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Let $R$ be a commutative ring, suppose that $I$ and $J$ are ideals of $R$. Suppose that $R/I\cong S_1$ and $R/J\cong S_2$. It is true that if $S_1\subset S_2$ if then $J\subset I$?

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Technically $R/I$ is never a subset of $R/J$ unless $I=J$. –  Brian M. Scott Aug 10 '12 at 16:23
    
To Brian: Considere $R$ the polinomial ring in one variable over the reals. Let $I$ be the ideal generated by $p(x)=x-1$ and $J$ the ideal generated by $q(x)=x^2+1$. Then $R/I$ is the real field and $R/J$ is the complex field. –  zacarias Aug 10 '12 at 16:39
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No, $R/I$ is isomorphic to the real field, and $R/J$ is isomorphic to the complex field, but $R/I\nsubseteq R/J$. –  Brian M. Scott Aug 10 '12 at 16:43
    
Brian, you're right.I have edited the question. –  zacarias Aug 10 '12 at 16:55
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Even with your changes, it still is problematic. You are choosing an isomorphism... –  M Turgeon Aug 10 '12 at 17:08

1 Answer 1

up vote 2 down vote accepted

Consider the following. Let $k$ be a field. For all $a\in k$, we have $$k[X]/(X-a)\cong k.$$ However, each ideal $(X-a)$ is maximal, and so none is contained in the other.

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