Tensor product of free module Let $R$ be a commutative ring with unity and $S$ a free $R$-module of rank $n$.
"Consider the real vector space $V:=S\otimes \mathbb{R}$."
I am trying to understand this particular setting: I know what a communtative ring with unity is and also what a free $R$-module of rank $n$ is. But what does one mean with the tensor product of $S$ and $\mathbb{R}$? There are so many definitions of tensor products ... Here, I seem to have a tensor product of a free module and a field?
 A: This has not a definite answer, that's why most of the time they put something below the $\otimes$ to let you know how you should identify the two spaces. Let me give you two different examples on what I would thing is natural.
So for instance you can have $S=\mathbb Z^n$ which is a module over $R=\mathbb Z$. Then the natural choice for $S\otimes \mathbb R$ would be expressions $s\otimes r$ but you have that $R=\mathbb Z$ acts on both $S,\mathbb R$ so you would like to identify the two elements $(ks)\otimes r$ and $s\otimes (kr)$ for any $s\in S \text{ and } r \in \mathbb R$. This will give you that $S\otimes \mathbb R=\mathbb R^n$.
However if you have $S=\mathbb Z_k^n$ and $R=\mathbb Z_k$ for some $k$ then you don't want to make any identification whatsover, so you just have linear combinations of elements $s\otimes r$ without any identification. i.e. the tensor product of $S$ and $\mathbb R$ as sets.
Of course for the first case one could also not demand any sort of identification and just get $\mathbb Z^n\otimes \mathbb R$ be just linear combinations of elements $s\otimes r$ without any relation. Usually one "knows" what the natural thing to consider is, and if there is place to confusion they put a little reminder in the product. For instance for the "usual" case of identifying the two multiplications I would write $\mathbb Z^n\otimes_{\mathbb Z} \mathbb R$ and for the tensor product without any identification I would put $\mathbb Z^n\otimes_0 \mathbb R$
