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Let $R \subset S \subset Q$ be three rings and let $N$ be a module over $Q$. Assume also that $S$ is flat over $R$. Then is it true that

$S \otimes_R N \cong N$?

I know that $R \otimes_R N=N$.

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    Can you please consider changing the title for something more specific? – 2017-02-08
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    So I think that there is no easy expression for $S \otimes_R N$ in general. You still have $S \otimes_R R[X] \cong S[X]$ when $R \subset S$, for instance. – 2017-02-08

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Let $R=\Bbb Z \subset S = \Bbb Z^2 = Q = N$. Then $S$ is free over $R$, so it is flat, and $$S \otimes_R N \cong N^2 \qquad\text{as $R$-modules}$$ but $$N^2 \not \cong N \qquad\text{as $R$-modules}$$ since $N^2 = \Bbb Z^4$ is not even isomorphic to $N=\Bbb Z^2$ as abelian groups (e.g. consider tensoring by $\Bbb F_2$ over $\Bbb Z$, or quotient by $(2\Bbb Z)^n$ ($n = 2$ or $4$)).

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    So is $S \otimes_R N \cong SN$ as $S$-modules? By $SN$ I mean all the elemets of the form $\sum_i a_in_i$ where $a_i \in S$ and $n_i \in N$? – 2017-02-08
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    @MathStudent : no because $SN$ is just $N$ itself. Here the notation $\sum a_in_i$ means that $a_i \in S$ acts on $n_i \in N$ as $N$ is a $S$-module (since it is a $Q$-module), so $a_i n_i \in N$. – 2017-02-08