Just a clarification about a notation used in a question (does $I^n$ mean $I\otimes\cdots\otimes I$?)

Question

In the following question (Direct Image by a Blow up) I have a question about the notation being used.

In this question, it is shown that $$\pi_{*}(\mathcal{O}_{\widetilde{X}}(-nE))= I_{Y/X}^{n}$$

where $\pi : \widetilde{X} \longrightarrow X$ is the blowing up of $X$ along a smooth subvariety $Y \subset X$ with exceptional divisor $E$.

Does the $I^{n}$ notation signify $\underbrace{I \otimes I \otimes \cdots \otimes I}_{n\ \textrm{times}}$ ?

Thank you very much.

score 2 · Answer 1 · answered Feb 03 '20 at 03:34

2

No, it means the product of ideal sheaves, $I\cdot I \cdot \ldots \cdot I$. This is not in general the same as the version with tensor products.

answered Feb 03 '20 at 03:34

KReiser

Ok. thank you very much, @KReiser. – Zlattan R.S Feb 03 '20 at 17:59
In what case is the direct image of a tensor product a tensor product? Again, Thank you very much. – Zlattan R.S Feb 03 '20 at 18:10
This is a large departure from your current question and an interesting enough problem to be asked on it's own, as you've now done here. – KReiser Feb 03 '20 at 19:09

1 Answers1