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

  • $\begingroup$ Why the tag "algebraic number theory"? $\endgroup$ Mar 4 '11 at 14:43
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    $\begingroup$ 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. $\endgroup$ Mar 4 '11 at 15:09
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    $\begingroup$ 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. $\endgroup$
    – Bhargav
    Mar 4 '11 at 15:14
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    $\begingroup$ 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. $\endgroup$
    – Lars
    Mar 4 '11 at 19:29
  • $\begingroup$ 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. $\endgroup$
    – Sasha
    Mar 9 '11 at 12:13

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.

  • $\begingroup$ Hi Mohan, I hope you don't but I edited it to fix the display. $\endgroup$ Mar 5 '11 at 20:48
  • $\begingroup$ Well, I sure don't mind. Thanks a lot, Donu. $\endgroup$ Mar 5 '11 at 21:12

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.


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.


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.

  • $\begingroup$ Shouldn't that be "I also think that the answer is yes"? $\endgroup$ Mar 5 '11 at 12:08
  • $\begingroup$ Yes to your comment answer also is the. $\endgroup$ Mar 5 '11 at 21:13

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