Studies in Rings generalised Unique Factorisation Rings

-53-
Recall that a
right (left) regular element in a
ring R is any element x such that xy = 0 implies y = 0
(yx = 0 implies y = 0).
Lemma 2.27.
Let R be a Noetherian ring in which every left
regular element is regular.
Suppose aR = Rb for some
a € Rand b regular in R. Then aR = Ra and a is regular.
Proof:
Since aR = Rb, we have a = ub and b = av, for some
u and v in h. Thus bv = av 2 E aR = Rb and
2
b = av = ubv = u(bv) = u(av ) E u(aR) = uRb, therefore
'th e r e exists an e Lcrne n t p Ln R such that b = upb,
i.e. (l-up)b = O.
By regularity of b, up = 1, so that
u is left regular and from hypothesis u is regular. Now
using the regularity of u and the equation up = 1, it
can be seen that up = pu = 1.
Consequently
aR = Rb = Rpub = (Rp)ub = Rub
= Ra
Now the regularity of a follows from the regularity
of b
and the fact that Ra = Rb.