## 88.9 Smooth or unobstructed categories

Let $p : \mathcal{F} \to \mathcal{C}_\Lambda$ be a category cofibered in groupoids. We can consider $\mathcal{C}_\Lambda$ as a category cofibered in groupoids over $\mathcal{C}_\Lambda$ using the identity functor. In this way $p : \mathcal{F} \longrightarrow \mathcal{C}_\Lambda$ becomes a morphism of categories cofibered in groupoids over $\mathcal{C}_\Lambda$.

Definition 88.9.1. Let $p : \mathcal{F} \to \mathcal{C}_\Lambda$ be a category cofibered in groupoids. We say $\mathcal{F}$ is smooth or unobstructed if its structure morphism $p$ is smooth in the sense of Definition 88.8.1.

This is the “absolute” notion of smoothness for a category cofibered in groupoids over $\mathcal{C}_\Lambda$, although it would be more correct to say that $\mathcal{F}$ is smooth over $\Lambda$. One has to be careful with the phrase “$\mathcal{F}$ is unobstructed”: it may happen that $\mathcal{F}$ has an obstruction theory with nonvanishing obstruction spaces even though $\mathcal{F}$ is smooth.

Remark 88.9.2. Suppose $\mathcal{F}$ is a predeformation category admitting a smooth morphism $\varphi : \mathcal U \to \mathcal{F}$ from a predeformation category $\mathcal{U}$. Then by Lemma 88.8.8 $\varphi$ is essentially surjective, so by Lemma 88.8.7 $p: \mathcal{F} \to \mathcal{C}_\Lambda$ is smooth if and only if the composition $\mathcal U \xrightarrow {\varphi } \mathcal{F} \xrightarrow {p} \mathcal{C}_\Lambda$ is smooth, i.e. $\mathcal{F}$ is smooth if and only if $\mathcal{U}$ is smooth.

Lemma 88.9.3. Let $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. The following are equivalent

1. $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth,

2. $\Lambda \to R$ is formally smooth in the $\mathfrak m_ R$-adic topology,

3. $\Lambda \to R$ is flat and $R \otimes _\Lambda k'$ is geometrically regular over $k'$, and

4. $\Lambda \to R$ 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

1. $R$ 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 88.8.6 and the fact that $\mathcal{C}_\Lambda$ is the same as $\underline{\Lambda }|_{\mathcal{C}_\Lambda }$ in the classical case. $\square$

Lemma 88.9.4. Let $\mathcal{F}$ be a predeformation category. Let $\xi$ be a versal formal object of $\mathcal{F}$ lying over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. The following are equivalent

1. $\mathcal{F}$ is unobstructed, and

2. $\Lambda \to R$ is formally smooth in the $\mathfrak m_ R$-adic topology.

In the classical case these are also equivalent to

1. $R \cong \Lambda [[x_1, \ldots , x_ n]]$ for some $n$.

Proof. If (1) holds, i.e., if $\mathcal{F}$ is unobstructed, then the composition

$\underline{R}|_{\mathcal{C}_\Lambda } \xrightarrow {\underline{\xi }} \mathcal{F} \to \mathcal{C}_\Lambda$

is smooth, see Lemma 88.8.7. Hence we see that (2) holds by Lemma 88.9.3. Conversely, if (2) holds, then the composition is smooth and moreover the first arrow is essentially surjective by Lemma 88.8.11. Hence we find that the second arrow is smooth by Lemma 88.8.7 which means that $\mathcal{F}$ is unobstructed by definition. The equivalence with (3) in the classical case follows from Lemma 88.9.3. $\square$

Lemma 88.9.5. There exists an $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$ such that the equivalent conditions of Lemma 88.9.3 hold and moreover $H_1(L_{k/\Lambda }) = \mathfrak m_ R/\mathfrak m_ R^2$ and $\Omega _{R/\Lambda } \otimes _ R k = \Omega _{k/\Lambda }$.

Proof. In the classical case we choose $R = \Lambda$. More generally, if the residue field extension $k/k'$ is separable, then there exists a unique finite étale extension $\Lambda ^\wedge \to R$ (Algebra, Lemmas 10.153.9 and 10.153.7) of the completion $\Lambda ^\wedge$ of $\Lambda$ inducing the extension $k/k'$ on residue fields.

