A small question related to Waring's Problem in number field I was reading "Generalization of Waring's Problem to Algebraic Number Fields"
by Carl Ludwig Siegel and there was this one bit I don't quite get and I'd appreciate
if some one could explain it to me.
We have $K$, an algebraic number field of degree $n$ and it's a totally real field.
Given a totally positive algebraic integer $v$, we let $A(v)$ to be
the number of solution to 
$$
\lambda_1^r + ..  + \lambda_m^r = v,
$$
where each $\lambda_i$ is a totally positive algebraic integer.
He says
"Since there are $n-1$ independent units in $K$, we may assume,
during the proof of the Theorem, that $v = O(M^{1/n})$".
Here $M$ is the norm of $v$ and the "Theorem" proves
the asymptotic formula for $A(v)$ 
under certain conditions as $M$ goes to infinity.
I just don't quite see the reason may we can assume $v = O(M^{1/n})$.
I'd appreciate if someone could explain me this bit.
Thanks!
 A: Let me quote the full paragraph from Siegel's paper first:

For any totally positive unit $\epsilon$ in $K$, the formula
  $A(\epsilon^r\nu)=A(\nu)$ holds good. Since there are $n-1$ indepenent
  units in $K$, we may assume, during the proof of the Theorem, that
$$ \nu= O(M^{1/n}),\ \ \ \ \ \ \ \  \nu^{-1} = O(M^{-1/n}).$$

So the point is to be able to assume $\nu / M^{1/n}$ is bounded by constants above and below. If $\epsilon$ is a totally positive unit and we let $\nu'=\epsilon^r \nu$, then $\nu'$ also has norm $M$, and
$$\nu'/M^{1/n} = \epsilon^r (\nu/M^{1/n}).$$
Then by a suitable choice of $\epsilon$, we can bound $\nu'/M^{1/n}$ by a value independent of $M$ (which is what will go to infinity). Since $A(\nu')=A(\nu)$, we can replace $\nu$ by $\nu'$, and so assume the bound holds for $\nu$.
Here's some more detail:
If $\sigma_1,...,\sigma_n$ are the embeddings of $K$ into $\mathbb{R}$, let $u_i = \nu/\sigma_i(\nu)$. Then $u_i$ is a totally positive unit, and we have
$$ M = \nu^n \prod_{i=1}^n u_i.$$
It follows that
$$ \log(M^{1/n}/\nu') = \frac{1}n\sum_{i=1}\log(u_i) - r\log(\epsilon)$$
Dirichlet's unit theorem says under the map $\xi: K \rightarrow \mathbb{R}^n$ given by
$$x \mapsto (\log(|\sigma_1(x)|),...,\log(|\sigma_n(x)|),$$
the units of $\mathcal{O}_K$ form a lattice of rank $n-1$ in a hyperplane $V\subset \mathbb{R}^n$. The totally positive units form a sublattice of finite index in this lattice, as does the sublattice $\Lambda$ of totally positive units of the form $\epsilon^{nr}$. Let $u=\prod_{i=1}^n u_i$. Choose positive numbers $c$ and $C$, and a fundamental domain $F\subset V$ for the action of $\Lambda$ (via translation in $\mathbb{R}^n$) such that $F$ is contained within the ball of radius $C$ centred at the origin, but outside the ball of radius $c$. Then we can pick $\epsilon^{nr}$ so that $\xi(u/\epsilon^{nr})$ lies in F. The expression
$$ \sum_{i=1}^n \log(u_i) - nr\log(\epsilon)$$
is the coordinate of $\xi(u/\epsilon^{nr})$ corresponding to the identity embedding of $K$, and is bounded above by $C$ and below by $c$. Thus we have obtained bounds:
$$ c < \log(M/(v')^n) < C$$
and so
$$e^{-C/n} < \nu'/M^{1/n} < e^{-c/n}$$
