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Add nontriviality assumption

Without this assumption, the statement is false. For no integer r >= 0
there is an injection from k[y_1,...,y_r] to the zero ring.
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iblech authored and aisejohan committed Jun 3, 2017
1 parent 51e21f4 commit 886da4ab466b262a4326636c343bd55109d47def
Showing with 1 addition and 1 deletion.
  1. +1 −1 algebra.tex
@@ -26602,7 +26602,7 @@ \section{Noether normalization}
Let $k$ be a field. Let $S = k[x_1, \ldots, x_n]/I$ for some ideal $I$.
There exist $r\geq 0$, and $y_1, \ldots, y_r \in k[x_1, \ldots, x_n]$
If $I \neq (1)$, there exist $r\geq 0$, and $y_1, \ldots, y_r \in k[x_1, \ldots, x_n]$
such that (a) the map $k[y_1, \ldots, y_r] \to S$ is injective,
and (b) the map $k[y_1, \ldots, y_r] \to S$ is finite.
In this case the integer $r$ is the dimension of $S$.

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