6
$\begingroup$

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.

Thanks in advance.

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$ – Pedro Tamaroff Feb 26 '18 at 19:21
1
$\begingroup$

$\require{begingroup}\begingroup\DeclareMathOperator{\Spec}{Spec}$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?

Well, what does it mean for a map to be closed? It means you have a closed subscheme. By Proposition II.5.9 we know that closed subschemes correspond to ideal sheafs. In particular, for affine ring homomorphisms, a closed subscheme corresponds to a surjective map and the corresponding ideal sheaf is the kernel of this map.

So there you go. The diagonal inside of $V \times_U V$ is the closed subscheme determined by the kernel of the map $b \otimes b' \mapsto bb'$. $\endgroup$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.