A study and contrast of polynomials and power series over commutative rings

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A study and contrast of polynomials and power series over commutative rings
Altman, Alan Paul
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v, 34 leaves : ; 29 cm


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Commutative rings ( lcsh )
Polynomial rings ( lcsh )
Power series rings ( lcsh )
Commutative rings ( fast )
Polynomial rings ( fast )
Power series rings ( fast )
bibliography ( marcgt )
theses ( marcgt )
non-fiction ( marcgt )


Includes bibliographical references (leaves 33-34).
General Note:
Submitted in partial fulfillment of the requirements for the degree of Master of Arts, Department of Mathematical and Statistical Sciences, 1986.
Statement of Responsibility:
by Alan Paul Altman.

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University of Colorado Denver
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17800820 ( OCLC )
LD1190.L62 1986m .A47 ( lcc )


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A STUDY AND CONTRAST OF POLYNOMIALS AND POWER SERIES OVER COMMUTATIVE RINGS by Alan Paul Altman B.A., University of Colorado, 1982 A thesis submitted to the Faculty of the School of the University of Colorado in partial fulfillment of the. requirements for the degree of Master of Arts Department of Mathematics 1986


,:f. This thesis for the Master of Arts degree by Alan Paul Altman has been approved for the Department of Mathematics by Richard r.:;undgren / Date II ;s-/ito


iii Altman, Alan Paul A Study and Contrast of Polytiomials and Power Series over Commutative Rings Thesis directed by Associate Professor Sylvia Lu Fundamental notions of study in ring theory include units, zero divisors, nilpotent elements, prime ideals, primary ideals, nilradical, Jacobson radicals, and radicals of ideals. Let A be a commutative ring with identity. The purpose of this paper is to study and compare two rings, polynomial rings A[x] and power series rings A[[x]] in one indeterminate over the same commutative ring A. In the first chapter, theories involving the above mentioned fundamental notions that are needed in this paper are discussed. Chapters two through four deal with these notions for polynomial and power series rings. For example, conditions for polynomials and power series to be units, zero divisors, or nilpotent elements are examined. McCoy's theorem, which states a simple condition for a polynomial to be a zero divisor, is presented with an alternative and elegant proof given by W. R. Scott. Ideal structures of those tw6 rings are also studied. In general, the conditions imposed on power series rings to satisfy certain properties are stronger than the conditions on polynomial rings to


iv satisfy the same properties. In the final chapter, we show that these conditions on power series rings can be weakened if we restrict A to be a ring called a Noetherian ring. For this purpose, Cohen's theorem is introduced showing a simplification of conditions for a ring to be Noetherian. It is applied in the proof of the famous Hilbert basis theorem and its generalization to power series rings. Finally, the analogue of McCoy's theorem for power series rings Noetherian rings is presented and proved with the aid of the Hilbert basis theorem.


v CONTENTS CHAPTER I. SOME FUNDAMENTAL NOTIONS IN THE THEORY OF RINGS ............. I II. UNITS OF POLYNOMIAL AND POWER SERIES RINGS . . . . . . . . . . . . . . . . 9 III. ZERO DIVISORS AND NILPOTENTS IN A[x] AND A [ [ X J J . . . . . . . . . . . . . . . 1 5 IV. PRIMARY IDEALS AND RADICALS OF A[x] AND A[[x]] ................................... 19 V. A[x] AND A[[x]] OVER A NOETHERIAN RING A ................................. 24 BIBLIOGRAPHY..................................... 33


Chapter I Some Fundamental Notions in the Theory of Rings Throughout this paper all rings are commutative with identity. An element a of a ring A is a unit if there is an element bin A such that ab=1. A zero divisor in a ring A is an element a in A such that ab=O for some nonzero element b in A. A nilpotent element in a ring A is an element a in A such that an=O for some positive integer n. The smallest positive integer n' such that an'=o is called the nilpotency of a. An ideal in a ring A is a set of elements I such that I is an additive Abelian group and A prime ideal is an ideal P such that if ab is in P then a is in P or b is in P. (2)={0,2} is a prime ideal of the ring Z4={0,1 ,2,3}. A maximal ideal is a proper ideal M which is not contained properly in any ideal I+A. (2) is also a maximal ideal of z4 as (0) is the only other ideal of z4 and (O)C(2). A local ring is a ring with exactly one maximal ideal. The ring Z4 is a local ring as (2) is the only maximal ideal of z4


