Studies in Rings generalised Unique Factorisation Rings

More documents

Recommendations

Info

-37- entries 'at is in M 2 ( Q) = N. Put I = X M 2 ( R) = M 2 ( R) X, -1 (' -1 then I .s N. Furthe rm0 re X € M 2 RpI, since Xis the scalar matrix/with non zero entries a~l which is -1 -1 () () -1 in Rp and thus I = X M 2 R = M 2 R X is contained () -1 -1 ( ) ( ) in M 2 Rp and 11 = I I = M 2 R. Therefore M 2 R is a GUFR which is not prime. Remark 2 05. (1) The principal ideal theorem for a right Noetherian ring asserts that the minimal prime ideals over any normal ideal has height atmost 10 Thus in a GUFR even though every non-mi~imal prime ideal contains normal ideals, each normal ideal is contained in either a minimal prime ideal or in a prime ideal of height 1. (2) If R is Noetherian ring satisfying descending chain condition on prime ideals, then R is a GUFR with the over ring 5 if and only if every height 1 prime ideal of R contains an S-invertible principal ideal. (3) By Proposition 1 0 16 , if R is a GUFR with over ring S, then every prime ideal contains a normal S-invertible ideal if and only if every minimal prime ideal contains a normal S-invertib1e ideal.
-38- Let R be a GUFR with over-ring S. We shall make use of a certain p a r t i a I quo t i en t ring of R. Let C =[a€ R/aR = Ra is S-invertjble}. We prove C consists of regular elements and C is a (right and left) Ore set. QUOTIENT RINGS Theorem 2 0 6 . Let R be a GUFR with the over-ring S. Then C contains only regulor elements and C is an Ore set. Proof Let a E: C, we provelR( a) = r R( a) = o. Sinc e a f e, aR = Ra is S-invertible and so there exists an R-subbimodule 1- 1 of 5 such that (aR)1- 1 = 1- 1(aR) = R. Thus we can find elements r i c R, si ( 1- l f 5 for i=1,2, •.. ,n, such that which implies n n L (ar.)s. = 1. iAe 1= · 1 1 1 n a l: r.s.=l · 1 l 1. fS(a E r.s.) = lS(l)=O and consequently . 11.1 1= n 1= lR (a) ~ r R (a) = o. ~ (a) < ls( a L r.s.) = I S ( l ) = o. Similarly - · 111 1= For the second part of the theorem, let a,b e C, then aR=Ra and bR = Rb. Now abR = a(Rb) = Rabo Since
Page 1 and 2: STUDIES IN RINGS GENERALISED UNIQUE
Page 3 and 4: \CKN0 1vVLEDGEWIENT Chapter 1 INTRO
Page 5 and 6: -2- REMIl'JDER OF Tt-iE COM~1UTAT l
Page 7 and 8: -4- non zero prime ideal of R conta
Page 9 and 10: -6- If both conditions hold, R is s
Page 11 and 12: -8- In non-commutative rings, it tu
Page 13 and 14: -10- The next two propositions guar
Page 15 and 16: -12- (a) I is a semiprime ideal (b)
Page 17 and 18: -14- Definition 1 029. An ideal P i
Page 19 and 20: -16- Proposition 1 033. 'equiva len
Page 21 and 22: -18- Proposition 1038. For a ring R
Page 23 and 24: -20- there exists d ' E:. 0 such th
Page 25 and 26: -22- expression. Then it can be int
Page 27 and 28: -24- Definition 1 0 4 9 . A regul~r
Page 29 and 30: -26- that the right ideal generated
Page 31 and 32: -28- Proposition 1.62. Let R be a r
Page 33 and 34: -30- It is found that the elements
Page 35 and 36: -32- The concept of a ring with few
Page 37 and 38: -34- universal enveloping algebra-
Page 39: -36- (3) We give an example of a GU
Page 43 and 44: -40- P.O C = ~ for i = 1 , 2 , ...
Page 45 and 46: -42- Noetherian order in an Artinia
Page 47 and 48: -44- Proof Since R is a Noetherian
Page 49 and 50: -46- In the proof of le~ma 2.17 we
Page 51 and 52: -48-- ( 1) CR(P) 5 CR(O) for each h
Page 53 and 54: -50- Remark 2.22. Prime GUFRs are t
Page 55 and 56: -52- Proof: As in [22, Proposition
Page 57 and 58: -54- Theorem 2.28. Let R be a Noeth
Page 59 and 60: -56- Lemma 2 032. Let R be a prirne
Page 61 and 62: Therefore, in both cases we have pr
Page 63 and 64: -60- RI~JGS yVITli [::NOUGH INVER~r
Page 65 and 66: -62- Th e o r ern 2.43. Let R be a
Page 67 and 68: -64- Theoren12.46. Let R be a pri.m
Page 69 and 70: Chapter-3 EXTENSIONS AND RINGS WITH
Page 71 and 72: -68- If S is a finite normalising e
Page 73 and 74: -70- regular element of R is regula
Page 75 and 76: -72- Lemma 3.110 Let P be a prime i
Page 77 and 78: -74- Now, if x ( P, then xS = Sx is
Page 79 and 80: -76- so that Sg = gS $ P. Since S i
Page 81 and 82: -78- is an r' E R such that Equa t
Page 83 and 84: -80- As in [12, lemma 1 03]. Lemma
Page 85 and 86: -82- Examples 3023 (1) In [13J, a c
Page 87 and 88: -84- each j which is not possible.
Page 89 and 90: -86- Thus the ideal T(z+ux)T is not
Page 91 and 92:
-88- in the generating set of I suc
Page 93 and 94:
-90- Proof First we prove that R is
Page 95 and 96:
Chapter 4 INTRODUCTION LOCALISATION
Page 97 and 98:
-94- Now i f [~~] ~ C( Q) a nd [~~
Page 99 and 100:
-9()- Remark 4 05. It can be seen t
Page 101 and 102:
-98- Remark 4 012. Let P be a class
Page 103 and 104:
-100- Lemma 4 019. An invertible id
Page 105 and 106:
-io>- Theorem 4 023. If R is a Noet
Page 107 and 108:
Lemma 4.28. Suppose P and Q are max
Page 109 and 110:
-106- and P is primeo Cons equen t
Page 111 and 112:
-108- M annihilated by I, I~(lV~) =
Page 113 and 114:
-1 lC)- Defini.tion 4 038. Let R be
Page 115 and 116:
Chapter-5 REMARKS In this concludin
Page 117 and 118:
-114- -1 -1 and so a I = R and a a
Page 119 and 120:
REFERENCES [1] AoW. Chatters (1984)
Page 121 and 122:
-118- [17J B.J. Muller Localisation
show all

Studies in Rings generalised Unique Factorisation Rings

Create successful ePaper yourself

Delete template?

Save as template?