In the general case we proceed as follows. Choose a smooth $\Lambda$-algebra $P$ and a $\Lambda$-algebra surjection $P \to k$. (For example, let $P$ be a polynomial algebra.) Denote $\mathfrak m_ P$ the kernel of $P \to k$. The Jacobi-Zariski sequence, see (88.3.10.2) and Algebra, Lemma 10.134.4, is an exact sequence

$0 \to H_1(\mathop{N\! L}\nolimits _{k/\Lambda }) \to \mathfrak m_ P/\mathfrak m_ P^2 \to \Omega _{P/\Lambda } \otimes _ P k \to \Omega _{k/\Lambda } \to 0$

We have the $0$ on the left because $P/k$ is smooth, hence $\mathop{N\! L}\nolimits _{P/\Lambda }$ is quasi-isomorphic to a finite projective module placed in degree $0$, hence $H_1(\mathop{N\! L}\nolimits _{P/\Lambda } \otimes _ P k) = 0$. Suppose $f \in \mathfrak m_ P$ maps to a nonzero element of $\Omega _{P/\Lambda } \otimes _ P k$. Setting $P' = P/(f)$ we have a $\Lambda$-algebra surjection $P' \to k$. Observe that $P'$ is smooth at $\mathfrak m_{P'}$: this follows from More on Morphisms, Lemma 37.34.1. Thus after replacing $P$ by a principal localization of $P'$, we see that $\dim (\mathfrak m_ P/\mathfrak m_ P^2)$ decreases. Repeating finitely many times, we may assume the map $\mathfrak m_ P/\mathfrak m_ P^2 \to \Omega _{P/\Lambda } \otimes _ P k$ is zero so that the exact sequence breaks into isomorphisms $H_1(L_{k/\Lambda }) = \mathfrak m_ P/\mathfrak m_ P^2$ and $\Omega _{P/\Lambda } \otimes _ P k = \Omega _{k/\Lambda }$.

Let $R$ be the $\mathfrak m_ P$-adic completion of $P$. Then $R$ is an object of $\widehat{\mathcal{C}}_\Lambda$. Namely, it is a complete local Noetherian ring (see Algebra, Lemma 10.97.6) and its residue field is identified with $k$. We claim that $R$ works.

First observe that the map $P \to R$ induces isomorphisms $\mathfrak m_ P/\mathfrak m_ P^2 = \mathfrak m_ R/\mathfrak m_ R^2$ and $\Omega _{P/\Lambda } \otimes _ P k = \Omega _{R/\Lambda } \otimes _ R k$. This is true because both $\mathfrak m_ P/\mathfrak m_ P^2$ and $\Omega _{P/\Lambda } \otimes _ P k$ only depend on the $\Lambda$-algebra $P/\mathfrak m_ P^2$, see Algebra, Lemma 10.131.11, the same holds for $R$ and we have $P/\mathfrak m_ P^2 = R/\mathfrak m_ R^2$. Using the functoriality of the Jacobi-Zariski sequence (88.3.10.3) we deduce that $H_1(L_{k/\Lambda }) = \mathfrak m_ R/\mathfrak m_ R^2$ and $\Omega _{R/\Lambda } \otimes _ R k = \Omega _{k/\Lambda }$ as the same is true for $P$.

Finally, since $\Lambda \to P$ is smooth we see that $\Lambda \to P$ is formally smooth by Algebra, Proposition 10.138.13. Then $\Lambda \to P$ is formally smooth for the $\mathfrak m_ P$-adic topology by More on Algebra, Lemma 15.37.2. This property is inherited by the completion $R$ by More on Algebra, Lemma 15.37.4 and the proof is complete. In fact, it turns out that whenever $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth, then $R$ is isomorphic to a completion of a smooth algebra over $\Lambda$, but we won't use this. $\square$

Example 88.9.6. Here is a more explicit example of an $R$ as in Lemma 88.9.5. Let $p$ be a prime number and let $n \in \mathbf{N}$. Let $\Lambda = \mathbf{F}_ p(t_1, t_2, \ldots , t_ n)$ and let $k = \mathbf{F}_ p(x_1, \ldots , x_ n)$ with map $\Lambda \to k$ given by $t_ i \mapsto x_ i^ p$. Then we can take

$R = \Lambda [x_1, \ldots , x_ n]^\wedge _{(x_1^ p - t_1, \ldots , x_ n^ p - t_ n)}$

We cannot do “better” in this example, i.e., we cannot approximate $\mathcal{C}_\Lambda$ by a smaller smooth object of $\widehat{\mathcal{C}}_\Lambda$ (one can argue that the dimension of $R$ has to be at least $n$ since the map $\Omega _{R/\Lambda } \otimes _ R k \to \Omega _{k/\Lambda }$ is surjective). We will discuss this phenomenon later in more detail.

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