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 By 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 muitipli- catively 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 3 j 4^ ... , 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.