The Stacks project

Lemma 47.8.4. Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $M$ and $N$ be $I$-power torsion modules.

  1. $\mathop{\mathrm{Hom}}\nolimits _{D(A)}(M, N) = \mathop{\mathrm{Hom}}\nolimits _{D({I^\infty \text{-torsion}})}(M, N)$,

  2. $\mathop{\mathrm{Ext}}\nolimits ^1_{D(A)}(M, N) = \mathop{\mathrm{Ext}}\nolimits ^1_{D({I^\infty \text{-torsion}})}(M, N)$,

  3. $\mathop{\mathrm{Ext}}\nolimits ^2_{D({I^\infty \text{-torsion}})}(M, N) \to \mathop{\mathrm{Ext}}\nolimits ^2_{D(A)}(M, N)$ is not surjective in general,

  4. (47.8.3.1) is not an equivalence in general.

Proof. Parts (1) and (2) follow immediately from the fact that $I$-power torsion forms a Serre subcategory of $\text{Mod}_ A$. Part (4) follows from part (3).

For part (3) let $A$ be a ring with an element $f \in A$ such that $A[f]$ contains a nonzero element $x$ annihilated by $f$ and $A$ contains elements $x_ n$ with $f^ nx_ n = x$. Such a ring $A$ exists because we can take

\[ A = \mathbf{Z}[f, x, x_ n]/(fx, f^ nx_ n - x) \]

Given $A$ set $I = (f)$. Then the exact sequence

\[ 0 \to A[f] \to A \xrightarrow {f} A \to A/fA \to 0 \]

defines an element in $\mathop{\mathrm{Ext}}\nolimits ^2_ A(A/fA, A[f])$. We claim this element does not come from an element of $\mathop{\mathrm{Ext}}\nolimits ^2_{D(f^\infty \text{-torsion})}(A/fA, A[f])$. Namely, if it did, then there would be an exact sequence

\[ 0 \to A[f] \to M \to N \to A/fA \to 0 \]

where $M$ and $N$ are $f$-power torsion modules defining the same $2$ extension class. Since $A \to A$ is a complex of free modules and since the $2$ extension classes are the same we would be able to find a map

\[ \xymatrix{ 0 \ar[r] & A[f] \ar[r] \ar[d] & A \ar[r] \ar[d]_\varphi & A \ar[r] \ar[d]_\psi & A/fA \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A[f] \ar[r] & M \ar[r] & N \ar[r] & A/fA \ar[r] & 0 } \]

(some details omitted). Then we could replace $M$ by the image of $\varphi $ and $N$ by the image of $\psi $. Then $M$ would be a cyclic module, hence $f^ n M = 0$ for some $n$. Considering $\varphi (x_{n + 1})$ we get a contradiction with the fact that $f^{n + 1}x_ n = x$ is nonzero in $A[f]$. $\square$


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