The Stacks project

Lemma 88.29.6. With notation and assumptions as in Situation 88.29.1. Let $\xi $ be a versal formal object for $\mathcal{F}$ lying over $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_{\Lambda , k})$. Then there exist

  1. an $S \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_{\Lambda , l})$ and a local $\Lambda $-algebra homomorphism $R \to S$ which is formally smooth in the $\mathfrak m_ S$-adic topology and induces the given field extension $l/k$ on residue fieds, and

  2. a versal formal object of $\mathcal{F}_{l/k}$ lying over $S$.

Proof. Construction of $S$. Choose a surjection $R[x_1, \ldots , x_ n] \to l$ of $R$-algebras. The kernel is a maximal ideal $\mathfrak m$. Set $S$ equal to the $\mathfrak m$-adic completion of the Noetherian ring $R[x_1, \ldots , x_ n]$. Then $S$ is in $\widehat{\mathcal{C}}_{\Lambda , l}$ by Algebra, Lemma 10.97.6. The map $R \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology by More on Algebra, Lemmas 15.37.2 and 15.37.4 and the fact that $R \to R[x_1, \ldots , x_ n]$ is formally smooth. (Compare with the proof Lemma 88.9.5.)

Since $\xi $ is versal, the transformation $\underline{\xi } : \underline{R}|_{\mathcal{C}_{\Lambda , k}} \to \mathcal{F}$ is smooth. By Lemma 88.29.5 the induced map

\[ (\underline{R}|_{\mathcal{C}_{\Lambda , k}})_{l/k} \longrightarrow \mathcal{F}_{l/k} \]

is smooth. Thus it suffices to construct a smooth morphism $\underline{S}|_{\mathcal{C}_{\Lambda , l}} \to (\underline{R}|_{\mathcal{C}_{\Lambda , k}})_{l/k}$. To give such a map means for every object $B$ of $\mathcal{C}_{\Lambda , l}$ a map of sets

\[ \mathop{\mathrm{Mor}}\nolimits _{\widehat{\mathcal{C}}_{\Lambda , l}}(S, B) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\widehat{\mathcal{C}}_{\Lambda , k}}(R, B \times _ l k) \]

functorial in $B$. Given an element $\varphi : S \to B$ on the left hand side we send it to the composition $R \to S \to B$ whose image is contained in the sub $\Lambda $-algebra $B \times _ l k$. Smoothness of the map means that given a surjection $B' \to B$ and a commutative diagram

\[ \xymatrix{ S \ar[r] & B \ar@{=}[r] & B \\ R \ar[u] \ar[r] & B' \times _ l k \ar[u] \ar[r] & B' \ar[u] } \]

we have to find a ring map $S \to B'$ fitting into the outer rectangle. The existence of this map is guaranteed as we chose $R \to S$ to be formally smooth in the $\mathfrak m_ S$-adic topology, see More on Algebra, Lemma 15.37.5. $\square$


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