Proposition 9

Suppose the sequence x regular for M. If N is an A-module such that $(x)N = 0$, one has $\mathrm{Ext}_A^i(N, M) = 0$ for $i < n$, and the mapping $\alpha \mapsto \theta_x \circ \alpha$ from Hom_A (N, M/(x) M) into Ext^n_A (N, M) (which is also the iterated connecting homomorphism associated with (19), cf. X, p. 127, Corollary 3) is bijective.

It remains to prove that the homomorphism $\psi^i : \alpha \mapsto \theta_x \circ \alpha$ from $\mathrm{Ext}_A^{i-n}(N, M/(x) M)$ into $\mathrm{Ext}_A^i(N, M)$ is bijective for $i \leq n$. Let us reason by induction on $n$, the assertion being trivial for $n = 0$. Put $M_1 = M/x_1 M$, $x' = (x_2, ..., x_n)$, so that $x'$ is $M_1$-regular. By the induction hypothesis, the homomorphism

from $\mathrm{Ext}_A^{i-n}(N, M/(x) M)$ into $\mathrm{Ext}_A^{i-1}(N, M_1)$ is bijective for $i < n$. On the other hand, consider the exact sequence

the connecting homomorphism $\mathrm{Ext}_A^i(N, M_1) \to \mathrm{Ext}_A^{i+1}(N, M)$ associated with it is $\varphi^i : \beta \mapsto \theta_{x_1} \circ \beta$ (X, p. 125, Prop. 5) ;

since one has $\mathrm{Ext}^i(1_N, (x_1)_M) = \mathrm{Ext}^i((x_1)_N, 1_M) = 0$, one deduces from this exact sequences

Since $\mathrm{Ext}_A^i(N, M_1) = 0$ for $i < n - 1$, it follows that $\mathrm{Ext}_A^{i+1}(N, M) = 0$ for $i < n - 1$, that is to say for $i + 1 < n$; it follows that $\varphi^i$ is bijective for $i < n$. Since $\psi^i(\alpha) = \theta_x \circ \alpha = \theta_{x_1} \circ \theta_{x'} \circ \alpha = \varphi^{i-1} \circ \overline{\psi}^i(\alpha)$ for $\alpha \in \mathrm{Ext}_A^{i-n}(N, M/(x) M)$, $\psi^i$ is indeed bijective for $i \leq n$.

EXERCISES