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

$\displaystyle \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

$\displaystyle 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

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

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

$\displaystyle \frac{c_i}{\vert\sigma_i(a)\vert}\leq A$    for $\displaystyle i\leq r$    and $\displaystyle \quad
\left(\frac{c_i}{\vert\sigma_i(a)\vert}\right)^2\leq A$    for $\displaystyle 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$.


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

Lemma 12.2.1   We have

$\displaystyle \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:

$\displaystyle \vert f(u) - s\vert$ $\displaystyle = \vert f(a) - f(b_j) - s\vert$    
  $\displaystyle \leq \vert f(b_j)\vert + \vert s - f(a)\vert$    
  $\displaystyle =\vert f(b_j)\vert + \vert z_1(\log(c_1) - \log(\vert\sigma_1(a)\vert)) + \cdots + z_{r+s}(\log(c_{r+s}) - \log(\vert\sigma_{r+s}(a)\vert))\vert$    
  $\displaystyle =\vert f(b_j)\vert + \vert z_1\cdot \log(c_1/\vert\sigma_1(a)\ver...
...ots + \frac{1}{2}\cdot z_{r+s} \log((c_{r+s}/\vert\sigma_{r+s}(a)\vert)^2)\vert$    
  $\displaystyle \leq \vert f(b_j)\vert + \log(A)\cdot\left(\sum_{i=1}^{r}\vert z_i\vert + \frac{1}{2}\cdot \sum_{i=r+1}^s\vert z_i\vert\right).$    

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

$\displaystyle c_1\cdots c_r\cdot (c_{r+1}\cdots c_{r+s})^2 = A$   and$\displaystyle \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

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

Proof. It is easier if we write

$\displaystyle z_1\log(c_1)$ $\displaystyle +\cdots +z_{r+s}\log(c_{r+s}) =$    
  $\displaystyle z_1\log(c_1)+\cdots + z_r\log(c_r)+ \frac{1}{2}\cdot z_{r+1}\log(c_{r+1}^2) + \cdots + \frac{1}{2}\cdot z_{r+s}\log(c_{r+s}^2)$    
  $\displaystyle =w_1\log(d_1)+\cdots + w_r\log(d_r)+ w_{r+1}\log(d_{r+1}) + \cdots +\cdot w_{r+s}\log(d_{r+s}),$    

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

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

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$

William Stein 2004-05-06