The Stacks project

[Lemma 4.2, Bhatt-Algebraize]

Lemma 15.97.3. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ be a limit of an inverse system $(A_ n)$ of rings. Suppose given $K_ n \in D(A_ n)$ and maps $K_{n + 1} \to K_ n$ in $D(A_{n + 1})$. Assume

  1. the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,

  2. $K_1$ is a perfect object, and

  3. the maps induce isomorphisms $K_{n + 1} \otimes _{A_{n + 1}}^\mathbf {L} A_ n \to K_ n$.

Then $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ is a perfect object of $D(A)$ and $K \otimes _ A^\mathbf {L} A_ n \to K_ n$ is an isomorphism for all $n$.

Proof. We already know that $K$ is pseudo-coherent and that $K \otimes _ A^\mathbf {L} A_ n \to K_ n$ is an isomorphism for all $n$ by Lemma 15.97.1. Thus it suffices to show that $H^ i(K \otimes _ A^\mathbf {L} \kappa ) = 0$ for $i \ll 0$ and every surjective map $A \to \kappa $ whose kernel is a maximal ideal $\mathfrak m$, see Lemma 15.77.3. Any element of $A$ which maps to a unit in $A_1$ is a unit in $A$ by Algebra, Lemma 10.32.4 and hence $\mathop{\mathrm{Ker}}(A \to A_1)$ is contained in the Jacobson radical of $A$ by Algebra, Lemma 10.19.1. Hence $A \to \kappa $ factors as $A \to A_1 \to \kappa $. Hence

\[ K \otimes _ A^\mathbf {L} \kappa = K \otimes _ A^\mathbf {L} A_1 \otimes _{A_1}^\mathbf {L} \kappa = K_1 \otimes _{A_1}^\mathbf {L} \kappa \]

and we get what we want as $K_1$ has finite tor dimension by Lemma 15.74.2. $\square$

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Your email address will not be published.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