k is an alegraically closed field and A is a commutative kalgebra. We also know that A is a Noetherian domain and its Krull dimension is one. Are there any necessary and sufficient conditions on A under which A becomes finitely generated module over a polynomial algebra k[c] for some c in A? Does anybody know any papers or books that discuss this? Thanks guys.

4$\begingroup$ The answer seems to be that it is necessary and sufficient for $A$ to be finitely generated as a $k$algebra. Necessity: finitely generated as a module implies finitely generated as an algebra, and finitely generated over finitely generated is finitely generated. Sufficiency: apply Noether normalization as suggested in the responses below (e.g. Eisenbud, Thm. 13.3). This has an evident generalization to any finite Krull dimension. $\endgroup$– Pete L. ClarkMar 10, 2010 at 13:16

$\begingroup$ +1 for catching my error, Prof. Clark. $\endgroup$– Harry GindiMar 10, 2010 at 13:29
3 Answers
Dear Amitsur,
It might help you to think geometrically. For example, $k[x,x^{1}]$ is the ring of functions on a hyperbola $xy = 1$, and the projection from this hyperbola to the line $x = y$ is a finite projection. This corresponds to the fact that $k[x,x^{1}]$ is finitely generated as a module over $k[x + x^{1}].$ (If we write $f = x + x^{1}$, then $x^2  f x +1 = 0$ and $x^{2}  f x^{1} + 1 = 0$.)
This is always the case, by Noether normalization. For a proof, see for examlple, Eisenbud  "Commutative algebra with a view towards algebraic geometry"  Theorem 13.3.
Edit: this is false, one need a finiteness assumption. See comments below.

$\begingroup$ It's clearly not always the case, because A could be L[T], a polynomial ring in one variable over a field L which is a huge algebraically closed field and a transcendental extension of k $\endgroup$ Mar 10, 2010 at 13:03

$\begingroup$ you are right off course. My mistake, I misread the question. $\endgroup$– the LMar 10, 2010 at 13:05

$\begingroup$ This is incorrect. If $A$ is a finitely generated module over $k[t]$, then in particular $A$ is a finitely generated $k$algebra, hence a HilbertJacobson ring. The Laurent series ring $k[[t]]$ is a PID (and not a field) with only finitely many prime ideals, hence not HilbertJacobson: see e.g. Section 8 of math.uga.edu/~pete/integral.pdf. $\endgroup$ Mar 10, 2010 at 13:06
This follows from a direct generalization of the Noether normalization lemma. It is covered in these notes from Mel Hochster. These notes prove it in a pretty general form (when the base ring is only an integral domain rather than a field).
Edit: A sufficient condition is that the algebra is finitely generated, but it is clearly not necessary.
Edit 2: I misread the question. I thought he was asking if A is finitely generated over some polynomial algebra (including infinitely generated polynomial algebras).

$\begingroup$ Thank you guys. I'll check the sources you gave me. So, the Laurent polynomial ring A=k[x,x^{1}] must be finitely generated k[f]module for some f in A then. This is kind of strange, isn't it? $\endgroup$– AmitsurMar 10, 2010 at 13:02

$\begingroup$ See the comments above. This is incorrect (note: being finitely generated over k is clearly necessary.) $\endgroup$ Mar 10, 2010 at 13:06

1$\begingroup$ Dear Amitsur, perhaps it seems less strange if you realize that you can take f=x+x^{1} and that x then satisfies the equation x^2fx+1=o, so that 1 and x generate A as a k[f]module . $\endgroup$ Mar 10, 2010 at 13:46

$\begingroup$ lol, of course! Thanks Georges Elencwajg. I guess I need to get some rest. Thank you all for your useful comments. $\endgroup$– AmitsurMar 10, 2010 at 13:57