Prime Ideals in Commutative Rings

Home , Ideal (ring theory), Minimal prime ideal, Prime element, Prime ideal, Radical of an ideal

PRIME IDEALS IN COMMUTATIVE RINGS

APPROVED:

Majbr Professor

Minor Professor r x Director of the Department of Mathematics PRIME IDEALS IN COMMUTATIVE RINGS

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

Marlene H. Clayton, B. S,

Denton, Texas

August, 1970 TABLE OP CONTENTS

Chapter Page I. MULTIPLICATIVELY CLOSED SETS AMD PRINCIPAL PRIMES 1 II. FACTORIZATION OF IDEALS 16 III. MAXIMAL AND MINIMAL PRIMES 22 IV. PRIME IDEALS IN RINGS WITH THE DESCENDING CHAIN CONDITION 56 BIBLIOGRAPHY ..... 40

1X1 CHAPTER I

MULTIPLICATIVELY CLOSED SETS AND PRINCIPAL PRIMES

This thesis is a study of some properties of prime ideals in commutative rings with unity. Throughout the thesis the symbols €, d , and < will mean is an element of/' set containment, and proper set containment, respectively. Capital letters will denote sets and lower case letters wi-11 denote elements of sets. . All rings will he commutative and have a multiplicative identity (i.e. unity element) unless otherwise stated.

Definition 1.1: The ideal P in a ring- R is a prime ideal if and only if for a, b e R, ab € P implies a e P or b € P.

Theorem 1.2; If I is an ideal in a ring R, then I is a prime ideal if and only if R/I is an integral domain.

Proof; Suppose I is a prime ideal. Let + I)(r2 + I) =0+1=1, where r^ + I and r^ + I belong to R/I. Now by multiplication of residue classes this implies *"^2 + ^ = 0 + !• Since a + I = b. + I if and only if a - b e I then r^g - 0 s I,

> or r 1r2 e I. But I is a prime ideal so either r^ e I which Implies r^ + I = I (because residue classes are either identi- cal or disjoint) or r? e I which implies r2 + I = I.

1 Suppose R/I is an integral domain. Let ab e I, ab

also belongs to ab + I and hence ab +1=0+1=1. This

says (a + l)(b + I) = I, but since R/I is an integral do-

main there are no non-zero divisors of zero and hence

a+I=Iorb+I=I. If a + I = I this implies a e I.

If b + I = I this implies b € I.

Definition 1.5; A subset S of a ring R is said to be

muitiplicatively closed if and only if S is not empty5 S does not contain the zero element of R, and if a, b e S, then ab e S.

Theorem 1.4: If I is an ideal in R such that I ^ R,

let S = R\lJ then I is prime if and only if S is muitiplicatively closed.

Proof; Suppose I is prime. The set S is not empty

because I 4 R implies 1 I which says 1 € S. The zero

element belongs to I and hence o | S. Let a, b e S, then

since S = r\i, a, b | I, Now either ab e S or ab | S.

Assume ab | S. Then ab e I and hence either a e I or b e I.

But this contradicts the fact that a, b | I, Therefore, the

assumption that ab | S is false and hence ab e S.

Suppose now that S is muitiplicatively closed. Let ab e 1, and hence ab | S. Since S is a muitiplicatively closed set then we consider these cases:

(1) a, b both do not belong to S which implies a, b e I, 3

(2) a e S and b | S which implies b e I,

(5) a | S and b e S which implies a e I, or (4) a € S and b € S which implies ab e S but that contradicts the fact that ab | S. Thus, in the three cases which can occur, either a e I or b e I.

Lemma 1.5: Let R be a commutative ring with unity.

Let S = {a-p a2, a^, ••• , an, ...3, where a^ e R for i = 1, 2, 5} •••. Denote by (s) the collection of all finite sums of the form S r^a^ where r^ e R and a^ e S then (S) is an ideal. In the event that S is a finite set, say

## a S C^l^ ^ ^ * s n}, the notation ^ a~j, a^>, ^ ^ ^n^ is sometimes used instead of (S). ft Proof: Let x € (S) and r e R, then x = Sr,a. where is| 1 1 r^ £ R and a^ e S, and n is a positive integer. Now rx = r i-Ir. a. =5Z?(r v . a.7 ) = 5Z(rrv y. ) a. and rr. e R for all i=/ 11 •_/ l i *.( i I l i = 1, 2, 3, n. The finite sum/lI ?(r. a. ) belongs to l-i 1 x (S), and hence, rx e (S). m Let z and y belong to (S), then z = ]EIr.a. where r. € R ;«i 1 1 1 and a. e S, and m is a positive integer. The element t _ y =S r.a. where r . e R and a. e S, and t is a positive i J J J J integer. Now let r. = -r . and a. = a . for j = 1, 2, 3, • J 4^3 j ... , t. Then z - y =ZZr.a. --Lr .a. i i js, j j = Sr. a, -iZ(~r ^0 a J_. m •=J i i +o m+j m -fc = 22 r. a. +5~'r , . a , . 1 1 m+J ra+J ,r.a.. I* i 1 1 Therefore, z - y e (S), hence (S) is an ideal in R.

Lemma 1.6: If R is a commutative ring with unity and I is an ideal of R such that I 4 and a is a fixed element of R\I, then the set S such that S = {x | x = i + ra where i e I, r e R and a is a fixed element of R\l) is an ideal in R. Proof: Let s e S and r e R, then s = i + r-^a for some

i e I and r^ e R. . Now rs = r(i + r1a) = ri + (rr.^) a, where ri e I and rr^ e R. Hence rs e S.

Let x, y e S, then x = i + ra and y = I + ra for some ±} i e I, r, r s R. Now x - y = (i + ra) - (i + ra)

= (i - I) 4- ra - ra

= (i - i) + a(r - r),

where i - I € I and r ~ r e R. Hence, x - y e Sa and thus S is an ideal in R.

Definition 1.7* If I is an ideal in R and a e R, a | I, let (I, a) denote the ideal generated by all the elements in I and by a.

