Lemma 3

There exists a homomorphism $g : E \to T$ whose restriction to $T$ is both a homothety and a pseudo-isomorphism.

Let $a$ be the annihilator of $T$; as $T$ is a finitely generated torsion $A$-module, $a \neq 0$. Let $p_i$ ($1 \leq i \leq k$) be the prime ideals of height 1 containing $a$ (which are finite in number ($§\ 1$, no. 6, Theorem 4)); if this number is 0, T is pseudo-zero (Proposition 9(a)) and we may take $g = 0$. Otherwise, let $S = \bigcap_i (A - p_i)$; by Lemma 2, $S^{-1}A$ is a principal ideal domain and hence $S^{-1}M$, which is a torsion-free finitely generated $S^{-1}A$-module, is free (Algebra, Chapter VII, §4, no. 3, Corollary 2 to Theorem 2) and, as $S^{-1}M = (S^{-1}E)/(S^{-1}T)$, $S^{-1}T$ is a direct factor of $S^{-1}E$ (Algebra, Chapter II, §1, no. 11, Proposition 21). Now,

(Chapter II, §2, no. 7, Proposition 19); hence there exist $s_0 \in S$ and $g_0 \in \operatorname{Hom}_A(E, T)$ such that $s_0^{-1}g_0$ is a projector of $S^{-1}E$ onto $S^{-1}T$. If $h_0 \in \operatorname{Hom}_A(T, T)$ denotes the restriction of $g_0$ to $T$, there therefore exists $s_1 \in S$ such that $s_1 h_0(x) = s_1 s_0 x$ for all $x \in T$; writing $s = s_1 s_0$, $g = s_1 g_0$, $h = s_1 h_0$, $h$ is therefore the homothety of ratio $s$ on $T$ and is the restriction of $g$ to $T$. It remains to verify that $h$ is a pseudo-isomorphism. Now, if $p = 0$ or if $p \in P$ is distinct from the $p_i$ ($1 \leq i \leq k$), $T_p = 0$ (Chapter II, §4, no. 4, Proposition 17) and $h,\ : T_p \to T_p$ is an isomorphism; if on the contrary $p$ is equal to one of the $p_i$ ($1 \leq i \leq k$), $s$ is invertible in $A$, and $h_{,,}$, the homothety of ratio $s$ on $T_{p_i}$, is also an isomorphism, which completes the proof of Lemma 3.

We now prove Theorem 4. Let $g : E \to T$ be a homomorphism satisfying the properties of Lemma 3; let $h$ be the restriction of $g$ to $T$ and let $x$ be the canonical projection of $E$ onto $M$. We show that the homomorphism $f = (g, x) : E \to T \times M$ solves the problem. There is the commutative diagram:

where the rows are exact. The snake diagram (Chapter I, §1, no. 4, Proposition 2) gives the exact sequence:

and hence $\operatorname{Ker}(f)$ is isomorphic to $\operatorname{Ker}(h)$ and $\operatorname{Coker}(f)$ to $\operatorname{Coker}(h)$. As $h$ is a pseudo-isomorphism, so is $f$.

We can say that "to within a pseudo-isomorphism" Theorem 4 reduces the study of finitely generated $A$-modules to that of torsion-free modules on the one hand and to that of torsion modules on the other. Moreover, we have seen above (Example 3) that a torsion-free module is pseudo-isomorphic to its bidual and hence to a reflexive module. As for torsion modules, there is the following result, which determines them to within a pseudo-isomorphism: