Let $I$ be an ideal of ring $R$ . If $K$ is ideal in $R/I$. Show that $K=J/I$ for some ideal $J$ which $I\subset J$ .
I was thinking $J=a+(-I)$ which $a\in K$ but I stuck to prove that $J/I\subset K$
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Let $\phi(a)= a+I$ be the natural map $R \to R/I$. Then for any ideal $J \subset R/I$ we have an ideal $\phi^{-1}(J) \subset R$ containing $I$ – reuns May 12 '17 at 16:23
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$J=K-I$? Are you talking about the set difference? $K$ is a set of cosets, not a set of elements of $R$, so that interpretation doesn't make any sense. – rschwieb May 12 '17 at 16:24
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Look up the fourth (or Lattice) isomorphism theorem for rings. – Soby May 12 '17 at 16:27
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@user1952009 then?please help me – lio May 12 '17 at 16:33
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I gave you the solution : for some ideal $J $ of $R/I$ look at $\phi^{-1}(J) = { a \in R \ \vert \ \phi(a) \in J}$. Show it is an ideal of $R$, obviously $J=\phi(\phi^{-1}(J)) = \phi^{-1}(J)/I $ – reuns May 12 '17 at 16:44