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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.

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Alex
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1 Answers1

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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.

Stahl
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  • Is there a reason that in this definition of 'Tensor Products' and the description of '$R$ modules' we discussed previously, it is preferable to work with formal linear combinations (as in $L(M \times N)$ above) rather than usual linear combinations (where one can simplify)? – Alex Nov 15 '16 at 18:39
  • $M\times N$ has ``too many relations:" in particular, the $R$-module structure on $M\times N$ has $r(m,n) = (rm,rn)$, which we don't want in our tensor product. $L(M\times N)$ is the free $R$-module on $M\times N$, which by taking quotients, we can impose the relations we want (to form the tensor, instead of $M\times N$). For any $R$-module $M$, you can view $M$ as a quotient of $L(M)$ by the submodule generated by the relations defining the $R$-module structure on $M$: ($r[m] - [r\cdot m]$, $[m + m'] - [m] - [m']$). – Stahl Nov 15 '16 at 21:40
  • If you have a chance see this post, I think I will better understand the case of modules if I understand the vector space case first. – Alex Nov 23 '16 at 17:36
  • Why can't we impose the relations by taking quotients on $V \times W$ directly, i.e. the vector space $(V \times W)/\mathcal{R} $, instead of imposing the relations on the free vector space $Free(V \times W)$ (space of formal linear combinations of $V \times W$) by taking quotients $(Free(V \times W)) / \mathcal{R}$? – Alex Nov 23 '16 at 22:33