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Is the kernel of a map from a $\mathbb{Z}$-free module to an abelian group $\mathbb{Z}$-free? It seems like the answer should be yes as the kernel is contained inside the free $\mathbb{Z}$-module but I can't quite put my finger on the precise reason why.

Rhoswyn
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    Submodules of free modules over a pid (like $\mathbb{Z}$) are free, so yes. – Randall Jan 23 '19 at 16:53
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    For example, see here: https://math.stackexchange.com/questions/162945/submodule-of-free-module-over-a-p-i-d-is-free-even-when-the-module-is-not-finit – Randall Jan 23 '19 at 16:54

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