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 $J\subset I$?
Factor Rings in Commutative Rings
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abstract-algebra
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3Technically $R/I$ is never a subset of $R/J$ unless $I=J$. – 2012-08-10
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0To 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. – 2012-08-10
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2No, $R/I$ is **isomorphic** to the real field, and $R/J$ is **isomorphic** to the complex field, but $R/I\nsubseteq R/J$. – 2012-08-10
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0Brian, you're right.I have edited the question. – 2012-08-10
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1Even with your changes, it still is problematic. You are choosing an isomorphism... – 2012-08-10
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0If by $S_1 \subset S_2$ you mean $S_1$ embeds in $S_2$ then this is easily seen to be false. Consider $R=k[x,y]$, $I=(x)$ and $J=(y)$. – 2012-08-10
1 Answers
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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.