Studies in Rings generalised Unique Factorisation Rings

-40-
P.O C = ~ for i = 1 , 2 , ... , n·. Then P.Ts are prime
1 ).
ideals of T for i=1,2, ... ,ne Let J be an ideal of T
such that P.T < J for each i = 1,2, ... ,n. Then
~
p. T n R < J n R = I for 1 ~ i ~ n ,
1
be the .mi.ni rna I primes over I.
Then it is obvious
that P. f P.! for i = 1,2, ... ,11, j = 1,2, •.. ,m and
). J
thus each P.' contains elements of Co Therefore the
J
product Pl'P2' ...P m'
also contains elements of C. But
Pl'P2' •.•P m
' S I, consequently I contains an element C,
i.e., I contains a unit of T. Also 'vve have IT = (J n R)T S J.
Hence J contains a unit of T. Thus J = T and we proved
that PIT, P 2T,
... ,PnT are maximal ideals of T.
Further, if M is any maximal ideal, then M = P.T
1
for some i = 1,2, •.• ,n. For, if M ~ P.T for all
1
i = 1,2, •.. ,n .
i
= 1,2, •.. ,n.
.Then Mn R is not contained in ·P. for any
1
Thus, as above,
it can be seen that
(~,nR)nc 1= y1, which implies that M contains a unit of T,
contradicting the maximality of M.This completes the proof.
In an NUFR,
the minimal prime ideal not containing
a normal Q(R)-invertible ideal is 0, and so, OT = 0 is
a ma x i ma 1 i ciea 1 0 f ToTh U 5 we 0 b t a in,