2 Proposition P is a prime ideal of A iff A/P is a domain. proof: Pis a prime ideal of A iff ab in P implies a is in P or b is in P iff (a+P)(b+P)=(ab+P)=P implies a+P=P or b+P=P iff A/P is a domain. Proposition 2 M is maximal ideal of A iff AIM is a field. proof: M is maximal iff M+(x)=(1) for all x in A-M iff for all x in A-M there is a yin A such that xy is in 1+M iff for all nonzero x+M in AIM there is a y+M in AIM such that (x+M)(y+M)=1+M iff A/M is a field. Corollary If I is a maximal ideal of A, then I is a prime ideal of A. proof: If I is a maximal ideal of A, then A/I is a field. So A/I is also a domain and I is prime by Proposition 1. Proposition 3 Let P 1 ... ,Pn be prime ideals and let I be an ideal contained in the union of these prime


3 ideals. Then I is contained in Pi for -some i. proof: The proof is by induction on n. Clearly the statement is true for n=1. Assume it is true for n-1. Suppose that I is not in the union of any n-1 of the prime ideals Pi, i=1 ,2, ... ,n. Then for each j, there exists an element bj of I which is in Pj but not in any Pi for Now b 2 b3 bn+b1 b3 bn+ ... +b 1 b 2 ... bn_1 is an element of I and therefore is in some Pi, say P 1 It follows that b 2 b3 bn=O(P 1 ) which is impossible since no one of the factors is in the prime ideal P 1 Hence I is in the union of some n-1 of the prime ideals Pi, i=1 ,2, ... ,n, and by the hypothesis of induction, must therefore be contained in some Pi. The set of all nilpotent elements of a ring A is called the nilradical of A. The nilradical of a a ring A, denoted by N(A), can be shown to be an ideal, as if xis in N(A), then ax is in N(A) for all a in A; and additive closure can be shown with use of the binomial theorem. The nilradical of A has the following relationship with the prime ideals of A.


4 Proposition 4 The -ni lradical N(A) of A is equal to the intersection of all the prime ideals of A. proof: Let {Pi: i I} be the set of all prime ideals of A Then N( A ) '= {) P i as x i n N (A ) i m p 1 i e s HI that xn=O is in Pi for all prime Pi in A. So xn is in (\ i'i..I Pi is prime. Pi and x is in (\ Pi as each H. I Now, suppose x is not nilpotent. Let S be the set of ideals I with the property that if n>O then xn is not in I. S is not empty because 0 is inS. OrderingS _by inclusion, S has a maximal element P by Zorn's lemma. To show P is a prime ideal let y,z not be in P. Then the ideals P+(y) and P+(z) strictly contain P and thus do not belong to S. So xm P+(y) and xn 'i. P+(z) for some m,n. Then xm+nz. P+(yz) and P+(yz) is not in S. So yz is not in P and P is prime. The Jacobson radical of ring A ,denoted J(A), is defined to be the intersection of all the.maximal ideals of A. Clearly, the nilradical is contained in the Jacobson radical.


5 Theorem x is an element of the Jacobson radical J(A) iff 1-xy is a unit of A for all y in A. proof: x is not in J(A) iff (x)+M=(1) for some maximal ideal M of A iff xy+m=1 for some min M, and some y in A iff 1-xy is not a unit in A as 1-xy is in M. Theorem 1 is the most basic theorem concerning the Jacobson radical. Corollary 1 If xis an element of the nilradical N(A) then 1+x is a unit of A. proof: N(A) is contained in the Jacobson radical and let Y=-1. The next corollary is a generalization of the previous corollary to Theorem 1. Corollary 2 If x is an element of the nilradical N(A), then u+x is a unit of A for all units u in A. proof: Let u+x=a where xis in N(A) and u is a unit in A. Then 1 +xu-1 =au-1 If we put x'=xu-1 and a'=au-1, then 1 +X'=a' is a unit by


