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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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
2 Answers
$\mathbb{Z} = 2\mathbb{Z} + 3\mathbb{Z}$. I am sure you can now answer your question yourself.
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0of course it's a silly question. But I/(I | J)=J/(I | J) implies I | J = IJ, where | means intersection? – 2010-12-09
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2What do you mean by the equality sign? Do you mean isomorphic or are you talking about equality of sets inside $R/I\cap J$? – 2010-12-09
You can almost never conclude that $R/J$ is free.
If $R$ is a nonzero commutative ring and $J$ is an ideal of $R$, then $R/J$ is a free $R$-module iff $J = 0$. Indeed, if $J \neq 0$, let $x$ be a nonzero element of $J$. Then for all $y \in R/J$, $xy = 0$. On the other hand, consider $\bigoplus_{i \in I} R$ for some nonempty index set $I$: taking $y$ to be an element of the form $(1,0,\ldots,0)$ -- i.e., which has one coordinate equal to $1$ and all the other coordinates equal to $0$ -- we find that $xy = 0$ implies $x = 0$, contradiction.
In other words, if $J \neq 0$, then $R/J$ has a nonzero annihilator, which is not the case for any nontrivial free module.