before $(H,\\{R_{h}\\}_{h})$ is a grading of $R$, and the inclusion $i:H\to G$
is a morphism of gradings $(H,\\{R_{h}\\}_{h})\to(G,\\{R_{g}\\}_{g})$ of $R$.
By universality there exists a map $u:G\to H$ of gradings in the opposite
direction. The composition $i\circ u$ is an endomorphism of
$(G,\\{R_{g}\\}_{g})$, which by universality is unique and thus equal to the
identity. Hence $i$ is surjective. It follows that $G=H$, as was to be shown.
∎

######  Definition 10.4.

Let $n\in\mathbb{Z}_{\geq 1}$. We write
$\mu_{n}\subseteq\overline{\mathbb{Z}}$ for the group of $n$-th roots of
unity. We equip $\overline{\mathbb{Z}}\otimes_{\mathbb{Z}}\mathbb{Z}[\mu_{n}]$
with a Hilbert lattice structure with the inner product given by

$\langle
x,y\rangle_{n}=\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L\otimes\mathbb{Z}[\mu_{n}])}\sigma(x)\cdot\overline{\sigma(y)}$

for $x,y\in\overline{\mathbb{Z}}\otimes_{\mathbb{Z}}\mathbb{Z}[\mu_{n}]$ and
any $\mathbb{Q}\subseteq L\subseteq\overline{\mathbb{Q}}$ such that
$[L:\mathbb{Q}]<\infty$ and $x,y\in L\otimes\mathbb{Z}[\mu_{n}]$.

The proof that this is a Hilbert lattice is completely analogous to the proof
for $\overline{\mathbb{Z}}$. One can show that for
$x,y\in\overline{\mathbb{Z}}$ we have $\langle
x,y\rangle_{n}=\varphi(n)\langle x,y\rangle$, where $\varphi$ is the Euler
totient function, using that we have a natural bijection
$\operatorname{X}(L)\times\operatorname{X}(\mathbb{Z}[\mu_{n}])\to\operatorname{X}(L\otimes\mathbb{Z}[\mu_{n}])$.

######  Lemma 10.5.

Suppose $R\subseteq\overline{\mathbb{Z}}$ is a subring and
$(G,\\{R_{g}\\}_{g})$ is a grading of $R$. Then for all $g,h\in G$ distinct we
have $\langle R_{g},R_{h}\rangle=0$.

###### Proof.

Let $H$ be as in Lemma 10.3. If $R_{g}=0$ or $R_{h}=0$ we are done, so suppose
$g,h\in H$ and let $n\in\mathbb{Z}_{\geq 1}$ such that $g^{n}=1=h^{n}$. As
$\varphi(n)\langle R_{g},R_{h}\rangle=\langle R_{g},R_{h}\rangle_{n}$ it
suffices to show that $\langle R_{g},R_{h}\rangle_{n}=0$. Since $g\neq h$
there exists by Pontryagin duality some morphism
$\chi\in\operatorname{Hom}(H,\mu_{n})$ such that $\chi(g)\neq\chi(h)$. One
then verifies that the induced $\mathbb{Z}[\mu_{n}]$-algebra automorphism
$\chi$ of $R\otimes\mathbb{Z}[\mu_{n}]$ that maps $x\otimes 1\in
R_{i}\otimes\mathbb{Z}[\mu_{n}]$ to $x\otimes\chi(i)$ for all $i\in H$ is an
isometry. Hence for $x\in R_{g}$ and $y\in R_{h}$ we have

$\langle x,y\rangle_{n}=\langle\chi x,\chi
y\rangle_{n}=\langle\chi(g)x,\chi(h)y\rangle_{n}=\langle
x,\chi(g)^{-1}\chi(h)y\rangle_{n}.$

Because $\chi(g^{-1}h)$ is a non-trivial $n$-th root of unity,
$1-\chi(g^{-1}h)$ divides $n$. Hence

$0=\langle R_{g},(1-\chi(g^{-1}h))R_{h}\rangle_{n}\supseteq\langle
R_{g},nR_{h}\rangle_{n}=n\langle R_{g},R_{h}\rangle_{n},$

and we are done. ∎

Theorem 1.7. Every subring of $\overline{\mathbb{Z}}$ has a universal grading.

