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Improve a lemma

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aisejohan committed Sep 2, 2019
1 parent 3f2a0d0 commit 0a7676cfe86d46888b4955d64438fc7d58755bd5
Showing with 11 additions and 6 deletions.
  1. +11 −6 algebra.tex
@@ -27063,8 +27063,8 @@ \section{Noether normalization}

\begin{lemma}
\label{lemma-Noether-normalization-over-a-domain}
Let $R \to S$ be an injective finite type map of domains. Then there exists an
integer $d$ and a factorization
Let $R \to S$ be an injective finite type ring map. Assume $R$ is a domain.
Then there exists an integer $d$ and a factorization
$$
R \to R[y_1, \ldots, y_d] \to S' \to S
$$
@@ -27083,15 +27083,20 @@ \section{Noether normalization}
Lemma \ref{lemma-finite-is-integral})
we can find monic $P_i \in K[y_1, \ldots, y_d][T]$ such that
$P_i(x_i) = 0$ in $S_K$. Let $f \in R$ be a nonzero element such that
$fP_i \in R[y_1, \ldots, y_d][T]$ for all $i$. Set $x_i' = fx_i$
and let $S' \subset S$ be the subalgebra generated by $y_1, \ldots, y_d$
and $x'_1, \ldots, x'_n$. Note that $x'_i$ is integral over
$fP_i \in R[y_1, \ldots, y_d][T]$ for all $i$. Then $fP_i(x_i)$
maps to zero in $S_K$. Hence after replacing $f$ by another
nonzero element of $R$ we may also assume $fP_i(x_i)$ is zero in $S$.
Set $x_i' = fx_i$ and let $S' \subset S$ be the $R$-subalgebra generated by
$y_1, \ldots, y_d$ and $x'_1, \ldots, x'_n$. Note that $x'_i$ is integral over
$R[y_1, \ldots, y_d]$ as we have $Q_i(x_i') = 0$ where
$Q_i = f^{\deg_T(P_i)}P_i(T/f)$ which is a monic polynomial in
$T$ with coefficients in $R[y_1, \ldots, y_d]$ by our choice of $f$.
Hence $R[y_1, \ldots, y_n] \subset S'$ is finite by
Lemma \ref{lemma-characterize-finite-in-terms-of-integral}.
By construction $S'_f \cong S_f$ and we win.
Since $S' \subset S$ we have $S'_f \subset S_f$ (localization is exact).
On the other hand, the elements $x_i = x'_i/f$ in $S'_f$ generate $S_f$
over $R_f$ and hence $S'_f \to S_f$ is surjective. Whence
$S'_f \cong S_f$ and we win.
\end{proof}


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