
Lemma 10.122.5. In Situation 10.122.4. 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 10.35.3. We have $\varphi (a_0 + a_1x + \ldots + a_ k x^ k) u t_ i \in \mathop{\mathrm{Im}}(\varphi )$. By Lemma 10.122.3, 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. $\square$

There are also:

• 3 comment(s) on Section 10.122: Zariski's Main Theorem

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).