Kernel of the product homomorphism $A\otimes A\rightarrow A$ Let $R$ be a commutative ring with $1$ and let $A$ be a commutative $R$-algebra. We view $A\otimes A$ also as $R$-algebra, the multiplication on generators being given by $(a\otimes b)(a'\otimes b')= aa'\otimes bb'$.

Denote by $m: A\otimes A\rightarrow A$ the product homomorphism, i.e. the map that maps a generator $a\otimes b$ to $ab$.

My question is: 


  Why is the kernel of $m$ generated by the elements $a\otimes 1 - 1\otimes a$ where $a$ runs through $A$?


The answer seems to be so easy that I could not find it anywhere in books. Alas, it is not clear to me :(

Thank you for your help!
Hint: Show that $a\otimes b - ab\otimes 1$ is in $I$, the ideal generated by elements of the form $x\otimes 1 - 1\otimes x$. This can easily be done by noting that $a\otimes b = (a\otimes 1)(1\otimes b)$, then substituting $1\otimes b = b\otimes 1 - (b\otimes 1 - 1\otimes b)$.

From here, show that every element of $A\otimes A/I$ can be written as $x\otimes 1 + I$.

Since $I$ is a sub-ideal of the kernel of your morphism, your morphism factors:

$$A\otimes A \to A\otimes A/I \to A$$

If $I$ is not the kernel, then $A\otimes A/I \to A$ is not $1-1$. But every element of $A\otimes A/I$ can be written as $x\otimes 1 + I$ and it must be sent to $x$. So this is necessarily $1-1$.
Easy: for $a\otimes b$ in the kernel, write $a\otimes b = (a\otimes 1-1\otimes a)(1\otimes b)$ and be done 

=)