Theorem 1.81 Let S be a multiplicatively closed set in a ring R, and let I be an ideal in R maximal with respect to the exclusion of S. Then I is prime. Proof» Let a, b £ R and ab € I, and assume a, b | I. Consider the ideals (I, a) and (I, b). Now (I, a)DI, (I, b)3I and hence (I, a) and I, b) each contain an element in S, since I is maximal with respect to the exclusion of S. 5

Let s^ be an element in (I, a) such that e S} and let Sg he an element in (I, b) such that Sg e S. Hence s^ = i^ + r^a where i^ € I and e R, and Sg = ig + r^b where ig e I and rg e R. Now s-^Sg belongs to S because S is a multiplicatively closed set, therefore s-^Sg = (i^ + r^a)(ig + rgb)

= + ^lr21:) + rla^"2 + rir2ak* But i-^ig € i]_(r2lD) € ^-2^rl8^ € al3^rlr2^ € ^ since I is closed under addition then s^Sg e I, but this contradicts our hypothesis and hence our assumption that a, b | I is false. Therefore, a € I or b € I.

Definition 1.9* A non-unit element p e R is called a principal prime if the principal ideal (p) is prime and non-zero.

Definition 1.10* An element u e R is called a unit if and only if there exists an element w e R such that uw = 1, where 1 is the unity in R.

Definition 1.11! An element a e R is called an irreducible element if and only if a is not a unit and a = be implies b is a unit or c is a unit.

Lemma 1.12: In an integral domain D with unity a principal prime element p. is an irreducible element.

Proof: Let p be a principal prime element in Da then (p) is a prime ideal in D. Assume p is not an irreducible element.

Then p = ab where neiths r a nor b is a unit in D. Now ab = p implies ab e (p). Since (p) is a prime ideal then either a e (p) or b € (p). 6

Suppose a e (p), then a = pk where k e D. Now ab = p so by substitution (pk)*b = P p•(kb) = P p-(kb) = p.l and by the cancellation law kb = 1. Hence b is a unit which contradicts our assumption. In a similar way if b € (p) then a would be a unit. Therefore, our assumption that p is not irreducible is false and, therefore, in an integral with unity a principal prime is an irreducible element.

Example 1.13 > The following example shows the existence of an irreducible element in an integral domain with unity that is not a principal prime. Consider the integral domain R = {a + a, b e I, where I denotes the integers. Now 21 - (1+2 ,T5)(1 - 2-^F) hence 21 e R. Also 21 = 3'1 • The following argument shows that the element 3 is irreducible in R. Assume 3 = (a + b ^ -$) (c + d V-5*) for some a, b, c, del. Now consider the norm function N such that 2 2 N(x + y ^ -5') = x + 5y . Since the norm function is multiplicative, then N(3) = N(a + bv/^) N(c + d V/-5"), hence 9 = 2 2 2 2 * (a + 5b )(c + 5d ). Now the possible cases for integer solutions are (1) a2 + 5b2 = 1, (2) a2 + 5b2 = 3 and (3) a2 + 5b2 = 9- 7

2 2 Case (1): a + 5"b =1. The possible Integer solutions are a = i 1 and b = o. But then a + b V -5' would be a unit. 2 2 Case (2): a + 5b =3. There are no integer solutions for a and b in this case. 2 2 Case (3): a + 5b =9. The possible integer solutions are (i) a = i 3j b = o and (ii) a = i 2, b = 1" 1. But in either case if a2 + 5b2 = 9 then c2 + 5d2 = 1 and by Case (1) c + d V-5 would be a unit. Since Case (2) cannot occur for any choice of integers a, b, c, d and since the only choices of integers for Cases (1) and (3) imply either a + b \J -5* or c + d V -5* is a unit, then 3 is irreducible in R. But 3 is not a principal prime for 21 e (3) and 21 = (1 + 2>TT5t)(l - 2-T^S) but 1 + 2 v/~^T| (3). For if 1 + e (3) then 1+2 \J -5' e (3) then 1 + 2 = 3 (a + b v/-5*) for some &, b e I. Hence 1 + 2 V™^5^= 3a + 3b V-ff, but this implies

3a = ls which implies a = 1/3 which contradicts our hypothesis that a be an Integer. Also, 3b = 2 which implies b = 2/3 which also contradicts our hypothesis. Therefore, 3 is an irreducible element in an integral domain with unity but 3 is not a principal prime. Example 1.14: The following example shows the'existence of a principal prime element in a ring (not an integral domain) that is not an irreducible element. Consider the ring 1/(6) = {0 + (6), 1 + (6), 2 + (6), 3 + (6), 4 + (6)j 5 + (6)). The element 3 + (6) is a principal prime since (3 + (6)) = {0 + (6), 3 + (6) ) is a prime ideal. But 3 + (6) is not irreducible element since [5 + (6)] = [3 + (6)][3 + (6)] and 3 + (6) is not a unit.

Lemma 1,15' In an integral domain D with unity, if a principal prime divides a product ah, then either p divides a or p divides b. Proof I If the principal prime p divides ab, then ab e (p) and hence a e (p) or b e (p) since (p) is prime. If a € (p), then a = kp for some k e D, hence p divides a. A similar argument holds if b € (p).

Corollary 1.16: If a principal prime divides b-^bgb^ * it divides some b^. Proof: The proof follows by induction.

Lemma 1.17* If P is a prime ideal in R a commutative ring with unity, and x a ring element such that x € P for k a positive integer then x e P. Proof: The proof will be by induction on k. If k = 1 then clearly x € P. Now suppose If x Is any ring element such that x^ e P where j 1, and j = k - 1, then x e P. Let xk e P, then x^" = xxk_1. Since P Is prime then x e P or x^"""^ e P. k-1 If x e P then by our induction hypothesis we have x e P. 1^. Therefore, if x e P for k a positive integer then x e P. Lemma 1.18: In a commutative ring with unity a unit times a principal prime is a principal prime. Proof: Let p be a principal prime element, then (p) is a prime ideal. Consider (up) where u is a unit. Since u is a unit there exists a w such that uw = 1. Let x e (p), then x = rp where r e R, then x = r»l*p x = r(wu) • p x = (rw)« up x = r-up where r e R. Hence x e (up) and thus (up) (p). Now let x e (up) then x = r'up where r'e R, and this implies x = rp where r = r'u e 'R. Hence x e (p) and (up)

