. Suppose G> (A) £ P, then there is some p e DP = np' and thus Def in .it ions 1.5 and 1.6 are equivalent. RC PeC Therefore, ail three definitions are equivalent, and (5? (A) = flP = np'so that (P(A) is the radical of the ideal A P<£ Pe C' in a ring R. Theorem 1.5 J If r e then there is some positive n integer n such that r e A. Proof; Let r e &(A), then M = {or"1" | i is a positive integer] is an m-system containing r so that M n A ^ (| by definition of Theorem 1.6; If A is an ideal in a commutative ring R,• then dP (A) = {r e R|rn e A for some positive integer n). Proof: From Theorem 1.5? • R which contains r meets A for rn e A. Therefore {r e R|rn e A for some positive integer n) € 0(A). Hence When only commutative rings are being studied, Theorem 1.6 is used as the definition of the radical of an ideal in a ring R. Definition 1.7: An ideal Q in a ring R is a semi-prime 2 ideal if for an ideal A in R, A c Q implies Acq. Theorem 1.7: If Q Is a semiprime ideal in a ring R n such that Acq for a positive integer n, then A c Q. Proof; Let Q "be a semiprime ideal in a ring R, and let A be an ideal in R such that An c Q, for a positive integer n. For n = 1 or n = 2, then An c Q Implies A c_ Q. Now suppose n y 2 and An c Q implies Acq for all n <( k. Suppose / N2 Ak+1 c Q. If k + 1 is even, then Ak+1 = VA / c_ Q. Since ) k+1 Q Is semiprime then A c Q and 2 <£ k since k) 2 so k+1 that by the hypothesis A 2' " c Q implies Ac=Q. If k + 1 k+2 (k+1) +1 is odd, then k +2s(k+l) + 1 is even, then A = A - / k+lN k+1 k+2 ( ^r~\2 k+2 \A )k c A c Q. Thus A c Q so that \A / = A c Q k+2 ~~ ~ p V 4-2 and Q is semiprime so that A c Q# But < k since k+2 k > 2, so that A < Q implies A c Q. Therefore, for any positive integer n, if An c Q, Q a semiprime ideal, then Acq. By definition of a semiprime ideal, every prime Ideal in a ring R is a semiprime ideal. 10 Theorem 1.81 The Intersecti on of any collection of prime ideals in a ring R is a cemiprime ideal. Proof: Let C be a nonempty collection of prime ideals P in a ring R. Then Q = HP is an ideal in R since the in- BEC tersection of any nonempty collection of ideals is an ideal. ? 2 Let A be an ideal in R such that A~ c Q. Then A c p for every P in C. Since each P e C is a prime Ideal, then each 2 P e C is a semiprime Ideal. Thus A 5. P implies A c p for all P e C. Therefore Ac Q so that, by Definition 1.7, Q is a semiprime ideal in R, Corollary; If A Is an ideal in a ring R, -0(A) is a semi- prime ideal. Definition 1.8: In a ring R, the radical of the ring is the radical of the zero ideal in R and is denoted by 0(R). Definition 1.9: Let r be an element in a ring R^ then r is nilpotent if there is some positive integer n such that r = 0. If A is an Ideal in R, A is a nil ideal If every element in A is nilpotent. A is a nilpotent ideal if there is some positive Integer n such that An = (0). Thus every nilpotent ideal is a nil ideal but not every nil ideal is a nilpotent ideal. Theorem 1.9? If r e then there is some positive integer n such that rn = 0. 11 Proof: By replacing A with (o), from Theorem 1.5? If r e £>(R) there Is some positive integer n such that rn e (o). Therefore rn ~ o for some positive integer n. Theorem 1.10* In a ring Ra 9(R) is a nil ideal which contains every nilpotent left (right) ideal in R. Proof: By Theorem 1.9* every element in #(R) is nil- potent so that by Definition 1.9* ^(R) is a nil ideal. Let n A he a left ideal in R such that A = (o) for some positive integer n, then An <= £>(R). Let a e A, then from the proof of Theorem 1.1, (a) c A + AR so that (a)n c (A + AR)n. For k = 1, (A + AR)k c Ak + A^R, Suppose k > 1 and (A + AR)kc Ak + AkR, then (A + AR)k+1= ( A+AR)(A+AR)k c (A+AR)(Ak+AkR) c A(Ak) + A(AkR) +(AR)Ak + (AR)(AkR). Thus A(Ak) + A(AkR) + (AR)Ak + (AR) (AkR) = k+1 k+1 k k k+1 A( ) + ^( ) R + A (RA ) + A.(RA )R c A^ ) + A(k+1) R + A^Ak^ + A(A^R = A(k+1)+ A(k+1)R + A(k+1) + k+1 k+1 A(k+1) R c A^ ) + A^ )R. Therefore, (A + AR)(k+1) c A^k+1^ + A^k+1^ R for any positive integer k. Hence (a)n 5 (A + AR)n c An + AnR = (o) so that (a)n= (o) and thus (a)n c £>(R). 12 Since ^(K) Is se:nlpriire„ (a)'P" c -£?(R) Implies (a) c £{R), by Theorem 1.7. Therefore, a s ^?(R). Since a was any element in A, then A c ^(R). Thus ^(R) contains every nilpotent left ideal in R. Similarly t?(R). contains every nilpotent right ideal in R and every nilpotent two sided ideal In R. Theorem2.HI If R Is a commutative ring^ then £>(R) = {r e R|rn = o for some positive integer n). The proof follows from Theorem 1.6 by replacing A with (o). Theorem 1.12: Iflis an Ideal in a ring R, the radical of the ring I is £*(R) H I. Proof: Let P(l) denote the radical of the ring I where I is an ideal in a ring R, then by definition P(l) c I. Let r e R such that r £>(R), then there is some m-system M in R such that r e M and 0 | 'M. Then 0 | M (1 I so that (o) D (Mill) = 0. Let S be the set of all ideals Q in I such that Qfl(MDI) = 0. Then S 4 0 since (o) e S and S can be partially ordered by containment. Then by Zorn's Lemma there Is a maximal element Q e S such that Qn(M(1l)=0 and Q is an ideal in I and Q is a prime ideal in I. Since Q c I then Q n I = Q so that Q n(Mfll) = QDM = 0. Thus for r g M, r | Q. Hence r is not in the intersection of all prime ideals in I so that r | P(I). Thus £fRj" c p{TJ implies PC1). 5 Since P(I) c z, then P(I) c Let r e 0(R) H I9 the/: r c •&•'•{ R) implies that every m-system in R that contains r contains o a.lso, If M is an m-system. in I such that r e M, then M is an m-system in R such that r e M so that o e M. Thus every m-system in I that contains r contains o also so that r e P(I). Therefore, 0(R) n I c P(l) so that P(I) = £>(R) n I. Examples: Let R be the ring of integers under ordinary addition and multiplication and let I = (2), then 0(1) = (2) since I itself is a prime ideal in this case. £>(R) = (o) in this case since the intersection of all prime ideals is (o). Thus £>(R) n I = (o), so that 0(1) 4 ^(R) n I. Thus the radical of an ideal I in a ring R is not necessarily the same as the radical of I considered as a ring. In this same ring of integers R; #((4) ) = (2) since (2) is the only proper prime ideal in R containing (4). Thus the radical of an ideal may be a prime ideal although the original ideal is not a prime ideal. Now ^((12))= (?) n (2) = (6) since (3), (2) are the only proper prime ideals containing (12). In this case the radical of (12) is a semiprime ideal but is not a prime ideal in R. 1 • A?.