Lemma 37.7.4. Let $Z \to X$ be a formally unramified morphism of schemes. Then the universal first order thickening $Z'$ is formally unramified over $X$.

Proof. There are two proofs. The first is to show that $\Omega _{Z'/X} = 0$ by working affine locally and applying Algebra, Lemma 10.149.5. Then Lemma 37.6.7 implies what we want. The second is a direct argument as follows.

Let $T \subset T'$ be a first order thickening. Let

$\xymatrix{ Z' \ar[d] & T \ar[l]^ c \ar[d] \\ X & T' \ar[l] \ar[lu]^{a, b} }$

be a commutative diagram. Consider two morphisms $a, b : T' \to Z'$ fitting into the diagram. Set $T_0 = c^{-1}(Z) \subset T$ and $T'_ a = a^{-1}(Z)$ (scheme theoretically). Since $Z'$ is a first order thickening of $Z$, we see that $T'$ is a first order thickening of $T'_ a$. Moreover, since $c = a|_ T$ we see that $T_0 = T \cap T'_ a$ (scheme theoretically). As $T'$ is a first order thickening of $T$ it follows that $T'_ a$ is a first order thickening of $T_0$. Now $a|_{T'_ a}$ and $b|_{T'_ a}$ are morphisms of $T'_ a$ into $Z'$ over $X$ which agree on $T_0$ as morphisms into $Z$. Hence by the universal property of $Z'$ we conclude that $a|_{T'_ a} = b|_{T'_ a}$. Thus $a$ and $b$ are morphism from the first order thickening $T'$ of $T'_ a$ whose restrictions to $T'_ a$ agree as morphisms into $Z$. Thus using the universal property of $Z'$ once more we conclude that $a = b$. In other words, the defining property of a formally unramified morphism holds for $Z' \to X$ as desired. $\square$

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