Lemma 15.26.4. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_ k(M)$. Asssume that $M_\mathfrak p$ is free of rank $k$ for every $\mathfrak p \not\in V(I)$. Then for every $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform
is locally free of rank $k$.
Proof. By Lemma 15.26.3 we have $\text{Fit}_ k(M') = R'$. By Lemma 15.8.7 it suffices to show that $\text{Fit}_{k - 1}(M') = 0$. Recall that $R' \subset R'_ a = R_ a$, see Algebra, Lemma 10.70.2. Hence it suffices to prove that $\text{Fit}_{k - 1}(M')$ maps to zero in $R'_ a = R_ a$. Since clearly $(M')_ a = M_ a$ this reduces us to showing that $\text{Fit}_{k - 1}(M_ a) = 0$ because formation of Fitting ideals commutes with base change according to Lemma 15.8.4 part (3). This is true by our assumption that $M_ a$ is finite locally free of rank $k$ (see Algebra, Lemma 10.78.2) and the already cited Lemma 15.8.7. $\square$
Your email address will not be published.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)
There are also: