The Stacks project

Proof. Choose an affine open $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$ such that $\mathop{\mathrm{Spec}}(k) \to S$ factors through it. By Artin's Axioms, Section 98.3 we obtain a predeformation category $\mathcal{F}_{\mathcal{X}, k, x_0}$ over the category $\mathcal{C}_\Lambda $. (As pointed out in locus citatus this category only depends on the morphism $\mathop{\mathrm{Spec}}(k) \to S$ and not on the choice of $\Lambda $.) By Artin's Axioms, Lemmas 98.6.1 and 98.5.2 $\mathcal{F}_{\mathcal{X}, k, x_0}$ is actually a deformation category. By Artin's Axioms, Lemma 98.8.1 we find that $T\mathcal{F}_{\mathcal{X}, k, x_0}$ and $\text{Inf}(\mathcal{F}_{\mathcal{X}, k, x_0})$ are finite dimensional $k$-vector spaces. $\square$

Proof. By Lemma 107.2.3 and Formal Deformation Theory, Lemma 90.13.4 (note that the assumptions of this lemma hold by Formal Deformation Theory, Lemmas 90.16.6 and Definition 90.16.8). By the uniqueness result of Formal Deformation Theory, Lemma 90.14.5 there exists a “minimal” versal ring $A$ of $\mathcal{X}$ at $x_0$ such that any other versal ring of $\mathcal{X}$ at $x_0$ is isomorphic to $A[[t_1, \ldots , t_ r]]$ for some $r$. This clearly implies the second statement. $\square$

Proof. Follows immediately from Artin's Axioms, Lemma 98.7.1 and Formal Deformation Theory, Lemma 90.29.6. (We also use that $\mathcal{X}$ satisfies (RS) by Artin's Axioms, Lemma 98.5.2.) $\square$

Proof. Since $U \to S$ is locally of finite type (as a composition of such morphisms), we see that $\mathop{\mathrm{Spec}}(k) \to S$ is of finite type (again as a composition). Thus the statement makes sense. The equivalence of (1) and (2) is the definition of $x$ being versal at $u_0$. The equivalence of (1) and (3) is Artin's Axioms, Lemma 98.12.3. Thus (1), (2), and (3) are equivalent.

If $x|_{U'}$ is smooth, then the functor $\hat x : \mathcal{F}_{U, k, u_0} \to \mathcal{F}_{\mathcal{X}, k, x_0}$ is smooth by Artin's Axioms, Lemma 98.3.2. Thus (4) implies (1), (2), and (3). For the converse, assume $x$ is versal at $u_0$. Choose a surjective smooth morphism $y : V \to \mathcal{X}$ where $V$ is a scheme. Set $Z = V \times _\mathcal {X} U$ and pick a finite type point $z_0 \in |Z|$ lying over $u_0$ (this is possible by Morphisms of Spaces, Lemma 67.25.5). By Artin's Axioms, Lemma 98.12.6 the morphism $Z \to V$ is smooth at $z_0$. By definition we can find an open neighbourhood $W \subset Z$ of $z_0$ such that $W \to V$ is smooth. Since $Z \to U$ is open, let $U' \subset U$ be the image of $W$. Then we see that $U' \to \mathcal{X}$ is smooth by our definition of smooth morphisms of stacks.

The final statement follows from the definitions as $\mathcal{O}_{U, u_0}^\wedge $ prorepresents $\mathcal{F}_{U, k, u_0}$. $\square$

Proof. The equivalence of (2) and (3) follows immediately from Formal Deformation Theory, Lemma 90.9.4.

Note that $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology if and only if $\mathcal{O}_{S, s} \to A' = A[[t_1, \ldots , t_ r]]$ is formally smooth in the $\mathfrak m_{A'}$-adic topology. Hence (2) does not depend on the choice of our versal ring by Lemma 107.2.4. Next, let $l/k$ be a finite extension and choose $A \to A'$ as in Lemma 107.2.5. If $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology, then $\mathcal{O}_{S, s} \to A'$ is formally smooth in the $\mathfrak m_{A'}$-adic topology, see More on Algebra, Lemma 15.37.7. Conversely, if $\mathcal{O}_{S, s} \to A'$ is formally smooth in the $\mathfrak m_{A'}$-adic topology, then $\mathcal{O}_{S, s}^\wedge \to A'$ and $A \to A'$ are regular (More on Algebra, Proposition 15.49.2) and hence $\mathcal{O}_{S, s}^\wedge \to A$ is regular (More on Algebra, Lemma 15.41.7), hence $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology (same lemma as before). Thus the equivalence of (2) and (1) holds for $k$ and $x_0$ if and only if it holds for $l$ and $x_{0, l}$.

