Lemma 91.8.1. Let $A \to A' \to B$ be ring maps such that $B = B \otimes _ A^\mathbf {L} A'$. Then $L_{B/A} = L_{B/A'}$ in $D(B)$.

## 91.8 Localization and étale ring maps

In this section we study what happens if we localize our rings. Let $A \to A' \to B$ be ring maps such that $B = B \otimes _ A^\mathbf {L} A'$. This happens for example if $A' = S^{-1}A$ is the localization of $A$ at a multiplicative subset $S \subset A$. In this case for an abelian sheaf $\mathcal{F}'$ on $\mathcal{C}_{B/A'}$ the homology of $g^{-1}\mathcal{F}'$ over $\mathcal{C}_{B/A}$ agrees with the homology of $\mathcal{F}'$ over $\mathcal{C}_{B/A'}$, see Lemma 91.6.1 for a precise statement.

**Proof.**
According to the discussion above (i.e., using Lemma 91.6.1) and Lemma 91.4.3 we have to show that the sheaf given by the rule $(P \to B) \mapsto \Omega _{P/A} \otimes _ P B$ on $\mathcal{C}_{B/A}$ is the pullback of the sheaf given by the rule $(P \to B) \mapsto \Omega _{P/A'} \otimes _ P B$. The pullback functor $g^{-1}$ is given by precomposing with the functor $u : \mathcal{C}_{B/A} \to \mathcal{C}_{B/A'}$, $(P \to B) \mapsto (P \otimes _ A A' \to B)$. Thus we have to show that

By Algebra, Lemma 10.131.12 the right hand side is equal to

Since $P$ is a polynomial algebra over $A$ the module $\Omega _{P/A}$ is free and the equality is obvious. $\square$

Lemma 91.8.2. Let $A \to B$ be a ring map such that $B = B \otimes _ A^\mathbf {L} B$. Then $L_{B/A} = 0$ in $D(B)$.

**Proof.**
This is true because $L_{B/A} = L_{B/B} = 0$ by Lemmas 91.8.1 and 91.4.7.
$\square$

Lemma 91.8.3. Let $A \to B$ be a ring map such that $\text{Tor}^ A_ i(B, B) = 0$ for $i > 0$ and such that $L_{B/B \otimes _ A B} = 0$. Then $L_{B/A} = 0$ in $D(B)$.

**Proof.**
By Lemma 91.6.2 we see that $L_{B/A} \otimes _ B^\mathbf {L} (B \otimes _ A B) = L_{B \otimes _ A B/B}$. Now we use the distinguished triangle (91.7.0.1)

associated to the ring maps $B \to B \otimes _ A B \to B$ and the vanishing of $L_{B/B}$ (Lemma 91.4.7) and $L_{B/B \otimes _ A B}$ (assumed) to see that

as desired. $\square$

Lemma 91.8.4. The cotangent complex $L_{B/A}$ is zero in each of the following cases:

$A \to B$ and $B \otimes _ A B \to B$ are flat, i.e., $A \to B$ is weakly étale (More on Algebra, Definition 15.104.1),

$A \to B$ is a flat epimorphism of rings,

$B = S^{-1}A$ for some multiplicative subset $S \subset A$,

$A \to B$ is unramified and flat,

$A \to B$ is étale,

$A \to B$ is a filtered colimit of ring maps for which the cotangent complex vanishes,

$B$ is a henselization of a local ring of $A$,

$B$ is a strict henselization of a local ring of $A$, and

add more here.

**Proof.**
In case (1) we may apply Lemma 91.8.2 to the surjective flat ring map $B \otimes _ A B \to B$ to conclude that $L_{B/B \otimes _ A B} = 0$ and then we use Lemma 91.8.3 to conclude. The cases (2) – (5) are each special cases of (1). Part (6) follows from Lemma 91.3.4. Parts (7) and (8) follows from the fact that (strict) henselizations are filtered colimits of étale ring extensions of $A$, see Algebra, Lemmas 10.155.7 and 10.155.11.
$\square$

Lemma 91.8.5. Let $A \to B \to C$ be ring maps such that $L_{C/B} = 0$. Then $L_{C/A} = L_{B/A} \otimes _ B^\mathbf {L} C$.

**Proof.**
This is a trivial consequence of the distinguished triangle (91.7.0.1).
$\square$

Lemma 91.8.6. Let $A \to B$ be ring maps and $S \subset A$, $T \subset B$ multiplicative subsets such that $S$ maps into $T$. Then $L_{T^{-1}B/S^{-1}A} = L_{B/A} \otimes _ B T^{-1}B$ in $D(T^{-1}B)$.

**Proof.**
Lemma 91.8.5 shows that $L_{T^{-1}B/A} = L_{B/A} \otimes _ B T^{-1}B$ and Lemma 91.8.1 shows that $L_{T^{-1}B/A} = L_{T^{-1}B/S^{-1}A}$.
$\square$

Lemma 91.8.7. Let $A \to B$ be a local ring homomorphism of local rings. Let $A^ h \to B^ h$, resp. $A^{sh} \to B^{sh}$ be the induced maps of henselizations, resp. strict henselizations. Then

in $D(B^ h)$, resp. $D(B^{sh})$.

**Proof.**
The complexes $L_{A^ h/A}$, $L_{A^{sh}/A}$, $L_{B^ h/B}$, and $L_{B^{sh}/B}$ are all zero by Lemma 91.8.4. Using the fundamental distinguished triangle (91.7.0.1) for $A \to B \to B^ h$ we obtain $L_{B^ h/A} = L_{B/A} \otimes _ B^\mathbf {L} B^ h$. Using the fundamental triangle for $A \to A^ h \to B^ h$ we obtain $L_{B^ h/A^ h} = L_{B^ h/A}$. Similarly for strict henselizations.
$\square$

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