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Let $\widehat{X}$ be the formal completion of $X=\mathbb{P}^N_k$ along a hypersurface for $N\geq 4$. The exercise is to prove $\operatorname{Pic}(\widehat{X})\rightarrow \operatorname{Pic}(Y)$ is an isomorphism.

The hint is to use Exercise II.9.6, Exercise III.4.6, and Exercise III.5.5. Let me explain the crux of each part that I understand so far.

For II.9.6, the main point of the application is that $\operatorname{Pic}(\widehat{X})\cong \varprojlim_n \operatorname{Pic}(\widehat{X})(X_n)$ where $X_n$ is the scheme $(\widehat{X},\mathcal{O}_{\widehat{X}}/\mathfrak{I}^n)$ where $\mathfrak{I}$ is the ideal of definition to this formal scheme. In other words, $X_n$ is precisely $(Y,\mathcal{O}_X/\mathscr{I}^n_Y)$ for $\mathscr{I}_Y$ the ideal sheaf of $Y$. The only thing one needs to check is that the inverse system of global sections groups is Mittag-Leffler and this should easy enough due to a Noetherian hypothesis.

The next part is to study the maps $\operatorname{Pic}(X_{n+1})\rightarrow \operatorname{Pic}(X_n)$. These induced maps clearly exist and these Picard groups can be identified with $H^1$ of their unit sheaves. This leads to the next hint -- I take a short exact sequence $$ 0\rightarrow \mathscr{I}_Y^n/\mathscr{I}_Y^{n+1}\rightarrow \mathcal{O}_{X_{n+1}}^*\rightarrow \mathcal{O}_{X_{n}}^*\rightarrow 0 $$ which exists for each $n$. More importantly, Exercise III.4.6 give rise to an exact sequence of abelian groups $$ \dots \rightarrow H^1(X,\mathscr{I}_Y^n/\mathscr{I}^{n+1}_Y)\rightarrow \operatorname{Pic}(X_{n+1})\rightarrow\operatorname{Pic}(X_{n})\rightarrow H^2(X,\mathscr{I}^{n}_Y/\mathscr{I}^{n+1}_Y)\rightarrow\dots $$ Now my thought would be that the last hint (Exercise III.5.5) should be applicable to this case to get some cohomology to vanish. However, the hypothesis that $N\geq 4$ is needed to show that $H^2(Y,\mathscr{I}^n,\mathscr{I}^{n+1})$ is always vanishing. In this case, we always have a surjective map $\operatorname{Pic}(X_{n+1})\rightarrow \operatorname{Pic}(X_n)$ for all $n\geq 1$. And thereby I get a surjection for the desired map above.

Crux of the Post / Question: How do we get injectivity? In general, I do not think we can get $H^1(\mathscr{I}^n/\mathscr{I}^{n+1})$ to vanish so I am stuck...

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The kernel $I^n/I^{n+1}$ is isomorphic to $\mathcal{O}_Y(-dn)$ where $d=\deg Y$. Now look at exercise III.5.5c again, where the vanishing result is $H^i(Y,\mathcal{O}_Y(n))=0$ for $0<i<\dim Y$ for $Y$ a complete intersection.

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  • $\begingroup$ Oh perfect! Just to be precise, the isomorphism for the kernel to that invertible sheaf follows from the short exact sequence associated to an effective divisor and induction right? $\endgroup$
    – Shrugs
    Mar 28, 2023 at 2:35
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    $\begingroup$ I'm not sure exactly what your argument there is, but I would say it like this: $I^n/I^{n+1}$ as a sheaf on $X$ is isomorphic to $i_*i^*I^n$, where $i:Y\to X$ is the closed immersion of $Y$. But $I^n\cong \mathcal{O}_X(-dn)$, as it's generated by the $n^{th}$ power of the degree-$d$ equation picking out $Y$. $\endgroup$
    – KReiser
    Mar 28, 2023 at 3:24
  • $\begingroup$ Oh I meant like noticing that $I^n/I^{n+1}=I^n\otimes O_X/I$ and then trying to do something from there. In hindsight that doesn't work and what you're saying was what I realize worked better... $\endgroup$
    – Shrugs
    Mar 28, 2023 at 3:31

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