###### Proof.

Let $R$ be a subring of $\overline{\mathbb{Z}}$, which is also a sublattice of
$\overline{\mathbb{Z}}$. Let $\textup{Y}=\\{\textup{Y}_{i}\\}_{i\in I}$ be a
universal decomposition of the lattice $R$, which exists by Theorem 5.4. We
obtain this decomposition by starting with the graph $\mathcal{G}$ on the
vertex set $\textup{indec}(R)$ with edges between $x,y\in\textup{indec}(R)$ if
and only if $\langle x,y\rangle\neq 0$, then taking $I$ to be the set of
connected components of $\mathcal{G}$ and $\textup{Y}_{i}$ the group generated
by $i\in I$. For $x=\sum_{i}\upsilon_{i}\in R$ with
$\upsilon_{i}\in\textup{Y}_{i}$ write $\textup{supp}(x)=\\{i\in
I\,|\,\upsilon_{i}\neq 0\\}$. Now consider the free abelian group
$\mathbb{Z}^{(I)}$ and let $U$ be the group obtained from it by dividing out

$N=\langle i+j-k\,|\,i,j\in
I,\,k\in\textup{supp}(\textup{Y}_{i}\cdot\textup{Y}_{j})\rangle.$

We have an induced map $f:I\to\mathbb{Z}^{(I)}\to U$ which induces a
decomposition $f(\textup{Y})$ of $R$, which is also a grading. That it is
universal follows from the fact that Y is a refinement of $\mathcal{R}$ for
any grading $(G,\mathcal{R})$ by Lemma 10.5. ∎

######  Example 10.6.

From Theorem 1.7 it follows in particular that $\overline{\mathbb{Z}}$ has a
universal grading. It follows from Lemma 10.5 that such a grading gives an
orthogonal decomposition of the lattice. By Theorem 1.1 such a decomposition
is trivial, and thus $\overline{\mathbb{Z}}$ has a trivial universal grading.

######  Example 10.7.

Every countable torsion group occurs as the group of a universal grading of a
subring of $\overline{\mathbb{Z}}$. Note that such a group is a subgroup of
$\Omega=\bigoplus_{\smash{p\in\mathcal{P}}}(\mathbb{Q}/\mathbb{Z})$, where
$\mathcal{P}$ is some countably infinite set. We choose $\mathcal{P}$ to be
the set of positive prime numbers. Fixing some embedding
$\overline{\mathbb{Z}}\to\mathbb{C}$ we have a well-defined $x$-th power of
$p$ in $\overline{\mathbb{Q}}\cap\mathbb{R}_{>0}$ for all $x\in\mathbb{Q}$.
Let $[\cdot]:\mathbb{Q}/\mathbb{Z}\to[0,1)\cap\mathbb{Q}$ be the (bijective)
map that assigns to each class its smallest non-negative representative. It is
then easy to verify that
$R=\mathbb{Z}[p^{x}\,|\,p\in\mathcal{P},\,x\in\mathbb{Q}_{\geq
0}]\subseteq\overline{\mathbb{Z}}$ has a grading
$(\Omega,(R_{(x_{p})_{p}})_{(x_{p})_{p}})$ with

$R_{(x_{p})_{p}}=\Big{(}\prod_{p\in\mathcal{P}}p^{[x_{p}]}\Big{)}\cdot\mathbb{Z}.$

In turn any subgroup $G\subseteq\Omega$ gives a grading $(G,(R_{g})_{g})$ of
the subring $\bigoplus_{g\in G}R_{g}\subseteq R$, and this grading must be
universal because a universal grading exists and all $R_{g}$ are of rank 1.

######  Lemma 10.8.

Suppose $R\subseteq\overline{\mathbb{Z}}$ is a subring and
$(G,\\{R_{g}\\}_{g})$ is a grading of $R$. If $K=\mathbb{Q}(A)$ for some
subset $A\subseteq\bigcup_{g\in G}R_{g}$, then $(G,\\{R_{g}\cap K\\}_{g})$ is
a grading of $R\cap K$.

###### Proof.

It is clear that $(G,\\{R_{g}\cap K\\}_{g})$ is a grading of $R\cap K$ once we
show $\bigoplus_{g}(R_{g}\cap K)=R\cap K$. For this it remains to show that
$R\cap K\subseteq\sum_{g}(R_{g}\cap K)$. Let $x\in R\cap K$. As $x\in R$ we
may uniquely write $x=\sum_{g}x_{g}$ for some $x_{g}\in R_{g}$. Without loss
of generality $A$ is closed under multiplication, so that $A$ generates $K$ as
a $\mathbb{Q}$-vector space. Then we may write $x=\sum_{a\in A}r_{a}a$ for
some $r_{a}\in\mathbb{Q}$ which are almost all equal to zero. Hence a positive
integer multiple $\lambda x$ of $x$ satisfies $\sum_{g}\lambda x_{g}=\lambda
x=\sum_{a\in A}\lambda r_{a}a$ with $\lambda r_{a}\in\mathbb{Z}$ for all $a$
and thus $\lambda r_{a}a\in R_{g_{a}}$ for some $g_{a}\in G$. It follows from
uniqueness of the decomposition that $\lambda x_{g}=\sum_{a\in
A,\,g_{a}=g}\lambda r_{a}a$ and thus $x_{g}\in K$. We conclude that $x_{g}\in
R_{g}\cap K$ and thus $x\in\sum_{g}(R_{g}\cap K)$, as was to be shown. ∎

We may also lift Theorem 1.4 of [12] to our more general setting.

