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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.

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

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  • $\begingroup$ thanks for the answer, but what if replace prime by principal? $\endgroup$ – A.Messab Feb 18 '16 at 10:40
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I do not know why you want to replace 'prime' by 'principal', since these properties do not really relate, but here is an example:

$R=k[x,y], K=(x), I=(x,y)$. $I$ is not principal but $I'=I/K=(y)$ is principal in $R'=R/K=k[y]$.

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