Lemma 10.168.5. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

$A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is unramified,

$C_0$ is of finite type over $B_0$.

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is unramified.

**Proof.**
Set $B_ i = A_ i \otimes _{A_0} B_0$, $C_ i = A_ i \otimes _{A_0} C_0$, $B = A \otimes _{A_0} B_0$, and $C = A \otimes _{A_0} C_0$. Let $x_1, \ldots , x_ m$ be generators for $C_0$ over $B_0$. Then $\text{d}x_1, \ldots , \text{d}x_ m$ generate $\Omega _{C_0/B_0}$ over $C_0$ and their images generate $\Omega _{C_ i/B_ i}$ over $C_ i$ (Lemmas 10.131.14 and 10.131.9). Observe that $0 = \Omega _{C/B} = \mathop{\mathrm{colim}}\nolimits \Omega _{C_ i/B_ i}$ (Lemma 10.131.5). Thus there is an $i$ such that $\text{d}x_1, \ldots , \text{d}x_ m$ map to zero and hence $\Omega _{C_ i/B_ i} = 0$ as desired.
$\square$

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