Lemma 37.9.10. Consider a commutative diagram of first order thickenings

$\vcenter { \xymatrix{ (T_2 \subset T_2') \ar[d]_{(h, h')} \ar[rr]_{(a_2, a_2')} & & (X_2 \subset X_2') \ar[d]^{(f, f')} \\ (T_1 \subset T_1') \ar[rr]^{(a_1, a_1')} & & (X_1 \subset X_1') } } \quad \begin{matrix} \text{and a commutative} \\ \text{diagram of schemes} \end{matrix} \quad \vcenter { \xymatrix{ X_2' \ar[r] \ar[d] & S_2 \ar[d] \\ X_1' \ar[r] & S_1 } }$

with $X_2 \to X_1$ and $S_2 \to S_1$ étale. For any $\mathcal{O}_{T_1}$-linear map $\theta _1 : a_1^*\Omega _{X_1/S_1} \to \mathcal{C}_{T_1/T'_1}$ let $\theta _2$ be the composition

$\xymatrix{ a_2^*\Omega _{X_2/S_2} \ar@{=}[r] & h^*a_1^*\Omega _{X_1/S_1} \ar[r]^-{h^*\theta _1} & h^*\mathcal{C}_{T_1/T'_1} \ar[r] & \mathcal{C}_{T_2/T'_2} }$

(equality sign is explained in the proof). Then the diagram

$\xymatrix{ T_2' \ar[rr]_{\theta _2 \cdot a_2'} \ar[d] & & X'_2 \ar[d] \\ T_1' \ar[rr]^{\theta _1 \cdot a_1'} & & X'_1 }$

commutes where the actions $\theta _2 \cdot a_2'$ and $\theta _1 \cdot a_1'$ are as in Remark 37.9.3.

Proof. The equality sign comes from the identification $f^*\Omega _{X_1/S_1} = \Omega _{X_2/S_2}$ of Lemma 37.9.9. Namely, using this we have $a_2^*\Omega _{X_2/S_2} = a_2^*f^*\Omega _{X_1/S_1} = h^*a_1^*\Omega _{X_1/S_1}$ because $f \circ a_2 = a_1 \circ h$. Having said this, the commutativity of the diagram may be checked on affine opens. Hence we may assume the schemes in the initial big diagram are affine. Thus we obtain commutative diagrams

$\vcenter { \xymatrix{ (B'_2, I_2) & & (A'_2, J_2) \ar[ll]^{a_2'} \\ (B'_1, I_1) \ar[u]^{h'} & & (A'_1, J_1) \ar[ll]_{a_1'} \ar[u]_{f'} } } \quad \text{and}\quad \vcenter { \xymatrix{ A'_2 & & R_2 \ar[ll] \\ A'_1 \ar[u] & & R_1 \ar[ll] \ar[u] } }$

The notation signifies that $I_1, I_2, J_1, J_2$ are ideals of square zero and maps of pairs are ring maps sending ideals into ideals. Set $A_1 = A'_1/J_1$, $A_2 = A'_2/J_2$, $B_1 = B'_1/I_1$, and $B_2 = B'_2/I_2$. We are given that

$A_2 \otimes _{A_1} \Omega _{A_1/R_1} \longrightarrow \Omega _{A_2/R_2}$

is an isomorphism. Then $\theta _1 : B_1 \otimes _{A_1} \Omega _{A_1/R_1} \to I_1$ is $B_1$-linear. This gives an $R_1$-derivation $D_1 = \theta _1 \circ \text{d}_{A_1/R_1} : A_1 \to I_1$. In a similar way we see that $\theta _2 : B_2 \otimes _{A_2} \Omega _{A_2/R_2} \to I_2$ gives rise to a $R_2$-derivation $D_2 = \theta _2 \circ \text{d}_{A_2/R_2} : A_2 \to I_2$. The construction of $\theta _2$ implies the following compatibility between $\theta _1$ and $\theta _2$: for every $x \in A_1$ we have

$h'(D_1(x)) = D_2(f'(x))$

as elements of $I_2$. We may view $D_1$ as a map $A'_1 \to B'_1$ using $A'_1 \to A_1 \xrightarrow {D_1} I_1 \to B_1$ similarly we may view $D_2$ as a map $A'_2 \to B'_2$. Then the displayed equality holds for $x \in A'_1$. By the construction of the action in Lemma 37.9.2 and Remark 37.9.3 we know that $\theta _1 \cdot a_1'$ corresponds to the ring map $a_1' + D_1 : A'_1 \to B'_1$ and $\theta _2 \cdot a_2'$ corresponds to the ring map $a_2' + D_2 : A'_2 \to B'_2$. By the displayed equality we obtain that $h' \circ (a_1' + D_1) = (a_2' + D_2) \circ f'$ as desired. $\square$

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