I am trying to understand and fill the gaps of Rotman's proof (in his homological algebra text). This approach is different from this post ("Post 1") and is more complete than this one ("Post 2").
I understand everthing about the proof except showing that multiplication is well defined.
1. Definition and preliminary notes about Rotman's text
Let $k$ be a commutative ring. $R$ is a k-algebra if $R$ is a $k$-module satisfying $$ a(rs) = (ar)s = r(as) $$ for all $a \in k$ and $r,s \in R$
Note that rings are assume to be unital and modules are assumed to have the property $1_R m = m$.
2. Result I'm trying to understand
3. My partial understanding
Instead of constructing the trilinear function suggested, it seems easier to me (as in the second linked post) to construct a bilinear function $f_m : R \times S^{op} \to M$ defined by $f_m(r,s) = rms$.
Post 1 suggests that we must assume that $am = ma$ with $m \in M$ and $a \in k$ when we regard M as a $k$-Module. But Rotman mentions this nowhere, and I didn't understand Post 1's comments about why this must be so.
Using this assumption, we clearly have $f_m$ bilinear for each $m$ so that $f_m$ extends to $\hat f_m : R \otimes S^{op} \to M$. We also have $f_{m + m'} = f_m + f_{m'}$, so we see that $(r \otimes x) m = rms$ is well defined.
4. Questions
As I mentioned above, I want a clearly understanding of why $am = ma$ for $a \in k$ and $m \in M$. Does this follow from Rotman's definition of a $k$-algebra, or is this an assumption he left out?
Rotman tries to show multiplication is well-defined by constructed a k-trilinear map. I assume what he's doing is using the universal property to show that $\phi : R \otimes S^{op} \otimes M \to M$ is well-defined. Then the desired multiplication function is
$$ (R \otimes S^{op}) \times M \xrightarrow{h} R \otimes S^{op} \otimes M \xrightarrow{\phi} M $$ where $h$ is the tensor product. Is this right?
