Vtyjkyuk

Recently I've been reading this paper, and in the first section on (differential calculus on associative algebras) they reference a theorem from Bourbaki's Algebra I Chapter 3. I have a problem with the proof of this theorem.

We consider a commutative ring $R$, and an associative $R$-algebra $A$.

Definition. If $Gamma$ is an $A$-bimodule, then we say a homomorphism $g:Arightarrow Gamma$ is a derivation of $A$ into $Gamma$ if $g(ab)=g(a)b+ag(b)$.

Definition. A first order calculus over $A$ is a pair $(B,d)$ where $d$ is a derivation of $A$ into $B$, and $B$ is the minimal left-$A$-module containing $operatornameimd$.

Then, we define an $A$-bimodule homomorphism $$
beginaligned
mu:Aotimes_R A&rightarrow A\
sum_iin Ia_iotimes b_i&mapsto sum_iin Ia_i b_i. \
endaligned
$$
We also define an $R$-linear map $$
beginaligned
d:A&longrightarrow operatornamekermusubset Aotimes_R A\
a&longmapsto 1otimes a - aotimes 1
endaligned
$$and it can be shown that $d$ is a derivation of $A$ into $operatornamekermu$ (we will call this kernel $I$ from now on).

Then the universal first order calculus over $A$ is the pair $(I,d)$.

What I would like to prove is this:

Theorem. For every first order differential calculus $(Gamma,g)$, there exists a unique epimorphism $p:Irightarrow Gamma$ such that $pcirc d=g$.

The paper references Bourbaki for this theorem, but I don't really follow the proof. The first thing I saw in Bourbaki was the claim "for any $A$-bimodule homomorphism $f:Irightarrow Gamma$, $fcirc d$ is a derivation of $A$ into $Gamma$" - which I don't understand at all. We would have that $$
beginaligned
fcirc d(ab)&=f(d(ab))\
&=f(d(a)b+ad(b))\
&=f(d(a)b)+f(ad(b))\
&=fcirc d(a)f(b)+f(a)fcirc d(b)
endaligned
$$but I don't see the logic behind this being a derivation unless we define the $A$ action using $f$.

Any help would be appreciated, thank you!

The last line of your calculation is incorrect: $f$ is not a map of algebras, it is a map of bimodules. This means that
$$
f(ax)=af(x);;;mboxand;;;f(xa)=f(x)a
$$
for all $xin I$ and $ain A$. Therefore, the last line in your calculation should be
$$
(fcirc d)(ab)=(fcirc d)(a)b+a(fcirc d)(b).
$$