# 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?

• Maybe one could use that $I/I^2$ is Kähler differentials, it must be well known how the number of generators of $\Omega^1$ as a module relates to the number of generators of $B$ as an algebra over $A$... Jul 20, 2014 at 22:10
• Yes, the elements $db_i\in \Omega^1$ form a generating set. But how to go from $I/I^2$ to $I$?
– AYK
Jul 20, 2014 at 22:23
• is finite type really enough? One has a map from the sequence $0 \to I_A \to A[\mathbf{x}] \otimes_A A[\mathbf{x}] \to A[\mathbf{x}] \to 0$ to the sequence $0 \to I \to B \otimes_A B \to B \to 0$. Then you have an exact sequence $I_A \to I \to \mathrm{coker}(I_A \to I)$ and also by the snake lemma a surjection $\mathrm{ker}(A[\mathbf{x}] \to B) \twoheadrightarrow \mathrm{coker}(I_A \to I)$. It's easy to see $I_A$ is finitely generated, but the condition that $\mathrm{ker}(A[\mathbf{x}] \to B)$ is precisely asks $B$ to be finitely presented and not just finitely generated. Jul 21, 2014 at 1:42
• @VivekShende: But we don't need $ker(A[x]\rightarrow B)$ to be finitely generated; we just need a certain quotient of it to be finitely generated, no? Jul 21, 2014 at 1:50

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 $s-\nabla(s)\otimes 1=s\in I’$.
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.
• The elements $b_i$ generate $B$ as an algebra over $A$, not as a module. The solution given above (by Rosen) is correct. Anyway, thanks for the effort!
• Oh, I missed that it's generated as an algebra. Fixed it. Still, I think it should be pointed out that the basic property of $I$ (for any $B$) is $I=(b\otimes 1 - 1 \otimes b \mid b \in B)$. The proof is also implicitly in Julian's answer.