6 Corollary 1 to Theorem 1 as x' is in N(A) which is contained in J(A). Therefore a'u=a=u+x is a unit of A as both a' and u are units in A. The r ad i c a 1 of an i de a 1 I of a r i n g A, w h i c h is denoted by r(I), is the set of all elements a in A such that an is in I for some positive integer n. For example, in the ring z8 the radical of the ideal (4)={0,4} is {0,2,4,6}. Note that r(I) is an ideal of A which contains I. From the definition we can see immediately that the nilradical is the radical of (0). Just as the nilradical of A was related to the prime ideals of A, the radical of an ideal has a similar relationship with the prime ideals of A. Note that f(r(I))=N(A/I) under the natural homophorism f:A-7A/I. Proposition 5 The radical of an ideal I is the intersection of all the prime ideals that contain I. proof: f(r(I))=N(A/I) and N(A/I) is the set of all elements a+I such that (a+I)n=I. This is the intersection of all prime ideals of A/I by Proposition 3; which are the prime ideals of


A that contain I. An ideal q of A is a primary ideal if ab is in q implies that a is in q or bn is in q for some positive integer n. Note that q is a primary ideal of A iff every zero divisor in A/q is nilpotent. Proposition 6 7 If q is a primary ideal of a ring A then r(q) is a prime ideal of A. proof: ab is in r(q) implies (ab)n=anbn is in q. So an is in q or (bn)m is in q for some positive integer m which means that a is in r(q) or b is in r(q). If r(q) is a prime ideal P then we say that q is p-primary. We will later introduce an example in which the converse of Propostition 6 is not true. That is, r(q) is prime does not necessarily imply that q is primary. However: Proposition 7 If r(q) is maximal then q is primary. proof: r(q)/q is the nilradical in A/q as a+q is in r(q)/q iff for some positive integer n an+q=q iff (a+q)n=q iff a+q is in


N(A/q). Thus A/q has only oneprime ideal and every element in A/q is a unit or nilpotent. Therefore every zero divsor in A/q is nilpotent. 8


9 Chapter II Units of Polynomial and Power Series Rings Let A be a ring and S the set of all sequences of the form (a0,a1 ,a2 ... ,am,O,O, ... ), denoted by (an) where ai are in A for all i. We define (an)+ (bn)= (an+bn) and (anHbn)=(cn) where Then: 1) (S + .) is a ring with 0=(0,0, ... ,0, ... ) and 1=(1 ,0,0, ... ,0, ... ). 2) A can be embedded into S by the monomorphism such that F(a)=(a,O,O, ... ). 3) Let x=(0,1,0, ... ), then xk=(0,0, ... ,1,0, ... ) and k ax =(a,O,O, ... )(0,0, ... ,1 ,0, ... )=(0,0, ... ,a,O, ... ) 4) For all f=(an) in S there exists a finite sequence a 0,a1 ... ,am in A such that "" f = a 0 + a 1 x + . +am xm = a i x 1 f i s c a 11 e d a i=O polynomial in x over A. The ring S is called the ring of polynomials over A in the indeterminate x, denoted by A[x]. If we let the sequence in 4) be finite or infinite then Sis called the ring of formal power series over A in


1 0 the indeterminate x, denoted by A[[x]]. Namely A[[x]] ea consists of those fs such that f= aixi which is i=O called a power series in x over A. The polynomial ring in indeterminates x 1 ,x2 ... ,xn over A, denoted by A[x1 ,x2 ... ,xn], is defined to be A[x1 ,x2 ... xn]=A[x1 ,x2 ... xn_1 ][xn] for each The power series ring A[[x1 ,x2 .. xn]J can be defined in a similar way. In Chapter I we remarked that the converse of Proposition 6 is not true in general. We now give an example of an ideal whose radical is prime but the ideal is not primary. Example: Let A=k[x,y,z,]/(xy-z2) and let x',y',z' denote the images of x,y,z respectively in A. Then P = ( x z ) i s p r i me s i n c e (A I p}; k [ y ] a do m a i n N ow we have x'y'=z'2 is in p2 but x' is not in P 2 and y' is not in r(P2 )=P; hence P 2 is not primary. In this paper we mainly consider polynomial and power series rings in one indeterminate. If I is 1\ an ideal of A then I[x]={f=L a-xi in A[x]; ai in I} 1 eC> i=O and I[[x]]={f= 2_ a-xi in A[[x]]; ai in I } I[x] and i=O 1 I[[x]] are ideals of A[x] and A[[x]]. On the other hand, IA[x] is an ideal 6ontained in I[x] andIA[[x]] is contained in I[[x]]. We can study more about the relationship between ideals of A and ideals of A[x]