Choose a scheme $U$ and a smooth morphism $U \to \mathcal{X}$ such that $\mathop{\mathrm{Spec}}(k) \times _\mathcal {X} U$ is nonempty. Choose a finite extension $l/k$ and a point $w_0 : \mathop{\mathrm{Spec}}(l) \to \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} U$. Let $u_0 \in U$ be the image of $w_0$. We may apply the above to $l/k$ and to $l/\kappa (u_0)$ to see that we can reduce to $u_0$. Thus we may assume $A = \mathcal{O}_{U, u_0}^\wedge $, see Lemma 107.2.6. Observe that $x_0$ is in the smooth locus of $\mathcal{X} \to S$ if and only if $u_0$ is in the smooth locus of $U \to S$, see for example Morphisms of Stacks, Lemma 101.33.6. Thus the equivalence of (1) and (2) follows from More on Algebra, Lemma 15.38.6. $\square$

Proof. Let $(\xi _ n, f_ n)$ be the versal formal object over $A$. By Artin's Axioms, Lemma 98.9.5 we know that $\xi = (A, \xi _ n, f_ n)$ is effective. By assumption $\mathcal{X}$ is locally of finite presentation over $S$ (use Morphisms of Stacks, Lemma 101.27.5), and hence limit preserving by Limits of Stacks, Proposition 102.3.8. Thus Artin approximation as in Artin's Axioms, Lemma 98.12.7 shows that we may find a morphism $U \to \mathcal{X}$ with source a finite type $S$-scheme, containing a point $u_0 \in U$ of residue field $k$ satisfying (1) and (2) such that $U \to \mathcal{X}$ is versal at $u_0$. By Lemma 107.2.6 after shrinking $U$ we may assume $U \to \mathcal{X}$ is smooth. $\square$

Proof. If the local ring of $S$ at the image point is a G-ring, then this follows immediately from Lemma 107.2.8 and the fact that the map from a Noetherian local ring to its completion is flat. In general we prove it as follows.

Step I. If $A$ and $A'$ are two versal rings to $\mathcal{X}$ at $x_0$, then the result is true for $A$ if and only if it is true for $A'$. Namely, after possible swapping $A$ and $A'$, we may assume there is a formally smooth map $\varphi : A \to A'$ such that the composition

is the morphism $\mathop{\mathrm{Spec}}(A') \to \mathcal{X}$, see Lemma 107.2.4 and Remark 107.2.9. Since $A \to A'$ is faithfully flat we obtain the equivalence from Morphisms of Stacks, Lemmas 101.25.2 and 101.25.5.

Step II. Let $l/k$ be a finite extension of fields. Let $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ be the induced morphism. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$ and let $A \to A'$ be as in Lemma 107.2.5. Then again the composition

is the morphism $\mathop{\mathrm{Spec}}(A') \to \mathcal{X}$, see Remark 107.2.9. Arguing as before and using step I to see choice of versal rings is irrelevant, we see that the lemma holds for $x_0$ if and only if it holds for $x_{l, 0}$.

Step III. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then we can choose a finite type point $z_0$ on $Z = U \times _\mathcal {X} x_0$ (this is a nonempty algebraic space). Let $u_0 \in U$ be the image of $z_0$ in $U$. Choose a scheme and a surjective étale map $W \to Z$ such that $z_0$ is the image of a closed point $w_0 \in W$ (see Morphisms of Spaces, Section 67.25). Since $W \to \mathop{\mathrm{Spec}}(k)$ and $W \to U$ are of finite type, we see that $\kappa (w_0)/k$ and $\kappa (w_0)/\kappa (u_0)$ are finite extensions of fields (see Morphisms, Section 29.16). Applying Step II twice we may replace $x_0$ by $u_0 \to U \to \mathcal{X}$. Then we see our morphism is the composition

The first arrow is flat because completion of Noetherian local rings are flat (Algebra, Lemma 10.97.2) and the second arrow is flat as a smooth morphism is flat. The composition is flat as composition preserves flatness. $\square$