The Stacks project

Proof. By Lemma 51.4.7 we know $\text{cd}(A, I) \leq d$. We will use Lemma 51.2.6 to show

is surjective which will finish the proof. Pick $\mathfrak p \in V(I) \setminus V(I')$. By our assumption on $I'$ we see that $\mathfrak p$ is not a maximal ideal of $A$. Hence $\dim (A_\mathfrak p) < d$. Then $H^ d_{\mathfrak pA_\mathfrak p}(A_\mathfrak p) = 0$ by Lemma 51.4.7. $\square$

Proof. By Lemma 51.4.7 we know $\text{cd}(A, I) \leq d$. Thus for any finite $A$-module $N$ we have $H^ i_{V(I)}(N) = 0$ for $i > d$. Let us say property $\mathcal{P}$ holds for the finite $A$-module $N$ if $H^ d_{V(I)}(N) = 0$. One of our assumptions is that $\mathcal{P}(M)$ holds. Observe that $\mathcal{P}(N_1 \oplus N_2) \Leftrightarrow (\mathcal{P}(N_1) \wedge \mathcal{P}(N_2))$. Observe that if $N \to N'$ is surjective, then $\mathcal{P}(N) \Rightarrow \mathcal{P}(N')$ as we have the vanishing of $H^{d + 1}_{V(I)}$ (see above). Let $\mathfrak p_1, \ldots , \mathfrak p_ n$ be the minimal primes of $A$ with $\dim (A/\mathfrak p_ i) = d$. Observe that $\mathcal{P}(N)$ holds if the support of $N$ is disjoint from $\{ \mathfrak p_1, \ldots , \mathfrak p_ n\}$ for dimension reasons, see Lemma 51.4.7. For each $i$ set $M_ i = M/\mathfrak p_ i M$. This is a finite $A$-module annihilated by $\mathfrak p_ i$ whose support is equal to $V(\mathfrak p_ i)$ (here we use the assumption on the support of $M$). Finally, if $J \subset A$ is an ideal, then we have $\mathcal{P}(JM_ i)$ as $JM_ i$ is a quotient of a direct sum of copies of $M$. Thus it follows from Cohomology of Schemes, Lemma 30.12.8 that $\mathcal{P}$ holds for every finite $A$-module. $\square$

Proof. The support of $M/fM$ has dimension $< d$ by our assumption on $f$. Thus $H^ d_{V(I)}(M/fM) = 0$ by Lemma 51.4.7. Thus $H^ d_{V(I)}(fM) \to H^ d_{V(I)}(M)$ is surjective. Since by Lemma 51.4.7 we know $\text{cd}(A, I) \leq d$ we also see that the surjection $M \to fM$, $x \mapsto fx$ induces a surjection $H^ d_{V(I)}(M) \to H^ d_{V(I)}(fM)$. $\square$

Proof. Set $d = \dim (A)$. Let $\mathfrak p_1, \ldots , \mathfrak p_ n \subset A$ be the minimal primes of dimension $d$. Recall that the finite $A$-module $H^{-i}(\omega _ A^\bullet )$ is nonzero only for $i \in \{ 0, \ldots , d\}$ and that the support of $H^{-i}(\omega _ A^\bullet )$ has dimension $\leq i$, see Lemma 51.9.4. Set $\omega _ A = H^{-d}(\omega _ A^\bullet )$. By prime avoidance (Algebra, Lemma 10.15.2) we can find $f \in A$, $f \not\in \mathfrak p_ i$ which annihilates $H^{-i}(\omega _ A^\bullet )$ for $i < d$. Consider the distinguished triangle

See Derived Categories, Remark 13.12.4. By Derived Categories, Lemma 13.12.5 we see that $f^ d$ induces the zero endomorphism of $\tau _{\geq -d + 1}\omega _ A^\bullet$. Using the axioms of a triangulated category, we find a map

whose composition with $\omega _ A[d] \to \omega _ A^\bullet$ is multiplication by $f^ d$ on $\omega _ A[d]$. Thus we conclude that $f^ d$ annihilates $H^ d_{V(I)}(\omega _ A)$. By Lemma 51.16.3 we conlude $H^ d_{V(I)}(\omega _ A) = 0$. Then we conclude by Lemma 51.16.2 and the fact that $(\omega _ A)_{\mathfrak p_ i}$ is nonzero (see for example Dualizing Complexes, Lemma 47.16.11). $\square$

Proof. Recall that the symbolic power $\mathfrak p^{(m)}$ is defined as the kernel of $A \to A_\mathfrak p/\mathfrak p^ mA_\mathfrak p$. Since localization is exact we conclude that in the short exact sequence

the support of $\mathfrak a_ n$ is contained in $\{ \mathfrak m\}$. In particular, the inverse system $(\mathfrak a_ n)$ is Mittag-Leffler as each $\mathfrak a_ n$ is an Artinian $A$-module. We conclude that the lemma is equivalent to the requirement that $\mathop{\mathrm{lim}}\nolimits \mathfrak a_ n = 0$. Let $f \in \mathop{\mathrm{lim}}\nolimits \mathfrak a_ n$. Then $f$ is an element of $A = \mathop{\mathrm{lim}}\nolimits A/\mathfrak p^ n$ (here we use that $A$ is complete) which maps to zero in the completion $A_\mathfrak p^\wedge$ of $A_\mathfrak p$. Since $A_\mathfrak p \to A_\mathfrak p^\wedge$ is faithfully flat, we see that $f$ maps to zero in $A_\mathfrak p$. Since $A$ is a domain we see that $f$ is zero as desired. $\square$

Proof. Since $A \to A^\wedge$ is faithfully flat we have $H^ d_{V(I)}(A) \otimes _ A A^\wedge = H^ d_{V(IA^\wedge )}(A^\wedge )$ by Dualizing Complexes, Lemma 47.9.3. Thus we may assume $A$ is complete.

Assume $A$ is complete. Let $\mathfrak p_1, \ldots , \mathfrak p_ n \subset A$ be the minimal primes of dimension $d$. Consider the complete local ring $A_ i = A/\mathfrak p_ i$. We have $H^ d_{V(I)}(A_ i) = H^ d_{V(IA_ i)}(A_ i)$ by Dualizing Complexes, Lemma 47.9.2. By Lemma 51.16.2 it suffices to prove the lemma for $(A_ i, IA_ i)$. Thus we may assume $A$ is a complete local domain.

Assume $A$ is a complete local domain. We can choose a prime ideal $\mathfrak p \supset I$ with $\dim (A/\mathfrak p) = 1$. By Lemma 51.16.1 it suffices to prove the lemma for $\mathfrak p$.

By Lemma 51.16.4 it suffices to show that $H^0_{V(\mathfrak p)}(\omega _ A^\bullet ) = 0$. Recall that

By Lemma 51.16.5 we see that the colimit is the same as

Since $\text{depth}(A/\mathfrak p^{(n)}) = 1$ we see that these ext groups are zero by Lemma 51.9.4 as desired. $\square$

Proof. By More on Algebra, Lemma 15.52.6 the completion $A^\wedge$ is normal and hence a domain. Thus the assumption of Proposition 51.16.6 holds and we conclude. The statement on affineness follows from Lemma 51.4.8. $\square$