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Hello,

Consider the following question. Let $A$ be a finitely generated reduced algebra over an algebraically closed field $k$. Consider the group of units of $A$, modulo the group $k^*$. Is this group always finitely generated?

Admittedly, I did not think of that question seriously, but I will be glad to hear the answer.

Thank you, Sasha

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Why the tag "algebraic number theory"? –  Martin Brandenburg Mar 4 '11 at 14:43
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The question is motivated by Dirichlet's unit theorem, I guess. At least when A is one-dimensional it is more or less the function field analogue. –  Qiaochu Yuan Mar 4 '11 at 15:09
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Here's a proof sketch. Reduce to A being normal first using reducedness, and then choose a normal projective compactification Spec(A) -> X. Note that by normality of X there is valuation map val:A* -> Z^S, where S is the (finite) set of Weil divisors contained in X - Spec(A), and val is given by sending an element of A* to its valuation at the corresponding dvr. Then ker(val) comprises meromorphic functions on X with no poles, so is simply k*, and img(val) is finitely generated as Z^S is so. Hence, A*/k* is finitely generated. –  Bhargav Mar 4 '11 at 15:14
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The most general version I know is Lemme 1 on p. 7 in this nice article: math.jussieu.fr/~kahn/preprints/picfini.pdf This paper has also other information related to your question. –  Lars Mar 4 '11 at 19:29
    
Thank you all, for answering. I can not decide which answer should be "the answer". And yes, the tag "algebraic number theory" was because without Dirichlet's unit theorem I would not have think of that problem. –  Sasha Mar 9 '11 at 12:13
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4 Answers

As always, the algebraic geometry version is easier than the number theoretic one. For convenience (which can be tweaked) let me assume that $A$ is the coordinate ring of an affine open set of a smooth projective variety $X$ and let $K$ be the rational functions on $X$. Then we have the natural map $K^*\to \mathrm{Div} X$ given by divisor of a function. Let $D_1,\ldots, D_n$ be the divisors at infinity. The kernel of the above map is just $k^*$ and an element of $A$ is a unit if and only if its divisor is supported wholly on the $D_i$'s. Thus, we see that $A^*/k^*$ is a subgroup of the finitely generated free abelian group generated by the $D_i$'s.

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Hi Mohan, I hope you don't but I edited it to fix the display. –  Donu Arapura Mar 5 '11 at 20:48
    
Well, I sure don't mind. Thanks a lot, Donu. –  Pete L. Clark Mar 5 '11 at 21:12
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I translate into English Lemma 6.5 from Sansuc's paper Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres, J. reine angew. Math. 327 (1981), 12-80, scan here.

Let $X$ be an algebraic variety over an arbitrary field $k$, i.e a geometrically integral algebraic $k$-scheme. We denote $ U(X) := k[X]^* / k^* $, the group of units of the ring of regular functions $k[X]^*$ modulo nonzero constants.

Rosenlicht's theorem: Let $X$ and $Y$ be two algebraic $k$-varieties and $G$ be a connected, smooth, linear algebraic $k$-group, .

(i) $U(X)$ is a finitely generated free abelian group;

(ii) $U(X\times_k Y)=U(X)\oplus U(Y)$;

(iii) $U(G)=\hat{G}(k)$ (the group of $k$-characters of $G$).

Reference: M. Rosenlicht, Toroidal algebraic groups, Proc. AMS 12 (1961), 984–988, scan here. For other references see Sansuc's paper.

(i) answers the question. Rosenlicht's proof of (i) is close to Mohan's answer.

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Yes. See the beginning of section 3 of "Compactifications of subvarieties of tori" by Jenia Tevelev. He has a finitely generated integral domain $A$ (he calls it $\mathcal O(X)$) over an algebraically closed field $k$, and states that it is "well-known" that $A^\ast/k^\ast$ is a finitely generated free group. I Googled a little and I found other statements of this result, but I couldn't find a better reference.

Note that if you accept the statement for integral domains, it's not hard to show it for reduced algebras more generally.

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I also think the answer is "yes" and I also haven't yet nailed down a precise reference. I found though a nice analysis of the relative unit group $R^{\times}/k^{\times}$ in the case $R = k[X]$ is the coordinate ring of a regular, integral affine curve over an arbitrary field $k$ in the following paper:


Rosen, Michael $S$-units and $S$-class group in algebraic function fields. J. Algebra 26 (1973), 98-–108.


It seems vaguely plausible that you could use this to prove the general (integral) case by some fibering argument, but I haven't really thought this through.

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Shouldn't that be "I also think that the answer is yes"? –  Qiaochu Yuan Mar 5 '11 at 12:08
    
Yes to your comment answer also is the. –  Pete L. Clark Mar 5 '11 at 21:13
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