Theorem I.191 If an element in an integral domain with unity is expressible as a product of ' ' '^n PrinciPal primes, then that expression is unique up to a permutation of the p*s and multiplication of them by unit factors. Proof: The proof will follow by using indue tion on the number n of principal prime factors of a. Let a = p where p is a principal prime. Assume a = xy where x and y are not units, but this contradicts Lemma 1.12 which says that a principal prime element in an integral domain is an irreducible element. Therefore, either x or y is a unit, say x is a unit. 10

Then(l/x)a = y, a principal prime since the product of a unit 1/x and a principal prime a is a principal prime. Hence the theorem is true for n = 1. Suppose the theorem is true for all a that can be expressed as a product of n - 1 principal primes.

p = q q q q where the Let a = PxP2P5 * **Pn-l n l 2 3 " ' k are principal primes. By Corollary 1.16 if a principal prime divides a product •••"b . it divides some b. . Now q, x. c. n ik. p 5 so q divides some divides q1q2 " ' q_k = PXP2 '*'Pn-l n k Pj_> say p . (If q^. did not divide pn we could renumber the p's).

Now qfc divides pn implies pn = uq^ where u is a unit (since q^ and pn are principal primes in an integral domain and

p hence irreducible elements). Now a = p^P2P-j *"'Pn l n

= qlq2 ••• hence a/pn = pxp2 • *'P^ = qxq2 • • • qk_1(l/u) and by the induction hypothesis n - 1 = k - 1 since &/pn is the product of n-1 principal primes. Therefore, n = k, and the p^s and q*s differ only by unit factors.

Theorem 1.20; Let R be an integral domain with unity and let S be the set of all elements in R expressible as a product of principal primes. Then if S is not empty it is a multiplicatively closed set such that if a> b are non-units in R and ab e S, then a e S and b e S.

Proof: Let S = {s|s e R and s = P^P2 ***Pn^ where the p*s are principal primes, and suppose S is not empty. The 11

zero element does not belong to S since we are in an integral domain and zero could not possibly be the product of principal

primes. Let s.^, Sg € S, then = P-j_P2 ' ' *Pk? &nd Sg

= q-^ •'*qm where the p^s and q>s are principal primes. Now

S1S2 = ^1^2 '**Pk^l^2 ' "'^m an(^ hence sis2 ~*-s a Pro(^uct of principal primes which implies s-^g e S. Therefore, S is a multiplicatively closed set.

Now suppose a, b are non-units in R and ab € S. We wish

to show a £ S and b e S. We will use induction on n} the number of principal primes in the product, to show this. Note ab cannot be the product of exactly one principal prime since a principal prime in an integral domain is irreducible.

Suppose ab = p-jPg, then p.^^ divides a or p1 divides b. With- out loss of generality suppose p^ divides a, then a = kp^,

hence kp-^b = p-jPg and by cancellation, kb = p0. Now either (1) b = dpg or (2) k = cpg.

Case (1)1 If b = dpg then kb = Pg

k (dp2) = p2 kd = 1.

Thus k and d are units. Hence kb = p2 implies b = (1/k) Pg and b is a product of principal primes and, therefore, b € S. Now

ab = p1p2 ab = p-^kb a = p-^k and since k is a unit then a e S. 12

Case (2): If k = cp2, then kb = p2

cp2b = p2 cb = 1. Thus c and b are units. But this contradicts our hypothesis that b is a non-unit. Hence Case (2) cannot occur. Therefore, the theorem is true if ab is a product of exactly two principal primes.

Suppose that the theorem is true for any non-unit elements a, t> e R such that a5 is a product of exactly n-1 principal primes.

Now suppose a and b are non-units in R such that ab is the product of exactly n principal primes. Let ab = p p ...p

then pn divides ab which implies pn divides a or pn divides b.

If pn divides a, then a = kpn, and either k is a unit or non- unit. if k is a unit then a = kp implies a = p where 5 is a principal prime since a unit times a principal prime is a principal prime. Hence, a e S in this case, and

ab = (kpn) b = P^Pg ••• Pn

Kb = pip2 ... Pn_1

b = (i/K) pip2 _ pn_ij

hence b e S.

If E is not a unit, then a = Ep^ and ab = (Kp ) b = p p 110 1 l 2 • • • Pn w* * implies Kb = P;LP2 . .. pn_i and hence, b e & and k e S by the induction hypothesis. Now E e S implies

K = ••• qr where the q>s are principal primes, thus a = kpn = q;Lq2 ... qr pn e S. 13

If p divides b then a e S and b e S by a similar n argument.

Definition 1.21: An integral domain R with unity is a Unique Factorization Domain (UFD) if it satisfies the following conditions: (UF 1.) Every non-unit of R is a finite product of irreducible factors. (UF 2.) The foregoing factorization is unique to within order and unit factors. More ex- plicitly, UF2 means the following: If

a = pip2 "*pm = qlq2 '' ' where the p^s and q's are irreducible, then m = n, and on renumbering the q*s, we have that p^ and q^ are associates for i = 1, 2, •**, m.

Theorem 1.22: In a UFD every irreducible element is a principal prime. Proof: Suppose R is a UFD, and let q be an irreducible element in R. Consider the principal ideal (q). Now (q) + (0) since q 4= 0. If a, b e R such that ab e (q), then ab = rq for some r e R. Now since R is a UFD, then a and b have unique factorizations except for the order of the factors and except for unit factors. Suppose a = p-^pg ***P^ anc* = p]_p2 **"pe

where the p*s and p 5s are irreducible elements in R. Also, r has a unique factorization, say r = Pjp2 '' *pm w^ere p*s are irreducible elements in R. Now ab = rq implies

plp2 "pkplp2 '•* pe = P1P2 14 and since we are in a UFD then q must "be an associate of some p. or some p' , hence q is one of the factors of a or b. X J If q is one of the factors of a then q divides a and a = rq where r e R and thus a e (q). If q is one of the factors of b then q divides b and b = rq where r e R and thus b e (q). Hence (q) is a prime ideal and, therefore, q is a principal prime.

Theorem If R is a ring such that every ideal in R (other than R) is prime, then R is a field. Proof: Since every ideal other than R is prime then the zero ideal (0) must be prime. Hence if ab e (0) then either a e (0) or b e (0), which implies either a = 0 or b = 0. Therefore, if ab = 0, a = 0 or b = 0. Hence R is an integral domain.

