Lemma 10.137.9. A composition of standard smooth ring maps is standard smooth.

Proof. Suppose that $R \to S$ and $S \to S'$ are standard smooth. We choose presentations $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and $S' = S[y_1, \ldots , y_ m]/(g_1, \ldots , g_ d)$. Choose elements $g_ j' \in R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ mapping to the $g_ j$. In this way we see $S' = R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]/ (f_1, \ldots , f_ c, g'_1, \ldots , g'_ d)$. To show that $S'$ is standard smooth it suffices to verify that the determinant

$\det \left( \begin{matrix} \partial f_1/\partial x_1 & \ldots & \partial f_ c/\partial x_1 & \partial g_1/\partial x_1 & \ldots & \partial g_ d/\partial x_1 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_ c & \ldots & \partial f_ c/\partial x_ c & \partial g_1/\partial x_ c & \ldots & \partial g_ d/\partial x_ c \\ 0 & \ldots & 0 & \partial g_1/\partial y_1 & \ldots & \partial g_ d/\partial y_1 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 0 & \partial g_1/\partial y_ d & \ldots & \partial g_ d/\partial y_ d \end{matrix} \right)$

is invertible in $S'$. This is clear since it is the product of the two determinants which were assumed to be invertible by hypothesis. $\square$

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