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 $A_ f \cong A_{f^ n}$. $\square$

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 $K$ is derived complete, and that this ideal is radical.

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