Now let a be any non-zero element in R. Consider the 2 2 2 2 ideal (a ). The element a e (a ) hence aa e (a ). Now 2 2 2 since (a ) is prime then a e (a ) which implies a = ra where r e R. Let x be any non-zero element in R. Now, ra2 = a xra 2 = xa

x(ra) a - xa = 0 a[x(ra) -x] = 0. We are in an integral domain and a |= 0 so x(ra) -x = 0. This implies x(ra) = x and ra is the unity in R. 15

2 Now let b be any non-zero element in R. Then (b ) 2 2 2 is a prime ideal and b e (b ), hence bb e (b ). And since 2 2 - 2 (b ) is prime then b e (b ) which implies b = rb where - 2 r € R. Hence, b = rb b - ?b2= 0 b ( r a ) - rbb = 0 b(ra - rb) = 0. The element b | 0 so ra - rb = 0, which implies rb = ra. Hence, r is the inverse for b. Therefore, R is a field. CHAPTER II

FACTORIZATION OF IDEALS

Definition 2.1: The product of two ideals A, B in a ring R is defined by

AB = { iba.b. |a. e A,b. e B, n any positive integer.} j JL _L JL X

Lemma 2.2: If R is a commutative ring with unity, and A and B are ideals of R, then AB is an ideal In R. Proof; Let x e AB and r e R. • Since x e AB, then h ^ x = 5_. a.b. where a, e A and b. e B, and n is a positive X X X X Ifi integer. Then rx = r 5Z a.b. ui 1 i ra b = i7T i i1

= 2_Z a.bi .l e AB.* where ra.l = a.l e A for each i = 1,2, •••, n. Therefore, rx e AB. m - — - If z € AB, then z = 21 a.b. where a. e A, b. e B and m Jsi J J J 3 is a positive integer. Let eL = - a^.+n and b^ = then

x - z = ^a.b. - Ea.b, l=t ii j-i J J = f"a.b. - &(-a . , ) (b . , ) us 1 1 J+n j+n' a b + a b = £ i i £( i+n)( i+n)

= J&a. b. + Hufa. b . i -< i i i=n»i i i (Tiaji = *L_ia.b. e AB. Thus x - z e AB for any' x, z e AB, hence, AB is azi ideal in R.

Theorem 2.3' If A is a prime ideal in a ring R and A = BC, where B and C are ideals in R, then either BCA or CCA.

16 17

Proof! Suppose B A, then there exists some b^ e B i ,21, such that b, Yc A. Now 2_jb.c. e A for all b.c. € C and any 1 ;s< ix 11 ^ positive integer n. In particular, t>-]_ci € ^ "^or ci € but this implies b-^ e A or c^ e A since A is a prime ideal. But b^ £ A implies c^ e A for all c^ in C and thus, CCA. In a similar manner it can be shown that if C <£ A then BCA,

Theorem 2.4: If A is an ideal in a ring R and A is not a prime ideal, then there exists ideals B and C in R such that ACB, Ace and BCCA. Proof: If A is not a prime ideal in R then there exist elements b, c e R such that be e A but b | A and c ^ A. Let B = (b,A) and C = (c,A). It was shown in Lemma 1.6 that B and C are ideals. Clearly A<^B and AdC.

Now (b, A) (c, A) = { ^(iVjb + a1)(s1c + a^lrJD + a± e (b,A), sic + ai e anc^ r±> si 6 ^ ^s an Let x e (b,A)(CjA), then x - + a. )(s.c + a.), where | iX. JL JLm r^, s^ € R, a^, iL e A and n is a positive integer; then x = yr, s.bc + r.ba. + s.a. C + a. a.. Notice that (r.s.) be e A TT* 1 1 11 11 11 v 1 1' since be e A and A is an ideal, (r^b) £L e A, (s^c) a^ e A and ai^i e ^ s^nce A ^-s an ideal. Now since ideals are closed with respect to addition, then x e A and, therefore, (b,A)(c,A) = BCCA. Definition 2.5« A commutative ring is called noetherian if it has an identity and if it satisfies the equivalent conditions (1), (2), and (3). [A noetherian domain is a noetherian ring without proper zero-divisors.} 18

(1) (^Ascending chain condition , or a.c.c.) Every strictly ascending chain { Ug ( ( ••• of ideals of R is finite. Or alternatively: Given an ascending chain ... of ideals in R, there exists an integer n such th± Un = Un+1 = Un+2 = ... .

(2) (^Maximum condition/' ) In every non-empty family of ideals of R, there exists a maximal element, that is an ideal not contained in any other ideal of the family. (Of course, such a maximal element need not "be a maximal ideal of R.) The maximum condition implies that every ideal U 4 R is contained in a maximal ideal, as is easily seen by con- sidering the family of all ideals 4 R containing U. (3) (Finite basis condition.) Every ideal U of R has a finite basis; this means that U contains a finite set of elements a^, ag, an such that U = Ra^ + Rag + ... + Ran

+ Ja-^ + Jag + ••• + Jan, where J is the set of integers (such a set is called a basis, or a set of generators of U.) If R has a unity, then U = Ra^ + Rag + ••• Ran.

Theorem 2.6: Every ideal A in a noetherian ring contains a product of prime ideals. Proof: Since R is a noetherian ring, then in every non- empty family of ideals of R, there exists a maximal element. Now suppose that there exists an ideal B such that B^R and B does not contain a product of prime ideals. Now consider the family F of all ideals not containing a product of prime ideals. 19 Then the family F has a maximal ideal, say A. Now A is not prime for if it were, then A would not belong to the family F, since one prime ideal can be considered a product and AciA. Now by Theorem 2.4 there exists ideals C and D in R such that ACC, A CD and CDC A. But since A is maximal, then C and D must contain a product of prime ideals, say C^P-^Pg ••• P^ and D3Q1Q2 • •• Q . Now A3 CD3 P1P2 ••• PkD=> P-jPg • •• ••• Q , hence, A contains a product of prime ideals, but this contradicts the fact that A € P and hence, our assumption is false and every ideal in a noetherian ring contains a product of prime ideals.

