commutative algebra, diagonal morphism can anyone help me with the following statement (it is part of a bigger proof where it is not explained). 
Let $B$ be a finite type commutative $A$-algebra (where $A$ is a commutative ring), and consider the kernel $I$ of the diagonal homomorphism $B\otimes_A B\to B$ (defined by $b\otimes b'\mapsto bb'$). Then $I$ is a finitely generated ideal.
My guess is the following: if $b_1,\ldots,b_n$ generate $B$ over $A$ (as an algebra), then the elements $b_i\otimes 1 - 1\otimes b_i$ are the desired generators of $I$. Is there an easy way to see this?
Thanks in advance.
 A: Let $I'\subset B\otimes_A B$ be the ideal generated by the elements $b_i\otimes 1-1\otimes b_i$, and define
$$
R=\{b\in B:b\otimes 1-1\otimes b\in I'\}.
$$
It’s not hard to check that $R$ is an $A$-subalgebra of $B$, so that $R=B$ (because the generators $b_i$ are in $R$ by construction). Now $b\otimes 1-1\otimes b\in I'$ for all $b\in B$ implies $b\otimes b'-bb'\otimes 1 = (b\otimes 1)(1\otimes b'-b'\otimes  1)\in I'$, so that $s-\nabla(s)\otimes 1\in I'$ for all $s\in B\otimes_A B$, where $\nabla:B\otimes B\to B$ is the codiagonal (i.e., the linear map sending every $b \otimes b'$ to $bb'$). Finally, if $s\in\ker(\nabla)$, then $s-\nabla(s)\otimes 1=s\in I’$.
A: It's a (pretty easy) standard exercise in algebra to show $I=(b\otimes 1 - 1 \otimes b\mid b \in B)$. 
Let $a\in A, b,c \in B$. Then 
$$ab\otimes 1 - 1 \otimes ab = (a\otimes 1)(b\otimes 1 - 1\otimes b)$$
$$bc\otimes 1 - 1 \otimes bc = (b\otimes 1)(c\otimes 1 - 1 \otimes c) + (b \otimes 1 - 1 \otimes b)(1 \otimes c)$$
Hence if $B$ is generated as $A$-algebra by $b_1,\ldots, b_n$, then $I$ is generated as ideal in $B\otimes_AB$ by $b_i \otimes 1 - 1 \otimes b_i\,\,(i=1,\ldots,n)$ as desired.  