Theorem 1.8. Every integrally closed subring of $\overline{\mathbb{Z}}$ has a
universal grading with a subgroup of $\mathbb{Q}/\mathbb{Z}$.

###### Proof.

Let $R$ be an integrally closed subring of $\overline{\mathbb{Z}}$ and let
$(G,\\{R_{g}\\}_{g})$ be a universal grading, which exists by Theorem 1.7. It
suffices to show that every finitely generated subgroup $H$ of $G$ is cyclic.

Let $H\subseteq G$ be finitely generated and thus finite by Lemma 10.3.
Moreover, by Lemma 10.3 we have $R_{h}\neq 0$ for all $h\in H$, so we may
choose some non-zero $a_{h}\in R_{h}$. Let $A=\\{a_{h}\mid h\in H\\}$ and
$K=\mathbb{Q}(A)$. Then by Lemma 10.8 we get a grading $(H,\\{R_{h}\cap K\\})$
of $S=R\cap K$. Since $K$ is a field and $R$ is integrally closed, the ring
$S$ is integrally closed. As $K$ is finite over $\mathbb{Q}$, the same holds
for the field of fractions of $S$. Hence we may apply Theorem 1.4 from [12] to
conclude that the universal grading of $S$ has a cyclic grading group $Y$. By
universality we get a morphism of gradings and thus a morphism of groups $Y\to
H$. The latter is surjective since $0\neq R_{h}\cap K\ni a_{h}$ for all $h\in
H$. Thus $H$ is cyclic, as was to be shown. ∎

######  Lemma 10.9.

Suppose $R\subseteq\overline{\mathbb{Z}}$ is a subring and
$(G,\\{R_{g}\\}_{g})$ is a grading of $R$. If the universal grading of $R_{1}$
is trivial and $R_{g}\neq 0$ for all $g\in G$, then $(G,\\{R_{g}\\}_{g})$ is
universal.

###### Proof.

Suppose $(H,\\{S_{h}\\}_{h})$ is a universal grading of $R$, which exists by
Theorem 1.7, and let $f:(H,\\{S_{h}\\}_{h})\to(G,\\{R_{g}\\}_{g})$ be the map
given by universality. Then $R_{1}=\bigoplus_{h\in\ker(f)}S_{h}$, which is a
grading of $R_{1}$. Since the universal grading of $R_{1}$ is trivial, it
follows that $R_{1}=S_{1}$. By Lemma 10.3 we have $S_{h}\neq 0$ for all
$h\in\ker(f)$, so it follows that $\ker(f)=1$ and that $f$ is injective. From
the fact that $R_{g}\neq 0$ for all $g\in G$ it follows that $f$ must be
surjective. Thus $f$ is an isomorphism of gradings and $(G,\\{R_{g}\\}_{g})$
is universal. ∎

######  Example 10.10.

We will show that every subgroup of $\mathbb{Q}/\mathbb{Z}$ occurs as the
group of a universal grading of an integrally closed subring of
$\overline{\mathbb{Z}}$.

Recall from Definition 7.12 the notation $\mu_{\infty}$ for the group of roots
of unity in $\overline{\mathbb{Z}}$. For a prime number $p$ write
$\mu_{p^{\infty}}=\\{\zeta\in\mu_{\infty}\,|\,(\exists\,n\in\mathbb{Z}_{\geq
0})\ \zeta^{p^{n}}=1\\}$ and
$\mu_{p}=\\{\zeta\in\mu_{\infty}\,|\,\zeta^{p}=1\\}$. The map
$\zeta\mapsto\zeta^{p}$ gives an isomorphism
$\mu_{p^{\infty}}/\mu_{p}\to\mu_{p^{\infty}}$. Taking the direct sum over all
$p$ we get an isomorphism $\mu_{\infty}/\mu_{0}\to\mu_{\infty}$, where
$\mu_{0}=\\{\zeta\in\mu_{\infty}\,|\,(\exists\,n\text{ square-free})\
\zeta^{n}=1\\}$. Thus it suffices to show that for every $\mu_{0}\subseteq
M\subseteq\mu_{\infty}$ the group $G=M/\mu_{0}$ occurs as a universal grading
group.

