Suppose that $R$ is a ring, $I$ and $J$ are ideals, and $R=I+J$. Can we conclude that $R/J$ is free?
If $R=I+J$, can we conclude that $R/J$ is free?
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abstract-algebra
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4Put $I=R$, so your question is if every quotient $R/J$ is free, which is absurd. If $R$ is an integral domain, this is not torsion free unless $J=0$. – 2010-12-09