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.



commutative algebra, diagonal morphism
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 $s-\nabla(s)\otimes 1=s\in I’$.


