# Comparing two different conditions for an ideal to correspond to a closed subscheme

Let $I$ be an ideal in the graded ring $S = A[x_0, \ldots, x_r]$. In Exercise II.5.10(a), Hartshorne defines the saturation $\bar I$ of $I$ to be the set $$\bar I = \{s \in S \mid \text{for all } i = 0, \ldots, r, \text{there exists an } n \text{ such that } x_i^n s \in I\}.$$ An ideal is saturated if $I = \bar I$.

Now let $\mathfrak{m} = (x_0, \ldots, x_r)$ be the irrelevant ideal in $S$ and define $H_{\mathfrak{m}}^0(I)$ by $$H_{\mathfrak{m}}^0(I) = \bigcup_{n \geq 0} (0 :_I \mathfrak{m}^n)$$ where $(0:_I \mathfrak{m}^n) = \{a \in I \mid \mathfrak{m}^n a = 0\}$.

Question 1: How do I show that, if $I$ is a saturated ideal of $S$, then $H_{\mathfrak{m}}^0(I) = 0$?

The two conditions in the question (namely "$I$ is saturated" and "$H_{\mathfrak{m}}^0(I) = 0$") are related to the bijection between saturated ideals of $S$ and closed subschemes of $\operatorname{Proj}(S)$. The first one is in part (d) of the Hartshorne exercise quoted above, while the second is described in this Mathoverflow answer, or from Theorem A4.2 in Eisenbud's book on Commutative Algebra, or in Theorem 13.21 in "Twenty-four hours of local cohomology" by Iyengar et al. (which one can stumble across on-line).

I am also interested in a second, more difficult, result that we should be able to prove from the assumption that $I$ is saturated, namely

Question 2: How does one show that, if $I$ is a saturated ideal of $S$, then $H_{\mathfrak{m}}^1(I) = 0$?

Here $H_{\mathfrak{m}}^1(I) = \lim_{n\to \infty} \operatorname{Ext}_R^i(R/\mathfrak{m}^n, I)$ is the first local cohomology group at $\mathfrak{m}$. It is known to be zero if and only if there exists an $I$-regular sequence in $\mathfrak{m}$ of length at least two.

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First, the ring $S$ (in Hartshorne) is a polynomial ring in $r+1$ indeterminates. Second, assume that $r\ge 1$.
Note that $H_{\mathfrak{m}}^0(S/I)=\overline{I}/I$. Then use the short exact sequence $$0\longrightarrow I \longrightarrow S \longrightarrow S/I\longrightarrow 0$$ which gives rise to a long exact sequence $$0\longrightarrow H_{\mathfrak{m}}^0(I) \longrightarrow H_{\mathfrak{m}}^0(S) \longrightarrow H_{\mathfrak{m}}^0(S/I)\longrightarrow H_{\mathfrak{m}}^1(I)\longrightarrow H_{\mathfrak{m}}^1(S)\longrightarrow \cdots$$ Since $\mathrm{depth}(S)\ge 2$, $H_{\mathfrak{m}}^0(S)=H_{\mathfrak{m}}^1(S)=0$ and this is it.