11 and A[[x]] using the following lemma. Lemma 1 If I is an ideal of A then A[[x]]/I[[x]]. (also A[x]/I[x]) proof: Let F be an epimorphism = ....... such that aixi)= 2. (ai+I)xi. Then f i=O i=O is in the kernel ofF iff aj+I=I for every j iff aj is in I for all j iff f is in I[[x]]. So by the fundamental theorem of homomorphisms (A/I)[[x]];A[[x]]/I[[x]]. A parallel argument can be made to show (A/I)[x]:A[x]/I[x]. Theorem 2 If Pis a prime ideal of A then P[[x]] (respectively P[x]) is a prime ideal of A[[x]] (respectively A[x]). proof : P i s a p r i me i de a I of A i m p 1 i e s that A I P is a domain by Proposition 1. So (A/P)[[x]] is also a domain; and A[[x]]/P[[x]] is a domain by Lemma 1. Therefore P[[x]] is a prime ideal of A[[x]] by Proposition 1. A parallel argument can made to show P[x] is a prime ideal of A[x].


1 2 Theorem 3 f is a unit of A[[x]J iff a 0 is a unit of A ...... proof: The necessity is clear. To show f=L_ aixi i=O is a unit in A[[x]J, we must produce a g= L bi xi such that fg= 1. Let b 0:a0 -1 and i=O having b 0,b1 ... ,bn appropriately chosen we need bn+ 1 Since fg=a, the coefficient of xn+1 of fg is aibj=O. Which means i+j=n+1 a 0 bn+ 1=-a1bn.. -anbo So let bn+ 1 = a 0 1 (-a1bn-an+1 b 0 ) as a 0 is a unit in A. For a polynomial fin A[x] to be a unit we must impose more restrictions on the coefficients of f. Theorem 4 proof: ,., Let A be a domain, then f= xi is a unit 1 1=0 in A[x] iff a 0 is a unit in A and a 1 =a 2 = ... =an=0; namely, the group of units of A[x] and that of A are identical. "' If f= L a. xi is a unit in A[x] then there is 1 i=O a g= .f. bixi such that fg=1. Since A has no i=O zero divisors anbmfo and hence deg(fg)=deg(f)+ deg(g). So since deg(fg)=O, deg(f)=O which


13 For rings A with zero divisors, the group of units of A[x] and that of A are not identical. More precisely, we have: Theorem 5 f is a unit in A[x] iff a 0 is a unit in A and a1 ,a2 ,an are nilpotent. proof: Let P be a prime ideal of A and let F be the natural homomorphism A-7A/P. We can extend F to F':A[x]......:r(A/P)[x] where F'(f)= a 0+P +(a1+P)x+ ... +(an+P)xn. Now f is a unit of A[x] implies F'(f) is a unit of (A/P)[x]. By Theorem 4 a 0 is a unit of A and ai are in P for all i>O. Then a 0 is a unit of A and ai are in the intersection of all prime ideals of A which is the nilradical of A asP was arbitrary. So a 0 is a unit of A and a1 ,a2 ,an are nilpotent. Now let a1 ,a2 ,an be nilpotent with maximum nilpotency t and let a 0 oe a unit of A. Then 2 n a1 x+a2x + ... +anx is nilpotent as we can find a sufficiently large t' such that ( 2 n)t' a1 x+a2x + ... +anx =0 for t'>nt. a 0+a1 x+ ... +anxn is a unit of A[x] by Corollary 2 to Theorem 1. So


1 4 If M is a maximal ideal in A, the most we can say about M[x] and M[[x]] is that M[x] and M[[x]] are prime ideals in A[x] and A[[x]] respectively. M[x] and M[[x]] need not be maximal, for if M[[x]] is maximal in A[[x]] then is a field by Proposition 2 and Lemma 1. However, in (A/M)[[x]] there are nonzero elements which are not units. For example, x is not a unit in (A/M)[[x]] by Theorem 3. A similar argument applies to M[x]. In fact, some basic properties of polynomial and power series rings over a field Fare that F[[x]] is a local ring with maximal ideal (x) and that F[x] is a principal ideal domain (a domain where every ideal is generated by one element) which is not a local rin-g.