• ^cx^aeA 136 of all nilpotent right ideals in a ring R, fJg}geB be the family of all nilpotent left ideals in R, and fKy}^eC be the family of all nilpotent ideals in R. if Wa =ag^I , W = J and W = S Ky then M p yeC wa = w = wy . Ik Proof: Since each nilpotent ideal is a nilpotent right ideal and a nilpotent left ideal, then ¥ c ¥ and ¥ c ¥ . Y — a y—. (3 t 01 T € f T 1 a aeA, then I + RI is a two sided ideal in R and I c I + RI. Then, as in the proof of Theorem 1,10 (I + RI)n c in + Rin = (o) for some positive integer n, since I is nilpotent. Then (I + RI) e ^KY^YeC ^ 1 — 1 + RI impliesag^Ia c so that ¥^ c ¥^ . Thus ¥a= ¥^ . Similarly, for any nilpotent left ideal J, so that L cl p — Y and hence ¥D = ¥ . Therefore, ¥ = ¥ = ¥ . 3 Y a 0 y Definition 1.10: The nil'radical of a ring R is R w =w =w ^( ) = a B Y where Wa,¥g,¥Y were defined in Theorem 1.15. Then from the construction of 7?(R), 7l(R) is a nil ideal which contains every nilpotent left or right ideal in. R. Theorem 1.14; if a ring R is left (right) Noetherian, then 72 (R) is a nilpotent ideal. Proof: Let S be the family of all nilpotent left ideals in 7?(R), then S 4 0 since (o) e S. Since R is left Noetherian. there is a maximal element I e S. If I = ^(R)a the theorem follows. Suppose I 4 ?2(R), then ??(R)\l 4 0- Let a e ^(R)\l, then, by Definition 1.10, there are nilpotent left ideals J such that n a e I1 + I2 + ... + i Let 1 = 1 + I + x + x then c ' 1 2 * * * n> I I ' but I 4 I ' since a € I ' and a | I. But this contradicts the maximality of I in S. Therefore I — ?2(R) so that 71 (R) is a nilpotent ideal. 15 • A similar proof for R being a right Noetherian ring shows S?(R) is a nilpotent ideal. Thereforej in the case R is left (right) Noetherian, S since "by Theorem 1.10, Theorem 1.15: If R is left (right) Artinian, then every nil left (right) ideal is nilpotent. Proof; Let R be a left Artinian ring, if R contains no non nilpotent ideals, the theorem follows. Suppose R contains non nilpotent ideals. Let I be a non nilpotent left ideal in R, then let F be the family of non nilpotent left ideals con- tained in I. Then F 4 0 since I 6 P. Sinccs R is left Artinian, P contains a minimal element, l_. Since I-,c. I and I is 2 ~~ 1 1 non nilpotent, then 1^ = i^, since 1^ is minimal in F. Let S be the family of all left ideals J in R such that I j 4 (0) and c then S 4 0 since 1^ e S. Therefore S contains a minimal element J^, since R is left Artinian. Since- I J 1 1 ^ there is some x e J1 such that A = {ixjiel^ 4 (o). Let r e R and a € A, then a = ix for some i e I so that ra = r(ix) = (ri) x e A since ri e I1 for ^ a left ideal. € Let a, b e A, then a = i^x and b = i2x for some i^ig Hence a - b = i^x - igX = (^ - ig) x e A since (^ - ig) e ^ for l1 a left ideal. Therefore A is a left ideal in R. Let r 6 '!]_(A), then r = ia where i e ^ and a e A. Thus a e A implies a = i^x for some ±1 e ^ so that r = ifi-jx) = (ii^) x e A so that I-^(A) c A and A c ^ construction 16 of A. Hence I.^A = A 4 (o) so that A £ S, Since x e J-^, then A cz so that A = by the minimality of Therefore, there is some i € 1-^ so that ix = x. By induction, . n . (n-1) . • n. 1 . n !• o, 1 x = 'x = •.. = ix = x. Since x 4 °> 1 4 0 f°r every positive integer n. Thus i is not nilpotent and i e 1^ c I implies I is not nil. Therefore, since any non nilpotent left ideal is not nil, then every nil left ideal is nilpotent. A similar proof for R a right Artinian ring shows every nil right ideal is a nilpotent. Theorem 1.16: If R is a left (right) Artinian ring, then •^(R) and 7?(R) are nilpotent and 0(R) = ??(R). Proof: By the definition of 71(H), 7{(R) is a nil ideal, so that if R is left Artinian, by Theorem 1.15, 71(H) is nil- potent. Then by Theorem 1.10, ??(R) c 0(R). Also, by Theorem 1.10, £>(R) is a nil ideal so that by Theorem 1.15, 0(R) is nilpotent. Thus, by Definition 1.10, 0(R) c ft(R). Therefore, £>(R) = 7?(R). Theorem 1.17: If R is a commutative ring, then £>(R) = 7?(R). Proof: Let a € -0(R) where R is a commutative ring. Then an = o for some positive integer n. Since R is a • commutative ring, then (a)n = (o) so that by Definition 1.10, (a) c fl(R). Thus a e 7?(R) so that 0(R) c 7?(R). Let x e ??(R), then since ??(R) is a nil ideal, xm - o for some positive integer m. Then by Theorem 1.11, since R is commutative, x e £>(R). Thus 7l(R) c £>(R) so that 7l(R) = £>(R). 17 Therefore, if R is a ^Onraiatax-ive ring or if R is a left (right) Artinian ring, then £(R) and »1(R) coincide as the radical of the ring R. Examples: Let TL "be the ring of integers and let R = Z/(8) = {0,1,2,3,^5,5/?]. Then R is a commutative ring with unity, where in general n = n + (8). Thus, in R, the nilpotent elements are 0,2,5,6. Therefore, "by Theorem l'.ll, £>(R) = jb,2,5,5j. Thus, since [£>(R)]^ = (5), -0(R) is a nil- potent ideal. Then, by Theorem 1.17, ??