Theorem 2.7* In a noetherian ring the zero ideal can be expressed as the product of prime ideals. Proof; By Theorem 2.6, every ideal in a noetherian ring contains a product of prime ideals. Let P-^Pg ••• P^. he the product of prime ideals such that (0)3 P^P^ ... P^. It has previously been shown that the product of ideals is an ideal. Since every ideal contains the zero ideal, then

P1P2 PkD(°)* Hence' (°) = pip2 Pk*

Theorem 2.8; Suppose A is an ideal in a ring R such that A is contained in a finite union U P. of prime ideals i 1 of R. Then A is contained in at least one of these prime ideals. 20 Proof: Let A "be an ideal such that Adfp., Without l-i 1 loss of generality we may assume that <£: P. for any k and j, rv n ^ for if P. cp., then U p. = UP.. k J i i*t i i*t Assume A is not contained in any of the prime ideals. Then there exists an a^ | P^, and since P^ P^ for

1 - 2, 3, n, then there exists a p2 e P2 such that p2 | P^ p-j e P-j such that p^ | P^

pn € Pn such that pn ^ Pl*

Now since P^ is a prime ideal then ^^2^^ ••• Pn I Pj* Now let = a2p2^J> ''' pn* a like manner for each a^ ^ P^ for i = 2, 3, .n, there exists an such that a. e A, e P. for J 4 i* otj_ ^ P^. Since each a. e A, then the sum otj_ € A, hut k Pj_ for i = 1, 2, • ••, n. Suppose C~t I * i £ Oj_ e Pjj then t=i + ctg + • • • + a,j + • • • + cx^ = p . where p . € P 3 3

aj + ax + • •. + aJ_1 + aj+1 + . • • a— p..

aj = Pj " (°1 + °2 + aj-l + aj+l + ••'+%) and since for each i =f j ol- € P ., then a. = p . - p . where J J J J Pj = (^ + Oq + • • • + ocj^ + a.+1 + • •• + a^). Hence, a. = p . - p . e P.. This contradicts the fact that a. | P., J J J J J J hence ^ P^ for i = 1, 2, .n and thus, there exists ah element of A that is not contained in any P^. But this contradicts our hypothesis and, therefore, A must be contained in some P^. 21

Theorem 2.9: Let P^, P^, ... Pn denote a finite r\ collection of prime ideals in a ring R. Then U P. is an 1 hi ideal in R if and only if t/ P. = P. for some j = 1, 2. ..., n. a k""' 1 J Proof: Suppose UP. = P. for some j = 1, 2, n. 1st X J n. Now P . is a prime ideal by the hypothesis and hence I/ P. is J 1 an ideal in R.

Suppose 0 P. is an ideal in R, call it A. Then n A a Up. and by Theorem 2.8 A is contained in at least one of I £ I X the prime ideals P,, Pp, ... P . And since each P. d Op. J- n 3J in i for j = 1, 2, ..., n, then Op. = P . for some j = 1, 2, • • n< l-l 1 J CHAPTER III MAXIMAL AND MINIMAL PRIMES

Definition 3.1: An ideal I in a ring R is said to be maximal if there exists no ideal properly contained in R in which I is itself properly contained.

Theorem 3.2: If R is a commutative ring with unity 1 4 0? and I is an ideal in R, I 4 Rj then I is a maximal ideal if and only if R/I is a field. • Proof: Consider the natural homomorphism cp: R-*R/I. Now cp(a) = a + I for all a e R, and since 1 is the unity in R, then 1 + I is the unity in R/l. For if a + I € R/I and 1 + I e R/I, then (1 + I)(a + I) = 1-a + I = a + I. In a similar manner, 0 + I is the additive identity element in R/l. Note that 0 + I =| 1 + I, for if they were equal, then 1-0 = 1 e I, which implies I = R. But this contradicts the hypothesis. Suppose I is a maximal ideal in R. Let a + I e R/I, where a + I 4 0 +1. (Such an element exists since 1 + I 4 0 + I). Now let N = (I,a). N is an ideal which contains I, hence N = R since I is a maximal ideal. There- fore, every element in R can be written in the form ar + i where r e R, and i e I. Since 1 e R, there exists r1 € R, i^ e I such that 1 = r.^a + i^. Then 1 - r^a = i^

22 23 which implies 1 - râ e I. Now 1 - râ e I implies 1 + I = râ + I, thus 1+1= (r^+I)(a+l). This shows + I is the inverse of a + I, and hence, R/I is a field. Conversely, suppose R/I is a field. Let M be an ideal in R such that I <( M. Then there exists an element m e M such that m ^ I. Now m + I 4 I because m - o = m | I. Since m + I 4 Ij then there exists an inverse for m + I since R/I is a field. Let m^ + I denote the inverse of m + I, hence, (m + l)(m-^ + I) = 1 + I. Now let r e R, then

(m + I)(m-L + I)(r + I) = mm-^r + I = (uim^ + I) (r + I) = (1 + I)(r + I) = r + I. Therefore, mn^r + I = r + I. This implies mm^r -re ic. M, and thus, mm^r - r = m for some m e M. From mm^r - r = i it follows that r = mm-^r - in, but mm^r e M and m e M hence, mm^r - m e M. Thus r e M and, therefore, RcM and hence, M = R. Since M was assumed to be an arbitrary ideal in R such that I <( M, it follows that I is a maximal ideal in R.

Theorem 3.31 If R is a commutative ring with identity, then maximal ideals different from R itself are prime. Proof: Let I be a maximal ideal of R, I 4 R. Now by Theorem 3.2, I is a maximal ideal if and only if R/I is a field, hence, R/I is a field. Since R/I is a field, R/I is an integral domain and by Theorem 1.2, I is a prime ideal. 24

Definition 3.4: A set S is said to "be partially ordered if and only if there exists a binary relation defined for certain ordered pairs (a, b) of elements of S such that:

(1) a £ a for all a e S (reflexive)

(2) if a £ b and b £ a5 then a = b (antisymmetric) (3) if a £ b and b £ c, then a 1 c ( transitive)

Definition 3«5: A subset of a partially ordered set S is called totally ordered if and only if for a, I) e either a £ b or b ( a,

Definition 3.6: An element u of S is called an upper bound of a subset S-^ of S if and only if a ^ u for all a e S^<

Definition 3.7* An element m of a partially ordered set

S is maximal if and only if there is no element s e S such that m <( s.

