Finishing the proof of Dirichlet's Unit Theorem

We begin by finishing Dirichlet's proof that the group of units $U_K$ of $\O _K$ is isomorphic to $\mathbf{Z}^{r+s-1}\oplus \mathbf{Z}/m\mathbf{Z}$, where $r$ is the number of real embeddings, $s$ is half the number of complex embeddings, and $m$ is the number of roots of unity in $K$. Recall that we defined a map $\varphi :U_K\to \mathbf{R}^{r+s}$ by

$$\varphi (x) = (\log\vert\sigma_1(x)\vert,\ldots, \log\vert\sigma_{r+s}(x)\vert).$$

Without much trouble, we proved that the kernel of $\varphi$ if finite and the image $\varphi$ is discrete, and in the last section we were finishing the proof that the image of $\varphi$ spans the subspace $H$ of elements of $\mathbf{R}^{r+s}$ that are orthogonal to $v=(1,\ldots, 1, 2, \ldots, 2)$, where $r$ of the entries are $1$'s and $s$ of them are $2$'s. The somewhat indirect route we followed was to suppose

$$z\not\in H^{\perp} = \Span (v),$$

i.e., that $z$ is not a multiple of $v$, and prove that $z$ is not orthogonal to some element of $\varphi (U_K)$. Writing $W=\Span (\varphi (U_K))$, this would show that $W^{\perp}\subset H^{\perp}$, so $H\subset W$. We ran into two problems: (1) we ran out of time, and (2) the notes contained an incomplete argument that a quantity $s=s(c_1,\ldots, c_{r+s})$ can be chosen to be arbitrarily large. We will finish going through a complete proof, then compute many examples of unit groups using .

Recall that $f:K^*\to \mathbf{R}$ was defined by

$$f(x) = z_1\log\vert\sigma_1(x)\vert + \cdots + z_{r+s}\log\vert\sigma_{r+s}(x)\vert = z\bullet \varphi (x)$$   (dot product)$$,$$

and our goal is to show that there is a $u\in U_K$ such that $f(u)\neq 0$.

Our strategy is to use an appropriately chosen $a$ to construct a unit $u\in U_K$ such $f(u)\neq 0$. Recall that we used Blichfeld's lemma to find an $a\in\O _K$ such that $1\leq \vert\Norm _{K/\mathbf{Q}}(a)\vert\leq A$, and

$$\frac{c_i}{\vert\sigma_i(a)\vert}\leq A i\leq r \quad \left(\frac{c_i}{\vert\sigma_i(a)\vert}\right)^2\leq A i=r+1,\ldots, r+s.$$ (12.5)

Let $b_1,\ldots, b_m$ be representative generators for the finitely many nonzero principal ideals of $\O _K$ of norm at most $A=A_K=\sqrt{\vert d_K\vert} \cdot \left( \frac{2}{\pi}\right)^s$. Modify the $b_i$ to have the property that $\vert f(b_i)\vert$ is minimal among generators of $(b_i)$ (this is possible because ideals are discrete). Note that the set $\{\vert f(b_i)\vert : i = 1,\ldots, m\}$ depends only on $A$. Since $\vert\Norm _{K/\mathbf{Q}}(a)\vert\leq A$, we have $(a)=(b_j)$, for some $j$, so there is a unit  $u\in \O _K$ such that $a=u b_j$.

Let

$$s=s(c_1,\ldots, c_{r+s}) = z_1\log(c_1)+\cdots +z_{r+s}\log(c_{r+s})\in \mathbf{R}.$$

Lemma 12.2.1   We have

$$\vert f(u) - s\vert \leq B = \max_i(\vert f(b_i)\vert) + \log(A)\... ...i=1}^{r}\vert z_i\vert + \frac{1}{2}\cdot \sum_{i=r+1}^s\vert z_i\vert\right),$$

and $B$ depends only on $K$ and our fixed choice of $z\in H^{\perp}$.

