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If R is a principal ideal domain, and b is a non-zero element in R, how can it be shown that there are only finitely many ideals in R/(b)? (i.e. R modulo the ideal generated by b)

Note: It is not established that R is commutative, so I'm not assuming that.

I think we should be able to say that since EVERY ideal in R is principal, then the number of ideals is less than or equal to the characteristic of R/(b) for b prime, i.e. less than or equal to b, and hence finite.

What about the case where b is not prime? Every Principal Ideal Domain is a Unique Factorization Domain, so we can express b (if it is not a unit) as a finite product of irreducibles (and hence primes since R is a P.I.D.).

Let b=p_1*p_2*...*p_n, where the p_i are not necessarily distinct.

Then can't we say by the Chinese Remainder Theorem that R/(b) is isomorphic to the direct product R/(p_1) x R/(p_2) x ... x R/(p_n)?

Since the number of ideals in each component R/(p_i) of the direct product is less than or equal to i, i=1,2,...,n, can we say that the number of ideals for R/(b) is less than or equal to min{p_1,...,p_n}?

Everything you're saying is correct. The most important fact here is, really, that if you have an ideal in R/(b), it must also be principle, meaning there should be some (a) such that I = (R/(b))/(a). So a must, among other things, be a divisor of b.

The remainder theorem makes this more explicit.

The number of ideals, though, is not min{p1...pn}. There are at least n ideals, one for each pi. You'd have to know more about R (or do something more complicated) to count the exact number.
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thanks for the verification and added insight, that's exactly what I needed.

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#### Clyde Oliver

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