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I have questions related to Hartshorne's Exercise II.5.14, set up as follows: Let $X$ be a connected normal closed subscheme of $\mathbb{P}^r_k$, where $k$ is an algebraically closed field, and write $S$ for the homogeneous coordinate ring of $X$. Also take $S' = \oplus_{n \ge 0} \Gamma(X,\mathcal{O}_X(n))$ and view it as a graded ring.

I'm trying to show $S$ is a domain and $S'$ is its integral closure. So far, I've proven $X$ is an integral scheme. My current questions:

  • It follows from the text (namely Exercise 3.12(b) and Corollary 5.16(a)) that $X$ can be identified with Proj $S$. If $S$ is a domain, surely Proj $S$ is integral; is the converse true?

  • Hartshorne's suggestion is to view $S'$ as $\Gamma(X,\mathcal{F})$ for a sheaf of rings $\mathcal{F} = \bigoplus_n \mathcal{O}_X(n)$, and in fact show $\mathcal{F}$ is a sheaf of integrally closed domains. Can anyone provide a hint on how this might be done?

Mr. Chip
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    For your first question, it is not true that $\operatorname{Proj} S$ integral implies $S$ is a domain. For example, consider any domain $S$, and $T = S[x]/(x^2)$, where $x$ has degree $1$. Then, $\operatorname{Proj} S \cong \operatorname{Proj} T$ by Exercise II.5.13, but $T$ is not a domain. – Takumi Murayama Jul 30 '16 at 04:14
  • @TakumiMurayama: Do you consider in this counterexample $S$ to be endowed with arbitrary graded ring structure or a specific one (...in order to construct the Proj)? The point which confuses me a bit is that for a nontrivial $s \in S_1$ the homogeneous localization of $T$ at that this $s$ if I'm not missing something contains nilpotent $\overline{x} $ which in $T_{(s)}$ is non zero, or not? – user267839 Apr 02 '24 at 21:39
  • @user267839 You are correct—the example I gave above is incorrect. One way to fix the example is to use $T = S[x]/(x^2)$, where $S$ is a graded domain (or more generally a reduced graded ring) and $x$ has degree 0. By [EGAII, Proposition 2.4.4(i)], $\operatorname{Proj}T$ is reduced and is isomorphic to $\operatorname{Proj}S$. (To address your second question, with this change, the element $x$ has image $0$ in $T_{(s)}$ since $x/1 = x^2/x = 0$ in $T_{(s)}$.) – Takumi Murayama Apr 04 '24 at 13:27
  • @TakumiMurayama: Thanks, yes. More generally for graded $S$ we can for arbitrary positive integers $r, s>0$ form another graded ring $R$ declared by (1) $R_0= B$ for any ring $B$ which arise from some ring map $B \to S_0$, (2) $R_d:=0$ for $d<r$, (3) $R_k:=S_{ks}$ such that the Proj's of $S$ and $R$ are isomorphic. I'm not sure if we can find even more "different" rings with still isomorphic proj's – user267839 Apr 04 '24 at 14:28
  • a trifle: I assume in your last sentence you have chosen $s:=x$? – user267839 Apr 04 '24 at 14:30
  • @TakumiMurayama: But wait, why is $x$ zero inside $T_{(s)}$, for any homogeneous $s\in S$ and not $s=t$ only? (...this we need to have an iso between the Proj's of $S$ and $T$; at least we need homog $s_1,..., s_d$ such that the $D_+(s_i)$ cover the projs of $S$ and $T$ and that $S_{(s_i)} \cong T_{(s_i)}$ hold) – user267839 Apr 04 '24 at 15:14
  • #Update: I think the problem has meanwhile been resolved: for homogeneous $s \in S$ inside $T_{(s)}$ the multiplication between $x$ and a $r/s^d \in T_{(s)}$ where $r \in S_{d \cdot \text{deg}(s)}$ is given by $x \cdot r/s^d := p(x) \cdot r/s^d$ (inside $S_{(s)}$ where $p:S[x]/(x^2) \to S$ canonical projection/"killing" of $x$, so $x$ acts indeed as multiplication by zero in $T_{(s)}$ and so is zero. And thus we have indeed $T_{(s)} \cong S_{(s)}$. Is the argumentation correct? – user267839 Apr 04 '24 at 15:34
  • @TakumiMurayama: but it is a priori not clear why $x/1$ should be zero in $T_{(s)}$. You argued that inside $T_{(s)}$ we can express $x/1$ as $x^2/x$ (...then indeed it would be zero, as by construction $x^2$ is zero and the proof in fine), but the problem here is why we can express $x/1$ as $x^2/x$ inside $T_{(s)}$? The point is that a priori I not see any reason why $x$in $T_s$ becomes invertible, in order to make expressions like $a/x$ inside $T_s$ welldefined. – user267839 Apr 04 '24 at 23:07
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    @user267839 Right, $S[x]/(x^2)$ does not seem to work. Instead, I believe the example I wrote in the answer below works. – Takumi Murayama Apr 05 '24 at 00:20

