Lemma 10.123.5. In Situation 10.123.4. Suppose $u \in S$, $a_0, \ldots , a_ k \in R$, $u \varphi (a_0 + a_1x + \ldots + a_ k x^ k) \in J$. Then there exists an $m \geq 0$ such that $u \varphi (a_ k)^ m \in J$.
Proof. Assume that $S$ is generated by $t_1, \ldots , t_ n$ as an $R[x]$-module. In this case $J = \{ g \in S \mid gt_ i \in \mathop{\mathrm{Im}}(\varphi )\text{ for all }i\} $. Note that each element $u t_ i$ is integral over $R[x]$, see Lemma 10.36.3. We have $\varphi (a_0 + a_1x + \ldots + a_ k x^ k) u t_ i \in \mathop{\mathrm{Im}}(\varphi )$. By Lemma 10.123.3, for each $i$ there exists an integer $n_ i$ and an element $q_ i \in R[x]$ such that $\varphi (a_ k^{n_ i}) u t_ i - \varphi (q_ i)$ is integral over $R$. By assumption this element is in $\varphi (R)$ and hence $\varphi (a_ k^{n_ i}) u t_ i \in \mathop{\mathrm{Im}}(\varphi )$. It follows that $m = \max \{ n_1, \ldots , n_ n\} $ works. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: