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Dirichlet's unit theorem states that (i) the group of units, $\mathscr{U}_K$, of the ring of integers of a number field $K$ is finitely generated, and (ii) the rank of $\mathscr{U}_K$ is equal to $r_1 + r_2 - 1$, where $r_1$ is the number of real embeddings and $r_2$ the number of conjugate pairs of complex embeddings of $K$.

Q. What about a reference to an elementary, short proof of (i)?

In principle, (i) is much weaker than (ii), so the question shouldn't be so implausible. The reason why I'm asking is that I seem to have an alternative, extravagant proof of a well-known result, but for the approach to be of any potential interest I need (a reference to) an elementary, short proof of (i).

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    $\begingroup$ Did you look in any algebraic number theory book that proves the Unit Theorem? Property (i) is nearly always just the first part of the proof! That and the upper bound $r_1 + r_2 -1$ are comparatively easy. The only really hard part of the full Unit Theorem is showing the rank is $r_1 + r_2 -1 $. $\endgroup$
    – KConrad
    Apr 16 '17 at 21:58
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    $\begingroup$ I did, but I am really hoping for something shorter and more elementary than the proof I've learned from Narkiewicz's Elementary and Analytic Theory of Algebraic Numbers (which is basically the same as the one in Ireland and Rosen's book suggested by Myshkin and Francesco Polizzi): After seeing Perdry's proof of Krull's intersection theorem, I convinced myself that there might have been published an equally slick proof of (i). After all, the bound implied by the proof in Narkiewicz's or Ireland and Rosen's book is still very tight. Does this sound completely implausible? $\endgroup$ Apr 16 '17 at 22:20
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    $\begingroup$ I've never seen an elementary proof of the group being finitely generated that doesn't essentially get $r_1+r_2-1$ as an upper bound as part of the argument. The proof in Narkiewicz is for $S$-units, so it has the appearance of being more complicated. Have you looked in Marcus's Number Fields, pp. 142-143? $\endgroup$
    – KConrad
    Apr 16 '17 at 22:52
  • $\begingroup$ It follows from the fact that every $abelian$ subgroup of $GL_n(\mathbb{Z})$ is finitely generated. $\endgroup$ Apr 17 '17 at 0:17
  • $\begingroup$ @KConrad I followed your suggestion and gave a look to Marcus' Number Theory. To my eyes, the proof is indeed neater than the ones I've seen so far, though this is mainly due to the way in which it's presented, not really to the circle of ideas on which it relies. In any case, it's getting definitely clear that my expectations were ungrounded: Since you've never seen an elementary proof of (i) that doesn't essentially yield, as part of the argument, the upper bound $r_1+r_2-1$ on the rank, it's highly probable that the kind of short and elementary proof I was looking for isn't (yet?) known. $\endgroup$ Apr 17 '17 at 13:14
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$O_K$ is a discrete cocompact subring in $K=\mathbf{R}^{r_1}\times\mathbf{C}^{r_2}$. So the group $U_K^\times$ of invertible elements of $O_K$ is a discrete subgroup in $K^\times$, the group of invertible elements of $K$, which is isomorphic to $\mathbf{R}^{r_1+r_2}\times W$ with $W$ compact (isomorphic to $(\mathbf{Z}/2\mathbf{Z})^{r_1}\times(\mathbf{R}/\mathbf{Z})^{r_2}$).

A discrete subgroup of such a Lie group is finitely generated and its $\mathbf{Q}$-rank is bounded above by $r_1+r_2$. Indeed the composite homomorphism $U_K\to \mathbf{R}^{r_1+r_2}$ is proper and hence maps injectively $U_K$ to a discrete subgroup of $\mathbf{R}^{r_1+r_2}$.

(Probably the upper bound by $r_1+r_2-1$ comes from an additional argument saying that $U_K$ goes into some subgroup of elements of norm one, but the point here is precisely that we don't need this to get the above weaker upper bound $r_1+r_2$.)

