0
$\begingroup$

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.)

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ Do you want to add some hypotheses on $R$? This can definitely fail if $\text{Spec} R$ is disconnected or reducible. $\endgroup$
    – Jason Starr
    Nov 28, 2013 at 10:45
  • $\begingroup$ 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. $\endgroup$
    – Andreas Maurischat
    Nov 28, 2013 at 12:13
  • 2
    $\begingroup$ 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 $A$. $\endgroup$
    – Jason Starr
    Nov 28, 2013 at 13:13
  • $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Nov 28, 2013 at 18:19
  • 1
    $\begingroup$ 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. $\endgroup$
    – Andreas Maurischat
    Jan 7, 2019 at 7:27

1 Answer 1

Reset to default
3
$\begingroup$

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...

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ This is in Nagata, Local rings, I.14.4. $\endgroup$
    – Cantlog
    Nov 28, 2013 at 22:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.