1 5 Chapter III Zero Divisors and Nilpotents in A[x] and A[[x]] An element fin A[x] is a zero divisor if there exists a nonzero element gin A[x] such that fg=O. A theorem by N.H. McCoy characterizes zero divisors in. polynomial rings. Theorem 6 [13] (McCoy's theorem) fin A[x] is a zero divisor iff there exists a nonzero a in A such that af=O. The traditional proof given by McCoy consists fl'\ of choosing a nonzero polynomial g= bixi in A[x] of i =0 ,, least degree m such that fg=O, where aix1 Then i=O anbm=O and ang=O as angf=O and deg(ang)

1 6 exist a nonzero a in A such that af=O and let g be a nonzero polynomial of least degree m 1\ (Y, such that fg=O. Let f= i:O a-xi and g= 2 b-xi 1 ; =" 1 where and Since aibmfO for some i. So Let r be the largest integer such that arg and then 0=fg=(a0 + ... +anxn)(b0 + ... +bmxm)= (_a0 + ... +arxr)(b0 + ... +bmxm). So arbm=O and deg(arg)c Example: Let K be a field withy and {zi}i=O = indeterminates. Let I be the ideal of K[y,{zi}i=O] oc generated by yz 0 and {zi+yzi+1 }i=O If we set oO A=K[y,{zi}i=O]/I, a 0:y+I and bi=zi+I for i>O then = (a0+x)( L bixi)=O in A[[x]] and clearly a 0 +x i=O cannot be annihilated by any constant in A. There is, however, an analogue to Theorem 5 for power series rings A[[x]] if A is some special ring. Theorem 7 [2] Let A be a ring with no nonzero nilpotent


17 elements; and let f,g in A[[x]] (respectively A[x]) such that fg=O. Then each coefficient of g annihilates f. proof: Let I and J be the ideals of A generated by all the coefficients of f and g respectively. If Pis a prime ideal of A then P[[x]] is prime b_y Theorem 2. Since fg=O in P[[x]], f is in P[[x]] or g is in P[[x]]. This means that aibj is in P for all ai in I and bj in J. So IJ is contained in every prime ideal so that IJ is contained in (0) by Proposition 4 as the nilradical of A is (0) by hypothesis. Next, we consider nilpotent polynomials and power series. Theorem 8 n f=L aixi in A[x] is nilpotent iff i=O a 0,a1 ... an are nilpotent in A. proof: Let P be a prime ideal of A and let F:A-7A/P be the natural homomorphism. Extend F to F': A[ x]---7(A/P )[ x] such that F' (f)= 2 (ai +P)xi. Now, f is nilpotent in i=O A[x] means that F'(f) is nilpotent in


(A/P)[x] and thus all ai are in P as (ai+P)=P implies ai is in P. Therefore ai are in the intersection of all the prime ideals of A, the nilradical of A, asP was 18 arbitrary. So a 0,a1 ... ,an are nilpotent in A by Proposition 4. Now assume a 0,a1 ... ,an are nilpotent. Then we can choose a positive n integer m so large that fm=(L aixi)m=O. So i=O f is nilpotent. Theorem 8 remains partially true if A[x] is replaced by A[[x]]. Theorem 9 oQ f= L aixi in A[[x]] is nilpotent implies that i=O an are nilpotent for all proof: Let P be a prime ideal of A and let Then the proof follows the same as in Theorem 8. The converse of Theorem 9 is not true in general. However, the converse is true for certain kinds of rings. We shall consider this problem in Chapter V.


1 9 Chapter IV Primary Ideals and Radicals of A[x] and A[[x]] With the aid of the previous theorems, we are now able to study the ideal structure of A[x] and A[[x]]. Theorem 10 If q is a P-primary ideal in A then q[x] is a P[x]-primary ideal in A[x]. proof: It suffices to show that every zero 4ivisor of A[x]/q[x] is nilpotent. (A I q ) [ x ] by Lemma 1 So i f f = 2._ (a i + q) xi i s i=O a zero divisor of (A/q)[x] then there exists an a+q in A/q such that and (a+q)f=q by Theorem 6. aai+q=q for all coefficients ai +q of f. This means that aai is in q for all i. Since q is primary and a is not in q as we have ain is in q. Thus ai+q are nilpotent in A/q for all i. So f is nilpotent in (A/q)[x] by Theorem 8 and f is also nilpotent in A[x]/q[x] by Lemma 1. Next