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    $\begingroup$ Precisely, the upper bound $r_1+r_2-1$ comes from the fact that the image of the logarithmic map $U_K\to \mathbf R^{r_1+r_2}$ is contained in the hyperplane defined by the equation $\sum x_i=0$. The precise determination of the rank of $U_K$, though, requires an additional argument and construction of independent units. $\endgroup$
    – ACL
    Apr 16 '17 at 21:55
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    $\begingroup$ @ACL the precise hyperplane depends on how the logarithmic map is defined (coefficients of 2 for complex places). $\endgroup$
    – KConrad
    Apr 16 '17 at 21:57
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    $\begingroup$ @KConrad Of course it does! I expected comments allowed for such an imprecision... $\endgroup$
    – ACL
    Apr 16 '17 at 22:00
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    $\begingroup$ I made the comment more for a student who might be passing by this page in the future. No big deal. $\endgroup$
    – KConrad
    Apr 16 '17 at 22:03
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This part of Dirichlet's unit theorem is the only one needed in the standard proof of the Weil-Mordell theorem over number fields, so you probably can find it in a few introductory books on elliptic curves.

In particular, "A Classical Introduction to Modern Number Theory" by Kenneth Ireland and Michael Ira Rosen contains a full proof in pages 326-327 (they call it "the weak Dirichlet unit theorem").

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I do not know if the following qualifies as "short" or "elementary": but it does not follow the usual pattern through Minkowski's Convex Body Theorem. Rather, it mimics the classical proof of Mordell–Weil: honestly, it copies it, but I write the details down to give a direct feeling of what comes into the game.

First of all, Hermite—Minkowski tells you that $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$ is finite, as follows. For every unit $u$, the extension $K(\sqrt{u})/K$ has discriminant whose norm equals a bounded power of $2$, because every integral element $r+s\sqrt{u}\in \mathcal{O}_{K(\sqrt{u})}$ (with $r,s\in K$) has minimal polynomial $f(X)=X^2-2rX+r^2-us^2$ with $2r,r^2-us^2\in\mathcal{O}_K$. This shows that $\mathcal{O}_{K(\sqrt{u})}/\mathcal{O}_K(\sqrt{u})$ is a finite group killed by $4$ and of rank bounded by $2[K\colon\mathbb{Q}]$. The discriminant of $K(\sqrt{u})/K$ is then $2\sqrt{u}\cdot[\mathcal{O}_{K(\sqrt{u})}\colon\mathcal{O}_K(\sqrt{u})]$ and thus we get a bound $\operatorname{Disc}_{K(\sqrt{u})/\mathbb{Q})}\leq \operatorname{Disc}_{K/\mathbb{Q}}^2\cdot 2^{a(K)}$ for a constant $a(K)$ depending on $K$ only. It follows that the quotient $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$ classifies quadratic extensions of $K$ (this is easy: Kummer theory works over any field, in this case of characteristic different from $2$) of degree bounded by $2[K\colon\mathbb{Q}]$ and discriminant bounded by $\operatorname{Disc}_{K/\mathbb{Q}}^22^{a(K)}$, and this is a finite set of extensions by Hermite–Minkowski. One might say that Hermite–Minkowski is not trivial, which is true: but it is easier than the full proof of Dirichlet's Unit Theorem, in the sense that it comes well before it in almost all books I know, and Hermite's original proof consists of elementary, although quite tedious, algebra not involving any topology.

Now comes the ''Mordell–Weil" part: consider the usual height $$ H(u)=\sqrt[{[K\colon\mathbb{Q}]}]{\prod_{\sigma\in\mathcal{M}_K^\infty}\max\{1,\vert\sigma(u)\vert\}^{\varepsilon(\sigma)}}\qquad\text{ for all }u\in\mathcal{O}_K^\times $$ where the product is over all infinite places and $\varepsilon(\sigma)=1$ if $\sigma(K)\subseteq\mathbb{R}$, and $2$ otherwise. It is immediate to see that $H(u)\geq 1$ for all $u$, that $H(uv)\geq H(u)H(v)$ for all $u,v$ and that $H(u^{m})=H(u)^m$ for all $u$ and $m\geq 1$; and it is a classical result that the Northcott property holds, namely that for every given bound $B$ there are only finitely many elements in $\mathcal{O}_K^\times$ such that $H(u)\leq B$: this is elementary again, using simply that algebraic integers of bounded degree and bounded height are roots of polynomials in $\mathbb{Z}[X]$ with bounded coefficients and degree, hence a finite number of polynomials. A proof of this can be found on page 503 of Northcott's original paper

