Let $A$ be a $k$-algebra, where $k$ is a commutative ring. A derivation $d:A \to M$, where $M$ is an $A$-bimodule is said to be universal, if for any derivation $d':A \to N$, $N$ an $A$-bimodule there exist a unique $A$-linear map $\phi: M \to N$ such that $\phi d= d'$.
Now, my exercise proposes to justify the existence of such $d$ and module $M$:
Consider $I$, kernel of multiplication $\mu : A \otimes A \to A$. What I can't see at the moment is the fact that $I$, as an $A$-module is generated by $1\otimes x - x\otimes 1$ for $x\in A$. Fact that $A$-span of elements of such type is in $I$ is trivial to verify, but why all elements of $I$ are combinations of elements of this type?
Further construction: If I prove this, then $I/I^2$ will be both left and right $A$ module and structures will be compatible (easy computation). $$d:A \to I/I^2 \\ x \mapsto [1\otimes x-x\otimes 1]$$ is the desired map, which is all rather easy to see, but I miss the detail above.
Any comment will be appreciated.
Let $J$ be the submodule of $A\otimes A$ generated by elements of the form $1\otimes x-x\otimes 1$. Consider an element of $A\otimes A$ of the form $a\otimes b$. This is $a(1\otimes b)$ in the $A$-module structure, and so modulo $J$ it is the same as $a(b\otimes 1)=ab\otimes 1=\mu(a\otimes b)\otimes 1$.
Since every element of $A\otimes A$ is a sum of elements of this form, we see that for any $t\in A\otimes A$, $t$ and $\mu(t)\otimes 1$ are equal mod $J$. In particular, if $t\in I$, then $\mu(t)\otimes 1=0$, so this says $t\in J$. Thus $I\subseteq J$, as desired.