In (1.0.3), Grothendieck states that, given non-commutative rings $A$ and $B$, a homomorphism $\varphi : A \to B$, and a left ideal $\mathfrak{J}$ of $A$, the left ideal $B\mathfrak{J}$ of $B$ generated by $\varphi (\mathfrak{J})$ is also the image of the canonical homomorphism $B\otimes _A \mathfrak{J} \to B$ of left $B$-modules. My questions are:
1: How is he tensoring two left $A$-modules? By '$B$' in $B\otimes _A \mathfrak{J}$ does he mean $B$ given the structure of a right $A$-module via $b.a \triangleq b \varphi (a)$?
2: Then how is he giving this the structure of a left $B$-module?
3: What is the canonical map from $B\otimes _A \mathfrak{J}$ to $B$? I imagine this will be obvious once it is clear what kind of modules he is actually tensoring.
1. "Yes" to the second question. More precisely, he is regarding $B$ as a $\left(B,A\right)$-bimodule, where the left $B$-module structure is obvious (i.e., given by multiplication) and where the right $A$-module structure is the one you define.
"No" to the first question. He is tensoring a $\left(B,A\right)$-bimodule with a left $A$-module. This yields a left $B$-module. This also answers question 2.
3. This is the left $B$-linear map which sends every pure tensor $b \otimes_A j$ to $b\varphi\left(j\right)$.