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Noether normalization tells us that a finitely generated $k$-algebra is an integral extension of a polynomial algebra over the field $k$.

My question is whether this still holds if we replace the field $k$ by a more general ring.

Question: Let $R$ be a commutative ring, and let $A$ be a finitely generated $R$-algebra which is flat over $R$. Are there $R$-algebraically independent elements $x_1,\dots, x_m\in A$ such that $A$ is integral over $R[x_1,\dots, x_m]$?

EDIT: Seeing that the general question above was already asked, I should change it to the case I am really interested in:

What is the answer in the case that $R$ is a local $k$-algebra of dimension zero over some field $k$? (So $R$ is nearly a field, but not reduced.)

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Andreas Maurischat
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    Do you want to add some hypotheses on $R$? This can definitely fail if $\text{Spec} R$ is disconnected or reducible. – Jason Starr Nov 28 '13 at 10:45
  • We can restrict to ${\rm Spec}R$ being connected. Maybe also that $A$ is even faithfully flat. The case I am troubled with is that of a nonreduced local ring $R$. Of course $R$ and $A$ should be commutative. – Andreas Maurischat Nov 28 '13 at 12:13
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    Unfortunately "connected" is not strong enough. Let $R$ be $k[u,v]/\langle uv \rangle$, and let $A$ be $R[s,t]/\langle t(u+v)-1,su \rangle$. Then $A$ is $R$-flat, but $A$ is not integral over $R[x_1,\dots,x_m]$ for any $R$-algebraically independent elements $x_1,\dots,x_m$ in $A$. – Jason Starr Nov 28 '13 at 13:13
  • My previous example is not faithfully flat over $R$, but that is easy to remedy. Just define $B$ to be $A\times R$ with its natural $R$-algebra structure. Really, you need to add some irreducibility hypotheses. – Jason Starr Nov 28 '13 at 18:19
  • Hey! Is that allowed? – Jason Starr Nov 28 '13 at 23:26
  • Dear @JasonStarr sorry for reviving this matter! I'm looking at Hochster's note (from Karl's answer) and I don't see where the domain assumption on D is used (neither for the normalization theorem nor the change-of-variable lemma preceding it). I understand why it's crucial that D→R be injective, but I don't see where the proof relies on properties of a domain (e.g on the fact localizations of a domain are injections). What am I missing? – Arrow Jan 07 '19 at 01:22
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    Dear @Arrow, in the proof in Hochster's note one localizes by more than one element. If D is not a domain their product could be zero. – Andreas Maurischat Jan 07 '19 at 07:27
  • Thanks to Andreas Maurischat for answering the question by @Arrow. – Jason Starr Jan 07 '19 at 13:52
  • Dear @AndreasMaurischat thank you! I just noticed Hochster explains this in the paragraph preceding the statement of the theorem. – Arrow Jan 08 '19 at 10:22

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Here's one generalization, taken from

Notes of Mel Hochster

Theorem: Let $D$ be an integral domain and let $R$ be any finitely generated $D$-algebra extension of $D$. Then there is a nonzero element $c \in D$ and elements $z_1, \ldots, z_d$ in $R_c$ algebraically independent over $D_c$ such that $R_c$ is module-finite over its subring $D_c[z_1, \ldots, z_d]$, which is isomorphic to a polynomial ring over $D_c$.

After looking at the proof very briefly, I don't see how to replace the inverting an element with a flatness hypothesis. Maybe you could blow something up and work on some charts though...

Karl Schwede
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