(R) = ^(R) since R is a commutative ring. Definition 1.111 An ideal IT in a ring R is called a nil radical of R if (i) S is a nil ideal and (ii) R/M contains no nonzero nilpotent ideal. Then, in a ring R, if t< = Sla where I ranges over all nil ideals in R, l{ is a nil radical of R "by Definition 1.11 and H contains every nil radical of R by construction of % From Definition 1.10, 7l{R) is a nil radical of R by Definition 1.11. Therefore, 7[(R) <= . Theorem 1.18: An ideal P in a ring R is semiprime if and only if R/P contains no nonzero nilpotent ideals. Proof: Let P be an ideal in R such that R/P contains no nonzero nilpotent ideals and let A be an ideal in R such that A2 e p. Then A/P is an ideal in R/P and (A/P)2= A2/P and since A2 c P, then A2/P = (o) in R/P so that (A/P)2= (o) implies A/P = (o) in R/P since R/P contains no nonzero nil- potent ideals. Hence A/P = (n"S in vt/v> 18 Therefore, by definition, P is a serniprime ideal. Now suppose P is a serniprime ideal and A is an ideal in R/P such that An = (o) in R/P, for some positive integer n. Then A an ideal in R/P implies A = A '/P where A7 is an ideal in R. Then A = (A'/P)n = (A')n/P = (o) in R/P. But (A')n/p . = (o) in R/P implies (A')n c p, and P is a serniprime in R, it follows by Theorem 1.7 that A' c p. Thus A'/P = (o) so that A = (o) in R/P. Therefore, R/P contains no nonzero nilpotent ideal. Therefore, from Definition 1.11, every nil radical in R is a serniprime ideal in R. Definition 1.12: In a ring R, let C be the collection of all nil radicalsfltfof R, then C 4 0 since n(R) € C. Let £ = fltf. mc Theorem 1.19: ^ is a nil radical of a ring R. Proof: Since st = ntf where C is the collection of all f/eC nil radicals in R and C is not empty, then £ is the intersection of a nonempty collection of ideals so that L is an ideal in R. Let a e then a for allN e C and each is a nil ideal so that there is some positive integer n such that an = o. Therefore, =£ is a nil ideal. Let A be an ideal in R such that A2 c =4 then A2 for all^e C. Since eachWis a serniprime ideal, by Theorem 1.18, A2 implies A ci^ for ' all V e C so that A c Thus A2 c ^ implies A c £ so that •£, is a serniprime ideal. Therefore, by Theorem 1.18, R/s£ contains no nonzero nilpotent ideals. Thus by Definition 1.11, =£ is a nil radical of R and is contained in every nil radical of R. 19 Theorem 1.20', In a ring R, r/. .Lb the intersection of" the ideals M in R such that R/M ha,s no nonzero nilpotent ideals. Proof: Let D 'be the collection of all ideals M in R such that R/M has no nonzero,nilpotent ideals, then from Theorem 1,19? R/s£ has no nonzero nilpotent ideals so that £ e D, Therefore, fl M <= sL Since s£ e D, n M is the in- MeD MeD tersection of a nonempty collection of ideals in R, so that fl M is an ideal in R. Let a e H M, then a e st and is a MeD MeD nil Ideal in R so that there is some positive integer n such that a11 = o. Thus n M is a nil ideal in R. Let A he an ideal MeD in R such that A 2 c n M. Then A"2 c M for all MeD. Since ""MeD R/M contains no nonzero nilpotent ideals for all MeD, each M e D is a semiprime ideal in R, by Theorem 1.18. Therefore, 2 A c M for all MeD implies A <£M for all MeD. Hence 2 A c n M. Then A c n M Implies A c n M so that fl M is a semi- ~MeD MeD ~MeD MeD prime Ideal. Thus R/fi M has no nonzero nilpotent ideals, by MeD Theorem 1.18. Therefore, D M Is a nil radical of R by MeD Definition 1,11. Therefore, aj c n M from Theorem 1.19. MeD Hence •£ - fl M. MeD If ©/is any nil radical in R, =£ c by Definition 1.12 and^ c %{ by definition of % Thus £ c c for every nil radicalM of R so that & is called the lower nil radical of R and H is called the upper nil radical of R. There is a 20 ring R in which *£ ip '4 in J&eobson- s Structure of Rings. page 255. Therefore, the distinction between lower and upper nil radical is necessary in such a ring. Theorem 1.21: In a ring R, £»(R) = Proof; From Theorem 1.8, ^(R) is a semiprime ideal so that by Theorem .1,18, R/-(R). Suppose ^(R) t then there is some p e 0(R) such that p Then, since £is a semiprime ideal (p)^ jr 4, then pRp jb Therefore, there is some c e R so that p = pep e £>(R) and Pj | L. Then in a similar manner there is some c^ e R so 0 that p2 = Pi !?! e ^(R) and p2 | ^ Let E = tpi=pi_l ci-l pi-l^pi-i G ^R),ci-i 6 R> Pj_ I then E n £= 0. Let Pn,Pm e E, then p^ e E so that P = P c p Then p e E so that n-+m n+(m-l) n+m~l (n-l)+m. n+m-l pn+(m-l) pn+(m~2)cn+(m-2)pn+(m-2) ~ ^pn+(m-5)Cn+m-5pn+(m-3)) cn+m-2^pn-HB~3Cn-+m-3pn+m$^ * Let k., = c n n 1 n+m-3 n+m-3 n+m-2 Pn+m-3Cn+m-3 SO that pn+m~l = pn+m-3klpn+m-3' Then pn+m-l _ (Pn+m-^Cn+m-4pn+m~4)k l (Pn-Hn- 4cn+m~4pn+m-4) so that k = c p k p c then p 83 p k p ^ 2 n+m-4 n.+m-4 l n+m-lf n-+m-4' n+m-l n+m~4 2 n+m-4. Since m is a positive integer, in m~2 steps we get p k n+m-l = Pn+(m-m)V2Pn+(m-m)= Pn ffi,SPn. similarly Pn-1 +m pmkn-2pm* Therefore p k n-4m ^^ ^m~2^nCn+m--l^m^:n---2' then jt - R so fchao p^xp^ e E. Therefore, E is an m-system in R such that E n L = 0 for =£ an ideal in R.- Hence by Theorem 1.2, there is a prime ideal P in R such that =£ c p and' P fl E = 0. But E c 0(R) by construction and E n P = 0 implies 0(R) Therefore, £>(R) is the lower nil radical of R so that 0(R) c 7?(R) for any ring R. if R is left (right) Noetherian, then fl(R) c &(-R) so that £>(R) = 7l(R) if R is left (right) Noetherian. Since 7l(R) is nilpotent if R is left (right) Noetherian, then £>(R) is nilpotent also. Since 0(R) = ^(r) if R is left (right) Artinian or if R is commutative, then 71(H) coincides with the lower nil radical of R if R is left (right) Noetherian, if R is left (right) Artinian, or. if R is commutative. If R is a left (right) Artinian ring, then by Theorem 1.15, since %( is a nil ideal in R, %{ is nilpotent. Therefore K c 0(R). But 0(R) = £ c %( so that K = £ if R is left (right) Artinian. If R is a commutative ring and a e % then there is some positive integer n such that a11 = o, since K is a nil ideal. But since R is a commutative ring a e -£?(R) by Theorem 1.11. Thus K e £>(r)? but ^>(R) = ^ c t( so that if R is commutative -0(R) = % Therefore if R is left (right) Artinian or if R is commutative, t( = =£ so that R has exactly one nil radical. CHAPTER II JACOBSOK RADICAL Definition 2.1: The additive Abelian group M in a ring R is said to "be an R-module if for every r € R and m e M, then rm e M such that (i) r(ra1+ nip) = rm^ + rm2.; (ii) (r + s)m. - rm^ + sm^; (iii) r(sm1) = (rsjm^; for all r, s e R and mmg e M. Definition 2.2: If M is an R-module in a ring R3 an element r e R is said to annihilate M if rm = a for all m e M. The annihilator of M is A(M) = {r e R|rm = o for all m e M}. Theorem 2.1'. If M is an R-module in a ring R, A(M) is a two-sided ideal in R. Proof: Let a e A(M) and r e R, then for every m e M, (ra)m = r(am) = r-o = o so that ra e A(M). Since M is an R-module^ rm e M for all m e M. Thereforef (ar)m = a(rm) = o since a € A(M). Thus ar e A(M). Let &, b e A(M), then (a-b)m = am-bm = o-o = o so that a-b e A(M). Therefore, A(M) is an ideal in R. • PQfi^ition 2.J>'. If M is an R-module in a ring Ra M is said to be an irreducible R-module if RM 4 (o) and the only submodules of M are (o) and M. 22 Definition 2,4: Tus Jacobson radical of a ring R, denoted by j?