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
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.
$\square$
\[ 0 \to R/I \xrightarrow {x} I/xI \to I' \to 0 \]
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