The Stacks project

Lemma 15.85.2. Let $A$ be a ring. Let $K \in D(A)$. The set $I$ of $f \in A$ such that $T(K, f) = 0$ is a radical ideal of $A$.

Proof. We will use the results of Lemma 15.85.1 without further mention. If $f \in I$, and $g \in A$, then $A_{gf}$ is an $A_ f$-module hence $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{gf}, K) = 0$ for all $n$, hence $gf \in I$. Suppose $f, g \in I$. Then there is a short exact sequence

\[ 0 \to A_{f + g} \to A_{f(f + g)} \oplus A_{g(f + g)} \to A_{gf(f + g)} \to 0 \]

because $f, g$ generate the unit ideal in $A_{f + g}$. This follows from Algebra, Lemma 10.23.2 and the easy fact that the last arrow is surjective. From the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits $ and the vanishing of $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{f(f + g)}, K)$, $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{g(f + g)}, K)$, and $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{gf(f + g)}, K)$ for all $n$ we deduce the vanishing of $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{f + g}, K)$ for all $n$. Finally, if $f^ n \in I$ for some $n > 0$, then $f \in I$ because $T(K, f) = T(K, f^ n)$ or because $A_ f \cong A_{f^ n}$. $\square$


Comments (2)

Comment #229 by on

There is a typo in the first sentence of the proof, furhter should be further.

Comment #5386 by suggestion_bot on

One could include a corollary that there is the largest ideal of A for which is derived complete, and that this ideal is radical.

There are also:

  • 2 comment(s) on Section 15.85: Derived Completion

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Your email address will not be published.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 091Q. Beware of the difference between the letter 'O' and the digit '0'.