Let \(A\) be a noetherian ring and \(\mfa\) an ideal. For a finite \(A\)-module \(M\) with \(\mfa M\ne M\), let \(\depth_\mfa M\) be the maximum length of an \(M\)-regular sequence in \(\mfa\). Use the convention \(\depth_\mfa M=\infty\) when \(M\) is finite and \(\mfa M=M\), including \(M=0\). This generalizes the notion of depth introduced in (II, §8).
For any \(A\)-module \(M\), show that the existence of an \(M\)-regular element in \(\mfa\) implies \(\Gamma_{\mfa}(M)=0\). For \(M\) finitely generated, show that \(\Gamma_{\mfa}(M)=0\) if and only if \(\depth_\mfa M\ge1\), with the stated convention.
Show inductively, for \(M\) finitely generated, that for any \(n\ge0\), the following conditions are equivalent:
\(\depth_{\mfa} M\ge n\);
- \(H_\mfa^i(M)=0\) for all \(0\le i<n\).
(The exceptional depth convention)
The maximum-length definition without the exceptional convention is insufficient for arbitrary ideals. For \(A=k\), \(\mfa=A\) and \(M=k\), the functor \(\Gamma_{\mfa}\) is zero on every module and all its derived functors vanish. But no nonzero scalar is an \(M\)-regular element with nonzero quotient, and zero does not act injectively on \(M\). Setting \(\depth_\mfa M=\infty\) for finite \(M\) with \(\mfa M=M\) resolves this case; see the definition of ideal depth. For a nonzero finite module over a local ring and a proper ideal, Nakayama’s lemma excludes this exceptional case.
Hint
When \(M\) is finitely generated, both conditions in (a) are equivalent to saying that \(\mfa\) is not contained in any associated prime of \(M\).
Solution
Write \(T^i(M)=H_{\mfa}^i(M)\) and \(T^0(M)=\Gamma_{\mfa}(M)\). By P-AGH333LOCALCOH, every element of every \(T^i(M)\) is killed by a power of \(\mfa\). Regular sequences have nonzero final quotient, and the finite-module depth convention is the one in D-DEFREGSQ.
1. If multiplication by some \(x\in\mfa\) is injective on \(M\), then \(T^0(M)=0\).
For \(m\in T^0(M)\), choose \(q\ge1\) with \(\mfa^qm=0\). Then \(x^qm=0\). A power of an injective endomorphism is injective, so \(m=0\). This proves the forward implication in (a) for arbitrary modules, and even when the injective multiplication map is surjective.
2. For a finite module \(M\), \(T^0(M)=0\) exactly when \(\mfa\) is contained in no associated prime of \(M\).When \(\mfa M\ne M\), this is equivalent to the existence of an \(M\)-regular element in \(\mfa\).
If \(\mathfrak p=\Ann_A(m)\) for some \(0\ne m\in M\) and \(\mfa\subseteq\mathfrak p\), then \(\mfa m=0\) and \(T^0(M)\ne0\). Conversely, if \(T^0(M)\ne0\), choose an associated prime \(\mathfrak p\) of this nonzero finite submodule. Such a prime exists for a nonzero finite module over a noetherian ring. It is the annihilator of a nonzero element \(m\in T^0(M)\), so it is also an associated prime of \(M\). Some power of \(\mfa\) annihilates \(m\), and primality gives \(\mfa\subseteq\mathfrak p\). This proves the first equivalence, including \(M=0\) with empty associated-prime set.
The associated-prime finiteness theorem and the zero-divisor description say that the zero divisors on a finite \(M\) form the union of its finitely many associated primes. Prime avoidance therefore supplies \(x\in\mfa\) outside that union when none of those primes contains \(\mfa\). It acts injectively on \(M\). If \(\mfa M\ne M\), the quotient \(M/xM\) maps onto the nonzero module \(M/\mfa M\), so \(M/xM\ne0\). Thus \(x\) is an \(M\)-regular element under the proper-quotient convention. Conversely, such an element implies \(T^0(M)=0\) by step 1.
