Quick question: SES where base ring is a PID

Question

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.

No need to assume that there is such $\phi$ because we always have it. — user26857, Jun 04 '14 at 07:26

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

Because any submodule of a free module over a PID is free.

You can learn about that here.

edited Apr 13 '17 at 12:21

Community

1

answered Jun 04 '14 at 04:02

rschwieb

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Quick question: SES where base ring is a PID

1 Answers1