The Stacks project

Lemma 90.15.4. Let $\mathcal{F}$ be a predeformation category satisfying (S2) which has a versal formal object. Then its minimal versal formal object satisfies (90.15.0.2).

Proof. Let $\xi $ be a minimal versal formal object for $\mathcal{F}$, see Lemma 90.14.5. Say $\xi $ lies over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. In order to parse (90.15.0.2) we point out that $T\mathcal{F}$ has a natural $k$-vector space structure (see Lemma 90.12.2), that $d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F}$ is linear (see Lemma 90.12.4), and that the action of $\text{Der}_\Lambda (k, k)$ is given by addition (see Lemma 90.12.6). Consider the diagram

\[ \xymatrix{ & \mathop{\mathrm{Hom}}\nolimits _ k(\mathfrak m_ R/\mathfrak m_ R^2, k) \\ K \ar[r] & \text{Der}_\Lambda (R, k) \ar[r]^{d\underline{\xi }} \ar[u] & T\mathcal{F} \\ & \text{Der}_\Lambda (k, k) \ar[u] \ar[ru] } \]

The vector space $K$ is the kernel of $d\underline{\xi }$. Note that the middle column is exact in the middle as it is dual to the sequence (90.3.10.1). If (90.15.0.2) fails, then we can find a nonzero element $D \in K$ which does not map to zero in $\mathop{\mathrm{Hom}}\nolimits _ k(\mathfrak m_ R/\mathfrak m_ R^2, k)$. This means there exists an $t \in \mathfrak m_ R$ such that $D(t) = 1$. Set $R' = \{ a \in R \mid D(a) = 0\} $. As $D$ is a derivation this is a subring of $R$. Since $D(t) = 1$ we see that $R' \to k$ is surjective (compare with the proof of Lemma 90.3.12). Note that $\mathfrak m_{R'} = \mathop{\mathrm{Ker}}(D : \mathfrak m_ R \to k)$ is an ideal of $R$ and $\mathfrak m_ R^2 \subset \mathfrak m_{R'}$. Hence

\[ \mathfrak m_ R/\mathfrak m_ R^2 = \mathfrak m_{R'}/\mathfrak m_ R^2 + k\overline{t} \]

which implies that the map

\[ R'/\mathfrak m_ R^2 \times _ k k[\epsilon ] \to R/\mathfrak m_ R^2 \]

sending $\epsilon $ to $\overline{t}$ is an isomorphism. In particular there is a map $R/\mathfrak m_ R^2 \to R'/\mathfrak m_ R^2$.

Let $\xi \to y$ be a morphism lying over $R \to R/\mathfrak m_ R^2$. Let $y \to x$ be a morphism lying over $R/\mathfrak m_ R^2 \to R'/\mathfrak m_ R^2$. Let $y \to x_\epsilon $ be a morphism lying over $R/\mathfrak m_ R^2 \to k[\epsilon ]$. Let $x_0$ be the unique (up to unique isomorphism) object of $\mathcal{F}$ over $k$. By the axioms of a category cofibred in groupoids we obtain a commutative diagram

\[ \vcenter { \xymatrix{ y \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ R'/\mathfrak m_ R^2 \times _ k k[\epsilon ] \ar[r] \ar[d] & k[\epsilon ] \ar[d] \\ R'/\mathfrak m_ R^2 \ar[r] & k. } } \]

Because $D \in K$ we see that $x_\epsilon $ is isomorphic to $0 \in \mathcal{F}(k[\epsilon ])$, i.e., $x_\epsilon $ is the pushforward of $x_0$ via $k \to k[\epsilon ], a \mapsto a$. Hence by Lemma 90.10.7 we see that there exists a morphism $x \to y$. Since $\text{length}_\Lambda (R'/\mathfrak m_ R^2) < \text{length}_\Lambda (R/\mathfrak m_ R^2)$ the corresponding ring map $R'/\mathfrak m_ R^2 \to R/\mathfrak m_ R^2$ is not surjective. This contradicts the minimality of $\xi /R$, see part (1) of Lemma 90.14.2. This contradiction shows that such a $D$ cannot exist, hence we win. $\square$


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 06T8. Beware of the difference between the letter 'O' and the digit '0'.