# An example of a prime quotien ideal where his corresponding ideal is not prime

If $R$ is a comutative ring with identity ring and $K$ is an ideal from it, let $R'=R/K$ and $I$ an ideal of $R$ satisfy $K\subseteq I$ and $I'$ is the coresponding ideal of $R'$ (we knew that correspondence theorem gives a certain one-to-one corespondence between the set of ideals of $R$ containing $K$ and the set of ideals of $R'$). can you give me some examples where $I'$ is prime then $I$ is not.

• There is no such example. In the correspondence between ideals in a ring and its quotient, prime ideals correspond to prime ideals. – Tobias Kildetoft Feb 18 '16 at 10:25

Note that we have by the third isomorphism theorem $$R/I \cong R/K\bigm/I'$$ hence $R/I$ is a domain iff $(R/K)/I'$ is, therefore $I$ is prime iff $I'$ is.
$R=k[x,y], K=(x), I=(x,y)$. $I$ is not principal but $I'=I/K=(y)$ is principal in $R'=R/K=k[y]$.