Consider $R=\mathbb{Z}[M]$, the smallest subring of $\overline{}\mathbb{Z}$
containing $M$, which is integrally closed. Define
$R_{\zeta\cdot\mu_{0}}=\zeta\cdot\mathbb{Z}[\mu_{0}]$ for all
$\zeta\cdot\mu_{0}\in M/\mu_{0}$ and note that this gives a grading
$(G,\\{R_{g}\\}_{g})$ of $R$. To prove this is a universal grading it suffices
by Lemma 10.9 to show that the universal grading of $\mathbb{Z}[\mu_{0}]$ is
trivial, or in turn, by Lemma 10.5, that $\mathbb{Z}[\mu_{0}]$ is
indecomposable. The elements of $\mu_{0}$ are indecomposable in
$\mathbb{Z}[\mu_{0}]$ because they are so in $\overline{\mathbb{Z}}$, and they
generate $\mathbb{Z}[\mu_{0}]$ as an additive group. From Proposition 8.5 we
may conclude that no pair $\zeta,\xi\in\mu_{0}$ is orthogonal, so from Theorem
5.4 it follows that $\mathbb{Z}[\mu_{0}]$ is indecomposable. Hence the grading
is universal.

##  11\. Enumeration of indecomposable algebraic integers of degree 2

The indecomposables of $\overline{\mathbb{Z}}$ of degree 1 are $1$ and $-1$.
In this section we compute the indecomposables of degree 2. The fields of
degree 2 over $\mathbb{Q}$ are $\mathbb{Q}(\sqrt{d})$ for
$d\in\mathbb{Z}\setminus\\{1\\}$ square-free. The following lemma is easily
verified by separating the cases $d$ negative and positive.

######  Lemma 11.1.

Let $d\in\mathbb{Z}\setminus\\{1\\}$ be square-free and let
$a,b\in\mathbb{Q}$. Then $q(a+b\sqrt{d})=a^{2}+|d|\cdot b^{2}$.∎

######  Lemma 11.2.

Let $\alpha\in\overline{\mathbb{Z}}$ and suppose one of the following holds:

  1. (i)

the real part of $\alpha^{2}$ is at least $2$ under every embedding
$\mathbb{Q}(\alpha)\to\mathbb{C}$;

  2. (ii)

the real part of $\alpha^{2}$ is at most $-2$ under every embedding
$\mathbb{Q}(\alpha)\to\mathbb{C}$;

  3. (iii)

$\alpha=(1+\sqrt{d})/2$ with $d\in\mathbb{Z}$ square-free such that $9\leq
d\leq 25$.

Then $\alpha$ has a non-trivial decomposition in a degree $2$ extension of
$\mathbb{Q}(\alpha)$.

###### Proof.

Let $K=\mathbb{Q}(\alpha)$ and $\gamma=\alpha^{2}/4$. Let $f=X^{2}-\alpha
X+1\in K[X]$, let $\beta\in\overline{\mathbb{Z}}$ be a root of $f$ and write
$L=K(\beta)$. Then
$(\beta-\alpha/2)^{2}=\beta^{2}-\alpha\beta+\alpha^{2}/4=\alpha^{2}/4-1=\gamma-1$.
For (i) and (iii) we will show $q(\beta-\alpha/2)\leq q(\alpha/2)$. Then
$(\beta,\alpha-\beta)$ is a decomposition of $\alpha$ by Lemma 4.3, and since
neither $0$ nor $\alpha$ is a root of $f$ we conclude that this decomposition
is non-trivial.

(i) Let $\sigma\in\operatorname{X}(L)$. For $\delta\in L$ write
$\textup{Re}_{\sigma}(\delta)$ and $\textup{Im}_{\sigma}(\delta)$ for the real
respectively imaginary part of $\sigma(\delta)$. By assumption
$\textup{Re}_{\sigma}(\gamma)\geq 1/2$. Thus
$\textup{Re}_{\sigma}(\gamma-1)^{2}=(\textup{Re}_{\sigma}(\gamma)-1)^{2}\leq\textup{Re}_{\sigma}(\gamma)^{2}$.
As $\textup{Im}_{\sigma}(\gamma-1)^{2}=\textup{Im}_{\sigma}(\gamma)^{2}$ we
may conclude that $|\sigma(\gamma-1)|\leq|\sigma(\gamma)|$. Then

$\displaystyle
q(\beta-\alpha/2)=\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L)}|\sigma(\gamma-1)|\leq\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L)}|\sigma(\gamma)|=q(\alpha/2),$

as was to be shown.

