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Clarify a proof

1 parent ed97289 commit 7788514bebf9c154bfef8eb8ab22643b08063a54 @aisejohan aisejohan committed Feb 17, 2017
Showing with 4 additions and 1 deletion.
  1. +4 −1 algebra.tex
@@ -11172,7 +11172,10 @@ \section{Valuation rings}
Assume that $x$ is not in $A$.
Let $A'$ denote the subring of $K$ generated by $A$ and $x$.
Since $A$ is a valuation ring we see that there is no prime
-of $A'$ lying over $\mathfrak m$. Hence we can write
+of $A'$ lying over $\mathfrak m$. Since $\mathfrak m$ is maximal
+we see that $V(\mathfrak m A') = \emptyset$. Then $\mathfrak m A' = A'$
+by Lemma \ref{lemma-Zariski-topology}.
+Hence we can write
$1 = \sum_{i = 0}^d t_i x^i$ with $t_i \in \mathfrak m$.
This implies that $(1 - t_0) (x^{-1})^d - \sum t_i (x^{-1})^{d - i} = 0$.
In particular we see that $x^{-1}$ is integral over $A$.

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