Lemma 37.6.7. Let $f : X \to S$ be a morphism of schemes. Then $f$ is formally unramified if and only if $\Omega _{X/S} = 0$.

**Proof.**
We recall some of the arguments of the proof of Morphisms, Lemma 29.32.5. Let $W \subset X \times _ S X$ be an open such that $\Delta : X \to X \times _ S X$ induces a closed immersion into $W$. Let $\mathcal{J} \subset \mathcal{O}_ W$ be the ideal sheaf of this closed immersion. Let $X' \subset W$ be the closed subscheme defined by the quasi-coherent sheaf of ideals $\mathcal{J}^2$. Consider the two morphisms $p_1, p_2 : X' \to X$ induced by the two projections $X \times _ S X \to X$. Note that $p_1$ and $p_2$ agree when composed with $\Delta : X \to X'$ and that $X \to X'$ is a closed immersion defined by a an ideal whose square is zero. Moreover there is a short exact sequence

and $\Omega _{X/S} = \mathcal{J}/\mathcal{J}^2$. Moreover, $\mathcal{J}/\mathcal{J}^2$ is generated by the local sections $p_1^\sharp (f) - p_2^\sharp (f)$ for $f$ a local section of $\mathcal{O}_ X$.

Suppose that $f : X \to S$ is formally unramified. By assumption this means that $p_1 = p_2$ when restricted to any affine open $T' \subset X'$. Hence $p_1 = p_2$. By what was said above we conclude that $\Omega _{X/S} = \mathcal{J}/\mathcal{J}^2 = 0$.

Conversely, suppose that $\Omega _{X/S} = 0$. Then $X' = X$. Take any pair of morphisms $f'_1, f'_2 : T' \to X$ fitting as dotted arrows in the diagram of Definition 37.6.1. This gives a morphism $(f'_1, f'_2) : T' \to X \times _ S X$. Since $f'_1|_ T = f'_2|_ T$ and $|T| =|T'|$ we see that the image of $T'$ under $(f'_1, f'_2)$ is contained in the open $W$ chosen above. Since $(f'_1, f'_2)(T) \subset \Delta (X)$ and since $T$ is defined by an ideal of square zero in $T'$ we see that $(f'_1, f'_2)$ factors through $X'$. As $X' = X$ we conclude $f_1' = f'_2$ as desired. $\square$

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