Every quotient of a PID by a prime ideal is again a PID

Prove that a quotient of a principal ideal domain by a prime ideal is again a principal ideal domain.

Let R be a principal ideal domain and let P \subseteq R be a prime ideal. Note that R/P is an integral domain.

Now by the lattice isomorphism theorem for rings, every ideal of R/P has the form I/P, where I is an ideal of R containing P. Since R is a principal ideal domain, I = (\alpha) for some \alpha \in R. Then I/P = (\alpha + P) is principal, and thus every ideal of R/P is principal.

Note that every ideal of a quotient of R is principal; we needed P to be prime so that R/P is also an integral domain.

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  • John McPerson  On July 21, 2011 at 9:52 am

    How about this. Assume I is non-zero. Since I is prime and R is a PID, I is maximal and so R/I is a field and hence a PID.

    • nbloomf  On July 21, 2011 at 10:56 am

      That works too.

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