(R)a is the set of all elements of R which annihilate all irreducible R-modules in R. If R contains no irreducible R-modules, then ^(R) ~ R. In the following theorems, let C denote the collection of all irreducible R-modules M, in a ring R. _Theore m 2,2: In a ring R, j?(R) =Me HA(M)C . Proof: If R has no irreducible R-modules, then G = 0. Since nA(M) £ R, let r e R such that r | fiA(M), then there MeC MeC is some MQ e C such that r "TyteC MeC Definition 2.5: If P is a left (right) ideal in R such that P 4 R and for any left (right) ideal I such that P 5 I implies P = I or I = R, then P is said to be a maximal left (right) ideal in R. If Q Is a left (right) ideal in R- such that Q 4 (o) and for any left (right) ideal I such that (o) c I c Q implies (o) = I or I = Q, then Q Is said to be a minimal left (right) ideal in R. 24 Definition 2.6: A left ideal I .in. J ring R is said to "be regular if there is seine a 6 R such that x - xa e I for all x e R. Definition 2.7: If M is an R-module in a ring R^ a mapping f of R on to M is an R - horn omo rp h. ism if, for any r, s e R, f(r) + f(s) = f(r -I- s) and f(rs) = rf(s). If such a mapping is one to one., it is called an R-isomorphism. Theorem 2.5: Let M be an irreducible R-module in .a ring R, then there is a maximal regular left ideal P .in R such that M = R/P under an R-isomorphism. Proof: Let m be a fixed element of M such that m 4 then Rm is a submodule of M and Rm 4 (o) since m 4 o. There- fore, Rm = M since M is irreducible. Let a mapping f from R to M be defined such that f(r) = rm for all r e R. Since Rm = M, f is an onto mapping so that f is an R^homomorphism. Let P = {x e R[xm = o}, then P is the kernel of f. Let p e P and r e R, then (rp)m = r(pm) « r.o = o so that rp e P for all r e R. Let p, q e P, then (p-q) m = pm-qm = o-o = o so that p-q e P. Thus P is a left ideal in R. Therefore, IV M = R/P under an R-isomorphism. Let P ' be a left ideal in R such that P P'/P c R/p = m so that Pr'/P = S where S c M. Since P c p 's then P'/P 4 (o) so that S 4 (o) in M. Since P '/P = S, then • S = {x e M|x = pm for some p e P'} is a subgroup of M. Let 7 a e S and r e R, then ra = r(pQm) for some pQ e P so that 7 ra =' (rpQ)m. Since P'is a left ideal, then rp e P and 25 thus ra e S. Let a/b e 3 and r e R, then r(a+b) - r(p^m + Pora) 7 for some p^jpp £ P so that r(a+b) = r(p^m) + r(pQm) = ra + rb. Let r,s e R and a e S, then (r + s)a - (r + s)(p m) for some pQ e P ' and thus (r + s)a = (rpQ + sp )m = r(p )m + (spQ)m - ra 4 sa. Similarly r(sa) = r [s(pom)j = r[(spo)m] = [r(spQ)]m -- [(rs)po]m = (rs)(p m) = (rs)a. Thus S is a sub- module of M and S 4 (°) so S = M since M is irreducible. Therefore, P '/P = M and R/P = M implies P ' = R. Thus P is a maximal left ideal in R. Since Rm - M, there is some a e R such that am = m. For any x e R, xam = xm so that xm - xam = (x-xa)m = o. Thus x-xa e P for all x e R so that P is regular. Therefore P is a maximal regular left ideal in R. Theorem 2.4: Every regular proper left ideal in a ring R is contained in a maximal regular left ideal in R. Proof: Let P be a regular proper left ideal in a ring R. If P is a maximal left ideal in R, then the theorem follows. If P is not a maximal left ideal, then there is some a € R such that x - xa e P for all x e R, since P is regular. For a e R, either a € P or a | P. Suppose a e P, then xa e P for all x e R. Thus xa + x-xa - x e P for all x € R and hence, R c p which contradicts P being a proper left ideal. There- fore a | P. Let F be the collection of all proper left ideals Q in R such that P c Q, then F 4 0 since P e F. Let Q e F, 26 then if a e Q, xa e Q for all x € R and since x-xa e P then x-xa e Q so that xa. + x--x& - x e Q. for all x e R. Thus Q = R which contradicts Q, a prope-r left ideal in R. Hence a | Q for all Q € C. F can be partially ordered by set inclusion. Let F ' be a chain In F, thenF7 Is totally ordered and / UQ' e F since F ' is a chain. For all Q e F Q' e F QiP' 0 o Q' c UQ' so that F' has an upper bound In F. Therefore, by ° ~q£F' Zorn's Lemma, there is a maximal element p' e F. If PQ is a left ideal in R such that P ' c P c R then P = R since + o - 5 o P I F for P c p' c p and P' maximal in F. Thus P ' is a oA T — + o maximal left ideal in R. Since a e R such that x-xa e P c p ' for all x e R, then P 7 is regular. Therefore P £P7 where p' ft is a maximal regular left ideal in R. Definition 2.8: If P is a left ideal In a ring R, then RIP = {x s R|xR c p}. in the following theorems, let D denote the collection of all maximal regular left ideals in a ring R. Theorem 2.5' In a ring R, $(R) = n(R:P). PeD Proof: From Theorem 2.2, $(R) = flA(M). By Theorem 2,3, M€C for each M e C there is some P'eD such that M = R/P . o o o o Therefore, DA(M)= flA(R/P). Thus for all PeD, A(R/P) = MeC PeD {xeR|xrp = o for all rp e R/P} = {xeR[xr e P for all r e R} {xeRlxR c p} = r:p. Therefore n(R:P) = nA(R/p) = oa(m) = $(r). ' Pe.D PeD MeC Theorem 2.6: In a ring R, j?(R) = np. PeD Proof: Let x e j?(R) in a ring R and let PQ e D, then there is some a e R such that x-xa e Pq since Pq is regular. From Theorem 2.5, £(R) = n(R:P) so that x e £(R) implies PeD x e (R:Pn) so that xR c p Hence xa e P . Thus u — O O xa + x-xa = x e PQ for each PeD. Thus x e HP and hence PeD J?(R) c np. ""PeD Let x e HP, then Q = {r + rx|r e R} is a regular left' BsD ideal in R. Suppose Q 4 then there is some p' e D such that Q c p Thus x e P 7 so that rx e P ' and hence • r + rx - rx = r e P ' for all r e R. Thus P ' = R which con- tradicts p.' e D. Therefore Q = R and since -x e R there is some y e R such that -x = y + yx. Hence x + y + yx = o. Suppose np ji j?