I have always wondered this statement is true.
Let $S$ be commutative algebra over $R$ and $M,N$ be modules over $S$. Then, $M\otimes_R N\simeq M\otimes_S N$ as $S$-modules. (Or can be as $R$-modules.)
Why I think it is true: A map between two modules can be induced by universal property of tensor product; $m\otimes n\mapsto m\otimes n$. Surjectivity is trivial and injectivity may be shown by the definition of tensor product. (Actually, it is a quotient module of a free module.)
So what I want to know is: is this known to be true? Or just this is my mistake.
Best Answer
This is false: the only result we have is the canonical homomorphism $\;M\otimes_R N\longrightarrow M\otimes_SN\; $ is surjective.
As a counter-example, consider $\mathbf C$ as a vector space on $\mathbf R$: then $\dim_\mathbf{R}(\mathbf C\otimes_\mathbf{R}\mathbf C) =4$, whereas $\dim_\mathbf{R}(\mathbf C\otimes_\mathbf{C}\mathbf C)=\dim_\mathbf{R}\mathbf C=2 $.