Is a projective module $P$ over a ring $R$ which is a finite direct product of fields free?
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Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community May 05 '22 at 12:40
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2Did you try $\mathbb Z/2$ and $\mathbb Z/2\oplus\mathbb Z/2$? – Pedro May 05 '22 at 19:29
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No. For example, $K\times\{0\}$ is a projective module over the ring $K\times L$ (being a direct summand of the free module $K\times L\cong (K\times \{0\})\oplus (\{0\}\times L)$). But it is not free.
More generally, over $K\times L$, any module of the form $K^n\times L^m$, with $n,m\in\mathbb{N}$, is projective. Such a module is free if and only if $n=m$.
Alex Kruckman
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