Lemma 10.15.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Let $\varphi : M \to M$ be an endomorphism such that $\varphi (M) \subset IM$. Then there exists a monic polynomial $P = t^ n + a_1 t^{n - 1} + \ldots + a_ n \in R[T]$ such that $a_ j \in I^ j$ and $P(\varphi ) = 0$ as an endomorphism of $M$.

**Proof.**
Choose a surjective $R$-module map $R^{\oplus n} \to M$, given by $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ for some generators $x_ i \in M$. Choose $(a_{i1}, \ldots , a_{in}) \in I^{\oplus n}$ such that $\varphi (x_ i) = \sum a_{ij} x_ j$. In other words the diagram

is commutative where $A = (a_{ij})$. By Lemma 10.15.1 the polynomial $P(t) = \det (t\text{id}_{n \times n} - A)$ has all the desired properties. $\square$

## 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Your email address will not be published.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)