Lemma 23.11.2. Let R be a Noetherian local ring. Let I \subset R be an ideal of finite projective dimension over R. If F \subset I/I^2 is a direct summand free of rank r, then there exists a regular sequence x_1, \ldots , x_ r \in I such that x_1 \bmod I^2, \ldots , x_ r \bmod I^2 generate F.
Local version of [Theorem 1.1, Vasconcelos]
Proof. If r = 0 there is nothing to prove. Assume r > 0. We may pick x \in I such that x is a nonzerodivisor and x \bmod I^2 generates a summand of F isomorphic to R/I, see Lemma 23.11.1. Consider the ring R' = R/(x) and the ideal I' = I/(x). Of course R'/I' = R/I. The short exact sequence
0 \to R/I \xrightarrow {x} I/xI \to I' \to 0
splits because the map I/xI \to I/I^2 sends xR/xI to a direct summand. Now I/xI = I \otimes _ R^\mathbf {L} R' has finite projective dimension over R', see More on Algebra, Lemmas 15.74.3 and 15.74.9. Hence the summand I' has finite projective dimension over R'. On the other hand, we have the short exact sequence 0 \to xR/xI \to I/I^2 \to I'/(I')^2 \to 0 and we conclude I'/(I')^2 has the free direct summand F' = F/(R/I \cdot x) of rank r - 1. By induction on r we may we pick a regular sequence x'_2, \ldots , x'_ r \in I' such that there congruence classes freely generate F'. If x_1 = x and x_2, \ldots , x_ r are any elements lifting x'_1, \ldots , x'_ r in R, then we see that the lemma holds. \square
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