Lemma 32.8.4. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is unramified, and

2. $f_0$ 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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