# Kernel of the diagonal homorphism $f:B\otimes_A B\rightarrow B$

I'm studying Hartshorne's Algebraic Geometry book, and in the remark 8.9.2 I understood everything, besides one detail that is bothering me.

He takes $U=SpecA\subset Y$, and $V=Spec B\subset X$, where $X$ and $Y$ are schemes, and a map $g:X\rightarrow Y$, such that $g(V)\subset U$. I know that $V\times_U V$ is isomorphic to $Spec(B\otimes_A B)$. But then he states that $\Delta(X)\cap (V\otimes_U V)$ is defined by the kernel of the diagonal morphism $f:B\otimes_A B\rightarrow B$, $f(b\otimes b')=bb'$, I can't see how to show this last part.

I know that the kernel of this map is $kerf=\{\sum a_{ij}b_i\otimes b_j|\sum a_{ij}b_ib_j=0 \}$, but I can't see what happens in the Spec.

• Try to forget about elements and consider categorical definitions of the diagonal, the kernel, and both the tensor and cartesian product. Then it should all be clear. – Pedro Tamaroff Feb 26 '18 at 19:21

$\require{begingroup}\begingroup\DeclareMathOperator{\Spec}{Spec}$Do you remember that in Proposition II.4.1 Hartshorne says that the diagonal map
$$\Delta : \Spec B \to \Spec B \times_{\Spec A} \Spec A = \Spec(B \otimes_A B)$$
is closed because the map $B \otimes_A B \to B$ defined by $b \otimes b' \mapsto bb'$ is surjective?
So there you go. The diagonal inside of $V \times_U V$ is the closed subscheme determined by the kernel of the map $b \otimes b' \mapsto bb'$. $\endgroup$