## 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.

• 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.