Northcott, D. (1949). An inequality in the theory of arithmetic on algebraic varieties. Mathematical Proceedings of the Cambridge Philosophical Society, 45(4), 502-509. doi:10.1017/S0305004100025202

By the first part, we know that there are finitely many units $\eta_1,\dots,\eta_r$ representing the elements in $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$; we also fix now a bound $B$ so that by the Northcott property there are only finitely many units $v_1,\dots,v_s$ of height bounded by $B$. Pick any $u\in\mathcal{O}_K^\times$: for every $n\geq 1$ we can write $$ u=\eta_{i_0}u_1^2,\quad u_1=\eta_{i_1}u_2^2,\quad\dots \quad u_{n-1}=\eta_{i_{n-1}}u_{n}^2 $$ and thus $$ u=\Big(\prod_{j=0}^n\eta_{i_j}^{2^j}\Big)u_{n+1}^{2^n}:=\eta(u;n)\cdot u_{n}^{2^n}. $$ with $\eta(u;n)$ belonging to the subgroup generated by $\eta_1,\dots,\eta_r$; moreover, the above equation shows $$ H(u)\geq H(u_n^{2^n})=H(u_n)^{2^n}\Rightarrow H(u_n)\leq H(u)^{1/2^n}\overset{n\to+\infty}{\longrightarrow} 1\quad(u\text{ is fixed}). $$ Therefore, if we choose $n$ to be big enough, we get $u_n\in\{v_1,\dots,v_r\}$ and therefore $u$ belongs to the subgroup generated by $\{\eta_1,\dots,\eta_r,v_1,\dots,v_s\}$, showing that $\mathcal{O}_K^\times$ is finitely generated.

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  • $\begingroup$ I'll need time to study your proof in detail, but the approach looks very interesting (though I still have to convince myself that it's the kind of short and elementary proof I was hoping for). $\endgroup$ Apr 26 '17 at 15:52
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    $\begingroup$ Well, if you want to "study" it, I'd rather advise to look at the beautiful presentation that Silverman offers in his Arithmetic of Elliptic Curves, chapter VIII, of the Mordell–Weil theorem. Mine is simply an adaptation. $\endgroup$ Apr 26 '17 at 16:01
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This follows, via the standard "logarithmic embedding", from the fact that a discrete subgroup of the additive group $\mathbb{R}^n$ is a lattice.

An elementary proof of (i) can be found in the book

K. Ireland, M. I. Rosen: A Classical Introduction to Modern Number Theory,

see in particular p. 326, $\S 3$: "The weak Dirichlet unit theorem".

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  • $\begingroup$ I guess it's a typo; a discrete subgroup of $\mathbf{R}^n$ need not be a lattice... or, you mean, is a lattice in the subspace it spans. $\endgroup$
    – YCor
    Apr 16 '17 at 22:48
  • $\begingroup$ Yes, of course you are right. Here $n$ must be intended as the dimension of the subspace spanned by the subgroup. $\endgroup$ Apr 17 '17 at 8:26
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William Stein gives a short argument (Section 8.1).

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  • $\begingroup$ I see section 4.3.1 is "inessential discriminant divisors." I never understood this phrase, which would make much more sense as essential discriminant divisors. In fact, Stein has another set of notes at wstein.org/129/ant/html/node29.html where he uses "essential discriminant divisor". In arxiv.org/pdf/1408.6251.pdf, Lagarias and Weiss write in a footnote on page 1 that prime divisors of the inessential discriminant divisor are the essential discriminant divisors. What nutty terminology! $\endgroup$
    – KConrad
    Apr 16 '17 at 23:59
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    $\begingroup$ Google tells me that I complained about this contradictory pair of terms before at mathoverflow.net/questions/239490/…. $\endgroup$
    – KConrad
    Apr 17 '17 at 0:04
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    $\begingroup$ @KConrad Well, you are consistent :) $\endgroup$
    – Igor Rivin
    Apr 17 '17 at 0:40
  • $\begingroup$ I do very much like the style of Stein's proof: He makes full justice, I think, of the elegance and simplicity of Dirichlet's idea(s). But, of course, this is, in essence, no different than the proofs mentioned so far (with the exception of the one provided by @FilippoAlbertoEdoardo). $\endgroup$ Apr 26 '17 at 16:39

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