Middle-East Journal of Scientific Research 15 (12): 1663-1665, 2013 ISSN 1990-9233 © IDOSI Publications, 2013 DOI: 10.5829/idosi.mejsr.2013.15.12.431

Non-nil Ideal Radical and Non-nil Noetherian R[x]

M. R. Alinaghizadeh and R. Modabbernia

Department of Mathematics, Shoushtar Branch, Islamic Azad University, Shoushtar, Iran

Abstract: In the part two of this paper we investigate Non-Nil ideal radical. In the part three we show that if R non-nil noetherian ring, Nil(R) divided prime ideal and nil(R[x])Í(f)"fÎR[x]\nil(R[x]). Then R[x] Non-nil noetherian Ring.

Key words : Non-nil radical ideal · non-nil noetherian · divided prime ideal · minimal prime ideal

INTRODUCTION (3)?(1) Suppose I be a non-nil ideal of R and F =

{(a1, a2,… am)| mÎN, aiÎI, (a1,… am) non-nil ideal}. We suppose that all of the rings in this article are Since I is non-nil thus $aÎl such that aÏnil(R), commutative with 1¹0. (a)ÎF, F¹Æ. So according to (3) F has maximal

Let R be a ring. Then nil(R) denotes the set of element (a1,..,am) we will show that I = (a1,..,am). nilpotent elements of R.I is called non-nil ideal of R if If I¹(a1,..,am) then $aÎI\(a1,..,am) thus I and {rÎR|$nÎN, rnÎI}. J is a non-nil (a1,..,am) ,…..,am,a) and (a1,..,am,a)ÎF we get to ideal radical of R if $ non-nil ideal I J P is divided a contradiction. prime ideal if pÍ(x)"xÎR\P. Min(R) is the Set of minimal prime ideals of R. If I be an ideal of R then Theorem 2.2: Let R and S be two rings and f:R®S an min(I) is the set of minimal prime ideals contains I.R is onto ring homomorphism. If R non-nil noetherian ring non-nil noetherian ring if every non-nil ideal of R be then S will be non-nil noetherian. finitely generated.

Proof: Assume that l1? l2? l3?……. be an increasing -1 -1 NON-NIL IDEAL RADICAL chain f non-nil ideals of S. f (l1)? f (l2)?……is a chain - of non-nil ideals of R and so (lk) = f Theorem 2.1: Let R be a ring then this following are 1 -1 -1 (ln) and thus "k³nlk = f(f (lk)) = f(f (ln)) = ln. equivalent: Corollary 2.3: Let R be a ring and I be an ideal of R. If 1) R non-nil noetherian. R non-nil noetherian then R/I is non-nil noetherian. 2) R satisfies in ascending chain condition on non-nil ideals of R. Theorem 2.4: Let R be non-nil noetherian ring and I be

3) Every non-empty set of non-nil ideals of R has non-nil ideal of R. There exists P1, P2,…, PnÎ min (R) maximal element. s.t. P1P2,…PnÍI.

Proof: (1)?(2) Let l1Íl2Íl3?……… be arbitrary Proof: Let F be the set of non-nil ideals of R which ascending chain of non-nil ideal of R. ÈiÎNli is a non-nil does not contain finite product of minimal prime ideal ideal of R.thus ÈiÎNli = ( , ,……, ) suppose that j of R and too F¹Æ. According to theorem 2.1. F has maximal element as P.P is not prime. Thus there exists = max {i1, i2,… ik} we have ÈiÎNliÍIj thus Ik = Ij "k³j. (2)?(3) Let F be non-empty set of non-nil ideal of ideals A,B of R s. t. ABÍP and A? P, B? P. Since R and l ÎF If l is not maximal element of F there 1 1 P? A+P, P? B+P so A+PÏF, B+PÏF and so $P1,…, Pn, exists l2 belong to F l1? l2. So if F does not have Q1…QmÎmin (R) s.t.P1…Pn?A+P, Q1Q2…Qm? B+P. maximal element then there will exist an infinite chain Thus P1…PnQ1,…Qm?(A+P)(B+P)?P is a of non-nil ideal of R and this is A contradiction. contradiction. Corresoponding Author: M.R. Alinaghizadeh, Department of Mathematics, Shoushtar Branch, Islamic Azad University, Shoushtar, Iran 1663 Middle-East J. Sci. Res., 15 (12): 1663-1665, 2013

Theorem 2.5: If R be non-nil noetherian ring then Lemma 3.1: Let R be a ring and f: R®R[x] denote the every non-nil ideal of R has primary decomposition. natural Ring homomorphism then nil(R[x]) = (nil(R))e = nil(R)R[x]. Proof: Suppose that S be the set of all non-nil ideal of R which does not have primary decomposition. We will Proof: By R.Y. Sharp (1990,exercise 1.36,page 24) show that S = Æ. If S¹Æ by theorem 2.1. S has n Nil(R[X])={c0+c1x+……+c0x |nÎN0, ciÎnil(R) for maximal element I.I is not primary and thus $a,bÎR n k all i = 0,….,n} and by R. Y. sharp (1990,exercise s.t.aÏl, "nÎN b Ïl. Let "kÎN lk = {xÎR|b xÎl} it is e n 2.47,page 44) (nil(R)) = {r0+r1x+…+rnx ÎR[x]|nÎN0, clear that "kÎN l is non-nil ideal and l ? l ? … k 1 2 r Înil(R) for all i=0,…..,n}thus we have nil(R[x]) = increasing chain of non-nil ideal of R. by. i (nil(R))e.

