Lemma 10.118.4. Let R \to S be a ring map. Let 0 \to M_1 \to M_2 \to M_3 \to 0 be a short exact sequence of S-modules. Then
U(R \to S, M_1) \cap U(R \to S, M_3) \subset U(R \to S, M_2).
Proof. Let u \in U(R \to S, M_1) \cap U(R \to S, M_3). Choose f_1, f_3 \in R such that u \in D(f_1), u \in D(f_3) and such that (10.118.3.1) holds for f_1 and M_1 and for f_3 and M_3. Then set f = f_1f_3. Then u \in D(f) and (10.118.3.1) holds for f and both M_1 and M_3. An extension of free modules is free, and an extension of finitely presented modules is finitely presented (Lemma 10.5.3). Hence we see that (10.118.3.1) holds for f and M_2. Thus u \in U(R \to S, M_2) and we win. \square
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