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As stated here, Noether's Normalization Theorem states:

Suppose that $R$ is a finitely generated integral domain over a field $K$. Then there exists an algebraically independent subset $\{y_1,y_2,y_3,\dots,y_r\}\in R$ such that $R$ is integral over $K[y_1,y_2,y_3,\dots,y_r]$.

If I correctly understand the statement, $R$ is a $K$-algebra, such that if $ab=0$, then $a=0$ and/or $b=0$. Let its generators be $\{a_1,a_2,\dots,a_n\}$. Then $K[a_1,a_2,\dots,a_n]=R$. $R$ is integral over itself. Hence, $R$ is integral over $K[a_1,a_2,\dots,a_n]$.

  1. Is my reasoning correct?
  2. Does this count as proof of Noether's Normalization Theorem?

Thanks in advance!

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  • $\begingroup$ The generators $\{a_1,\ldots,a_n\}$ need not be algebraically independent. $\endgroup$ – Zev Chonoles Jul 22 '13 at 5:24
  • $\begingroup$ @ZevChonoles- sorry this might be a stupid question, but my logic was: if $a_k=$ some combination of the other generators, then every element in $R$ can anyway be constructed by $\{a_1,a_2,\dots,a_n\}\setminus \{a_k\}$. So $\{a_1,a_2,\dots,a_n\}\setminus \{a_k\}\subset \{a_1,a_2,\dots,a_n\}$ can also be the set of generators.. Is it possible to construct an algebraically independent set of generators this way? $\endgroup$ – fierydemon Jul 22 '13 at 5:30
  • $\begingroup$ Ok sorry found a counterexample to my own question. If $\{a,b,c\}$ are the generators, and $a^2=b^3c^4$, then $a$ can't be removed from the list of generators. $\endgroup$ – fierydemon Jul 22 '13 at 5:33
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The theorem requires the $y_i$ to be algebraically independent. Your "proof" doesn't even mention this condition. So no, it is not correct.

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  • $\begingroup$ @ChrisEagle- Thanks for your answer! $\endgroup$ – fierydemon Jul 22 '13 at 5:34

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