# Hartshorne Algebraic Geometry Exercise III.11.5 (On Picard groups of Formal Completions)

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...

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 $$0 for $$Y$$ a complete intersection.
• 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 $Y$. Mar 28, 2023 at 3:24
• 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... Mar 28, 2023 at 3:31