Ordinary and derived patching support

Proof of Theorem 6.2

Let \({\mathfrak p}\in{\operatorname{Spec}} R\). If \({\operatorname{Ann}}_R(M)\not\subseteq{\mathfrak p}\), choose \(a\in{\operatorname{Ann}}_R(M)\setminus{\mathfrak p}\). The element \(a\) becomes a unit in \(R_{{\mathfrak p}}\), so \(M_{{\mathfrak p}}=0\).

Conversely, suppose \(M_{{\mathfrak p}}=0\). Choose generators \(m_1,\ldots,m_d\) of \(M\). For each \(i\), there is an element \(s_i\notin{\mathfrak p}\) such that \(s_im_i=0\). The product \(s=s_1\cdots s_d\) lies outside \({\mathfrak p}\) and annihilates every generator. Thus \(s\in{\operatorname{Ann}}_R(M)\setminus{\mathfrak p}\). This proves the equality.

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Proof of Theorem 6.3

If \(M=0\), the statement follows from the usual convention \({\operatorname{Fitt}}_0(0)=R\). Assume \(M\neq0\), and choose a finite presentation

with \(n\geq1\). The zeroth Fitting ideal is generated by the \(n\) by \(n\) minors of \(A\). After localization at a prime \({\mathfrak p}\), the module \(M_{{\mathfrak p}}\) vanishes exactly when \(A_{{\mathfrak p}}\) is surjective. A map onto the free module \(R_{{\mathfrak p}}^n\) is surjective exactly when one maximal minor is a unit. Thus \(M_{{\mathfrak p}}=0\) exactly when \({\operatorname{Fitt}}_0^R(M)\not\subseteq{\mathfrak p}\).

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Proof of Theorem 6.4

Use a finite presentation with target rank \(n\). If \(r>n\), every fiber has dimension at most \(n\), so the left hand locus is empty. The convention \({\operatorname{Fitt}}_{r-1}(M)=R\) for \(r-1\geq n\) makes the right hand locus empty as well. It remains to consider \(1\leq r\leq n\). At \({\mathfrak p}\), the fiber dimension is

It is at least \(r\) exactly when the rank is at most \(n-r\), which is equivalent to the vanishing of every minor of size \(n-r+1\). Those minors generate \({\operatorname{Fitt}}_{r-1}^R(M)\).

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Proof of Proposition 6.5

Base change carries a finite presentation matrix for \(M\) to a presentation matrix for \(M\otimes_RS\). Every defining minor is carried to the corresponding minor over \(S\).

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Proof of Theorem 6.7

By Theorem 6.2, full support is equivalent to

This holds exactly when \({\operatorname{Ann}}_R(M)\) is contained in every prime ideal, hence in the nilradical. Because \(R\) is Noetherian, the ideal \({\operatorname{Ann}}_R(M)\) is finitely generated. An ideal generated by finitely many nilpotent elements is nilpotent. This proves the equivalence of the first three conditions. If \(R\) is reduced, its nilradical is zero.

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Proof of Corollary 6.8

Every generic point is a prime of \(R\), and full support contains every prime.

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Proof of Theorem 6.9

Let \({\mathfrak p}\supseteq I\). Localizing gives

If \(M_{{\mathfrak p}}=0\), the quotient is zero. If \(M_{{\mathfrak p}}\neq0\), then \(I_{{\mathfrak p}}\) lies in the maximal ideal of the local ring \(R_{{\mathfrak p}}\). Nakayama's lemma shows that \(M_{{\mathfrak p}}/I_{{\mathfrak p}}M_{{\mathfrak p}}\neq0\). The stated equality follows.

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Proof of Theorem 6.11

Apply Theorem 6.9 with \(I={\mathfrak a}\). The support of the quotient is

which identifies with \({\operatorname{Spec}} R\). The remaining statements follow from Theorem 6.7.

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Proof of Corollary 6.12

The module \(M\) has full support by Theorem 6.11.

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Proof

The two nonassociate prime factors \(x\) and \(x-t\) occur with multiplicity one, so the principal quotient is reduced. The displayed isomorphism makes \(M\) free over \(S\). Its annihilator as an \(R\) module is \((x)\), hence

by Theorem 6.2. The generic point \((x-t)\) of the other component does not contain \(x\), so it is not in the support.

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Proof of Theorem 6.16

Choose a bounded complex \(P\) of finite projective \(R\) modules representing \(C\). Scalar extension to a commutative \(R\) algebra \(S\) is computed degree by degree as \(S\otimes_RP\); the canonical symmetry of tensor products gives the displayed order \(C\otimes_R^{\mathbf L}S\). Scalar extension preserves the bounds, finite generation, and projectivity of the terms. A derived isomorphism between two such projective models lifts to a homotopy equivalence, which scalar extension preserves. The resulting derived object and its residue field fibers are therefore independent of the chosen model.

Let \(q\in{\operatorname{Spec}}(R/I)\), and let \({\mathfrak p}\) be its inverse image in \(R\). The composite point \(R\to R/I\to\kappa(q)\) has kernel \({\mathfrak p}\), so it factors through a field embedding \(\kappa({\mathfrak p})\to\kappa(q)\). The ordinary tensor associator in every degree gives an isomorphism of complexes

Extension of fields is faithfully flat. It preserves and reflects exactness in each degree, hence also acyclicity. The left hand complex is thus nonzero in the derived category exactly when the residue field fiber of \(C\) at \({\mathfrak p}\) is nonzero. This proves the support identity on corresponding primes. The image of \({\operatorname{Spec}}(R/I)\to{\operatorname{Spec}} R\) is exactly \(V(I)\), giving the stated ambient spectrum equality, including the case \(I=R\).

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Proof of Theorem 6.18

Apply Theorem 6.16 to the patched complex and use the actual ring identification \(R\cong R_\infty/{\mathfrak a}\). Full patched support restricts to \(V({\mathfrak a})\), which corresponds to all of \({\operatorname{Spec}} R\). The supplied derived isomorphism with \(C\) preserves this support by the projective model argument in that theorem.

For a field valued point \(x:R\to E'\), put \({\mathfrak p}=\ker x\). The point factors as \(R\to\kappa({\mathfrak p})\to E'\), where the second map is a field embedding. Tensor association on a projective model identifies the fiber at \(x\) with the extension to \(E'\) of the fiber at \({\mathfrak p}\). The latter is nonzero by full support. Faithful flatness of the field extension reflects exactness in every degree, so the fiber at \(x\) is nonzero as well. This applies to every field valued point, with no closed point restriction.

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Proof of Corollary 6.19

The derived fiber is nonzero by Theorem 6.18. If every cohomology object vanished, a cochain model of that fiber would be acyclic. Its image in the derived category would then be zero, a contradiction. Hence some integer degree has nonzero cohomology.

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