(ii) Let $\textup{i}\in\overline{\mathbb{Q}}$ be a primitive fourth root of
unity. Then $\textup{i}\alpha$ satisfies the conditions to (1), hence it has a
non-trivial decomposition $(\beta,\textup{i}\alpha-\beta)$, where $\beta$ is a
root of $X^{2}-\textup{i}\alpha X+1$. In turn,
$(-\textup{i}\beta,\alpha+\textup{i}\beta)$ is a non-trivial decomposition of
$\alpha$, where $-\textup{i}\beta$ is a root of $X^{2}-\alpha X-1$. In
particular $-\textup{i}\beta$ is of degree at most $2$ over
$\mathbb{Q}(\alpha)$.

(iii) Since $d>0$ the field $K$ is totally real. Let
$\gamma_{1},\gamma_{2}\in\mathbb{R}$ be the images of $\gamma$ under
$\operatorname{X}(K)$ such that $\gamma_{1}<\gamma_{2}$. Because $9\leq d\leq
25$ we have $\gamma_{1}=(\sqrt{d}-1)^{2}/16\leq 1$ and
$\gamma_{2}=(\sqrt{d}+1)^{2}/16\geq 1$. Hence

$q(\beta-\alpha/2)=\frac{1}{2}\big{(}|\gamma_{1}-1|+|\gamma_{2}-1|\big{)}=\frac{1}{2}\big{(}(1-\gamma_{1})+(\gamma_{2}-1)\big{)}=\frac{2\sqrt{d}}{16}\leq\frac{1+d}{16}=q(\alpha/2),$

as was to be shown. ∎

######  Theorem 11.3.

The indecomposable elements of $\overline{\mathbb{Z}}$ of degree $2$ up to
conjugacy and sign are $\sqrt{-1}$, $\frac{1+\sqrt{-7}}{2}$,
$\frac{1+\sqrt{-3}}{2}$ and $\frac{1+\sqrt{5}}{2}$, for a total of $14$
indecomposables.

###### Proof.

First note that the $4$ listed elements indeed are indecomposable: We treated
$(1+\sqrt{-7})/2$ in Example 8.3, and the remaining $3$ have square-norm less
than $2$, so Proposition 8.1 applies. Since conjugation and multiplication by
$-1$ are isometries by Lemma 7.13, all $14$ are indecomposable.

Let $\alpha\in\overline{}\mathbb{Z}$ be of degree $2$ over $\mathbb{Q}$. It
remains to show that $\alpha$, up to conjugation and sign, admits a non-
trivial decomposition or is one of the $4$ listed indecomposables. Since
$\alpha$ is of degree 2 over $\mathbb{Q}$ it is an element of
$\mathbb{Q}(\sqrt{d})$ for some square-free $d\in\mathbb{Z}\setminus\\{1\\}$.
Then we may write $\alpha=(a+b\sqrt{d})/2$ for some $a,b\in\mathbb{Z}$ with
$a+b\in 2\mathbb{Z}$ and by conjugating and changing sign we may assume
$a,b\geq 0$. If $a\geq 2$ we have

$q(\alpha/2-1)=\Big{(}\frac{a}{4}-1\Big{)}^{2}+|d|\Big{(}\frac{b}{4}\Big{)}^{2}=\Big{(}\Big{(}\frac{a}{4}\Big{)}^{2}+|d|\Big{(}\frac{b}{4}\Big{)}^{2}\Big{)}+\Big{(}1-\frac{a}{2}\Big{)}\leq
q(\alpha/2),$

so $(1,\alpha-1)$ is a decomposition of $\alpha$. Since $\alpha\neq 1$, this
decomposition is non-trivial. Similarly we get a decomposition
$(\sqrt{d},\alpha-\sqrt{d})$ if $b\geq 2$, so either this decomposition is
non-trivial or $\alpha=\sqrt{d}$.

First suppose $\alpha=\sqrt{d}$. If $|d|<2$ we have $d=-1$ and $\sqrt{-1}$ is
listed. Otherwise $\alpha^{2}=d$ satisfies the conditions to Lemma 11.2.i or
ii, so $\sqrt{d}$ is not indecomposable. For $\alpha\neq\sqrt{d}$ the
remaining cases are $\alpha=(1+\sqrt{d})/2$, which is integral only if
$d\equiv 1\bmod 4$. If $d\leq-9$ the real part of $\alpha^{2}$ is
$(1+d)/4\leq-2$ under either embedding, so $\alpha$ satisfies the conditions
to Lemma 11.2.ii. If $-9<d<9$ we have $d\in\\{-7,-3,5\\}$ and thus
$\alpha=(1+\sqrt{d})/2$ is listed. For $9\leq d\leq 25$ we may apply Lemma
11.2.iii. The remaining case is $25<d$, where we have that
$\sigma(\alpha^{2})=[(1\pm\sqrt{d})/2]^{2}\geq(\sqrt{d}-1)^{2}/4\geq 2$ for
all $\sigma\in\operatorname{X}(\mathbb{Q}(\sqrt{d}))$, so Lemma 11.2.i
applies. Hence $\alpha$ is either listed or not indecomposable. ∎

Interesting to note is that all non-trivial decompositions of
$\alpha\in\overline{\mathbb{Z}}$ of degree 2 over $\mathbb{Q}$ that are
produced in Theorem 11.3 live in a field extension of degree at most 2 over
$\mathbb{Q}(\alpha)$.

