Well-defineness of scalar extension on $R$-module

Question

Let $R$ be a subring of $S$ containing $1$. $N$ an $R$-module. Then the construction of $S \otimes _R N$ (D&F p360) is by quotienting the free abelian group $(S \times N)^{ab}$ by the group $H$ with generating elements \begin{align*} & (s_1+s_2,n)-(s_1,n) -(s_2,n), \\ & (s,n_1+n_2) - (s,n_1) - (s,n_2), \\ & (sr,n) - s,rn). \quad \quad \quad \quad \quad \quad \quad (*) \end{align*} Let $s \otimes n$ be coset of $(s,n)$, then we can define left $S-$module by $$s ( \sum _{finite} s_i \otimes n_i ) = \sum_{finite} (ss_i) \otimes n_i . $$

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To show the above left action is well defined, Dummit&Foote argues we also quotient out \begin{align*} &(s'(s_1+s_2),n)-(s's_1,n) -(s's_2,n), \\ &(s's,n_1+n_2) - (s's,n_1) - (s's,n_2), \\ & (s'(sr),n) - s's,rn). \quad \quad \quad \quad \quad \quad \quad (**) \end{align*} so

$\sum s_i \otimes n_i - \sum s_i' \otimes n_i' = \bar{0} \Rightarrow \sum ss_i \otimes n_i - \sum ss_i' \otimes n_i' = \bar{0}.$

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I don't follow the boxed line. Yes, the LHS can be written as sums of elements of $H$, then we can replace those with $s$ within - but I couldn't write this out rigorously.

We have $LHS = \sum n_i h_i \in H$, $n_i \in \mathbb{Z}$, $h_i$ are elements of form $(*)$. Hence $\sum n_i h'_i \in H$ where $h'_i$ are elements of form $(**)$. Then why can we deduce $\sum n_i h'_i = RHS$ ? Here $LHS$ and $RHS$ denote left and right side of implication.

Answer 1

I will do this for two summands, it's the same for more, but it can get cumbersome to write out.

So, we want to show that the definition:

$s\cdot(s_1\otimes n_1 + s_2\otimes n_2) = (ss_1)\otimes n_1 + (ss_2)\otimes n_2$ is well-defined, that is, if:

$[(s_1,n_1)+(s_2,n_2)] + H = [(s_1',n_1') + (s_2',n_2')] + H$, then:

$[(ss_1,n_1) + (ss_2,n_2)] + H = [(ss_1',n_1') + (ss_2',n_2')] + H$, as well.

By assumption, we are given that:

$(s_1,n_1) + (s_2,n_2) - (s_1',n_1') - (s_2',n_2') \in H$.

Now, let's look a bit closer at what happens to generators of $H$, when you stick an arbitrary $s \in S$ in the "first coordinate":

If we take:

$(s_1+s_2,n_1) - (s_1,n_1) - (s_2,n_2) \in H$, we have:

$(s(s_1+s_2),n_1) - (ss_1,n_1) - (ss_2,n_1) = (ss_1 + ss_2,n_1) - (ss_1,n_1) - (ss_2,n_1)$

which is also a generator of $H$ (with $ss_1$ and $ss_2$ taking the place of $s_1$ and $s_2$, respectively).

Similarly, with $(s_1,n_1+n_2) - (s_1,n_1) - (s_1,n_2)$, if we replace $s_1$ with $ss_1$, we get the generator (which is, of course, also in $H$):

$(ss_1,n_1+n_2) - (ss_1,n_1) - (ss_1,n_2)$.

Finally, it is easy to see that for any $r \in R$:

$(s(s_1r),n_1) - (ss_1,rn_1) = ((ss_1)r,n_1) - (ss_1,rn_1)$, which is a generator of $H$ of the third type in your $(\ast)$.

Therefore, if a sum:

$\sum\limits_j (s_j,n_j) \in H$ (being some finite sum of generators), it follows that:

$\sum\limits_j (ss_j,n_j) \in H$ (for any $s \in S$), as well (being the same finite sum of generators with an $s$ left-multiplied to the first entry of each, which as we have seen above is also a sum of generators, so in $H$ by closure).

But this means, that:

$(s_1,n_1) + (s_2,n_2) - (s_1',n_1') - (s_2',n_2') \in H$ then implies:

$(ss_1,n_1) + (ss_2,n_2) - (ss_1',n_1') - (ss_2',n_2') \in H$ also, so that the two cosets:

$[(ss_1,n_1) + (ss_2,n_2)] + H$ and $[(ss_1',n_1') + (ss_2',n_2')] + H$ are equal and the (left) $S$-action is well-defined on $S\otimes_R N \stackrel{\text{def}}{=} (S \times N)^{ab}/H$.