Lemma 51.17.1. Let $p$ be a prime number. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with $p = 0$ in $A$. Let $M$ be a finite $A$-module such that $M \otimes _{A, F} A \cong M$. Then $M$ is finite free.

**Proof.**
Choose a presentation $A^{\oplus m} \to A^{\oplus n} \to M$ which induces an isomorphism $\kappa ^{\oplus n} \to M/\mathfrak m M$. Let $T = (a_{ij})$ be the matrix of the map $A^{\oplus m} \to A^{\oplus n}$. Observe that $a_{ij} \in \mathfrak m$. Applying base change by $F$, using right exactness of base change, we get a presentation $A^{\oplus m} \to A^{\oplus n} \to M$ where the matrix is $T = (a_{ij}^ p)$. Thus we have a presentation with $a_{ij} \in \mathfrak m^ p$. Repeating this construction we find that for each $e \geq 1$ there exists a presentation with $a_{ij} \in \mathfrak m^ e$. This implies the fitting ideals (More on Algebra, Definition 15.8.3) $\text{Fit}_ k(M)$ for $k < n$ are contained in $\bigcap _{e \geq 1} \mathfrak m^ e$. Since this is zero by Krull's intersection theorem (Algebra, Lemma 10.51.4) we conclude that $M$ is free of rank $n$ by More on Algebra, Lemma 15.8.7.
$\square$

## 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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: