When $a_1\otimes b_1=a_2\otimes b_2$

Question

Background Information

For the purposes of this question we shall let $R$ be a commutative ring and let $A$ and $B$ be $R$ modules (just treat them as bimodules). There exists an explicit construction of a tensor product $A\otimes B$ of $A$ and $B$, although strictly speaking tensor products are characterised by their universal property and this explicit construction is usually presented to prove the existence of tensor products.

In general, let $C$ be the $R$ module generated by all 2 tuples $(a,b)$ with $a\in A$ and $b\in B$. Then we define a submodule $D$ of $C$ to be the $R$ module generated by elements of the following forms:

(1) $(a_1+a_2,b) - (a_1,b)-(a_2,b)$

(2) $(a,b_1+b_2) - (a, b_1) - (a,b_2)$

(3) $r(a,b)-(ra,b)$

(4) $r(a,b)-(a,rb)$

with any $a,a_1,a_2\in A,b,b_1,b_2\in B$ and $r\in R$. The tensor product (up to isomorphism), $A\otimes B$, is given by the quotient $C/D$. (Please correct me if I am wrong about this construction as I did not refer to any sources while typing this post but rather I (hopefully) managed to recall this correctly.

The question

Is it true that $a_1\otimes b_1 = a_2\otimes b_2$ if and only if there exists collection of elements $\{k_j\}_{j\in J}, k_j\in C$ for a finite set $J$ where $k_j$ takes one of the above 4 forms such that when adding $\sum_{j\in J} r_j k_j$ to $(a_1,b_1)$ with $r_j\in R$ such that the terms in the summand annihilate each other till we ultimately yield $(a_2,b_2)$.

I am not trying to find a method to prove two tensors $a_1\otimes b_1, a_2\otimes b_2$ are equal for even if the above claim holds true the number of terms in the summand may be arbitrarily, albeit finitely, many. My question focuses more on the existence of such a finite set $\{k_j\}$, and not an algorithm to deduce the explicit elements of this set.

I apologise in advance if this question turns out to be silly as I learned this by myself and am not fully comfortable nor familiar with tensor product of modules yet. Thank you.

Answer 1

The answer to your question is yes (and you don't need the $r_j$'s).

First your construction is correct (see Wikipedia).

Now an element $x\in C/D$ is trivial if and only if $x\in D$, since $D$ is the subgroup generated by elements which takes one of the 4 options in your question (in fact the 3rd and 4th option could be written as $(ra,b) = (a,rb)$). This means that $x=y_1+y_2+...+y_k$ where each of the $y_i$ is one of the $4$ options in your question.

You don't need to multiply by $r_j$.

Answer 2

The question $a_1\otimes b_1 = a_2\otimes b_2$ is equivalent to $(a_2,b_2) \in (a_1,b_1) + D$ and $(a_1,b_2)$ can be transformed into $(a_2,b_2)$ by a finite number of relations given by $D$.