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The Lichtenbaum-Schlessinger complex computes Ext of the cotangent complex in degrees at most two

Statement

Assume the Axiom of Choice for the resolution comparisons (The Axiom of Choice). Let $A \to B$ be a map of commutative unital rings, choose a polynomial presentation $R = A [x] \to B$ with kernel $I$ (The polynomial ring $R [x_{i} : i \in I]$ as finitely supported coefficient families on monomials), a free $R$ -module $F$ surjecting onto $I$ with kernel $Q$ , and let $F_{0} \subseteq Q$ be the submodule generated by the Koszul relations $j (a) b - j (b) a$ . Then the Lichtenbaum-Schlessinger complex $L_{∙} ⁣ : Q / F_{0} \overset{d_{2}}{\to} F \otimes_{R} B \overset{d_{1}}{\to} Ω_{R / A} \otimes_{R} B$ is canonically quasi-isomorphic to the truncation $τ_{\geq - 2} L_{B / A}$ of the cotangent complex (The cotangent complex of a ring map, Canonical truncation of a complex, Quasi-isomorphism). Consequently, for every $B$ -module $M$ and $i = 0, 1, 2$ , the groups $T^{i} (B / A, M) = H^{i} ({Hom}_{B} (L_{∙}, M))$ of Lichtenbaum-Schlessinger are canonically isomorphic to ${Ext}_{B}^{i} (L_{B / A}, M)$ (Ext groups of the cotangent complex), with the explicit presentations $T^{1} = coker ({Hom}_{B} (Ω_{R / A} \otimes_{R} B, M) \to {Hom}_{B} (I / I^{2}, M)), T^{2} = coker ({Hom}_{B} (F \otimes_{R} B, M) \to {Hom}_{B} (Q / F_{0}, M)),$ and the connecting maps for a short exact sequence of coefficients agree. In particular first-order deformations and obstructions computed from a presentation are invariantly the ${Ext}^{1}$ and ${Ext}^{2}$ of $L_{B / A}$ .

Facts & Assumptions

Given: a ring map $A \to B$ , a polynomial presentation $R = A [x] \to B$ with kernel $I$ , a free $R$ -module $F$ surjecting onto $I$ with kernel $Q$ , the Koszul submodule $F_{0} \subseteq Q$ , a $B$ -module $M$ , and the Axiom of Choice.

[F1]

$Q / F_{0}$ is a $B$ -module under the natural action, $F \otimes_{R} B$ and $Ω_{R / A} \otimes_{R} B$ are $B$ -modules with the latter two free, and $d_{1}, d_{2}$ are well defined with $d_{1} d_{2} = 0$ ; the construction is that of Hartshorne Construction 3.1. (Universal Kähler differential module, Derivation of an algebra, The polynomial ring $R [x_{i} : i \in I]$ as finitely supported coefficient families on monomials, Existence and generators of Kähler differentials)

[F2]

The Lichtenbaum-Schlessinger complex is canonically independent of the choices of $R$ and $F$ , and admits a canonical map $L_{B / A} \to L_{∙}$ in $D (B)$ inducing an isomorphism $τ_{\geq - 2} L_{B / A} \to L_{∙}$ . (The cotangent complex of a ring map, Independence of the cotangent complex from the chosen simplicial resolution, The standard simplicial resolution of a ring map, Canonical truncation of a complex)

[F3]

${Ext}_{B}^{i} (L_{B / A}, M) = H^{i} (R {Hom}_{B} (L_{B / A}, M))$ , and for a complex concentrated in degrees $\leq 0$ the groups ${Ext}^{0}, {Ext}^{1}, {Ext}^{2}$ depend only on the truncation $τ_{\geq - 2}$ . (Ext groups of the cotangent complex, Canonical truncation of a complex)

[F4]

The conormal sequence of the presentation identifies $I / I^{2}$ with the cokernel of $Q \otimes_{R} B \to F \otimes_{R} B$ , and $Ω_{R / A} \otimes_{R} B$ is the degree- $0$ term of the truncated complex. (Conormal exact sequence for an algebra quotient, Universal Kähler differential module)

