Lemma 51.16.1. Let $A$ be a Noetherian ring of dimension $d$. Let $I \subset I' \subset A$ be ideals. If $I'$ is contained in the Jacobson radical of $A$ and $\text{cd}(A, I') < d$, then $\text{cd}(A, I) < d$.

## 51.16 Hartshorne-Lichtenbaum vanishing

This vanishing result is the local analogue of Lichtenbaum's theorem that the reader can find in Duality for Schemes, Section 48.34. This and much else besides can be found in [CD].

**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. $\square$

Lemma 51.16.2. Let $A$ be a Noetherian ring of dimension $d$. Let $I \subset A$ be an ideal. If $H^ d_{V(I)}(M) = 0$ for some finite $A$-module whose support contains all the irreducible components of dimension $d$, then $\text{cd}(A, I) < d$.

**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.
$\square$

Lemma 51.16.3. Let $A$ be a Noetherian local ring of dimension $d$. Let $f \in A$ be an element which is not contained in any minimal prime of dimension $d$. Then $f : H^ d_{V(I)}(M) \to H^ d_{V(I)}(M)$ is surjective for any finite $A$-module $M$ and any ideal $I \subset A$.

**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 $H^ d_{V(I)}(M) \to H^ d_{V(I)}(fM)$.
$\square$

Lemma 51.16.4. Let $A$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Let $I \subset A$ be an ideal. If $H^0_{V(I)}(\omega _ A^\bullet ) = 0$, then $\text{cd}(A, I) < \dim (A)$.

**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 avoidence (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). $\square$

Lemma 51.16.5. Let $(A, \mathfrak m)$ be a complete Noetherian local domain. Let $\mathfrak p \subset A$ be a prime ideal of dimension $1$. For every $n \geq 1$ there is an $m \geq n$ such that $\mathfrak p^{(m)} \subset \mathfrak p^ n$.

**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. $\square$

Proposition 51.16.6. Let $A$ be a Noetherian local ring with completion $A^\wedge $. Let $I \subset A$ be an ideal such that

for every minimal prime $\mathfrak p \subset A^\wedge $ of dimension $\dim (A)$. Then $\text{cd}(A, I) < \dim (A)$.

**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. $\square$

Lemma 51.16.7. Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset A$ be an ideal. Assume $A$ is excellent, normal, and $\dim V(I) \geq 1$. Then $\text{cd}(A, I) < \dim (A)$. In particular, if $\dim (A) = 2$, then $\mathop{\mathrm{Spec}}(A) \setminus V(I)$ is affine.

**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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)