(R) = nA(M)a then there is some MeC PeD MeC ° such that np £ A(M ). Thus for some p e np and some m e M u PeD PeD * pm 4 o. Since (np)m is a submodule of Mcand (np)m 4 (o), EfeD PeD then (np)m = since M is irreducible. Therefore, there is PeD some p^ e np such that p_m = -m since -m e M* Since d e np PcTl -J- 9 ¥ *"] ^ IX 3 reu peD there is some s e R such that p^ + s + sp^ = o. Thus o — o»m — (pj+s+sp^)m = p^ra + sm + sp-^iri — -m + sm -sm = ~m 28 which implies (flP)m ~ (o) which contradicts (np)m 4 (°)» PeD ' P.eD Therefore HP c: ^(R) so that $(R) - (IP. PeD ' PeD Thus far the characterization of $(R) has "been in terms of left ideals In R so that j?(R) might he called the left Jacobson radical of R. If a right Jacobson radical $ '(R) were constructed similarly in terms of right ideals instead of left, the same results would be obtained for right ideals. Thus, since $(R) = flA(M) is a two-sided ideal in R,for A(M) MeD a two-sided ideal in R for all M € D, then $'(R) is similarly a two-sided ideal in R and $(R) = Hence, the dis- tinction between right and left Jacobson radicals is unnecessary, Definition 2.9: In a ring R, let x e R, then (i) If there is some y e R such that x+y+xy = o, then x is right quasi-regular and y is the right quasi-inverse of x; (ii) If there is some z e R such that x+z+zx = o, then x is left quasi-regular and z is the left quasi-inverse of x; (iii) If x is left quasi-regular and right quasi-regular, then x is quasi-regular; (iv) If w e R is a right quasi-inverse of x and a left quasi-inverse of x, then w is a quasi-inverse of x; (v) If A is a left or right ideal in R such that every element of A Is right (left) quasi-regular, then A Is said to be right (left) quasi-regular. Theorem 2.7: In a ring R, let x e R. If y e R is a right quasi-inverse of x and z e R Is a left quasi-inverse of x, then y = z. ?9 Proof: From Definition 2.9. x+y+xy - o and x+z+zx = o, then x = -z-zx. Thus o = x+y+xy = (— z—zx) + y + (-z-zx)y - -z-zx + y - zy ~ zxy = -zx-zy~zxy~z+y = -z(x+y+xy) -z + y ~ —z*o " z + y =-z + y = o. Therefore z = y. Theorem 2.8: Every nilpotent element of a ring R is quasi-regular. Proof: Let x e R such that xn = o for some positive -I n-1 n-1 inceger n and. let y = S (—x) , then x+y+xy = x + S (—x) ^+x( 2 (—x)^) = o k=l k=i k=1 n-l , n-1 , and x+y+yx = x + S (-x) + (Z) (~x) )x = o. Therefore x+y+xy k=l k=l = x+y+yx = o. implies x is quasi-regular. Theorem 2.9* If a right (left) ideal A in a ring R is right (left) quasi-regular, then A is quasi-regular. Proof: Let A be a right quasi-regular right ideal in R and let x e A, then there is some y e R such that x+y+xy = o. Thus y = -x-xy and -x € A for x € A so that -xy e A since A is a right ideal. Therefore, -x~xy = y e A implies there is some- z e R such that y+z+yz = o. Thus y+x+xy = o and y+z+yz = o implies x = z by Theorem 2.7. Therefore x+y+xy = x+y+yx = o implies x is quasi-regular so that A is a quasi- regular right ideal in R. A similar proof if A is a left 30 quasi-regular left ideal in K shows A is a quasi-regular left ideal in R. Theorem 2,10: In a ring R, ^?(R) is a left 'quasi-regular left ideal in R and $(R) contains every left quasi-regular left ideal in R. Proof: From Theorem 2.2, $(R) = HA(M) and for each M e C, MeC A(M) is a left ideal in R, by Theorem 2.1. Thus ^(R) is a left ideal in R. From Theorem 2.6, $(R) = HP, so that, .by PeD the proof of Theorem 2.6, for x e $(R)5 there is some y e R such that x+y+yx = o. Therefore j?(R) is a left quasi-regular left ideal in R. Let I be a left quasi-regular left ideal in R and let MeC. Suppose IM 4 (°)> then there is some m e M such• that Im = M since M is irreducible. Thus there is some a € I such that am = -m. Since a e I and I is left quasi-regular, there is some b e R such that a+b+ba=o, thus b = -a-ba e I since I is a left ideal. Therefore, o = o*m = (a+b+ba)m = am+bm+bam = -m+bm-bm = -m. Thus o = -m implies Im = (o) which contradicts Im = M. Therefore IM = (o) so that I c A(M) for all MeC. Hence I c j?(R) so that every left quasi-regular left ideal in R is contained in j?(R). Corollary: j?(R) is a quasi-regular left ideal in R and contains every quasi-regular left ideal in R. Theorem 2.11: In a ring R, $(R) is a right quasi-regu- lar right ideal in R and j?(R) contains every right quasi-regular right ideal in R. Proof: As has already been stated, j?(R) is a right ideal in R, since $(R) is a tiAro-sided ideals and .from the preceding theorem, j?(R) is quasi-regular. Thus j?(R) is a right quasi-regular right ideal in R. In a similar proof to the proof of Theorem 2.10, $(R) contains every right quasi- regular right ideal in R. Corollary{ <$(R) is a quasi-regular ideal in R that, con- tains every quasi-regular right or left ideal in R. Theorem 2.12: In a ring R, $(R) = fxeR|axb is quasi- regular for all a,b € R}. Proof: Let x e ^(R), then axb e j?(R) for all a,b e R since j?(R) is an ideal in R. Therefore axb is quasi- regular by the Corollary to Theorem 2.11. Therefore, $(R) 5 € R|ax"b is quasi-regular for all a,b* e R}. Let z e {x e R|axb is quasi-regular for all a,b e R} and let M e C, then suppose zbQ ^ A(M) for some bQ e R. Thus there is some m | o in M such that zbQm 4 o. Since M Is an R-module, zbQm € M. Thus R(zbQm) = M since M is Irreducible. Therefore, there is some r e R such.that r(zbQm) = (rzbQ)m = -m. Since azb is quasi-regular for all a,b e R, then rzbQ is quasi-regular so that there is some c e R such that (rzbQ) + c + c(rzbQ) =• o. Therefore, o = o»m = (rzbQ + c + crzbQ)m = (rzbQ)m + cm + (crzbQ)m = -m + cm-cm = -m. Thus R(zbQm) = (o) which contradicts R(zb m) = M. "^9 Therefore, zb m = o for all m e M and thus zb e A(M). o o Therefore, zb e A(M) for all b e R. If m^ e M such that m]_ 4 (°)» then Rm^ = M. Thus zM = zRm^ = (o) since zb e A(M) for all b e R. Therefore, z e A(M) for all M e C so that z e $(R)• Therefore, {x e R|axb is quasi-regular for all a,To e R} <= j?