The Stacks project

Proposition 88.6.3. Let $I$ be an ideal of a Noetherian G-ring $A$. Let $B$ be an object of ( If $B$ is rig-smooth over $(A, I)$, then there exists a finite type $A$-algebra $C$ and an isomorphism $B \cong C^\wedge $ of $A$-algebras.

Proof. Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Write $P = A[x_1, \ldots , x_ r]^\wedge $. Choose generators $g_1, \ldots , g_ m \in J$. Choose generators $k_1, \ldots , k_ t$ of the module of relations between $g_1, \ldots , g_ m$, i.e., such that

\[ P^{\oplus t} \xrightarrow {k_1, \ldots , k_ t} P^{\oplus m} \xrightarrow {g_1, \ldots , g_ m} P \to B \to 0 \]

is a resolution. Write $k_ i = (k_{i1}, \ldots , k_{im})$ so that we have
\begin{equation} \label{restricted-equation-relations-straight-up} \sum \nolimits _ j k_{ij}g_ j = 0 \end{equation}

for $i = 1, \ldots , t$. Denote $K = (k_{ij})$ the $m \times t$-matrix with entries $k_{ij}$.

Let $A[x_1, \ldots , x_ r]^ h$ be the henselization of the pair $(A[x_1, \ldots , x_ r], IA[x_1, \ldots , x_ r])$, see More on Algebra, Lemma 15.12.1. We may and do think of $A[x_1, \ldots , x_ r]^ h$ as a subring of $P = A[x_1, \ldots , x_ r]^\wedge $, see More on Algebra, Lemma 15.12.4. Since $A$ is a Noetherian G-ring, so is $A[x_1, \ldots , x_ r]$, see More on Algebra, Proposition 15.50.10. Hence we have approximation for the map $A[x_1, \ldots , x_ r]^ h \to A[x_1, \ldots , x_ r]^\wedge = P$ with respect to the ideal generated by $I$, see Smoothing Ring Maps, Lemma 16.14.1. Choose a large enough integer $n$ as in Lemma 88.6.1. By the approximation property we may choose $g'_1, \ldots , g'_ m \in A[x_1, \ldots , x_ r]^ h$ and a matrix $K' = (k'_{ij}) \in \text{Mat}(m \times t, A[x_1, \ldots , x_ r]^ h)$ such that $\sum \nolimits _ j k'_{ij}g'_ j = 0$ in $A[x_1, \ldots , x_ r]^ h$ and such that $g_ i - g'_ i \in I^ nP$ and $K - K' \in I^ n\text{Mat}(m \times t, P)$. By our choice of $n$ we conclude that there is an isomorphism

\[ B \to P/(g'_1, \ldots , g'_ m) = \left(A[x_1, \ldots , x_ r]^ h/(g'_1, \ldots , g'_ m)\right)^\wedge \]

This finishes the proof by Lemma 88.6.2. $\square$

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Your email address will not be published.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 0GAT. Beware of the difference between the letter 'O' and the digit '0'.