Lemma 50.15.7. Let $X \to T \to S$ be morphisms of schemes. Let $Y \subset X$ be an effective Cartier divisor. If both $X \to T$ and $Y \to T$ are smooth, then the de Rham complex of log poles is defined for $Y \subset X$ over $S$.
Proof. Let $y \in Y$ be a point. By More on Morphisms, Lemma 37.17.1 there exists an integer $0 \geq m$ and a commutative diagram
\[ \xymatrix{ Y \ar[d] & V \ar[l] \ar[d] \ar[r] & \mathbf{A}^ m_ T \ar[d]^{(a_1, \ldots , a_ m) \mapsto (a_1, \ldots , a_ m, 0)} \\ X & U \ar[l] \ar[r]^-\pi & \mathbf{A}^{m + 1}_ T } \]
where $U \subset X$ is open, $V = Y \cap U$, $\pi $ is étale, $V = \pi ^{-1}(\mathbf{A}^ m_ T)$, and $y \in V$. Denote $z \in \mathbf{A}^ m_ T$ the image of $y$. Then we have
\[ \Omega ^ p_{X/S, y} = \Omega ^ p_{\mathbf{A}^{m + 1}_ T/S, z} \otimes _{\mathcal{O}_{\mathbf{A}^{m + 1}_ T, z}} \mathcal{O}_{X, x} \]
by Lemma 50.2.2. Denote $x_1, \ldots , x_{m + 1}$ the coordinate functions on $\mathbf{A}^{m + 1}_ T$. Since the conditions (1) and (2) in Definition 50.15.1 do not depend on the choice of the local coordinate, it suffices to check the conditions (1) and (2) when $f$ is the image of $x_{m + 1}$ by the flat local ring homomorphism $\mathcal{O}_{\mathbf{A}^{m + 1}_ T, z} \to \mathcal{O}_{X, x}$. In this way we see that it suffices to check conditions (1) and (2) for $\mathbf{A}^ m_ T \subset \mathbf{A}^{m + 1}_ T$ and the point $z$. To prove this case we may assume $S = \mathop{\mathrm{Spec}}(A)$ and $T = \mathop{\mathrm{Spec}}(B)$ are affine. Let $A \to B$ be the ring map corresponding to the morphism $T \to S$ and set $P = B[x_1, \ldots , x_{m + 1}]$ so that $\mathbf{A}^{m + 1}_ T = \mathop{\mathrm{Spec}}(P)$. We have
\[ \Omega _{P/A} = \Omega _{B/A} \otimes _ B P \oplus \bigoplus \nolimits _{j = 1, \ldots , m} P \text{d}x_ j \oplus P \text{d}x_{m + 1} \]
Hence the map $P \to \Omega _{P/A}$, $g \mapsto g \text{d}x_{m + 1}$ is a split injection and $x_{m + 1}$ is a nonzerodivisor on $\Omega ^ p_{P/A}$ for all $p \geq 0$. Localizing at the prime ideal corresponding to $z$ finishes the proof. $\square$
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