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Let $(X, O_X)$ be a ringed space, $I$ be a sheaf of ideals on $(X, O_X)$ and let $F$ be an $O_X$-module. I have two questions.

a) Is there an example where the presheaf $P : U \mapsto I(U) \cdot F(U) \leq F(U)$ is not a sheaf on $X$ ?

b) Is it correct that we define the product $I \cdot F$ as being the image of the sheaf morphism $$I \otimes_{O_X} F \to O_X \otimes_{O_X} F \cong F,$$ and if so, do we necessarily have that $I \cdot F$ is isomorphic (as sheaf of $O_X$-module) to the sheafification of $P$ ? According to this question, this is true but the comment has no further explanation as to why this is true.

For a), if I consider a collection of pairwise compatible sections $s_U \in P(U)$, we can write $$s_U = \sum_{i=1}^{n_U} a_i^U b_i^U$$ with $a_i^U \in I(U), b_i^U \in F(U)$, but I don't see how the conditions $s_U\vert_{U \cap V} = s_V\vert_{U \cap V}$ could help to glue the various $s_U$ together.

Thank you for your help.

Alphonse
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  • $I\otimes_{R} F$ is canonically isomorphic to $I\cdot F$ for any ideal $I$ of $R$ and $R$-module $F$ via the map $i\otimes m\mapsto i\cdot m$ if that is what you are asking in b). – Levent Mar 06 '18 at 13:35
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    @Levent : this is wrong, I think. See https://math.stackexchange.com/questions/298195/tensor-product-of-a-module-with-an-ideal-is-isomorphic-to-their-standard-product – Alphonse Mar 06 '18 at 13:37
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    Yes, you are right. I did not know that. Thank you :) – Levent Mar 06 '18 at 13:40

1 Answers1

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Let me answer b) first, because it helps with answering a): The image of the sheaf morphism $$I \otimes F \to O_X \otimes F \cong F$$ is by definition the sheafification of the presheaf $$U \mapsto Im(I(U) \otimes_{O_X(U)} F(U) \to F(U))$$ which is just your presheaf $P$. So the image is by definition the sheafification of your presheaf.

Answering a): Now I can talk about presheaves of tensor products being or not being sheaves, where an extremely important and instructive example is given by the projective line $X = \mathbb{P}^1_k$ over a field $k$. The ideal sheaf $I$ corresponding to a closed, $k$-rational point is isomorphic to $O(-1)$. Considering $F = O(1)$ we know $$I \otimes F \cong O_X,$$ but $I(X) = 0$, so the tensor product presheaf is not equal to its sheafification which satisfies $O_X(X) = k$.

Note that $F$ is a flat $O_X$-module, so the tensor product presheaf is actually isomorphic to the product, so this also gives an example for your situation.

Johann Haas
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    Your answer to b) is not entirely correct : the image of $I\otimes F\rightarrow F$ is the sheafification of $(I\otimes_{\mathcal{O}X}F)(U)\rightarrow F(U)$, but $(I\otimes{\mathcal{O}X}F)(U)\neq I(U)\otimes{\mathcal{O}X(U)} F(U)$ in general. There is a morphism : $I(U)\otimes{\mathcal{O}X(U)} F(U)\to(I\otimes{\mathcal{O}_X}F)(U)$ and in fact this morphism might not be surjective. In fact, they might not have the same image in $F(U)$. Luckily, both image have the same sheafification. You can see this by looking at stalks. – Roland Mar 06 '18 at 14:10
  • Thanks, I should have made that more explicit. The point is that sheafification commutes with colimits, so you can calculate the image using the presheaf-tensor product instead of the sheafified one. – Johann Haas Mar 06 '18 at 14:33