(R) so that $(R) = (x £ R | axb is quasi-regular for all a,b e R}. Theorem 2.1j?' In a ring R, $(R) = {x e RjxR is right quasi-regular}. Proof: Let x e $(R), then xr e $(R) for all r e R so that xR is right quasi-regular. Thus $(R) 5 {x e R|xR is right quasi-regular}. Let z e {x e R|xR is right quasi-regular}, then zr is right quasi-regular for all r e R. Since xR is a right ideal in R, then xR is a right quasi-regular right ideal so that xR <= j?(R), "by Theorem 2.11, Let M e C, then xR c A(M) so that xr e A(M) for all r e R. Let m e M such that m 4 °> then Rm = M since M is irreducible. Thus xM = xRm = (o) since xr e A(M) for all r € R. Hence x e A(M) so that x e j?(R) and thus fx e R|xR is right quasi-regular} c j?(R) so that j?(R) = {x e R|xR is right quasi-regular}. Theorem 2.14: In any ring R, j?(R) contains every nil ideal in R. 33 Proof; Let N "be a ail Ideal in R and let x e N. Since N is an ideal in R, axb e N for all a,b e R. Thus axb is nilpotent for all a/D e R and therefore axb is quasi-regular for all a,b e R by Theorem 2.8. Therefore, x e 'j?(R) by Theorem 2.12. Thus N c $(R) so that $(R) contains every nil ideal in R. Corollary: In a ring R, l(c ^(r) where % is the upper nil radical of R. Theorem 2.13! If R is a left (right) Artinian ring, then j?(R) is a nilpotent Ideal in R. Proof: Let R be a left Artinian ring and let J = j?(R)i 2 3 then J =>jr)j=)... is a decreasing chain of ideals in R so that there is some positive integer n such that J 3 j-^zd ... => Jn = Jn+1= ... = j2n= .... For any x e R, if J2nx = (o) then Jnx = (o) since J2n= Jn. Let I = {x e R|jnx = (o)}. If Jn c I, then J2n= (o) so that Jn = (o) and thus J is nilpotent. Suppose Jn Thus x e I. If J 4 (o) in R/l, then Jn contains a minimal left ideal P in R/l since R/l is left Artinian for R being left Artinian. Thus either J11!? = (o) or J^"p 4 (o). If ^ ^ J E =f (o) then (R/l) P 4 (o) so that P is an irreducible R/l module since P Is a minimal left ideal in R/l. Since ' J c J then J p e j P = (o) since anything in J annihilates all irreducible R/l modules. Therefore, jn p = (o) in R/l. 5 J] e»* Let P be the Ideal in R corresponding to P in R/l, then "*=• pn ri Ja P = (o) in R/I implies Jr~n p = jnp = (o) in R, hence P ^ I so that P = (o) in R/I which contradicts the selection of P being a minimal left ideal in R/l. Therefore Jn = (o) in R/I which implies Jn c I so that J2n = (o) and hence Jn=(o) Therefore, J = j?(R) is nilpotent if R is left Artinian. A similar proof for R being right Artinian shows $(R) is nil- potent . Corollary % If R is a left (right) Artinian ring, then j?(R) = ^ Therefore, In any ring R, j?(R) contains every nil radical of R. If R is left (right) Artinian, then j?(R) coincides with the nil radical of R. Theorem 2.16: If I is a two-sided ideal in R, then »{1) = m^(R). Proof: Let a e in^(R), then a e j?(R) so that ^arg is left quasi-regular for all rx,r2 e R. In particular, xay is left quasi-regular for all x, y e I, so that a e j?(I). Therefore, Ifl$(R) c ^(i), By definition, j?(l) c I. Suppose j?(R) = (o) in R and let P = (x e R[lx = (o)}, then P is a right Ideal R. Since I is an ideal in R, 1^(1) is a left Ideal in R and Ij?(l) <= ^(i) so that I#(I) is left quasi-regular. Thus lj?(l) c j?(R) = (0) by Theorem 2.12. Therefore, j?(l) c P and j?(l) c I implies £(I) <= pni. Let x e PDI, then x e P and x e I so that x2 = o. 35 Thus pni is a nil right ideal so that Pfll c $(~L), Therefore, pni = j?(l) so that j?(l) is a nil right ideal of R, Therefore, $(I) c'^(R) from Theorem 2,lb. Thus j?(l) ^ j?(R)flI so that J?(I) = ^(R)m for j?(R) = (o). Suppose $(R) 4 °> then (l+^(r) )/^(R) is an ideal in R/J?(R). Then (I+^(R) )/j?(R) ] = (o) since ^(R/^R)) = (o). For if P is an ideal in R/j?(R) and P is an ideal in R corresponding to P in R/$(R) and if P is a left quasi-regular left ideal in R/^(R), let a e P such that a = a + $(R) e P. Then there is some b e P such that o = a + "5 + Fa =? a + b + ha so that a + b + ha e £(R) and thus is left quasi-regular. Thus there is some c e R such that a + b + ba + c + c(a+b + ba) = o. Therefore a + (b + c + cb) +(b+c+cb)a=o implies a is left quasi-regular so that a e j?(R), therefore P c j?(R). Thus P = (o). Since $(R/$(R)) is a left quasi-regular left ideal in R/$(R), then J?(R/$(R)) = (o). Then from the pre- ceding part of the proof, (I+^(R)/j?(R) ] c ^(R/^(R)) = (o). Sincefl + i?(R)]/^(R) = I/IDJ?(R) by the Dedekind-Noether isomorphism theorems, then j?[l/(in^(R)) ] = (o) so that J?(I) £ in^(R). Therefore, ^(1) = Theorem 2.17', if $(R) is the Jacobson radical of a ring R, the Jacobson radical of the matrix ring over R n is $(R) n* ProofLet A e R^ be of the form A ^ Sil where left quasi—regular, then there is some ^ ^ such that ai;i + a^ + a^a.^ = o. Thus {x + xa-^-Jx € R} c_ R so that if y e R, then y = y+ o = y+y.o = y+ y(alx+ a^+ = y + ya^ + ya^ + ya^a^ = (y + ya^) + (y + ya^)^. SQ tha(. y e {x + xa11|x e R} and thus R = {x + xaxi|x e R}. Tnus for i = 1,2,..., n, there is an a^ e R such that a. 'li + aliall ~ ~ali' ^' e suc^ that ^"1 ^ o o » • • o^ • t A '= • • J * • a„n l.. o o * • • o j then A+A + A A — o so that A is left quasi-regular in R Let J^. = {A e Rn|aik = o for k 4 j and a±J € j?(R)}, then J is a left ideal in Rn for all j = 1,2,..., n. By a similar argument to the above for J = 1, each J is a left quasi- regular left ideal in R for J = 1,2,...,n. Thus J. C a(R ) for all j and hence Q(R) _ T . T , / K >n - J± + J2 + ...+ Jn c ^(Rn). Let c e J?(R ) and for any a e R, let A be the matrix with 0 • n' - pq a in the (p,q) position and zeros elsewhere,, Let a_,b e R, then ° . . . o V > p-D _ | ^ ^ ^ • # O ^\pCBgk I ' •P% • i : : * . : ° O • • - a R But C e ^(Rn) implies SA^B e K n)- Let C ' be the quasi- inverse of £A,k p CB q,k so that acp qb + c'1 1 + c.'.1,1 a.c pqb = o = acp qb + c/1,1 + acp qbe/. 