## 89.18 Schlessinger's theorem on prorepresentable functors

We deduce Schlessinger's theorem characterizing prorepresentable functors on $\mathcal{C}_\Lambda$.

Lemma 89.18.1. Let $F, G: \mathcal{C}_\Lambda \to \textit{Sets}$ be deformation functors. Let $\varphi : F \to G$ be a smooth morphism which induces an isomorphism $d\varphi : TF \to TG$ of tangent spaces. Then $\varphi$ is an isomorphism.

Proof. We prove $F(A) \to G(A)$ is a bijection for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_\Lambda )$ by induction on $\text{length}_ A(A)$. For $A = k$ the statement follows from the assumption that $F$ and $G$ are deformation functors. Suppose that the statement holds for rings of length less than $n$ and let $A'$ be a ring of length $n$. Choose a small extension $f : A' \to A$. We have a commutative diagram

$\xymatrix{ F(A') \ar[r] \ar[d]_{F(f)} & G(A') \ar[d]^{G(f)} \\ F(A) \ar[r]^{\sim } & G(A) }$

where the map $F(A) \to G(A)$ is a bijection. By smoothness of $F \to G$, $F(A') \to G(A')$ is surjective (Lemma 89.8.8). Thus we can check bijectivity by checking it on fibers $F(f)^{-1}(x) \to G(f)^{-1}(\varphi (x))$ for $x \in F(A)$ such that $F(f)^{-1}(x)$ is nonempty. These fibers are precisely $\text{Lift}(x, A')$ and $\text{Lift}(\varphi (x), A')$ and by assumption we have an isomorphism $d\varphi \otimes \text{id} : TF \otimes _ k \mathop{\mathrm{Ker}}(f) \to TG \otimes _ k \mathop{\mathrm{Ker}}(f)$. Thus, by Lemma 89.17.5 and Remark 89.17.6, for $x \in F(A)$ such that $F(f)^{-1}(x)$ is nonempty the map $F(f)^{-1}(x) \to G(f)^{-1}(\varphi (x))$ is a map of sets commuting with free transitive actions by $TF \otimes _ k \mathop{\mathrm{Ker}}(f)$. Hence it is bijective. $\square$

Note that in case $k' \subset k$ is separable condition (c) in the theorem below is empty.

Theorem 89.18.2. Let $F: \mathcal{C}_\Lambda \to \textit{Sets}$ be a functor. Then $F$ is prorepresentable if and only if (a) $F$ is a deformation functor, (b) $\dim _ k TF$ is finite, and (c) $\gamma : \text{Der}_\Lambda (k, k) \to TF$ is injective.

Proof. Assume $F$ is prorepresentable by $R \in \widehat{\mathcal{C}}_\Lambda$. We see $F$ is a deformation functor by Example 89.16.10. We see $\dim _ k TF$ is finite by Example 89.11.11. Finally, $\text{Der}_\Lambda (k, k) \to TF$ is identified with $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (R, k)$ by Example 89.11.14 which is injective because $R \to k$ is surjective.

Conversely, assume (a), (b), and (c) hold. By Lemma 89.16.6 we see that (S1) and (S2) hold. Hence by Theorem 89.15.5 there exists a minimal versal formal object $\xi$ of $F$ such that (89.15.0.2) holds. Say $\xi$ lies over $R$. The map

$d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F}$

is bijective on $\text{Der}_\Lambda (k, k)$-orbits. Since the action of $\text{Der}_\Lambda (k, k)$ on the left hand side is free by (c) and Lemma 89.12.6 we see that the map is bijective. Thus we see that $\underline{\xi }$ is an isomorphism by Lemma 89.18.1. $\square$

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