Let $M$ be an $R$-module, where $R$ is a PID. Assume there is a free $R$-module $F$ and a surjective map $\phi :F\rightarrow M$. Then why is $\ker(\phi)$ also free?
Thank you.
2026-03-27 23:57:28.1774655848
Quick question: SES where base ring is a PID
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Because any submodule of a free module over a PID is free.
You can learn about that here.