##  12\. Geometry of numbers

In this section we gather some known results about the geometry of numbers.

######  Definition 12.1.

Let $K$ be a number field. We write
$K_{\mathbb{R}}=\mathbb{R}\otimes_{\mathbb{Q}}K$ and
$K_{\mathbb{C}}=\mathbb{C}\otimes_{\mathbb{Q}}K$.

Recall for a number field $K$ the definition of $\operatorname{X}(K)$, the set
of ring homomorphisms from $K$ to $\mathbb{C}$.

######  Lemma 12.2.

We have an isomorphism of $\mathbb{C}$-algebras
$\Phi_{K}:K_{\mathbb{C}}\to\mathbb{C}^{\operatorname{X}(K)}$ given by

$\Phi_{K}(z\otimes\alpha)=(z\cdot\sigma(\alpha))_{\sigma\in\operatorname{X}(K)}.$

We have a natural inclusion $K_{\mathbb{R}}\to
K_{\mathbb{C}}\to\mathbb{C}^{\operatorname{X}(K)}$, and its image is given by
the subspace of elements invariant under the involution
$(x_{\sigma})_{\sigma}\mapsto(\overline{x_{\overline{\sigma}}})_{\sigma}$.
This inclusion induces an isomorphism of $\mathbb{R}$-algebras
$K_{\mathbb{R}}\cong\mathbb{R}^{r}\times\mathbb{C}^{s}$ for integers $r,s\geq
0$ such that
$r=\\#\\{\sigma\in\operatorname{X}(K)\,|\,\sigma[K]\subseteq\mathbb{R}\\}$ and
$r+2s=[K:\mathbb{Q}]$. ∎

######  Definition 12.3.

We equip $K_{\mathbb{C}}$ with the inner product induced by the standard
hermitian inner product on $\mathbb{C}^{\operatorname{X}(K)}$ and
$K_{\mathbb{R}}$ with its restriction, turning $K_{\mathbb{R}}$ into a real
inner product space. Since $K_{\mathbb{R}}$ is an inner product space we have
an induced measure on $K_{\mathbb{R}}$ we denote $\operatorname{vol}$.

######  Remark 12.4.

For a number field $K$ and $\alpha\in K$ we have
$q(\alpha)=\frac{1}{[K:\mathbb{Q}]}\|\Phi_{K}(\alpha)\|^{2}$. Note that the
norm on $K_{\mathbb{R}}$ is not the ‘standard’ norm on
$\mathbb{R}^{r}\times\mathbb{C}^{s}$. In terms of the latter vector space it
is given by

$(x_{1},\dotsc,x_{r},z_{1},\dotsc,z_{s})\mapsto\sqrt{|x_{1}|^{2}+\dotsm+|x_{r}|^{2}+2|z_{1}|^{2}+\dotsm+2|z_{s}|^{2}}.$

######  Theorem 12.5 (Proposition 4.26 in [14]).

Let $R$ be an order in a number field $K$. Then $\Phi_{K}[R]$ is a full rank
lattice in $K_{\mathbb{R}}$ with determinant $|\Delta(R)|^{1/2}$, where
$\Delta(R)$ is the discriminant of $R$. ∎

######  Definition 12.6.

For a commutative ring $R$ and $d\in\mathbb{Z}_{\geq 0}$ we write
$R[X]_{d}=\\{f\in R[X]\,|\,\deg(f)<d\\}$.

######  Lemma 12.7.

For a number field $K$ we have an isomorphism of real vector spaces
$K_{\mathbb{R}}[X]_{d}\cong(\mathbb{R}[X]_{d})^{r}\times(\mathbb{C}[X]_{d})^{s}$
for all $d\in\mathbb{Z}_{\geq 0}$ induced by the isomorphism
$K_{\mathbb{R}}\cong\mathbb{R}^{r}\times\mathbb{C}^{s}$ of Lemma 12.2. ∎

######  Theorem 12.8 (Minkowski, Theorem 4.19 in [14]).

Let $n\in\mathbb{Z}_{\geq 0}$, let $\Lambda\subseteq\mathbb{R}^{n}$ be a full
rank lattice and let $S\subseteq\mathbb{R}^{n}$ be a symmetric convex body. If
$\operatorname{vol}(S)>2^{n}\det(\Lambda)$, then there exists a non-zero
element in $\Lambda\cap S$. ∎

##  13\. Szegő capacity theory

In this section we will give a proof of Theorem 1.2, which is due to T.
Chinburg.

