Lemma 15.85.1. Let R \to S and S \to S' be ring maps. The canonical map \mathop{N\! L}\nolimits _{S/R} \otimes _ S^\mathbf {L} S' \to \mathop{N\! L}\nolimits _{S/R} \otimes _ S S' induces an isomorphism \tau _{\geq -1}(\mathop{N\! L}\nolimits _{S/R} \otimes _ S^\mathbf {L} S') \to \mathop{N\! L}\nolimits _{S/R} \otimes _ S S' in D(S'). Similarly, given a presentation \alpha of S over R the canonical map \mathop{N\! L}\nolimits (\alpha ) \otimes _ S^\mathbf {L} S' \to \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' induces an isomorphism \tau _{\geq -1}(\mathop{N\! L}\nolimits (\alpha ) \otimes _ S^\mathbf {L} S') \to \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' in D(S').
15.85 The naive cotangent complex
In this section we continue the discussion started in Algebra, Section 10.134. We begin with a discussion of base change. The first lemma shows that taking the naive tensor product of the naive cotangent complex with a ring extension isn't quite as naive as one might think.
Proof. Special case of Lemma 15.84.6. \square
Lemma 15.85.2. Let R \to S and R \to R' be ring maps. Let \alpha : P \to S be a presentation of S over R. Then \alpha ' : P \otimes _ R R' \to S \otimes _ R R' is a presentation of S' = S \otimes _ R R' over R'. The canonical map
NL(\alpha ) \otimes _ S S' \to \mathop{N\! L}\nolimits (\alpha ')
is an isomorphism on H^0 and surjective on H^{-1}. In particular, the canonical map
\mathop{N\! L}\nolimits _{S/R} \otimes _ S S' \to \mathop{N\! L}\nolimits _{S'/R'}
is an isomorphism on H^0 and surjective on H^{-1}.
Proof. Denote I = \mathop{\mathrm{Ker}}(P \to S). Denote P' = P \otimes _ R R' and I' = \mathop{\mathrm{Ker}}(P' \to S'). Suppose P is a polynomial algebra on x_ j for j \in J. The map displayed in the lemma becomes
\xymatrix{ \bigoplus _{j \in J} S' \text{d}x_ j \ar[r] & \bigoplus _{j \in J} S' \text{d}x_ j \\ I/I^2 \otimes _ S S' \ar[r] \ar[u] & I'/(I')^2 \ar[u] }
where the left column is \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' and the right column is \mathop{N\! L}\nolimits (\alpha '). By right exactness of tensor product we see that I \otimes _ R R' \to I' is surjective. Hence the bottom arrow is a surjection. This proves the first statement of the lemma. The statement for \mathop{N\! L}\nolimits _{S/R} \otimes _ S S' \to \mathop{N\! L}\nolimits _{S'/R'} follows as these complexes are homotopic to \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S' and \mathop{N\! L}\nolimits (\alpha '). \square
Lemma 15.85.3. Consider a cocartesian diagram of rings
\xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] }
If B is flat over A, then the canonical map \mathop{N\! L}\nolimits _{B/A} \otimes _ B B' \to \mathop{N\! L}\nolimits _{B'/A'} is a quasi-isomorphism. If in addition \mathop{N\! L}\nolimits _{B/A} has tor-amplitude in [-1, 0] then \mathop{N\! L}\nolimits _{B/A} \otimes _ B^\mathbf {L} B' \to \mathop{N\! L}\nolimits _{B'/A'} is a quasi-isomorphism too.
Proof. Choose a presentation \alpha : P \to B as in Algebra, Section 10.134. Let I = \mathop{\mathrm{Ker}}(\alpha ). Set P' = P \otimes _ A A' and denote \alpha ' : P' \to B' the corresponding presentation of B' over A'. As B is flat over A we see that I' = \mathop{\mathrm{Ker}}(\alpha ') is equal to I \otimes _ A A'. Hence
I'/(I')^2 = \mathop{\mathrm{Coker}}(I^2 \otimes _ A A' \to I \otimes _ A A') = I/I^2 \otimes _ A A' = I/I^2 \otimes _ B B'
We have \Omega _{P'/A'} = \Omega _{P/A} \otimes _ A A' because both sides have the same basis. It follows that \Omega _{P'/A'} \otimes _{P'} B' = \Omega _{P/A} \otimes _ P B \otimes _ B B'. This proves that \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B' \to \mathop{N\! L}\nolimits (\alpha ') is an isomorphism of complexes and hence the first statement holds.
We have
\mathop{N\! L}\nolimits (\alpha ) = I/I^2 \longrightarrow \Omega _{P/A} \otimes _ P B
as a complex of B-modules with I/I^2 placed in degree -1. Since the term in degree 0 is free, this complex has tor-amplitude in [-1, 0] if and only if I/I^2 is a flat B-module, see Lemma 15.66.2. If this holds, then \mathop{N\! L}\nolimits (\alpha ) \otimes _ B^\mathbf {L} B' = \mathop{N\! L}\nolimits (\alpha ) \otimes _ B B' and we get the second statement. \square
Lemma 15.85.4. Let A \to B be a local complete intersection as in Definition 15.33.2. Then \mathop{N\! L}\nolimits _{B/A} is a perfect object of D(B) with tor amplitude in [-1, 0].
Proof. Write B = A[x_1, \ldots , x_ n]/I. Then \mathop{N\! L}\nolimits _{B/A} is represented by the complex
I/I^2 \longrightarrow \bigoplus B \text{d}x_ i
of B-modules with I/I^2 placed in degree -1. Since the term in degree 0 is finite free, this complex has tor-amplitude in [-1, 0] if and only if I/I^2 is a flat B-module, see Lemma 15.66.2. By definition I is a Koszul regular ideal and hence a quasi-regular ideal, see Section 15.32. Thus I/I^2 is a finite projective B-module (Lemma 15.32.3) and we conclude both that \mathop{N\! L}\nolimits _{B/A} is perfect and that it has tor amplitude in [-1, 0]. \square
Lemma 15.85.5. Consider a cocartesian diagram of rings
\xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] }
If A \to B and A' \to B' are local complete intersections as in Definition 15.33.2, then the kernel of H^{-1}(\mathop{N\! L}\nolimits _{B/A} \otimes _ B B') \to H^{-1}(\mathop{N\! L}\nolimits _{B'/A'}) is a finite projective B'-module.
Proof. By Lemma 15.85.4 the complexes \mathop{N\! L}\nolimits _{B/A} and \mathop{N\! L}\nolimits _{B'/A'} are perfect of tor-amplitude in [-1, 0]. Combining Lemmas 15.85.1, 15.74.9, and 15.66.13 we have \mathop{N\! L}\nolimits _{B/A} \otimes _ B B' = \mathop{N\! L}\nolimits _{B/A} \otimes _ B^\mathbf {L} B' and this complex is also perfect of tor-amplitude in [-1, 0]. Choose a distinguished triangle
C \to \mathop{N\! L}\nolimits _{B/A} \otimes _ B B' \to \mathop{N\! L}\nolimits _{B'/A'} \to C[1]
in D(B'). By Lemmas 15.74.4 and 15.66.5 we conclude that C is perfect with tor-amplitude in [-1, 1]. By Lemma 15.85.2 the complex C has only one nonzero cohomology module, namely the module of the lemma sitting in degree -1. This module is of finite presentation (Lemma 15.64.4) and flat (Lemma 15.66.6). Hence it is finite projective by Algebra, Lemma 10.78.2. \square
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