Lemma 93.15.1. In Example 93.8.1 let $P$ be a $k$-algebra. Assume that $k \to P$ is of finite type and that $\mathop{\mathrm{Spec}}(P) \to \mathop{\mathrm{Spec}}(k)$ is smooth except at the maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ n$ of $P$. Let $P_{\mathfrak m_ i}$, $P_{\mathfrak m_ i}^ h$, $P_{\mathfrak m_ i}^\wedge $ be the local ring, henselization, completion. Then the maps of deformation categories
\[ \mathcal{D}\! \mathit{ef}_ P \to \prod \mathcal{D}\! \mathit{ef}_{P_{\mathfrak m_ i}} \to \prod \mathcal{D}\! \mathit{ef}_{P_{\mathfrak m_ i}^ h} \to \prod \mathcal{D}\! \mathit{ef}_{P_{\mathfrak m_ i}^\wedge } \]
are smooth and induce isomorphisms on their finite dimensional tangent spaces.
Proof.
The tangent space is finite dimensional by Lemma 93.8.5. The functors between the categories are constructed in Lemmas 93.8.7, 93.8.8, and 93.8.10 (we omit some verifications of the form: the completion of the henselization is the completion).
Set $J = \mathfrak m_1 \cap \ldots \cap \mathfrak m_ n$ and apply Lemma 93.12.5 to get that $\mathcal{D}\! \mathit{ef}_ P \to \mathcal{D}\! \mathit{ef}_{P^\wedge }$ is smooth and induces an isomorphism on tangent spaces where $P^\wedge $ is the $J$-adic completion of $P$. However, since $P^\wedge = \prod P_{\mathfrak m_ i}^\wedge $ we see that the map $\mathcal{D}\! \mathit{ef}_ P \to \prod \mathcal{D}\! \mathit{ef}_{P_{\mathfrak m_ i}^\wedge }$ is smooth and induces an isomorphism on tangent spaces.
Let $(P^ h, J^ h)$ be the henselization of the pair $(P, J)$. Then $P^ h = \prod P_{\mathfrak m_ i}^ h$ (look at idempotents and use More on Algebra, Lemma 15.11.6). Hence we can apply Lemma 93.14.3 to conclude as in the case of completion.
To get the final case it suffices to show that $\mathcal{D}\! \mathit{ef}_{P_{\mathfrak m_ i}} \to \mathcal{D}\! \mathit{ef}_{P_{\mathfrak m_ i}^\wedge }$ is smooth and induce isomorphisms on tangent spaces for each $i$ separately. To do this, we may replace $P$ by a principal localization whose only singular point is a maximal ideal $\mathfrak m$ (corresponding to $\mathfrak m_ i$ in the original $P$). Then we can apply Lemma 93.13.3 with multiplicative subset $S = P \setminus \mathfrak m$ to conclude. Minor details omitted.
$\square$
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