Lemma 15.90.2. 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 15.90.1 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 10.24.2 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.

Comment #5620 by on

@suggestion_bot: but this is exactly what the lemma is saying, by our definition of being derived complete in Definition 15.90.4. So that would be just a reformulation of this lemma with the language which is being introduced after the lemma. In most cases we don't do this, in other words, we mostly don't just reformulate things after introducing the notation. If others agree with your suggestion, then I will go ahead and do this.

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