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k is an alegraically closed field and A is a commutative k-algebra. 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.

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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. –  Pete L. Clark Mar 10 '10 at 13:16
    
+1 for catching my error, Prof. Clark. –  Harry Gindi Mar 10 '10 at 13:29
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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 $x^{-2} - f x^{-1} + 1 = 0$.)

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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.

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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 –  Kevin Buzzard Mar 10 '10 at 13:03
    
you are right off course. My mistake, I misread the question. –  Liran Shaul Mar 10 '10 at 13:05
    
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 Hilbert-Jacobson ring. The Laurent series ring $k[[t]]$ is a PID (and not a field) with only finitely many prime ideals, hence not Hilbert-Jacobson: see e.g. Section 8 of math.uga.edu/~pete/integral.pdf. –  Pete L. Clark Mar 10 '10 at 13:06
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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).

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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? –  Amitsur Mar 10 '10 at 13:02
    
See the comments above. This is incorrect (note: being finitely generated over k is clearly necessary.) –  Pete L. Clark Mar 10 '10 at 13:06
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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^2-fx+1=o, so that 1 and x generate A as a k[f]-module . –  Georges Elencwajg Mar 10 '10 at 13:46
    
lol, of course! Thanks Georges Elencwajg. I guess I need to get some rest. Thank you all for your useful comments. –  Amitsur Mar 10 '10 at 13:57
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