Example 15.100.10. Let $k$ be a field. Let $A = k[[x, y]]/(xy)$. By abuse of notation we denote $x$ and $y$ the images of $x$ and $y$ in $A$. Let $I = (x)$. Let $M = A/(y)$. There is a free resolution

$\ldots \to A \xrightarrow {y} A \xrightarrow {x} A \xrightarrow {y} A \to M \to 0$

We conclude that

$\mathop{\mathrm{Ext}}\nolimits ^2_ A(M, N) = N[y]/xN$

where $N[y] = \mathop{\mathrm{Ker}}(y : N \to N)$. We denote $A_ n = A/I^ n$, $M_ n = M/I^ nM$, and $N_ n = N/I^ nN$. For each $n$ we have a free resolution

$\ldots \to A_ n^{\oplus 2} \xrightarrow {y, x^{n - 1}} A_ n \xrightarrow {x} A_ n \xrightarrow {y} A_ n \to M_ n \to 0$

We conclude that

$\mathop{\mathrm{Ext}}\nolimits ^2_{A_ n}(M_ n, N_ n) = (N_ n[y] \cap N_ n[x^{n - 1}])/xN_ n$

where $N_ n[y] = \mathop{\mathrm{Ker}}(y : N_ n \to N_ n)$ and $N[x^{n - 1}] = \mathop{\mathrm{Ker}}(x^{n - 1} : N_ n \to N_ n)$. Take $N = A/(y)$. Then we see that

$\mathop{\mathrm{Ext}}\nolimits ^2_ A(M, N) = N[y]/xN = N/xN \cong k$

but

$\mathop{\mathrm{Ext}}\nolimits ^2_{A_ n}(M_ n, N_ n) = (N_ n[y] \cap N_ n[x^{n - 1}])/xN_ n = N_ n[x^{n - 1}]/xN_ n = 0$

for all $r$ because $N_ n = k[x]/(x^ n)$ and the sequence

$N_ n \xrightarrow {x} N_ n \xrightarrow {x^{n - 1}} N_ n$

is exact. Thus ignoring some kind of $I$-power torsion is necessary to get a result as in Lemma 15.100.8.

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