######  Definition 13.1.

We define for $\overline{\mathbb{Q}}$ and in turn any subfield of
$\overline{\mathbb{Q}}$ the max-norm

$|\alpha|_{\infty}=\max_{\sigma\in\operatorname{X}(\overline{\mathbb{Q}})}|\sigma(\alpha)|.$

One easily shows that the max-norm satisfies the following.

######  Lemma 13.2.

For all $\alpha\in\overline{\mathbb{Q}}$ we have
$q(\alpha)\leq|\alpha|_{\infty}^{2}$ and for all $n\in\mathbb{Z}_{\geq 1}$
also $|\alpha^{n}|_{\infty}=|\alpha|^{n}_{\infty}$. ∎

In this section we will identify a number field $K$ with its image in
$K_{\mathbb{R}}$. We extend $\sigma\in\operatorname{X}(K)$ to $K_{\mathbb{C}}$
by $z\otimes\alpha\mapsto z\cdot\sigma(\alpha)$ or equivalently projecting
$\mathbb{C}^{\operatorname{X}(K)}$ onto the component at $\sigma$. We
similarly equip $K_{\mathbb{C}}$ with a max-norm

$\lvert x\rvert_{K}=\max_{\sigma\in\operatorname{X}(K)}|\sigma(x)|.$

and $K_{\mathbb{R}}$ with the induced norm. Note that the max-norm
$|\,\cdot\,|_{\infty}$ of $K$ as subfield of $\overline{\mathbb{Q}}$ and the
max-norm $\lvert\,\cdot\,\rvert_{K}$ of $K$ as subfield of $K_{\mathbb{R}}$
agree.

######  Definition 13.3.

Let $X$ be a metric space with metric $d$ and let $S\subseteq X$ be a subset.
A rounding function from $X$ to $S$ is a map $\lfloor\cdot\rceil:X\to S$ for
which there exists some constant $c\in\mathbb{R}_{\geq 0}$ such that for all
$x\in X$ we have $d(x,\lfloor x\rceil)\leq c$. We call such a $c$ an error
constant for $\lfloor\cdot\rceil$.

For $\mathbb{Z}$ in $\mathbb{Q}$ with the metric induced by the usual absolute
value we may round to a nearest integer, giving a rounding function with error
constant $1/2$. For a naive rounding map for an arbitrary order $R$ with basis
$(\alpha_{i})_{i}$ of a number field $K$ with metric induced by $q$ we may
simply send $\sum_{i}x_{i}\alpha_{i}\in K$ with $x_{i}\in\mathbb{Q}$ to
$\sum_{i}\lfloor x_{i}\rceil\alpha_{i}\in R$. An error constant for this
rounding function is for example
$\frac{1}{2}\sum_{i=1}^{n}|\alpha_{i}|_{\infty}$. However, other rounding
functions exist, for example Babai’s nearest plane algorithm, which try to
achieve beter error constants.

######  Lemma 13.4.

Let $K$ be a $\mathbb{Z}$-algebra with a norm $|\cdot|$ and let $R$ be a
subring of $K$. Let $\alpha\in K$ and suppose there is a rounding function
$K\to R$ with respect to this norm and with error constant $c$. Consider
$K[X]$ and let $Y=X-\alpha$. Then $K[X]=K[Y]$ equipped with the max-norm
$\|\sum_{i}f_{i}Y^{i}\|\mapsto\max_{i}|f_{i}|$ has a rounding function
$\lfloor\cdot\rceil:K[X]\to R[X]$ with error constant $c$ and the following
additional property: If for some $n\geq 0$ and $f=\sum_{i}f_{i}X^{i}\in K[X]$
we have $f_{k}\in R$ for all $k\geq n$, then $f-\lfloor f\rceil$ has degree
less than $n$.

###### Proof.

Let $\lfloor\cdot\rceil_{1}:K\to R$ be the rounding function with error
constant $c$. We may assume without loss of generality that
$\lfloor\cdot\rceil_{1}$ restricted to $R$ is the identity. With the notation
as in Definition 12.6 we identify $K$ with $K[X]_{1}$ and $R$ with $R[X]_{1}$
and inductively define for $n\in\mathbb{Z}_{\geq 1}$ a rounding function
$\lfloor\cdot\rceil_{n}:K[X]_{n}\to R[X]_{n}$. For $f\in K[Y]_{n+1}$ write
$f=aY^{n}+g$ with $a\in K$ and $g\in K[Y]_{n}$. Now define $\lfloor
f\rceil_{n+1}=\lfloor a\rceil_{1}X^{n}+\lfloor g-\lfloor
a\rceil_{1}(X^{n}-Y^{n})\rceil_{n}$. Note that this is well-defined. Suppose
$\lfloor\cdot\rceil_{n}$ has error constant $c$. Then

