Remark 60.13.1. In Situation 60.7.5. Let $(U, T, \delta )$ be an object of $\text{Cris}(X/S)$. Write $\Omega _{T/S, \delta } = (\Omega _{X/S})_ T$, see Lemma 60.12.3. We explicitly describe a first order thickening $T'$ of $T$. Namely, set
with algebra structure such that $\Omega _{T/S, \delta }$ is an ideal of square zero. Let $\mathcal{J} \subset \mathcal{O}_ T$ be the ideal sheaf of the closed immersion $U \to T$. Set $\mathcal{J}' = \mathcal{J} \oplus \Omega _{T/S, \delta }$. Define a divided power structure on $\mathcal{J}'$ by setting
see Lemma 60.3.1. There are two ring maps
The first is given by $f \mapsto (f, 0)$ and the second by $f \mapsto (f, \text{d}_{T/S, \delta }f)$. Note that both are compatible with the divided power structures on $\mathcal{J}$ and $\mathcal{J}'$ and so is the quotient map $\mathcal{O}_{T'} \to \mathcal{O}_ T$. Thus we get an object $(U, T', \delta ')$ of $\text{Cris}(X/S)$ and a commutative diagram
of $\text{Cris}(X/S)$ such that $i$ is a first order thickening whose ideal sheaf is identified with $\Omega _{T/S, \delta }$ and such that $p_1 - p_0 : \mathcal{O}_ T \to \mathcal{O}_{T'}$ is identified with the universal derivation $\text{d}_{T/S, \delta }$ composed with the inclusion $\Omega _{T/S, \delta } \to \mathcal{O}_{T'}$.
Comments (2)
Comment #8734 by Eiki on
Comment #9342 by Stacks Project on