20 we prove that the radical r(q[x]) of q[x] is ,., P[x]. f= L aixi is in r(q[x]) iff there is i=O an m such that aixi)m is in q[x] iff i=O there is a k such that ai k is in q for all i iff aiis in p for all i as r(q)=p iff f is in P[x]. Next, we consider a generalization of Theorem 10 to power series rings. We find that this is possible under some restrictions on the base ring A. We are appreciative to Professor Paul O'Meara for a usefull suggestion on the proof of the following lemma. Lemma 2 Let x be an element of a ring A and let y be a nilpotent element of A. Then if x-y is not a zero divisor of A then xis not a zero divisor of A. proof: Let n be the nilpotency of y and assume that x is a zero divisor, that is, ax=O for some a!O. Next, let k be the least positive integer such that ayk=O. Such a k must exist and because ayn:o. Then and ayk-1(x-y)=O, which is a contradiction to the


21 fact that x-y is not a zero divisor. Hence, x is not a zero divisor. Theorem 11 [2] If q is a p-primary ideal of A and q contains a power of p then q[[x]] is a p[[x]]-primary ideal of A[[x]]. proof: p[[x]] is a prime ideal of A[[x]] and contains q[[x]]. Let f and g be two elements of A[[x]] such that fg is in q[[x]]. We need to show if f is not in p[[x]] then g is in q[[x]]. Working modulo q[[x]], we may suppose that q=(O) and show (0) is p[[x]]-primary provided that (0) is p-primary in A. We need to show if fg=O and f is not in p[[x]], then f is not a zero divisor so g=O. Let f=a0+a1 x+ ... +amxm+ ... where ai is in p for and am is not in p. (0) is pprimary implies ai is nilpotent for :"1\-1 So h= ai xi is ni 1 potent by theorem 8. i:O Since !m is not in p it is not a zero divisor; and thus f-h is not a zero divisor. By Lemma 2 f is not a zero divisor. Now, since pk is contained in q, (p[[x]])k is contained in q[[x]]. So p[[x]])=r(q[[x]]).


22 Recall that for any ring the nilradical is contained in the Jacobson radical. Theorem 12 In A[x] the Jacobson radical J(A[x]) is equal to the nilradical N(A[x]). proof: If f= 2: a1xi is in J(A[x]) then 1+fg is a i=O unit in A[x] for all g in A[x] by Theorem 1. f 2 n+1 In particular 1 + X= 1 +a 0 x+a 1 x + ... +anx is a unit. This means that a 0,a1 ... ,an are nilpotent in A by Theorem 5. So f is nilpotent in A[x] by Theorem 8. Thus f is in N(A[x]). Theorem 12 is not true for power series. Example: F[[x]] over a field F is a local ring with maximal ideal (x). Hence J(F[[x]])=(x). Since F[[x]] is an integral domain, (0) is a prime ideal. Therefore N(F[[x]])=(O) and hence N(F[[x]J)'iJ(F[[x]]). Theorem 13 ..,., f= 2:. ai xi is in the Jacobson radical i=O J(A[[x]]) iff a 0 is in the Jacobson radical J (A).


23 proof: By Theorem 1 we have that f is in J(A[[x]]) e>C> i f f 1 + f g i s a u n i t of A [ [ x ] ] for a 1 1 g =.L b i x i .: c in A [ [ x]] iff 1 +a 0 b 0 is a unit of A for a 11 bo in A iff a 0 is in J(A). 1'\ In the polynomial ring A[x], i=O in J(A[x]), then a 0 is in J(A). However, the converse of this statement is not true since 1+a0 b 0 is a unit of A does not imply 1+fg is a unit in A[x], r'Y' where g= .L_ bi xi, for the constant term of an element i=O of A[x] being a unit does not imply the element of A[x] is a unit.


