Does $S\subset R$ imply $A\otimes_{R} A \subset A\otimes_{S} A$? I'm currently trying to prove Proposition 3.2 from the paper "Galois correspondence for Hopf Bigalois Extensions" by Peter Schauenburg, focusing on the last two implications it declares as "clear". The details of the proof are irrelevant, I just assumed something that now I'm not sure if holds:

Let $A$ be an $R$-module, and let $S\subset R$ ($A$ is also an
$S$-module). Then $A\otimes_RA\subset A\otimes_S A$.

Does that "$\subset$" make sense? Or should it be "$A\otimes_RA$ is a quotient of $A\otimes_S A$"? If not, is there any relation I can find between two ordered sets $S\subset R$ and their corresponding $\otimes_S$, $\otimes_R$? Any help will be appreciated, thanks in advance.

Let $R$ be a ring, $M$ a right $R$-module, and $N$ a left $R$-module. The usual construction of the tensor product $M \otimes_R N$ is that we take the free abelian group on $M \times N$ and quotient by the minimum relations necessary for bilinearity: for all $m, m' \in M$, $r \in R$, and $n, n' \in N$,

*

*$(m, n + n') - (m, n) - (m, n') = 0$,

*$(m + m', n) - (m, n) - (m', n) = 0$,

*$(m \cdot r, n) - (m, r \cdot n) = 0$.

If $S \subseteq R$ is a subring, then $M \otimes_S N$ is constructed as the quotient of the same free abelian group by the same relations, except the last set of relations is smaller because we are only quantifying over all $r \in S$ rather than all $r \in R$. Thus, $M \otimes_R N$ is a quotient of $M \otimes_S N$. (You can also deduce this from the universal properties, but I think it's a little easier to see this from the construction.)
