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 $F$.
[Vasconcelos]
Proof.
By assumption we may choose a finite free resolution
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 $J/I^2 \cong R/I$.
$\square$
\[ 0 \to R^{\oplus n_ e} \to R^{\oplus n_{e-1}} \to \ldots \to R^{\oplus n_1} \to R \to R/I \to 0 \]
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)