Definition 3.8! A partially ordered set S is said to be inductive if and only if every totally ordered subset of S has an upper bound in S.

Lemma 3.9^ (Zorn-'s Lemma) If a partially ordered set

S is inductive, then there exists a maximal element in S.

Theorem 3.10: Let A be an ideal in R, and let M denote a multiplicatively closed set in R such that A n M = f, then M is contained in a maximal multiplicatively closed set M such that M n A = (J). 25

Proof: Let S "be the set of all multiplicatively closed sets in R which contain M but do not contain any element of A. Note M e S so S is not empty. Now clearly S is partially ordered by set containment. Let T be any totally ordered subset of S. There exist such subsets for M € S and the set whose only element is M, {M}, is totally ordered. Now let U = U M.. The set U can be shown to be a 1 h-LtT multiplicatively closed set by the following argument. Let x, y e U, then x and y are elements in R, each of which is contained in a multiplicatively closed set in T, say x e and y e Mg where and Mg e T. Now since T is totally ordered, then M-^Mg or Mg*^ M^ without loss of generality suppose M-^Mg. Hence, x e Mg and y e Mg implies xy € Mg since Mg is multiplicatively closed. Since xy s Mg, then xy e U and hence, U is a multiplicatively closed set. Clearly U contains M and U n A = (J), and since U is a multiplicatively closed set then U e S, and U is an upper bound for T. Hence, S is inductive and by Zorn^s Lemma there exists a maximal element M in S, that is M is a maximal multiplicatively closed set such that M^M- and, M n A =

Theorem 3*11' Let M be a multiplicatively closed set in R, and let A be an ideal in R such that A fl M = (j). Then A is contained in an ideal P such that if B is an ideal with the property P'<( BCR, then B contains an element of M. 26

Also, the Ideal P is a prime ideal. Proof: Let S be the set of all ideals in R which contain A but do not contain any element of M. Note A e S so S is not empty. S is partially ordered by set containment. Let T be any totally ordered subset of S. There exist such subsets for A e S and the set {A) is totally ordered. Now let U = U A., The set U can be shown to be an Aitl 1 ideal in R by the following argument. Let x, y e U and r be any element of R. Since x, y e U then x and y are each contained in some ideal A^, say x e A-^, and y e Ag. Now since T is totally ordered, then A^Ag or A^A^, without loss of

generality say A-^Ag. Then x e Ag and y e Ag and since A2

is an ideal, then x-y e Ag and rx e A^, and since A^U, then x-y and rx belong to U. Therefore, U is an ideal. Clearly U contains A and U (1 M = (j), and since U is an ideal, then U e S, and U is an upper bound for T. Hence S is inductive and by Zorn5s Lemma there exists a maximal element P in S, that is P is an ideal containing A and maximal with respect to the exclusion of M. That P is a prime ideal has been shown in Theorem 1.8.

Definition 3.12; A non-zero element a in a ring R is called a nilpotent element if there exists a positive integer n such that a11 = 0.

Theorem 3.13: In any ring R, the intersection of all prime ideals of R is the set of all nilpotent elements in R. 27

Proof: Let P be the intersection of all prime ideals in R and let N be the set of all nilpotent elements in R, i.e. N = [x|x e R and there exists a positive integer n such that xn = 0. } If x e If and x 4 0, then there exists a positive integer m such that xm = 0. Now xm e P since all ideals must contain the zero element. Hence, xm belongs to every prime ideal in R, and by Lemma 1.17 x belongs to every prime ideal and hence, x e P. Therefore, N<^P. Now suppose x | N, then x is not a nilpotent element and there is no positive power of x such that x11 = 0. 2 Hence, the set S = {1, x, x , ••• } does not contain the zero element and is a multiplicatively closed set. The zero ideal is an ideal in R that does not intersect S, and by Theorem 3.11 there exists a maximal ideal in R with respect to the exclusion of S, call it P, and P is a prime ideal. Since P is a prime ideal and maximal with respect to the exclusion of S, then x | P, and hence, x | P. Therefore, PCN, and since Ncp, then P = N. Theorem 3.1^' An integral domain with unity is a UFD if and .only if every proper non-zero prime ideal in R contains a principal prime. Proof: Suppose R is a UFD. If R is a field, then R contains no proper non-zero prime ideals, hence, the theorem is true. Now suppose R is not a field. Let I be a proper non-zero prime ideal in R. Let a be a non-zero element of I, 28

then a is not a unit since I =[ R. Either a is irreducible or a is not irreducible. If a is irreducible, then a is a principal prime by Theorem 1.22. If a is not irreducible, then since R is a UFD, a is a finite product of irreducible factors. Thus a = ^2.^2 **" ^n 6 Since I is prime, it follows that € I for some i = 1, 2, ..., n. Now since irreducible elements in a UPD are principal primes, then I contains a principal prime.

Suppose every proper non-zero prime ideal in R contains a principal prime. If R is a field, then R is already a UPD. Suppose then that R is not a field. Then R contains a non- unit b 4 0- The set of units in R forms a multiplicatively closed set M such that M fl (b) = ({). By Theorem 3.11 there exists a prime ideal Q^(b) such that Q, fl M. = ((), hence, Q is a proper non-zero prime ideal in R. Therefore, R contains at least one principal prime.

Let S be the set of all elements that are products of principal primes. Assume there exists a non-zero, non-unit element x € R such that x ^ S. S is a multiplicatively

closed set for if s, s e S, then s = p-^p0 • • • pn and

s = P^P2 **• Pm where the p*s and p*s are principal primes,

and ss = P]_P2 ••• PnP]_P2 **" ^m* Hence, ss e S. Clearly, the zero element does not belong to S since R is an integral domain. Consider the ideal (x). The ideal (x) and the multiplicatively closed set S have no element in common, for 29 suppose z e (x). Now z = xr where r e R. Now either r e S or r <£ S. If r | S, then z is the product of two elements, each of which cannot he written as the product of principal primes, and hence, z S. IfreS, then r =

V-JJPQ *'* wêre the € ôw z = xPiP2 '** ût x cannot be written as the product of principal primes and hence, z cannot he written as the product of principal primes, thus z £ S. Therefore, (x) n S = <$. Now, by Theorem 3.11* there exists a prime ideal P such that P^> (x) and P is maximal with respect to the exclusion of S. But since P is prime, it contains a principal prime "by our hypothesis, say p. Hence, p € P, and p e S, but this contradicts the fact that P fi S = (J). Therefore, every non-zero and non-unit element belongs to S, and hence, R is a UFD.

