Lemma 90.9.3. Let $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. The following are equivalent
$\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth,
$\Lambda \to R$ is formally smooth in the $\mathfrak m_ R$-adic topology,
$\Lambda \to R$ is flat and $R \otimes _\Lambda k'$ is geometrically regular over $k'$, and
$\Lambda \to R$ is flat and $k' \to R \otimes _\Lambda k'$ is formally smooth in the $\mathfrak m_ R$-adic topology.
In the classical case, these are also equivalent to
$R$ is isomorphic to $\Lambda [[x_1, \ldots , x_ n]]$ for some $n$.
Proof.
Smoothness of $p : \underline{R}|_{\mathcal{C}_\Lambda } \to \mathcal{C}_\Lambda $ means that given $B \to A$ surjective in $\mathcal{C}_\Lambda $ and given $R \to A$ we can find the dotted arrow in the diagram
\[ \xymatrix{ R \ar[r] \ar@{-->}[rd] & A \\ \Lambda \ar[r] \ar[u] & B \ar[u] } \]
This is certainly true if $\Lambda \to R$ is formally smooth in the $\mathfrak m_ R$-adic topology, see More on Algebra, Definitions 15.37.3 and 15.37.1. Conversely, if this holds, then we see that $\Lambda \to R$ is formally smooth in the $\mathfrak m_ R$-adic topology by More on Algebra, Lemma 15.38.1. Thus (1) and (2) are equivalent.
The equivalence of (2), (3), and (4) is More on Algebra, Proposition 15.40.5. The equivalence with (5) follows for example from Lemma 90.8.6 and the fact that $\mathcal{C}_\Lambda $ is the same as $\underline{\Lambda }|_{\mathcal{C}_\Lambda }$ in the classical case.
$\square$
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