Given that $R$ is a commutative ring with identity $1$. In the definition of a tensor product of two $R$-modules given below, what is $Y(S)$? Is $Y(S)$ the submodule of $Y$ consisting of all formal linear combinations of $(v,w)$ which can be expressed in the form of 1,2,3 or 4? If it is, then is it true that $(v_2,w_2) \in [(v_1,w_1)]$ iff $(v_1, w_1) - (v_2,w_2) \in Y(S)$ iff $(v_1, w_1)-(v_2,w_2)$ can be written in the form of 1,2,3 or 4?
Thanks.
As Slade said in their comment, $Y(S)$ is the submodule of $Y$ generated by expressions of the form $(rv,w) - r(v,w)$, $(v,rw) - r(v,w)$ (that's a typo in the book), $(v_1 + v_2, w) - (v_1,w) - (v_2, w)$, and $(v, w_1 + w_2) - (v,w_1) - (v,w_2)$. In general, $Y(S)$ will be bigger than $S$; there's no reason that an expression like $(rv,w) - r(v,w) + (v,rw) - r(v,w)$ should be precisely equal to an expression of one of those forms, but it is the sum of two such forms, so if you want a module containing all the forms, you must allow sums and multiples (i.e., $R$-linear combinations) of expressions of those forms. $(v_1,w_1)$ and $(v_2, w_2)$ are in the same equivalence class if and only if $(v_1,w_1) - (v_2,w_2)\in Y(S)$ (by definition), but as above, $(v_1,w_1) - (v_2,w_2)$ need not be exactly equal to an expression of one of the four forms listed.