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I'm doing part (a) of Exercise 5.14 in Chapter II of Hartshorne: Let $k$ be an algebraically closed field, and $X\subseteq\mathbb{P}_k^r$ a connected, normal and closed subscheme. Let $S$ be the homogeneous coordinate ring, and $S'=\bigoplus_{n\geq 0}\Gamma(X,\mathcal{O}_X(n))$. Then we have to show that $S$ is a domain and $S'$ is its integral closure.

Following the hint of Hartshorne, I proved that $X$ is integral, and that $S$ is a domain. He then proposes to consider the sheaf of rings $\mathcal{S}=\bigoplus_{n\geq 0}\mathcal{O}_X(n)$, and show that it is a sheaf of integrally closed domains. And this is the point where I'm failing to see how to proceed.

In this post, Brian Ng is using the following argument: for $\mathfrak{p}\in\operatorname{Proj S}$, we have $$\mathcal{S}_\mathfrak{p}\cong\bigoplus_{n\geq0}S(n)_{(\mathfrak{p})}\cong\{\frac{s}{b}\ |\ b\notin\mathfrak{p}\text{ homogeneous, every non-zero homogeneous part of $s$ has degree $\geq\deg b$}\}\subseteq S_{\mathfrak{p},\text{ hom.}}$$ where $S_{\mathfrak{p},\text{ hom.}}$ is the localization of $S$ at the homogeneous elements not in $\mathfrak{p}$. With this I agree. However, he then sais that this is integrally closed as $S_{\mathfrak{p},\text{ hom.}}$ is (because $X$ is normal), and that therefore that $S'=\Gamma(X,\mathcal{S})$ is integrally closed to, and I don't see how exactly we may conclude this. First of all, $X$ being normal only implies that $X_\mathfrak{p}\cong S_{(\mathfrak{p})}=(S_{\mathfrak{p},\text{ hom.}})_0$ is integrally closed. Furthermore, how does one conclude from the stalks being integrally closed that every ring of sections of the sheaf is integrally closed?

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Suppose $x\in S_1$ is a linear form on $\Bbb P^n$ not in $\mathfrak{p}$. I claim $\mathcal{S}_\mathfrak{p}\cong S_{(\mathfrak{p})}[x]$: as $\mathfrak{p}\in D(x)$, we can write $(\mathcal{O}_X(n))_\mathfrak{p} = (\mathcal{O}_X(n)|_{D(x)})_\mathfrak{p}$ since restriction commutes with stalks. By the identification of $D(x)$ with $\Bbb A^n$, we have $\mathcal{O}_X(n)|_{D(x)}$ is the sheaf associated to the $S_{(x)}$-module $x^nS_{(x)}$, and the prime ideal $\mathfrak{p}$ is associated to the prime ideal $\mathfrak{p}'=\mathfrak{p}_{(x)}$. So $$(\mathcal{O}_X(n)|_{D(x)})_\mathfrak{p} = (\widetilde{x^nS_{(x)}})_{\mathfrak{p}'}=x^nS_{(\mathfrak{p})}$$ and the direct sum $\mathcal{S}_\mathfrak{p}$ is $\bigoplus_{n\geq0} x^nS_{\mathfrak{p}}=S_\mathfrak{p}[x]$. As $S_{\mathfrak{p}}$ is integrally closed, $\mathcal{S}_\mathfrak{p}$ is too.

As for why integrally closed in all stalks means integrally closed for every open, define a sheaf $\mathcal{T}$ by the rule $\mathcal{T}(U)$ is the integral closure of $\mathcal{S}(U)$, which works because integral closure commutes with localization. Then $\mathcal{S}$ is naturally a subsheaf of $\mathcal{T}$ with the same stalks, which implies they must be equal.

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  • $\begingroup$ Thanks a lot, this is really clever and insightful! $\endgroup$ Commented Dec 15, 2020 at 15:40
  • $\begingroup$ Sorry for bothering, but could you in addition show $\mathcal S$ is a sheaf of domains? This is the condition you need for the second paragraph. Besides how can we know the global section of it is exactly the direct sum of global sections?(I mean sheafification of direct sum may add more elements to $\mathcal S(X)$) $\endgroup$
    – Richard
    Commented Oct 25, 2021 at 13:41
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    $\begingroup$ @Richard That seems relatively immediate to me: $\mathcal{S} = \mathcal{O}_X[t]$ where $t$ has degree one, and $\mathcal{O}_X$ is a sheaf of domains since $X$ is normal and connected hence integral. For your second question, everything works right over a noetherian topological space, which is enough for us because we're working in $\Bbb P^n_A$ for some noetherian ring $A$. $\endgroup$
    – KReiser
    Commented Oct 25, 2021 at 19:07
  • $\begingroup$ @KReiser in the concluding statement for the proof of your claim, did you make a typo? (Should be $\mathcal{S}_{\mathfrak{p}}=\bigoplus_{n\geq 0}x^nS_{(\mathfrak{p})}$ instead)? $\endgroup$
    – Ivan So
    Commented Jul 5, 2022 at 2:47
  • $\begingroup$ Meanwhile, in proving $\mathcal{S}_{\mathfrak{p}}$ is integrally closed, did you use the algebra fact that suppose $R$ is integrally closed over it's quotient field $K$, then $R[x]$ is integrally closed as a subring of $K[x]$? $\endgroup$
    – Ivan So
    Commented Jul 5, 2022 at 2:50

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