11, . Thus acp q b is *quasi - regular for all a,b e R so that c € $(R) ;md, therefore, C e JK*n) = *(*0n. Theorem 2.18'. If £>(R) is the prime radical of a ring R, then the matrix ring Rn has prime radical £>(R)n. Proof; Suppose R has a unity, then there is one-to-one correspondence between the ideals I c R and the ideals R and R//I = "^n - n ( )n ^n^n* Therefore, In is a prime ideal in Rn if and only if I is a prime ideal in R. Therefore, if F is the set of all prime ideals P in R, then R ^( n) - np = (np)_ = m). P ^(Rn) = €»(Rp> Rn - (^(R')riR)n - ^(R)n. Therefore, from the corollary to Theorem 2.l4, for a ring R, if Rn is the matrix ring over R, ^(^)n S ^(^)n« ^ R E = R R is left (right) Artinian, then <£( n) = ^( n) ^( n)> since £(R) = 0(R) = ??(R). Example: Let R be the ring of two-by-two matrices over the ring of integers Z, then R is a noncommutative ring with unity. Since £>(Z) = (o) then 0(R) = (o). Since 0(R) contains all nilpotent ideals in R, then (o) is the only nilpotent ideal in R. Let Z be the ring of integers, then Z is an integral with unity and &(Z) = (o). Since Z has the property that every prime ideal is a maximal ideal then $(Z) = HP where P ranges over all maximal ideals in Z and hence over all prime ideals in Z. Therefore ^(Z) = 9{Z ) = (o). Now consider Z/(8) = R, then, from a previous example, 6>(R) = {0,2,5,5} is the set of all nilpotent elements in R. Hence 0(R) c ^(r) and in this the only quasi-regular elements of R are {5,2,5,6} so that £(R) = e(R). Therefore j?(R) 4 (o) in Z/(8). CHAPTER III BROWN-McCOY RADICAL In Chapter II, the Jacobson radical of a ring R was shown to consist of elements of R that were left or right quasi-regular. In a somewhat similar manner, the Brown- McCoy radical of R can be defined in terms of elements of R with a specific property. Definition 3«1' In a ring R, let a e R and let G(a) = {ar-r + 2(x.ay. - x.y.)|r,x.,y. € R}, then a is said 3^ X 1 1 IL to be G-regular if a e G(a). A left (right) ideal A in R is said to be G-regular if every element of A is G-regular. Theorem 3.11 In a ring R, let a e R, then G(a) Is a two-sided ideal In R. Proof: Let g eG(a), then there is some r e R such that g = ar-r + 2(x.ay.-x.y.) for some x.,y. e R. Let p e R, then -i. -L i i i i pg = par - pr + pS^ay^ x^) = par-pr + ^(px^y^px y.^) = (par + 2pxiay1) - (pr + Zpx^) = a-o - o + 2(x/a. x.'y/) K X X 1 > where xf, y^ e R for all i, so that pg e G(a). Likewise gp = arp - rp + (Z(x - x^y ))p = arp - rp + S(xiay1p - x-y^p) so that gp e G(a). 39 Let g,h e G(a), then there are some e R such that g = ar - r + £(x..ay^ - x^y^) and h ;= as~s + 2(x.ay. - x y ). Thus g - h = (ar-r + Z(x, ay. - x.y. )) - J J J x == ar - as - r + s + 2(xiay1 - x^) - 2(x^ayj - jyj) a(r - s) - (r - s) + Z[ (x±ayi - x^ay ) - (x±y± - x^)] so that g-h e G(a). Therefore, G(a) is a two-sided ideal in R. Definition 3.2: In a ring R, let the Brown-McCoy radi- cal of R he denoted by ^(R), then %(R) = {a e R|(a) is G-regular) where (a) is the principal ideal generated by a e R. Lemma 3.1: In a ring R, let a, c e R, then if a - c is G~regular and c e G(a), then a is G-regular. Proof: Let a, c € R such that a - c is G-regular and c € G(a), then there exist r, x^y^^ e R such that a-c = (a-c)r - r + Z[x.(a-c)y. - x^] ar cr - r + 2[(xjLay1 - x^y^ - x±yi] = ar - r + 2(xiayi - xiyi) - cr - 2xj.cyi. Thus a = ar - r + S(x.ay - x.y.) + (c - cr - Sx.cy.) and -L ~L X 11 since c e G(a) then (c - cr - 2xicy^) e G(a) since G(a) is an ideal in R and ar - r + 2(x^ay^ - x^y^) e G(a). Therefore, a e G(a) since G(a) is an ideal. 4l Theorem 3.2: In a ring R, 7/l('R) Is a two-sided ideal. Proof: Let a € 5??(R) and r e R, then ar e (a) so that (ar) c (a) so that (ar) is G-regular since every element of (ar) is in (a) and thus G-regular. Similarly ra e (a) so that (ra) c (a) and thus G-regular. Therefore ar, ra e #?(R). Let a, b e 7!\{R) and let z £ (a-b), then z = a^ - b^ for -some a^ e (a) and b^ e (b). For a e 7/?(R) and a^ e (a), then there is some r, x. ,y. e R such that an - a, r - r + S(x.a_y. - x.y.). i 1 1 v l l^i u/i' Now z = a^ - b^, then a^ = z + b^ so that z 1 = z + °1 ( + tx)r - r + Z[x±(z + b1)y1 - Thus = zr - r + S(x±zyi - + t^r + Zx^.^. z - [zr - r + 2(xizyi - x±y. ) ] = - + b^r + Zx^y . Since b-L e (b) and b e /^(R), then b^ e G(b1) so that b + b r ~ l i + 2x1b1yjL is G-regular. Therefore, by Lemma. 3.1, since z - [zr - r + 2(x1zyi - x^)] is G-regular and zr - r + S(xizy1 - x±y±) e G(z) then z e G(z). Therefore, (a - b) is G-regular since every element in (a - b) is G-regular. Thus a - b e 57?(R) so that %(R) is a two-sided ideal in R. Theorem 3.3: In a ring R, 57?(R/^(R)) = o. 42 Proof: By Theorem 3.2., 7?i(R) is a two-sided ideal in H, so that there is a one-to-one correspondence between the ideals of R which contain %(R) and the ideals of R/^(R). Let b e R such that b = b + flf(R) e ^(R/5?|(R)) and let a e (b) in R. If (b) denotes the ideal in R/#?(R) which corresponds to (b) in R and (b) denotes the ideal in R/%(R) generated by b e R/^(R), then fb) = (b). Therefore a e (b) implies a e (b). Since b" e ^(R/59?(R)S (b) is G-regular in R/#?(R) so that a e G(a) = G(a) in R/^(R). Therefore, there is some c € G(a) such that c e G(a) so that a - c = (&-c) = o in R/%(R). Therefore, a - c e 55f(R) so that a - c is G-regular and c e G(a). Thus a e G(a) by Lemma 3.1. Hence (b) is G-regular so that b e 7l\(R). Thus b = o in R/^(R). Therefore, s?i(rA{R) = Co) - Definition 3.3* In a ring R, let a e R, then H(a) = - x±y±)|x±,yi e R) so that if a e H(A), then a is said to be H-regular. A left (right) ideal A in R is H-regular if every element of A is H-regular. Lemma 3.2: In a ring R, let a € R. If RaR c H(a)" then H(a) = R2. roof 2 - P * From construction, H(a) c R f0.r any a e R. r) p Suppose RaR c: H(a), then R^" c H(a) so that H(a) - R~. If R ' is a subring of R, then H?(a) is the ideal H(a) restricted to R' and H"'(a) cH(a), Hregularity is defined analogously to H-regularity. Theorem 3.4; In a ring R, the following are equivalent: (i) a € ^(R); (ii) RaR is G-regular; (iii) RaR is H-regular. Proof; Let a e 5^(R) in a ring R, then RaR e (a) and a e 5P?