Lemma 32.8.4. Notation and assumptions as in Situation 32.8.1. If
f is unramified, and
f_0 is locally of finite type,
then there exists an i \geq 0 such that f_ i is unramified.
Lemma 32.8.4. Notation and assumptions as in Situation 32.8.1. If
f is unramified, and
f_0 is locally of finite type,
then there exists an i \geq 0 such that f_ i is unramified.
Proof. Choose a finite affine open covering Y_0 = \bigcup _{j = 1, \ldots , m} Y_{j, 0} such that each Y_{j, 0} maps into an affine open S_{j, 0} \subset S_0. For each j let f_0^{-1}Y_{j, 0} = \bigcup _{k = 1, \ldots , n_ j} X_{k, 0} be a finite affine open covering. Since the property of being unramified is local we see that it suffices to prove the lemma for the morphisms of affines X_{k, i} \to Y_{j, i} \to S_{j, i} which are the base changes of X_{k, 0} \to Y_{j, 0} \to S_{j, 0} to S_ i. Thus we reduce to the case that X_0, Y_0, S_0 are affine
In the affine case we reduce to the following algebra result. Suppose that R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i. For some 0 \in I suppose given an R_0-algebra map A_ i \to B_ i of finite type. If R \otimes _{R_0} A_0 \to R \otimes _{R_0} B_0 is unramified, then for some i \geq 0 the map R_ i \otimes _{R_0} A_0 \to R_ i \otimes _{R_0} B_0 is unramified. This follows from Algebra, Lemma 10.168.5. \square
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