The Stacks project

Lemma 107.2.7. In Situation 107.2.1. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism such that $\mathop{\mathrm{Spec}}(k) \to S$ is of finite type with image $s$. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$. The following are equivalent

  1. $x_0$ is in the smooth locus of $\mathcal{X} \to S$ (Morphisms of Stacks, Lemma 101.33.6),

  2. $\mathcal{O}_{S, s} \to A$ is formally smooth in the $\mathfrak m_ A$-adic topology, and

  3. $\mathcal{F}_{\mathcal{X}, k, x_0}$ is unobstructed.

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DZS. Beware of the difference between the letter 'O' and the digit '0'.