Confused about Tensor Product of R-modules

Question

In Tu's book on Differential Geometry he first defines $Free(V\times W)$ as:

$$\sum r_i(v_i, w_i), r_i \in R, (v_i, w_i) \in V \times W$$ where the sum is finite.

The way I understand it is the above construction is of formal combinations and forgets the actual structure of the modules. In other words, if $v_1+v_2 = v_3$, it isn't true that in $Free(V\times W)$ that $(v_1, 0) + (v_2, 0) = (v_3, 0)$

Now to form the tensor product we quotient by the submodule, $S$ spanned by elements of the form: $$ (v_1 + v_2, w) - (v_1, w) - (v_2, w)\\ (v,w_1 + w_2) - (v, w_1) - (v, w_2)\\ (rv,w) - r(v,w)\\ (v, rw) - r(v,w)$$ Then we have a map from the product to the tensor product, $$(v,w) \rightarrow v\otimes w$$

However, if $v_3 = v_1 + v_2$, then I can't show that $$v_3\otimes w = v_1 \otimes w + v_2 \otimes w$$ which should be the case if $\otimes$ is a module homomorphism bilinear map.

Answer 1

Since $V \otimes W := \operatorname{Free}(V \times W)/S$ and $\otimes : V \times W \to V \otimes W$ is defined by $$(v,w) \mapsto v \otimes w := (v,w)+S,$$ the condition $(v_1 + v_2, w) - ((v_1, w) + (v_2, w)) \in S$ tells us that $$(v_1 + v_2, w)+S = ((v_1, w) + (v_2, w))+S = ((v_1,w)+S) + ((v_2,w)+S)$$ which is the same as $(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w$. Also observe that the other relations which define $S$ gives us \begin{align} v \otimes (w_1+w_2) &= v \otimes w_1 + v \otimes w_2, \\ (rv) \otimes w &= r(v \otimes w), \\ v \otimes (rw) &= r(v \otimes w).\end{align}

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Recall that if $M$ is an $R$-module and $S$ is a submodule of $M$, the quotient $M/S$ is defined by $M/\!\sim$, where $$(\forall u_1,u_2 \in U) \quad u_1 \sim u_2 \iff u_1-u_2 \in S.$$ In this case, the equivalence class of $m \in M$ is given by $m+S := \{m+s : s \in S\}$ (hence $m+S = m'+S \iff m-m' \in S$), and we define an $R$-module structure in $M/S$ by $$(\forall m,m' \in M)(\forall r \in R) \quad r(m+S)+(m'+S) := (rm+m')+S.$$

Answer 2

So for posterity, I want to write up an answer for others who may have the same confusion. As @KCd clarified, elements of $Free(V\times W)$ are of the form,

$$\sum r_i(v_i, w_i)$$

However if we write a particular element of $Free(V\times W)$ as $r_1(v_1 + v_2, w_1)$ and $v_3 = v_1 + v_2$ then $$r_1(v_1 + v_2, w_1) = r_1(v_3, w_1)$$ In other words, inside of our parentheses in our notation we are not taking formal sums, but instead combining elements of the module as we normally would.