Proof. Say $X \to S$ is unramified and flat. Then $\Delta : X \to X \times _ S X$ is an open immersion, see Morphisms, Lemma 29.35.13. We have to show that $\mathcal{C}_{X/S}$ is zero. Consider the two projections $p, q : X \times _ S X \to X$. As $f$ is formally unramified (see Lemma 37.6.8), $q$ is formally unramified (see Lemma 37.6.4). As $f$ is flat, $p$ is flat, see Morphisms, Lemma 29.25.8. Hence $p^*\mathcal{C}_{X/S} = \mathcal{C}_ q$ by Lemma 37.7.7 where $\mathcal{C}_ q$ denotes the conormal sheaf of the formally unramified morphism $q : X \times _ S X \to X$. But $\Delta (X) \subset X \times _ S X$ is an open subscheme which maps isomorphically to $X$ via $q$. Hence by Lemma 37.7.8 we see that $\mathcal{C}_ q|_{\Delta (X)} = \mathcal{C}_{X/X} = 0$. In other words, the pullback of $\mathcal{C}_{X/S}$ to $X$ via the identity morphism is zero, i.e., $\mathcal{C}_{X/S} = 0$. $\square$

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