Definition 3.15: Let A be an ideal in a ring R. A prime ideal P in R is said to be a minimal prime ideal belonging to A if A P and there is no prime ideal P in R such that Ac P <( P.

Theorem 3.16: In the ring of integers I, the minimal prime ideals belonging to a non-zero ideal (n) 4 I are the prime ideals (p) for which the integer p is a prime divisor of n.

Proof: Since (n) 4 I* then n 4 1, and we may also assume n is a positive integer. Let M be the set of minimal prime ideals belonging to a non-zero ideal (n). Let D be the set 50 of prime ideals (p) for which the integer p is a prime divisor of n. Let p-^, Pg* ••• P^- be the distinct prime factors of n. Let n = p-j^l ••• where the b*s are positive integers, Let x e D, then x = (p.) for 1-1, 2, ..., k. Let a e (n), then a = En where E is some integer. Hence, the

b, b0 b, \ element a = k (p-j^1l p2 2 •.. •. •p P^kk k))

p b., b~ p„ b.-l b a £ ( l i p2 2 ... i i ... Pk k) p±. This implies a € (p^), and hence, (n)c(p^). Suppose there exists a prime ideal (q) such that (n)c.(q) <( (Pj_)« Now q e (q) but q | (Pj_) since q is a prime number, and q = tp^ for some integer t implies t = 1, but this contradicts the choice of (q) such that (q) <( (p^). Therefore, there can exist no such (q). Hence, x € M which implies D^m, Now let Y e M. Then Y is some minimal prime ideal belonging to (n). Hence (n)CY and there is no prime ideal Y in the ring of integers such that (n)CY M and M* n A = (j). Now by Theorem 1.4, m''= r\p"' where P* is a prime ideal. But since m' is a maximal multiplicatively closed set which does not meet A, then P* is a minimal prime ideal belonging to A by the first part of the proof. 32

Now since m'd M, then p"o p. But Acp*cp and P is a minimal prime ideal "belonging to A, hence P = P which implies M ' = M. Therefore, M = R\P is a maximal multiplicatively closed set which does not meet A.

Definition 3.18* If A is an ideal in a ring R, then the set {x € R | some positive integral power of x "belongs to A) is called the radical of A.

Theorem 5.19* The radical of an ideal is an ideal. Proof: Let A be an ideal in a ring R. The radical of A is not empty, since A is an ideal and A has an element, say a. Now a^~ e A implies a e VT. Let x e /A, then there exists a positive integer n such that xn e A. Let y e >/A^ then 'there exists a positive integer n such that yn e A. In order to show x - y e A, it is necessary to find an integer m such that (x-y)m e A. Let m = 2nri, then the r + 1 term of (x-y)m is 1" (r) xm~ryr =

n na r T (^9 ) x ^ ~ j t jf 2nn - r then 2nn - r = n + n'

n n where n' ^ 0. Hence, x2nn-r _ xn+n _ xnx ^ x e A

n r n n r r and hence, x^ ^~ = x x e A. Thus, x2nn- y g ^ If 2nn - r n, then r )> 2nn-n r )> n(2n-l) and hence, r )> n. Since

>— X n, then y e A "by a similar argument as above, and hence, + <2F>) x2n5~V e A. Now since an ideal is closed with respect to addition and (x-y)m is the sum of terms of the form 1" (^Pn) x^nn_ryr 33 which belong to A, then (x-y)ra e A, which implies x - y e >T~A. Let r e R, then (rx)n = rnxn e A since xn 6 A. Hence, (rx)n e A and, therefore, rx €XTX

Theorem 3.20: If P is a prime ideal in a.ring R, then P is its own radical. Proof: The radical of P contains P since for any element , p e P, it follows that p e P hence, p evP. Let x £ JT then there exists a positive integer n such that xn € P where n ^ 1. If n = 1, then x e P. If n ) 1, then "by Lemma 1.17 x e P. Hence, NTPCP. Therefore, >fp = P if P is a prime ideal. Theorem 3.21; If A is an ideal in a ring R and P is a prime ideal in R, such that P^A, then P also contains the radical of A. Proof: Suppose P is a prime ideal in R such that P3A.

Let x €>rr, then there exists a positive integer m such that xm e A. Since P^A, then xm e P, and "by Lemma 1.17 x e P. Therefore, P^> ST.

Theorem 3.22: The radical of an ideal A in a ring R is the intersection of all minimal prime ideals belonging to A. • Proof: Let I be the intersection of all minimal prime ideals belonging to A. By Theorem 3.21 If A is contained in a prime ideal P, then P also contains A. Now the intersection of all the minimal prime ideals belonging to A would at 34 least contain A and hence,-the /x would "be contained in the intersection. Thus I It must now "be shown that JTai' or that if r is any element such that r € R and r krr, then r £ 1 . Let r e R and r * /a: Now consider the multiplicatively 2 3 closed set M = {r, r , r , •••). Now this set does not contain any element of A since r ^ implies r1 ^ A for every positive integer i. Now by Theorem 3.10 M is contained in a maximal multiplicatively closed set, say M which does not contain any element of A. Now r e M since M3 M and r e M. Also, M = R\P where P is a prime ideal "by Theorem 1.4. But since M is a maximal multiplicatively closed set that does not meet A then by Theorem 3.17 P is a minimal prime ideal belonging to A. Now since r e M, then r P and hence, r | i' , the intersection of all minimal prime ideals belonging to A. Thus i' . Since v/lTcl' and l'c\/A, therefore, n/TT = I Theorem 3.23; Let A denote an ideal in R. If P is a prime ideal in R such that P3A, then P contains a minimal prime ideal belonging to A. Proof; Let P be a prime ideal in R such that P3A, Then by Theorem 1.4 R\p is a multiplicatively closed set, and since PDA, then R\P does not meet A. Now by Theorem 3.10, R\P is contained in a maximal multiplicatively closed set M • such that M n A = (j). Now since M is a multiplicatively closed set then by Theorem 1.4 M = R\P where P is a prime 35 ideal. Obviously Pi>A since M n A = (J). But since M is a maximal multiplicatively closed set then by Theorem 3.17 P is a minimal prime ideal belonging to A. Now Mi> R\P- implies R\PDR\.P since M = R\P, and hence, PC P. Therefore, P contains a minimal prime ideal P belonging to A. CHAPTER IV.

