If $0\rightarrow K\rightarrow S$ is an injective ring homomorphism of commutative rings and if $M$ is an $S$-Module am I right that $M\otimes _K S\cong M$?
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What is $k$? Do you mean to say $\otimes_K$? – Arthur Mar 05 '20 at 09:30
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I mean tensor product over k, since both M and S are K-module because of the ring homomorphism. – 4780 Mar 05 '20 at 09:37
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Even I can suppose that S is flat as an K-module. – 4780 Mar 05 '20 at 09:39
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1You seem to use $k$ and $K$ as though they are the same. They are not. – Arthur Mar 05 '20 at 09:50
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No, this is false, even in very nice cases, for instance $M=S$, $K$ and $S$ are fields, and $S/K$ is finite (even Galois).
As an example, $\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C}\simeq \mathbb{C}^2$.
Captain Lama
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My problem in very particular case is this: If M is an K[x]-module(K is a field), I still cannot say that $ M\otimes_K{K[X]}$ is isomorphic to M. because in this case we know that M->M[x] is injective. – 4780 Mar 05 '20 at 10:53