Timeline for Pullback of ideal sheaf under base change by completion of base ring
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15 events
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| Dec 10, 2019 at 4:26 | comment | added | Kestutis Cesnavicius | Firstly, the compatibility I am using does not use that $f^*(I)$ is an ideal sheaf, just that it's image in $O_{X}$ cuts out the base change of the subscheme. Of course, $f^*$ is not exact for a general nonflat $f$, but that is not a problem (also, your $\pi$ is flat, so this point, even if irrelevant as I mentioned, does not occur in your situation). | |
| Dec 9, 2019 at 5:24 | comment | added | user267839 | this is exactly the problem I explained above. or do I make an error in my reasonings? | |
| Dec 9, 2019 at 5:22 | comment | added | user267839 | This is unfortunately exactly the obstruction I see to finish the proof that $f^*I_0=I$. because, if I understood the idea correctly, then by property of schematic image and the fact that $\pi$ is an iso restricted to $\hat{Y}/t^n \to Y/t^n$, then we see that $V(I)=Z$ as well as $V(f^*I_0)= V(f^{-1}((Z')$(<-problem) induce the same schematic image $Z'=V(I_0)$ and thus are alrady equal as they are contained in subsets where $\pi$ is an isomorphism. but to make this argument we need equality $V(f^*I_0)= V(f^{-1}((Z')$, and | |
| Dec 9, 2019 at 5:20 | comment | added | user267839 | not assumed $V$ to be Cartier divisors. thus I'm not sure why the compatibility $f^{-1}(V)=V(f^*I)$ should be true. or do you mean a "compatibility " between pull backs of quasi-coherent sheaves of ideals and their associated closed subschemes in another way? | |
| Dec 9, 2019 at 5:19 | comment | added | user267839 | as $f^*$ right exact we obtain exact sequence $f^*I \to f^*O_Y=O_X \to f^*i_*O_V=O_{f^{-1}(V)} \to 0$ (btw about $f^*i_*O_V=O_{f^{-1}(V)}$ I pretty sure that it's true, but haven't actually an argument; do you see a quick one?) on the other hand we have an exact sequence $0 \to I' \to O_X \to O_X/I' = i_*O_{f^{-1}(V)} \to 0$ where $I'$ is defined by $f^{-1}(V)$. by universal property of kernel we obtain an epi $e:f^*I \to I'$. but there is no reason why $e$ has to be in general an isomorphism. this holds for example if $I, I'$ are invertible, but we | |
| Dec 9, 2019 at 5:18 | comment | added | user267839 | @KestutisCesnavicius:about the statement that pulling back quasi-coherent sheaves of ideals and their associated closed subschemes are compatible I'm not sure. let me explain where I see the problem: by compatible you mean that for $f:X \to Y$ nice enough and closed $V \subset Y$ defined by ideal sheaf $I$, i.e. $V=V(I)$, then $f^{-1}(V)=V(f^*I)$, right? you claim that follows from right exactness of tensor product resp $f^*$ as it is a tp. we start with exact sequence defining $V$: $0 \to I \to O_Y \to O_Y/I = i_*O_V \to 0$ with $i: V \to Y$ canonical inclusion. | |
| Dec 8, 2019 at 13:45 | comment | added | Kestutis Cesnavicius | By the very definition of schematic image, it is the smallest closed subscheme of $Y$ through which $Z$ factors, so it is just $Z$ viewed inside $Y$ instead of $\widehat{Y}$. To conclude, I indeed use that pulling back quasi-coherent sheaves of ideals and their associated closed subschemes are compatible; this basically amounts to the right exactness of the tensor product. | |
| Dec 8, 2019 at 3:23 | comment | added | user267839 | Essentially, I need to know if and why following is true: Let $f: X \to Y$ morphism with "nice enough properties (like qcqs and and and...) between nice enough schemes as in our case" and $V \subset Y$ is reduced closed subscheme defined by ideal sheaf $J$. then $f^{-1}(V)$ is defiend by pulled back ideal $f^*J$. is this true & why? if we know it, then we are done, since $\widehat{Y}/t^n \cong Y/t^n$ is induced by $\pi$ and the whole story between $Z$ and $Z'$ take place inside $\widehat{Y}/t^n$ resp. $ Y/t^n$. | |
| Dec 7, 2019 at 23:21 | comment | added | user267839 | your last step, that "... by general definition, the schematic image $Z′=V(I_0)$ is $Z$ itself viewed inside $Y/t^n$..." translates to $\mathcal{I}= \pi^*\mathcal{I}_0$ I not understand. by assumption $Z$ is contained in $V(t) =\widehat{Y}/t^n$ and as $\widehat{Y}/t^n \cong Y/t^n$ and keeping in mind that this iso is given by restriction of $\pi$, we conclude that direct image is contained $Z' \subset Y/t^n$. that's fine. why does this already imply $\mathcal{I}= \pi^*\mathcal{I}_0$? | |
| Dec 7, 2019 at 23:21 | comment | added | user267839 | @KestutisCesnavicius:the last step I not understand: firstly, we indeed assume $\mathcal{I}$ be qc and thus it is uniquely determined by closed subscheme $Z = V(I) \subset \widehat{Y}$. futhermore by qcqs as you said we can explicitely determine the qc ideal sheaf which determine the schematic image $Z'$, namely it's explicitly kernel $K$ of $\mathcal{O}_{Y} \to g_*\mathcal{O}_Z$ where $g$ is the composition $Z \subset \hat{Y} \to Y$. direct calculation gives indeed $K=\mathcal{I}_0= \pi_* \mathcal{I} \cap \mathcal{O}_Y$. how | |
| Dec 7, 2019 at 22:08 | comment | added | Kestutis Cesnavicius | You probably want to assume that $\mathcal{I}$ is quasi-coherent. Given this, such an $\mathcal{I}$ is determined by its corresponding closed subscheme $Z \subset \widehat{Y}$, which by your assumption is a closed subscheme of some $\widehat{Y}/t^n \cong Y/t^n$. The ideal $\mathcal{I}_0$ is also quasi-coherent and its corresponding closed subscheme $Z' \subset Y$ is the schematic image of $Z$ in $Y$ (basically by definition, or because the morphism $\widehat{Y} \rightarrow Y$ is qcqs). By general definition, the schematic image $Z'$ is $Z$ itself viewed inside $Y/t^n$, hence the result. | |
| Dec 7, 2019 at 19:22 | history | edited | user267839 | CC BY-SA 4.0 |
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| Dec 7, 2019 at 19:14 | history | edited | user267839 | CC BY-SA 4.0 |
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| Dec 7, 2019 at 18:57 | history | edited | user267839 | CC BY-SA 4.0 |
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| Dec 7, 2019 at 18:48 | history | asked | user267839 | CC BY-SA 4.0 |