PRIME IDEALS IN RINGS WITH THE DESCENDING CHAIN CONDITION

Definition 4.11 A ring R is said to satisfy the descending chain condition (d.c.c.) if every strictly descending chain of ideals N-^ )> Ng )> N^ ••• is finite. (Or if N^D ••• is an infinite descending sequence of ideals, then there exists an integer n such that N^ = Nn for i = n + 1, n + 2, ••••)

Theorem 4.2: If R is an integral domain satisfying the d.c.c. for ideals, .then R is a field. Proof: Since R satisfies the d.c.c., then for every infinite descending sequence of Ideals, N^ONg^^'j there exists an integer n such that N^ = Nn for i = n + 1, n + 2, 2 So consider the descending chain of ideals (a)3(a ) ZD • O (an)D where a e R and a 4 0. Now by the d.c.c. there exists an integer m such that (am) = (a1*1*"*"). There-

m m+ fore, a = ra "'" where m+r 1e R. Now let x be any non-zero xra = xam m xraa = xam xraam xam a (xra-x) = 0. Now am 4 0 so xra - x = 0. This implies xra = x and ra is the unity in R.

36 37

Now let b "be any non-zero element in R. Consider the O 1*- descending chain of ideals (b):?(b )d. .. (b ) • • • % by the d.c.c. there exists an integer n such that (bn) = (bn+1). n — n+1 — Therefore, b = rb , where r e R. Now rbn+1 - bn = 0 rbnb - bn = 0

n rb b - bn(ra) = 0 bn(rb - ra) = 0.

Since bn 4 0* then rb - ra = 0. Hence, rb = ra, and r is the inverse of b. Therefore, R is a field.

Theorem 4.3I If R is a ring satisfying the d.c.c. for ideals, then prime ideals of R different from R are maximal. Proof; Let P be an arbitrary prime ideal such that P ( R. Now by Theorem 1.2, R/P is an integral domain. Let cp be the natural homomorphism from R onto R/P, and let N be an ideal in R/P. Now let N = cp~"'"(N) = {x e R|cp(x) e N} If x e P, then cp(x) = 5 e N which implies that PdcpT^N) = N. So N= cp"1 (N) 4 $ because P ^ f, and PCN.

If a, b € N, then cp(a-b) = cp(a) - cp(b) e N since N is an ideal, but this implies a - b e cp-1(N) = N. If r e R, then cp(ra) = cp(r) cp(a) e N since N is an ideal, and, it follows that ra e cp"^(N) = N. for all r e R. Hence, cp~1(N) = N is an ideal in R.

1 Suppose N-^ y Ngj, where cp" (N1) = N^ and cp'^Ng) = Ng. Let ng e Ng then cp(ng) e Ng N-^ which implies cp(ng) € and 38 hence, e N-^. Now suppose n e N-^, n | Ng and let n be an element of R such that cp(n) = ri € Np which Implies n € N^, and cp(n) = n | Ng which implies n £ Ng. Therefore, N-^ )> Ng.

Now assume there exists an infinite chain of ideals in R/P such that N-^ )> Ng )> ••*, then there exists a corresponding chain of ideals N^ /" Ng ^ • • • in R. But this contradicts the fact that R satisfies the d.c.c. and, thus, our assumption is false, therefore, R/P satisfies the d.c.c. Now "by Theorem 4.2, since R/P is an integral domain satisfying the d.c.c., then R/P is a field, and by Theorem 3.2, P is a maximal prime ideal.

Theorem 4.4: A ring R with identity satisfying the d.c.c. for ideals has only a finite number of prime ideals. Proof; Let R be a ring with identity that satisfies the d.c.c., then by Theorem 4.3, all prime ideals different from R are maximal. Now suppose that there are an infinite number of distinct prime ideals in R.

Consider the descending chain of ideals such that K-, 3 K ^ ... 3 K o ... where 1 <=0 l K = P ^1 *1

Kg = P1Pg

K3 ~ P1P2P3

Ki - P1P2 Pi where the P>s are distinct prime ideals. Now since R satis-

-P* n -f- V-\ r-i 4- 4-V\/^sV-i 4~ V* ^ V-* QV1 pf O QV"» 1 nf QfTQ7» V"l nil n Vt "f~ O *f* 39

Pn+l;:>Pn * Pn+lDPn-l Pn Pn+1D * * * 3 P1 P2 Pn+1> so

P = P P P Hence P P P Pn+p * * * n+l 1 2 n* > n+P l 2 -"V By Theorem 2.3, if a prime ideal A = BC, then B or C.

Now suppose that if A = P-^ ... Pn-1, then ADP. for some i = 1, 2, •••> n-1. Let A = ^±^2 •••Pns then A — ***Pn-l)Pn.

or P P Hence, A^(PjPg ••• Pn_i) n* n* then, of course, A contains a prime ideal. Suppose A^ P^Pg ,**^>n_i-' then hy the induction hypothesis bp. for some i = 1, 2, ..., n-1.

p then 3 P for SOm6 Now since Pn+1=> P^ ••• n> ^l i i = 1, 2, ..., n. But all prime ideals are maximal so that P , = P.. But this contradicts the fact that the P^s are n+1 i distinct prime ideals. Hence, the assumption is false and there are only a finite number of prime ideals. BIBLIOGRAPHY

Kaplansky, Irving, Commutative Rings, Boston, Allyn and Bacon, Inc., 1970.

McCoy, Neal H., Rings and ideals, The Mathematical Association of America, 19^8.

Miller, Kenneth S., Elements of Modern Abstract Algebra, New York, Harper and Brothers Publishers, 1958.

Waerden, Bertel Leendert von der, Modern Algebra, New York, Ungar Publishing Co., 19^9.

Zariski, Oscar and Pierre Samuel, Commutative Algebra, Vol. I (2 volumes), Princeton, Van Nostrand Company, Inc., 1958.