Lemma 10.123.6. In Situation 10.123.4. Suppose $u \in S$, $a_0, \ldots , a_ k \in R$, $u \varphi (a_0 + a_1x + \ldots + a_ k x^ k) \in \sqrt{J}$. Then $u \varphi (a_ i) \in \sqrt{J}$ for all $i$.

Proof. Under the assumptions of the lemma we have $u^ n \varphi (a_0 + a_1x + \ldots + a_ k x^ k)^ n \in J$ for some $n \geq 1$. By Lemma 10.123.5 we deduce $u^ n \varphi (a_ k^{nm}) \in J$ for some $m \geq 1$. Thus $u \varphi (a_ k) \in \sqrt{J}$, and so $u \varphi (a_0 + a_1x + \ldots + a_ k x^ k) - u \varphi (a_ k x^ k) = u \varphi (a_0 + a_1x + \ldots + a_{k-1} x^{k-1}) \in \sqrt{J}$. We win by induction on $k$. $\square$

## Comments (2)

Comment #4512 by Noah Olander on

Looks like in the second to last sentence in the proof you should subtract by $u\phi (a_k x^k)$ not $u \phi (a_k)$.

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• 3 comment(s) on Section 10.123: Zariski's Main Theorem

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