Studies in Rings generalised Unique Factorisation Rings

Therefore, in both cases we have proved that P
contains a normal invertible ideal and so R[x] is a prime
GUFR.
Remark 2.34.
Let P be a prime ideal of R. Then PR[x] is a prime
ideal of R[x]. We write
E p
= [f E: R[XV(f+PR[X])(R[X]/PH[X]) = (R[X]/PR[X])(f+PR[X]~
Since R[xJ/PR[x]
is prime,
(f+PR[x]) (R[x]/PR[x]) = (R[x]/PR[xJ) (f+PR[xJ) implies
that f+PR[x] is regular in R[xJ/PR[x] and that
f £ CR[x] (PR[x]). Therefore E p
~ CR[x] (PR[x]) for each
prime ideal P of R.
Also we write
Cl = Lf E R[x] / fR[x] = R[X]f} •
If R["1is a GUFI1, clearly C ~ Cl.
Theorem 2035.
Let R ~e a GUFR and suppose E p ~ CR[x](O) n Cl,
for every minimal prime ideal P of R. Then R[x] is a GUFR.
Proof:
Sinc e f, i saGUFR., }{ has a n Arti n i an quotien t r i n9
and so R[x] has an Artinian quotient ring [23, theorem 306J.
We denote the quotient ring of R[x] by Q(R[xJ).