Studies in Rings generalised Unique Factorisation Rings

-86-
Thus
the ideal T(z+ux)T is not contained in any
maximal ideal of T and hence (z+ux) is weakly I-invertible.
Theorem 3 0 30 .
Let R be a Noetherian ring with many normal elements
and I be a one sided ideal of R containing a normal element.
Then I can be generated by a set of weakly T-invertible
elements.
Proof:
Suppose I is a left ideal. Let (Zl'Z2 p • • ,zJ be
ageneratin9 set 0 f I a nd x be anar ma 1 e 1 erne ntin I.
Consider zl' by
theorem 3.29, there exists an element
U 1
,
such tha .~
zl+u1'x is weakly T-invertibleo Since
I -1
u = u1c for "i € R 1
some anti Cl € e, we have
.l
-1
u 1 = u1l =
not containing U
1
u1'c l
does not belong to the maxima 1 ideals
I
and u 1
belongs to all maximal ideals,
which contains u 1
' . Thus as in the proof of theorem 3 0 2 9 ,
Zl + u1x is a weakly I-invertible element in R. Similarly
we get a collection (Zi+Uix]
of weakly T-invertible
elements for each zi. Since x is normal, x £ C and so
x is invertible in T. Thus [zi+uix,x} is a collection
of weakly T-invertible elements in 10 We prove this is
generating set for 10