24 Chapter V A[x] and A[[x]] over a Noetherian Ring A A ring A is said to be Noetherian if it satisfies the following three equivalent conditions. 1) Every nonempty set of ideals of A has a maximal element. 2) Every ascending chain of ideals in A is stationary. This is known as the ascending chain condition (A.C.C.). 3) Every ideal in A, including A itself, is finitely generated. The equivalence of the three conditions is well known [1] p.74. It is clear that if A is a Noetherian ring then A/I is a Noetherian ring for any ideal I. The ring z of integers is an example of a Noetherian ring as every ideal in Z is generated by one element. On the other hand, A[x1 ,x2 ... ] is not a Noetherian ring. However, as we shall shortly see, if A is Noetherian then A[x] and A[[x]] are Noetherian. A very useful theorem in commutative


25 algebra was given by I. S. Cohen which simplifies condition 3 for a ring to be Noetherian. Theorem 14 (Cohen's theorem) A ring A in which every prime ideal is finitely generated is Noetherian. proof: Assume that every prime ideal of a ring A is finitely generated. and A is not Noetherian. Then there exists ideals in A that are not finitely generated. LetS be the set of all ideals of A that are not finitely generated. S i s not empty s o, b y Z or n 's 1 emma, S has a maximal element I. To show I is a prime ideal let ab be in I such that a is not in I and b is not in I. Then the ideal (I,a) is -properly larger than I and therefore is finitely generated. Let (I ,a) be generated by elements of the form i 1 +x 1a, ... ,in+xna where x 1 ... _,xn are in A and i1 ... ,in are I. Now let J be the set of all y in A such in that ya is in I. Then J contains both I and b and hence is finitely generated. We claim that I=Ci1 ... ,in,Ja). Take an arbitrary element z in I. Then z is in (I,a), so we have an expression


26 z=u 1Ci1 +x 1a)+ ... +un(in+xna). Here we see .that u 1 x 1 + ... +unxn lies in J. We have verified I=Ci1 ... ,in,Ja), which implies that I is finitely generated. Contradiction! Thus I is a prime ideal. Consequently, if all prime ideals of a ring A are finitely generated then A is Noetherian. As we stated earlier, if a ring A is Noetherian then A[x] and A[[x]] are also Noetherian. The fact that A[x] is Noetherian was first proved by David Hilbert (1862-1943) and the theorem is appropriately named the Hilbert basis theorem. The analogue of the Hilbert basis theorem also holds true for power series rings. Theorem 15 (Hilbert basis theorem) If A is Noetherian then A[x] (respectively A[[x]]) is Noetherian. proof: Let p be a prime ideal of A[[x]] and let p' be the image of p under the homomorphism of A[[x]J onto A obtained by mapping x to 0. Then we need to show if p' is finitely generated so is p. Let P'=(a1 ,a2 ... ,ar). We have two cases according if xis in p or


27 not. If x is in p then P=(a1 ,a2 ... ,ar,x). Assume x is not in p. Let f 1 ,f2 .. ,fr in p be series with constant terms a 1 ,a2 .. ,ar. We claim that P=(f1 ... ,fr). Take any g in p and let the constant term of g be b. Then b= Lbiai and gbi f i can be written xg1 and xg1 is in p. Therefore g1 is in p as x is not in p. In this manner write with g2 in p. Continuing this process leads us to h 1 h 2 ... hr in A [ [ X] ] hi = b i + C i X+ ... satisfying g=h 1 f 1 + ... +hrf r And finally by Cohen's theorem A[[xJ] is Noetherian. A parallel argument shows A[x] is Noetherian. When A is Noetherian, Theorem 6 (McCoy's theorem) has an analogue in A[[x]]. To prove this theorem we need the concept of primary decomposition. A primary decomposition of an ideal I in a ring A is an expression of I as a finite intersection of primary ideals qi. If I= n qi, where all r(qi)=Pi i=1 are distinct and each qi does not contain fl qJ., 1 then the primary decomposition is said j:H to be minimal or irredundant. In general, a primary decomposition of an ideal need not exist. However,


28 in a Noetherian ring every ideal has a primary decomposition as it can be shown that every ideal in a ring satisfying the ascending chain condition is a finite intersection of irreducible ideals and that every irreducible ideal in the ring is primary cf.[18]. Lemma 3 If x is an element of any ring A such that x is not in a p-primary.ideal q, then (q:x) is p-primary and therefore r(q:x)=p. proof: Let y be an element of (q:x). Then xy is in q, hence y is in p (as x is not in q). So q is contained in (q:x) which is contained in p. Thus r(q) =Pis contained in r(q:x) which is contained in r(p)=P Therefore r(q:x)=p. Lemma 4 Let A1 ... ,An be ideals and let p be a prime ideal such that p:(J i=1 A i. Then P=Ai for some i. n proof: If p= n Ai then p is contained in every Ai. i=1 If Ai is not in p for every i, then there exists an xi in Ai such that xi is not in p