2 Answers2

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To show $S$ is a domain, it suffices to show that $I_X$ is prime, which is equivalent to showing that $X$ is irreducible. Note that $X$ is reduced. Else then some local ring $\mathcal{O}_{x, X}$ contains nilpotents and then $\mathcal{O}_{x, X}$ is not integrally closed, since it is not an integral domain. If $X$ were reducible, then some point $x$ would be contained in two irreducible components, so the local ring at the point would have zero divisors. So since $X$ is normal, $X$ is irreducible and $S$ is a domain.

Consider the sheaf$$\mathscr{L} = \bigoplus_{n \ge 0} \mathcal{O}_X(n).$$Then$$\mathscr{L}_\mathfrak{p} = \bigoplus_{n \ge 0} S(n)_{(\mathfrak{p})} = \left\{ {s\over f} \in S_\mathfrak{p} : \deg s \ge \deg f\right\}.$$Any element integral over $\mathscr{L}_\mathfrak{p}$ is of course integral over $S_\mathfrak{p}$, and thus is in $S_\mathfrak{p}$ since $X$ is normal. However, nothing with total negative degree can be integral over $\mathscr{L}_\mathfrak{p}$, so $\mathscr{L}_\mathfrak{p}$ is integrally closed. Thus,$$\Gamma(X, \mathscr{L}) = \bigoplus_{n \ge 0} \Gamma(X, \mathcal{O}_X(n)) = S'$$is integrally closed. $S'$ is contained in the integral closure of $S$ by pages 122-123 of Hartshorne, so $S'$ is the integral closure of $S$.

Hi Erin, thanks for your answer. Can you elaborate on why in this setting (more general than the context of varieties) the primality of $I_X$ is equivalent to the irreducibility of $X$? To elaborate on what I mean: I find it hard to see the equivalence using the construction of $I_X$ in 5.16.

The ring is an algebraically closed field in this exercise, so it is basically the usual case. Perhaps that sentence should say "$I_X$ is prime if and only if $X$ is irreducible and reduced," but I handle reducibility in the next sentences, so I think my solution is fine as is.

Brian Ng
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  • Hi Erin, thanks for your answer. Can you elaborate on why in this setting (more general than the context of varieties) the primality of $I_X$ is equivalent to the irreducibility of $X$? – Mr. Chip Jul 30 '16 at 04:12
  • To elaborate on what I mean: I find it hard to see the equivalence using the construction of $I_X$ in 5.16. – Mr. Chip Jul 30 '16 at 04:30
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In the interest of clarifying the example I gave in the comments, I am writing an answer here.

The example $T = S[x]/(x^2)$ with $\deg(x) = 0$ is not quite correct. Instead, a corrected version is as follows. Consider the nilradical $\mathfrak{N}_+$ of $T_+$, that is, the set of nilpotent elements in $T_+$. Then, there is an isomorphism $$\operatorname{Proj}(T/\mathfrak{N}_+) \overset{\sim}{\longrightarrow} \operatorname{Proj}(T)_{\mathrm{red}}$$ by [EGAII, Proposition 2.4.4(i)]. Note that $T/\mathfrak{N}_+$ is non-reduced since $x \ne 0$ in $T/\mathfrak{N}_+$. Also, if $t \in T_d$ for $d > 0$, then $x/t = tx/t^2 = 0$ in $(T/\mathfrak{N}_+)_{(\bar{t})}$ since $tx \in \mathfrak{N}_+$.

Takumi Murayama
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