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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