29 f'IAi, but xis not in pas pis prime. Thus Ai is in p for some i, so that Ai=P for some i. Proposition 8 [1] p.52 theorem 4.5 Let I be a decomposable ideal with minimal (\ primary decompostion n qi. Let Pi=r(qi), i=1 Then the Pi are precisely the prime ideals which occur in the set of ideals r(I:x), where xis in A, and hence are independent of the particular decomposition of I. proof: For any x in A, (I:x)=(nqi:x)=n(qi:x), !\ hence r (I : X)= .f\ r ( q i : X)= n p j by Lemma 3. i = 1 xi q Suppose r(I:x) is prime, then r(I:x)=Pj for some j by Lemma 4. Hence every prime ideal of the form r(I:x) is one of Pj Conversely, for each i there must exist xi such that xi is in fl qJ but not in qi, since i;\:j the decomposition is minimal; and we have The prime ideals associated with a minimal primary decomposition of I in Proposition 8 is called the prime ideals belonging to I.


30 Proposition 9 ([1] p.53 proposition 4.7) If the zero ideal is decomposable, the set D of zero-divisors of A is the union of the prime ideals belonging to (0). proof: D= \_) r(O:x) and if (0) has an minimal primary X f. A decomposition n qi then as we have seen in i=1 the proof of Proposition 8 r(O:x)= f\ Pj' xtq which is contained in all Pj Hence D is (\ contained in U Pi. But also from i = 1 Proposition 8 each pi is of form r(O:x) fdr some x in A; hence ,, D. Thus D= V Pi i=1 Proposition 10 1\ V Pi is contained in i=1 Let A be a Noetherian ring and let (0)=q1 (\ q2{'J ... n qn be a minimal primary decomposition where r(qi)=Pi Then there exists nonzero elements a1 ,an in A such that aiPj=(O), proof: Consider the set S of all annihilators of nonzero elements of A. S has maximal elements Mj as A is Noetherian. The set of zero divisors of A is the union of these Mj's., So Y Mj= ,\.) Pi where r(qi)=Pi by


31 Proposition 9. Now to show that the Mj are prime, let Mj be the annihilator of x in A, denoted ann(x), and let ab be in Mj and a not in Mj. Then ax=O and ann(ax) contains Mj as a is not in Mj. But since Mj is maximal in S, ann(ax)=Mj; and since b annihilates ax, b is in Mj. So Mj is prime. By Proposition 3, each pi must be contained in an Mj. Therefore there exists nonzero elements Now we have the analogue to McCoy's theorem in A[[x]]. Theorem 16 [2] Let A be a Noetherian ring. Then there exists nonzero e 1 ements a 1 ... an in A such that f is a zero divisor in A[[x]] iff aif=O for some i. proof: Let (O)=q1nq2n ... /lqn be a minimal primary decomposition of (0) where r(qi)=Pi Then by Propostion 9 the set of zero divisors of A is ,... \) Pi and by Proposition 10 there are i=1 elements a 1 ... ,an in A such that aipi=(O) for every i=1 ,2, ... ,n. Since A is Noetherian


32 so is A[[x]] by Theorem 15 and hence (0)=q1 [[x]] n q2[[x]] n ... n qm[[x]] is a minimal primary decomposition of (0), for """' f= ;?; ai xi be longs to the intersection iff all ai belong to q 1n q2n ... /)qn=O, that is, f=O By Theorem 11 qi[[x]] is Pi[[x]]-primary. So, the set of zero divisors of A[[x]] is U pi[[x]] by Proposition 9. i = 1 Hence, f is a zero divisor of A[[x]] iff f is in pi[[x]] for some i iff a1f=O since aipi=(O). In the Noetherian case we also have the converse of Theorem 9. Theorem 17 [2] """'. Let A be Noetherian and f=L a-xi in A[[x]]. 1 i=O If there is a positive integer m such that aim=O for all i20 then f is nilpotent. proof: Let I be the ideal generated by the coefficients of f. Since A is Noetherian I is finitely generated. So, if aim=O for all then there is a positive integer k such that Ik=(O); which means that fk=O.


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