Proof. By properties of logarithms, $f(u)=f(a/b_j)=f(a)-f(b_j)$. We next use the triangle inequality $\vert a+b\vert\leq \vert a\vert+\vert b\vert$ in various ways, properties of logarithms, and the bounds (12.2.1) in the following computation:

$$\vert f(u) - s\vert = \vert f(a) - f(b_j) - s\vert$$

$$\leq \vert f(b_j)\vert + \vert s - f(a)\vert$$

$$=\vert f(b_j)\vert + \vert z_1(\log(c_1) - \log(\vert\sigma_1(a)\vert)) + \cdots + z_{r+s}(\log(c_{r+s}) - \log(\vert\sigma_{r+s}(a)\vert))\vert$$

$$=\vert f(b_j)\vert + \vert z_1\cdot \log(c_1/\vert\sigma_1(a)\ver... ...ots + \frac{1}{2}\cdot z_{r+s} \log((c_{r+s}/\vert\sigma_{r+s}(a)\vert)^2)\vert$$

$$\leq \vert f(b_j)\vert + \log(A)\cdot\left(\sum_{i=1}^{r}\vert z_i\vert + \frac{1}{2}\cdot \sum_{i=r+1}^s\vert z_i\vert\right).$$

The inequality of the lemma now follows. That $B$ only depends on $K$ and our choice of $z$ follows from the formula for $A$ and how we chose the $b_i$. $\qedsymbol$

The amazing thing about Lemma 12.2.1 is that the bound $B$ on the right hand side does not depend on the $c_i$. Suppose we could somehow cleverly choose the positive real numbers $c_i$ in such a way that

$$c_1\cdots c_r\cdot (c_{r+1}\cdots c_{r+s})^2 = A$$   and$$\qquad \vert s(c_1,\ldots, c_{r+s})\vert>B.$$

Then the facts that $\vert f(u)-s\vert\leq B$ and $\vert s\vert>B$ would together imply that $\vert f(u)\vert>0$ (since $f(u)$ is closer to $s$ than $s$ is to 0), which is exactly what we aimed to prove. We finish the proof by showing that it is possible to choose such $c_i$. Note that if we change the $c_i$, then $a$ could change, hence the $j$ such that $a/b_j$ is a unit could change, but the $b_j$ don't change, just the subscript $j$. Also note that if $r+s=1$, then we are trying to prove that $\varphi (U_K)$ is a lattice in $\mathbf{R}^0=\mathbf{R}^{r+s-1}$, which is automatically true, so we may assume that $r+s>1$.

Lemma 12.2.2   Assume $r+s>1$. Then there is a choice of $c_1,\ldots, c_{r+s} \in \mathbf{R}_{>0}$ such that

$$\vert z_1\log(c_1)+\cdots +z_{r+s}\log(c_{r+s})\vert > B.$$

Proof. It is easier if we write

$$z_1\log(c_1) +\cdots +z_{r+s}\log(c_{r+s}) =$$

$$z_1\log(c_1)+\cdots + z_r\log(c_r)+ \frac{1}{2}\cdot z_{r+1}\log(c_{r+1}^2) + \cdots + \frac{1}{2}\cdot z_{r+s}\log(c_{r+s}^2)$$

$$=w_1\log(d_1)+\cdots + w_r\log(d_r)+ w_{r+1}\log(d_{r+1}) + \cdots +\cdot w_{r+s}\log(d_{r+s}),$$

where $w_i=z_i$ and $d_i=c_i$ for $i\leq r$, and $w_i=\frac{1}{2}z_i$ and $d_i=c_i^2$ for $r<i\leq s$,

The condition that $z\not\in H^{\perp}$ is that the $w_i$ are not all the same, and in our new coordinates the lemma is equivalent to showing that $\vert\sum_{i=1}^{r+s} w_i \log(d_i)\vert>B$, subject to the condition that $\prod_{i=1}^{r+s} d_i = A$. Order the $w_i$ so that $w_1\neq 0$. By hypothesis there exists a $w_j$ such that $w_j\neq w_1$, and again re-ordering we may assume that $j=2$. Set $d_3=\cdots=d_{r+s}=1$. Then $d_1 d_2 = A$ and $\log(1)=0$, so

$$\left\vert\sum_{i=1}^{r+s} w_i \log(d_i)\right\vert = \vert w_1\log(d_1) + w_2\log(d_2)\vert$$

$$= \vert w_1 \log(d_1) + w_2\log(A/d_1)\vert$$

$$= \vert(w_1-w_2)\log(d_1) + w_2\log(A)\vert$$

Since $w_1\neq w_2$, we have $\vert(w_1-w_2)\log(d_1) + w_2\log(A)\vert\to\infty$ as $d_1\to \infty$. $\qedsymbol$