Theorem 2.6: $mÎN s.t. "i³m li = lm. We define E={bmy+c|yÎR, cÎl} E is a non-nil Lemma 3.2: Let R be a ring and f:R®R[x] denote natural Ring homomorphism then ideal of R and I? lm, I? E,I=EÇlm because aÎlm, a m m m m ÏI&b ÎE&b ÏI."xÎEÇlmx=b y+c s.t. yÎR, cÎl&b 2m m xÎl. Thus b yÎl and so yÎl2m = lm hence b yÎI & = = thus xÎI. E and lm have primary decomposition so I has primary decomposition and this is a contradiction. Proof: By R. Y. Sharp (1990, exercise 2.47, page 44)

Corollary 2.7: Let R be a non-nil noetherian ring and I be a non-nil ideal of R. = There exist prime ideals P1,……., Pn such that = P Ç..…ÇP . 1 n and by lemma 3.1. nil(R[x])=nil(R)R[x] thus

Proof: Since R non-nil noetherian by theorem 2.5. I has primary decomposition so there exists primary ideals Q1..…Qn such that I = Q1Ç..…ÇQn and for every i,j,1 Theorem 3.3: (R.Y. Sharp, 1990, theorem 8.7) Let R be a noetherian ring and let x be an indeterminate. Then and the ring R[x] of polynomials is again a noetherian ring.

? Qi thus Theorem 3.4: (Badawi, A., 2003, theorem 2.2.) Let R = . be a ring and nil(R) divided prime ideal of R.then R is a non-nil noetherian ring if and only if is a Corollary 2.8: Let R be a non-nil ideal of R. There noetherian domain. exists P1,.., PnÎMin(I) such that = P1Ç..…ÇPn. Theorem 3.5: Let R be a ring and nil(R) divided prime Corollary 2.9: If R be non-nil noetherian ring and I ideal of R and "fÎR[x]nil(R[x])Í(f). Then R non-nil non-nil ideal of R. Then Min(I) is a finite set. noetherian ring if and only if R[x] non-nil noetherian ring. Proof: By corollary 2.7. there exists P1,…, PnÎ min (l) s.t. = P1Ç..…ÇPn suppose that PÎMin(I) since = Proof: Let R[x] be non-nil noetherian since t: R[x]®R, n ÇpÎMin(l)P thus P1Ç..…ÇPn?P hence P = Pi$1£i£n. t(r0+r1x+…+rnx ) = r0 onto homomorphism by theorem 2.2. R is non-nil noetherian. NON-NIL NOETHERIAN R[X] WITH NON-NIL NOETHERIAN RING R Converse: Let R be non-nil noetherian since nil(R) prime ideal by lemma 3.1. nil(R[x]) prime ideal. Thus In this section we will show that if nil(R) divided nil(R[x]) divided prime ideal by lemma 3.1. and prime ideal of R and "fÎR[x]\nil(R[x])nil(R[x]Í)(f) theorems 3.3. and 3.4. noetherian ring and so and R non-nil noetherian ring then R[x] is non-nil noetherian. R[x] is non-nil noetherian. 1664 Middle-East J. Sci. Res., 15 (12): 1663-1665, 2013

REFERENCES 6. Dobbs, D.E., 1976. Divided rings and going-down. Pacific J. Math., 67: 353-363. 1. Atiyah, M.F. and I.G. Macdonald, Introduction to 7. Huckaba, J., 1998. Commutative Rings with zero commutative. divisors. New York, Basel: Marcel Dekker. 2. Van Derhall, A. 1969. Algebra reading, mass.: 8. Kaplansky, I. 1974. Commutative Rings. Chicago: Addison-Wesley Publishing Company, Inc. The University of Chicago Press. 3. Badawi, A. 1999b. On divided commutative rings. 9. Princeton, N.J. and D. Van Nostrand, 1958. Comm. Algebra, 27: 1465-1474. Company, Inc. 4. Badawi, A. 2003. On NON nil-Noetherian Rings. 10. Sharp, R.Y., 2000. Steps in commutative Algebra. Comm. Algebra, 31: 1669-1677. Cambridge University Press. 5. Divinsky, N.J., 1965. Rings and radicals. Toronto: 11. Zariski, O. and P. Samuel, 2002. Commutative University of Toronto Press. Algebra, Vol: 1-2.

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