The Stacks project

Lemma 89.8.6. Let $R \to S$ be a ring map in $\widehat{\mathcal{C}}_\Lambda $. Then the induced morphism $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth if and only if $S$ is a power series ring over $R$.

Proof. Assume $S$ is a power series ring over $R$. Say $S = R[[x_1, \ldots , x_ n]]$. Smoothness of $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ means the following (see Remark 89.8.4): Given a surjective ring map $B \to A$ in $\mathcal{C}_\Lambda $, a ring map $R \to B$, a ring map $S \to A$ such that the solid diagram

\[ \xymatrix{ S \ar[r] \ar@{..>}[rd] & A \\ R \ar[u] \ar[r] & B \ar[u] } \]

is commutative then a dotted arrow exists making the diagram commute. (Note the similarity with Algebra, Definition 10.138.1.) To construct the dotted arrow choose elements $b_ i \in B$ whose images in $A$ are equal to the images of $x_ i$ in $A$. Note that $b_ i \in \mathfrak m_ B$ as $x_ i$ maps to an element of $\mathfrak m_ A$. Hence there is a unique $R$-algebra map $R[[x_1, \ldots , x_ n]] \to B$ which maps $x_ i$ to $b_ i$ and which can serve as our dotted arrow.

Conversely, assume $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth. Let $x_1, \ldots , x_ n \in S$ be elements whose images form a basis in the relative cotangent space $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2)$ of $S$ over $R$. Set $T = R[[X_1, \ldots , X_ n]]$. Note that both

\[ S/(\mathfrak m_ R S + \mathfrak m_ S^2) \cong R/\mathfrak m_ R[x_1, \ldots , x_ n]/(x_ ix_ j) \]


\[ T/(\mathfrak m_ R T + \mathfrak m_ T^2) \cong R/\mathfrak m_ R[X_1, \ldots , X_ n]/(X_ iX_ j). \]

Let $S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ be the local $R$-algebra isomorphism given by mapping the class of $x_ i$ to the class of $X_ i$. Let $f_1 : S \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ be the composition $S \to S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to T/(\mathfrak m_ R T + \mathfrak m_ T^2)$. The assumption that $\underline{S}|_{\mathcal{C}_\Lambda } \to \underline{R}|_{\mathcal{C}_\Lambda }$ is smooth means we can lift $f_1$ to a map $f_2 : S \to T/\mathfrak {m}_ T^2$, then to a map $f_3 : S \to T/\mathfrak {m}_ T^3$, and so on, for all $n \geq 1$. Thus we get an induced map $f : S \to T = \mathop{\mathrm{lim}}\nolimits T/\mathfrak m_ T^ n$ of local $R$-algebras. By our choice of $f_1$, the map $f$ induces an isomorphism $\mathfrak m_ S/(\mathfrak m_ R S + \mathfrak m_ S^2) \to \mathfrak m_ T/(\mathfrak m_ R T + \mathfrak m_ T^2)$ of relative cotangent spaces. Hence $f$ is surjective by Lemma 89.4.2 (where we think of $f$ as a map in $\widehat{\mathcal{C}}_ R$). Choose preimages $y_ i \in S$ of $X_ i \in T$ under $f$. As $T$ is a power series ring over $R$ there exists a local $R$-algebra homomorphism $s : T \to S$ mapping $X_ i$ to $y_ i$. By construction $f \circ s = \text{id}$. Then $s$ is injective. But $s$ induces an isomorphism on relative cotangent spaces since $f$ does, so it is also surjective by Lemma 89.4.2 again. Hence $s$ and $f$ are isomorphisms. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 89.8: Smooth morphisms

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