Ideal sheaves of diagonal maps

Question

Let $X\subset A$ be the inclusion of a scheme $X$ into an ambient space with ideal sheaf $J \subset \mathcal{O}_A$ and let $\Delta_A,\Delta_X$ be the images of the diagonal embeddings $A \rightarrow A\times A$ and $X\rightarrow X\times X$. I was wondering about an explicit description of the kernel (ideal sheaves) of the two maps $\mathcal{O}_{A\times A} \twoheadrightarrow \mathcal{O}_{X\times X}$ and $\mathcal{O}_{\Delta A}\twoheadrightarrow \mathcal{O}_{\Delta X}$ (I omitted the push-forwards from the notation).
My naive guess would have been that the kernel of the first map should be something like $J\otimes J$ and the second something like $J$ (as ${\Delta}A, \Delta X$ look like $A,X$) and I'd love some insight on that thanks.

Answer 1

First, applying the pushforward along $\Delta$ to the exact sequence $$ 0 \to J \to \mathcal{O}_A \to \mathcal{O}_X \to 0,\tag{*} $$ one obtains $$ 0 \to \Delta_*J \to \Delta_*\mathcal{O}_A \to \Delta_*\mathcal{O}_X \to 0, $$ because the pushforward for a closed embedding is exact. This identifies one of the kernels with $\Delta_*J$.

On the other hand, pulling back (*) along the projections $p_1,p_2 \colon A \times A \to A$, one obtains $$ 0 \to p_1^*J \to \mathcal{O}_{A \times A} \to \mathcal{O}_{X \times A} \to 0 $$ and $$ 0 \to p_2^*J \to \mathcal{O}_{A \times A} \to \mathcal{O}_{A \times X} \to 0, $$ respectively. Now, since $X \times X = (X \times A) \cap (A \times X)$, it follows that $$ J_{X \times X} = p_1^*J + p_2^*J. $$ Note, by the way, that this is definitely distinct from their tensor product, for instance because, when $X$ is a Cartier divisor in $A$, so that $J$ is a line bundle, the tensor product is also a line bundle, while the ideal of $X \times X$ is not a line bundle, because the codimension of $X \times X$ is $2$.