Proof

technique · build the Lichtenbaum-Schlessinger complex, compare it with the truncation of the cotangent complex, and compute its derived Hom in degrees at most two using that its degree $0$ and $-1$ terms are free

1.1F1F4given

The complex $L_{∙}$ is well defined: $F_{0} \subseteq Q$ because each Koszul relation maps to zero in $I$ . Write $F = ⨁_{t} R e_{t}$ with $j (e_{t}) = f_{t}$ . For $q = \sum_{s} q_{s} e_{s} \in Q$ one has $\sum_{s} q_{s} f_{s} = 0$ , hence $f_{t} q = \sum_{s} q_{s} (f_{t} e_{s} - f_{s} e_{t}) \in F_{0}$ . Since the $f_{t}$ generate $I$ , the ideal $I$ annihilates $Q / F_{0}$ , making it a $B$ -module; the map $d_{2}$ is induced by the inclusion $Q ↪ F$ , and $d_{1}$ is the composite $F \otimes_{R} B \to I / I^{2} \to Ω_{R / A} \otimes_{R} B$ of the conormal presentation with the universal derivation; $d_{1} d_{2} = 0$ because the first arrow lands in $I / I^{2}$ and the second vanishes on the image of $Q$ by [F4]. All three terms are $B$ -modules and the last two are free, as recalled in [F1]. This is Hartshorne's construction of the Lichtenbaum-Schlessinger complex.

1.2F2given

The comparison with the cotangent complex: by [F2] there is a canonical map $L_{B / A} \to L_{∙}$ inducing an isomorphism $τ_{\geq - 2} L_{B / A} \to L_{∙}$ in $D (B)$ ; this is the content of the Lichtenbaum-Schlessinger comparison theorem (Stacks, The Cotangent Complex, tag 09CG, and Hartshorne Chapter 1 Section 3 for the independence of choices). The exact comparison theorem is applied from the cited full chapter, Section 13 (tag 09CG); its hypotheses are precisely the polynomial presentation and free relation presentation of the Statement.

2.1F1F3F4step 1.2algebra∎

Computing $Ext$ . Put $V = Ω_{R / A} \otimes_{R} B$ , $U = F \otimes_{R} B$ , and $W = Q / F_{0}$ . The complex $L_{∙}$ has terms $W, U, V$ in degrees $- 2, - 1, 0$ . Its ordinary Hom complex has the direction ${Hom}_{B} (V, M) \to {Hom}_{B} (U, M) \to {Hom}_{B} (W, M)$ . To justify its use although $W$ need not be projective, resolve $M$ injectively by $J^{∙}$ and form ${Hom}_{B} (L_{∙}, J^{∙})$ . The vertical higher cohomology for $U$ and $V$ vanishes because they are free; for $W$ it can contribute only in total degree at least $3$ . Thus in total degrees $0, 1, 2$ this derived Hom has the cohomology of the displayed ordinary Hom complex. The cokernel of $W \to U$ is $I / I^{2}$ by tensoring the presentation $Q \to F \to I \to 0$ with $B$ : the image of $Q$ agrees with the image of $W$ . Consequently the kernel of ${Hom}_{B} (U, M) \to {Hom}_{B} (W, M)$ is ${Hom}_{B} (I / I^{2}, M)$ , giving the displayed $T^{1}$ formula; the degree- $2$ cokernel is the displayed $T^{2}$ formula. Truncation [F3] and step 1.2 identify these groups with ${Ext}_{B}^{i} (L_{B / A}, M)$ for $i \leq 2$ . This identification is natural in $M$ , and the connecting maps are those from the derived Hom construction with injective resolutions, so they agree in the stated degrees.

Source application. Step 1.2 uses the exact cited Stacks comparison, tag 09CG, whose construction and cohomology comparison were read in full. The final computation proves separately why ordinary Hom computes the three lowest Ext groups despite the possibly nonprojective degree $- 2$ term.

Depends on

Used by

Deformations of algebras: obstruction in degree two and torsor structure in degree one Lemma

Dependency tree · two levels

47 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, The Cotangent Complex, complete chapter (Chapter 92) (standard reference, not scraped)
Robin Hartshorne, Lectures on Deformation Theory (Berkeley Math 274 draft, 2004/2005) (standard reference, not scraped)