Lemma 23.11.1. 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 isomorphic to R/I, then there exists a nonzerodivisor x \in I such that the image of x in I/I^2 generates F.
Proof. By assumption we may choose a finite free resolution
0 \to R^{\oplus n_ e} \to R^{\oplus n_{e-1}} \to \ldots \to R^{\oplus n_1} \to R \to R/I \to 0
Then \varphi _1 : R^{\oplus n_1} \to R has rank 1 and we see that I contains a nonzerodivisor y by Algebra, Proposition 10.102.9. Let \mathfrak p_1, \ldots , \mathfrak p_ n be the associated primes of R, see Algebra, Lemma 10.63.5. Let I^2 \subset J \subset I be an ideal such that J/I^2 = F. Then J \not\subset \mathfrak p_ i for all i as y^2 \in J and y^2 \not\in \mathfrak p_ i, see Algebra, Lemma 10.63.9. By Nakayama's lemma (Algebra, Lemma 10.20.1) we have J \not\subset \mathfrak m J + I^2. By Algebra, Lemma 10.15.2 we can pick x \in J, x \not\in \mathfrak m J + I^2 and x \not\in \mathfrak p_ i for i = 1, \ldots , n. Then x is a nonzerodivisor and the image of x in I/I^2 generates (by Nakayama's lemma) the summand J/I^2 \cong R/I. \square
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