3. If \(M\) is finite and \(\mfa M=M\), then \(T^i(M)=0\) for every \(i\ge0\).
For \(M=0\), every derived functor is zero. Otherwise choose generators \(m_1,\ldots,m_s\) of \(M\) and coefficients \(c_{ij}\in\mfa\) with \(m_i=\sum_j c_{ij}m_j\). The adjugate identity for the matrix \(1-(c_{ij})\) shows that its determinant annihilates each generator. This determinant has the form \(1-t\) with \(t\in\mfa\). Thus multiplication by \(t\) on \(M\) is the identity.
The endomorphism induced by multiplication by \(t\) on \(T^i(M)\) is also multiplication by \(t\): multiply every term of an injective resolution by \(t\) to obtain a chain map lifting the original scalar endomorphism. Because the original map is the identity, functoriality makes this induced endomorphism the identity. For any \(z\in T^i(M)\), however, some \(\mfa^q\) kills \(z\), so \(t^qz=0\). Since \(t\) acts identically, this implies \(z=0\). This proves every vanishing in the exceptional case, consistently with \(\depth_\mfa M=\infty\). Together with steps <1>1–<1>2 it also completes (a).
4. For finite \(M\) with \(\mfa M\ne M\), a regular sequence of length \(n\) in \(\mfa\) implies \(T^i(M)=0\) for \(i<n\).
1.Let \(n>1\) and write \(x\) for the first entry of a regular sequence of length \(n\).Put \(N=M/xM\).It is finite and \(N/\mfa N\cong M/\mfa M\ne0\).The other entries form an \(N\)-regular sequence of length \(n-1\), so induction gives \(T^j(N)=0\) for \(j<n-1\).
Apply the derived-functor long exact sequence to \[ 0\longrightarrow M\xrightarrow{x}M\longrightarrow N\longrightarrow0. \] For \(1\le i<n\), its segment \[ T^{i-1}(N)\longrightarrow T^i(M)\xrightarrow{x}T^i(M) \] makes multiplication by \(x\) injective, since the left term is zero. But every element of \(T^i(M)\) is killed by a power of \(x\), by the ideal-torsion property. Injectivity then forces \(T^i(M)=0\). Step 1 gives the missing degree zero, proving the induction. The multiplication maps here are scalar multiplication by the chain-map argument in step 3.
5. For finite \(M\) with \(\mfa M\ne M\), vanishing of \(T^i(M)\) for \(i<n\) implies the existence of an \(M\)-regular sequence of length \(n\) in \(\mfa\).
Again induct on \(n\). For \(n=0\) use the empty sequence; \(M\) is nonzero because \(\mfa M\ne M\). For \(n\ge1\), the hypothesis gives \(T^0(M)=0\). Step 2 supplies an \(M\)-regular element \(x\in\mfa\). Put \(N=M/xM\); this is finite and \(N/\mfa N\cong M/\mfa M\ne0\).
For \(0\le j<n-1\), the long exact sequence from step 4 has a segment \[ T^j(M)\longrightarrow T^j(N)\longrightarrow T^{j+1}(M) \] whose outer terms vanish. Thus \(T^j(N)=0\) for all \(j<n-1\). The induction hypothesis gives an \(N\)-regular sequence of length \(n-1\) in \(\mfa\). Prepending \(x\) gives an \(M\)-regular sequence of length \(n\), with the same nonzero final quotient. This proves the reverse induction without assuming a depth-drop formula.
6. Q.E.D.
Steps <1>1–<1>3 prove (a), with its arbitrary-module forward direction and its finite-module converse. For \(\mfa M\ne M\), steps <1>4–<1>5 prove (b) for every \(n\). For \(\mfa M=M\), step 3 proves all the required vanishings and depth is infinite by convention. Thus the equivalence in (b) holds for every finite module in the statement.