(R) implies (a) is G-regular. Therefore, RaR is G-regular. Thus (i) implies (ii). TR 8 Now let a e such that RaR is G-regular and let b e RaR then b is G-regular so that b e G(b)flR. Thus there is some r x y e R such that b = br r + ' i» i ~ 2(x±byi - x1yjL). There- fore, for any s, t e R, sbt = sbrt - srt + Z(sx by±t - sx1y1t)eH(b). Therefore, RbR c H(b) so that H(b) = R2 from Lemma 3.2. Thus 2 RaR CR = H(b) implies b € H(b). Therefore,RaR is H-regular and (ii) implies (iii). Let a e R such that RaR is H-regular, and let b e (&).» then b € RaR. Thus b^ is H-regular which implies b5 is G-regular. Let s = -(b + b2), then b -(bs - s) = b - (_b2_b5+b+b2) and b5 = b-(bs - s) is G-regular and bs - s e G(b) so that b e G-(b) from Lemma 3.1a and thus b is G-regular. Therefore, (a) is G-regular so that a e ^(R). Thus (iii) implies (i). Therefore, all three, are equivalent. 44 Theorem 3.5* If C In a;; ideal In a ring R, then ^(i) = $(R) n i. Proof; By construction Ws I) 5 I. Let "b e 7l\(l) and let a e (b), then a is G-regular with respect to I. Therefore,, a Is G-regular in R. Thus (b) is G-regular in R so that b e 5>??(R) Therefore, ^(1) £ 57?(R) so that Wl) c ty(R) D I. Now let b e $(R) n I, then I b I c ^(R) n I. Let a e Ibl, then a e G(a) HI, since Ibl <= %(R) (1 I is G— regular. Then 2 from the proof of Theorem 5.4, a e G(a) fl I Implies H>'(a) = I . 2 Then a e Ibl c I so that a e H (a) which implies 'Ibl is H-regular. Thus b € %l(l), by Theorem 5.4, so that ^(R) D I c Therefore I) = ^(R) fl I. Theorem 5.6; In a ring R with R^ denoting the matrix ring over R, then ^(R^) = 57?(R) . xj. n The proof follows similarly to the proof for the Jacobson radical of a matrix ring R over a ring R. Theorem 2.7* In a ring R, $(R) 5 %(R). t Proof: Let a e ^(R)? then (a) c j?(R) so that (a) is right quasi-regular. Let b e (a), then there is some r e R such that b + r + br = o. Thus b = -br-r so that b is G-regular. Therefore (a) Is G-regular and thus a e %(R) so that £(R) c ^(R). " Corollary 3.1: In a ring R, ^(R) contains every nil ideal In R. Corollary 3.2: In a ring "R, l( c $(R) and thus 7l(R) c ^(R) and £>(R) 5?/<(R). Therefore, In any ring R, -&>(P) c ??(R) c $(R) 5 ^(R). The following theorem is stated without proof. Theorem 3.8: If R is a left (right) Artinian ring, then Jfl(R) = j?(R) = 7?(R) = 0(R). Therefore, in a ring R with the descending chain condi- tion for right or left ideals in R, the Brown-McCoy radical coincides with the nil radical of R. From Chapter II, the Jacobson radical of R coincides with nil radical when R is a ring with the descending chain•condition for right or left ideals. The original concept of the radical of a ring R was defined as the nil radical of R where R is a ring with the descending chain condition for right or left ideals. The original concept of the radical of a ring R was defined as the nil radical of R where R is a ring with the descending chain condition for right or left Ideals in R. Other radicals were later defined for arbitrary rings which would coincide with the nil radical in the presence of the des- cending chain condition for right or left ideals. The Jacobson radical and the Brown-McCoy radical were two such radicals. Other radicals were defined by Levitzki, Baer, and others to coincide with the nilradical in the presence of the descending chain condition for right or left ideals. For each radical defined on an arbitrary ring, the radical of the residue class ring modulo Its radical is the zero' ideal in the residue class ring. Therefore, in a ring with 46 the descending chain condition for right or left ideals, all radicals coincide with the nil radical so that in such a ring, we may speak of the radical of the ring. BIBLIOGRAPHY Brown, B. and McCoy, N.H„, "The 'Radical of a Ring", Duke Mathematical Journal, -IS, (1948), pp. 495-499* ~ Gray, M., A Radical Approach to Algebra, Addison-Wesley Publishing Company, ReadTng7~Ma s s ac hu s e 11 s, (1970). Herstein, I. N., Noncommutative Rings, The Mathematical Association of America, (^T9"68y. Jacobson, N., Structure of Rings, American Mathematical Society, ~Providence7~~Rhode Island, (1956). Jacobson, N., The Theory of Rings, American Mathematical Society, New York," New York, (1943). McCoy, N. W., The Theory of Rings, The MacMillan Company, New York,"New York, "{"1964). Nagata, M., Local Rings, Interscience Tracts in Pure' and Applied MathematicsrioT' 15, Interscience Publishers, New York, New~York, (1962). 47 (A) is contained in every prime ideal that contains A. Let r e R such that r ^ A), then there is some m-system M in R such that r e M and A D M = (j), from Definition 1.4. Then, by Theorem 1.5, there Is a prime ideal p' in R contain- ing A such that p' fl M = |, Then r i p/, therefore r I D P. P€C Thus £>X7ry £ HP which implies HP c£^(A), Therefore PeC <$>(A) - OP. Hence Definitions 1.4 and 1.5 are equivalent, PeC Assume Definition 1.5 for the radical of the ideal A in a ring R. Then, if C = {P|P is a prime ideal in R and A c P} and if C ' = {P7|P7 is a minimal prime ideal in R containing A}, let r e DP. Then r is in every prime ideal PeC containing A, so that r is in every minimal prime ideal con- taining A. Thus r e DPso that HPenP7. P 7eC ' PeC "Pe'C 7 Let P be a prime ideal in R such that ACP, and let S be the family of prime ideals P 7 in R such that Ac'P'gP. Then S 4 ^ since P e S. Let T = {P'7|P7 e S}, then T 4 0 since P e S implies P e T. Then T can be partially ordered by set inclusion, and every element of T is an m-system by Theorem 1.2. Let T ' be a chain in T, then M7 = UP ' is an element of T and M7 P 7eT ' contains every element of TTherefore, by Zorn?s Lemma, T contains a maximal element M. Then by the construction of T, M is a prime ideal and M e S. Then M is a minimal element of S since the ordering of S is opposite to that of T. Thus M is a minimal prime ideal in R containing A. Since P was an arbitrary prime ideal in R containing A, then every prime ideal containing A contains a minimal prime ideal containing A. Let r e R such that r | HP, then there is some PeC such that PeC r | P. Since P is a prime ideal in R containing A, there is a minimal prime ideal P 7 such that Agp'cp. Then r <£ P im- plies r | P7. Thus r HP 7 and hence HPsnP 7 so that np«np. Hence PeC7 PeC PeC7 PeCTPeC 8