$\|f-\lfloor f\rceil_{n+1}\|=\max\Big{\\{}\big{|}a-\lfloor
a\rceil_{1}\big{|},\,\big{\|}(g-\lfloor a\rceil_{1}(X^{n}-Y^{n}))-\lfloor
g-\lfloor a\rceil_{1}(X^{n}-Y^{n})\rceil_{n}\big{\|}\Big{\\}}\leq c,$

so $\lfloor\cdot\rceil_{n+1}$ has error constant $c$. Note that $\lfloor
f\rceil_{m}=\lfloor f\rceil_{n}$ for $m\geq n$ and $f\in K[Y]_{n}$. Hence this
defines a rounding function $K[X]\to R[X]$ with error constant $c$ as
required. With $f$ as before of degree $n$, if $a\in R$ then $a-\lfloor
a\rceil_{1}=0$ and $f-\lfloor f\rceil$ has degree less than $n$. One then
inductively proves the final property of $\lfloor\cdot\rceil$. ∎

######  Lemma 13.5.

Let $d,k\in\mathbb{Z}_{>0}$. Then there exist $m\in\mathbb{Z}_{>0}$ such that
$d\mid\binom{am}{k}$ for all $a\in\mathbb{Z}_{>0}$.

###### Proof.

Note that $\binom{X}{k}=X(X-1)\dotsm(X-k+1)/(k!)\in\mathbb{Q}[X]$ is a
polynomial for fixed $k$. For $X$ a multiple of $d\cdot(k!)$ we have that
$d\mid\binom{X}{k}$. ∎

The following is a special case of a theorem of Szegő, adapted from [16].

######  Theorem 13.6 (Szegő).

Let $R$ be an order of a number field $K$ and let $r>1$. Then for each
$\alpha\in K$ there exists a monic non-constant $g\in R[X]$ such that for all
$z\in K_{\mathbb{C}}$ satisfying $\lvert g(z)\rvert_{K}<r$ we have $\lvert
z-\alpha\rvert_{K}<r$.

###### Proof.

We will write $Y=X-\alpha$ and $Y^{n}=\sum_{i=0}^{n}f_{n,i}X^{i}$ for
$n\in\mathbb{Z}_{\geq 0}$. Let $d_{0}=1$, and for $k\in\mathbb{Z}_{>0}$ let
$d_{k}\in\mathbb{Z}_{>0}$ be such that $d_{k}\alpha^{k}\in R$. For
$k\in\mathbb{Z}_{\geq 0}$ let $m_{k}\in\mathbb{Z}_{>0}$ be such that for all
$a\in\mathbb{Z}$ we have $d_{k}\mid\binom{am_{k}}{k}$, which exists by Lemma
13.5. Then $f_{n,n-k}=\pm\binom{n}{k}\alpha^{k}\in R$ for all
$k\in\mathbb{Z}_{\geq 0}$ and positive multiples $n$ of $m_{k}$ and such that
$k\leq n$. Write $M_{b}=\text{lcm}(m_{1},\dotsc,m_{b})$. Then for all $b$ and
$n$ such that $M_{b}\mid n$ we have $f_{n,n-k}\in R$ for all $0\leq k\leq b$.

Let $\lfloor\cdot\rceil_{1}:K\to R$ be a rounding function, which exists
because $R$ is a full rank lattice in $K_{\mathbb{R}}$ by Theorem 12.5. Let
$\lfloor\cdot\rceil:K[Y]\to R[X]$ be the rounding function given by Lemma 13.4
and let $c$ be its error constant. For $n\geq 0$ let $g_{n}=\lfloor
Y^{n}\rceil$ and $e_{n}=\sum_{i=0}^{n}e_{n,i}Y^{i}=Y^{n}-g_{n}$. Moreover, we
may assume that if $M_{b}\mid n$ for some $0\leq b\leq n$ then
$\deg(e_{n})<n-b$, as $f_{n,i}\in R$ for $i\geq n-b$. Since $r>1$ we may
choose $b\in\mathbb{Z}_{>0}$ such that $2cr^{-b}\leq r-1$. Next we may choose
$a\in\mathbb{Z}_{>0}$ such that $n=aM_{b}$ satisfies $n>b$ and $r^{n-1}\geq
2$. We claim $g=g_{n}$ satisfies the requirements.

Clearly $g_{n}$ is monic by the bound on the degree of $e_{n}$ and non-
constant as $n>0$. Let $\operatorname{X}(K)$ act on $K[X]$ coefficient-wise
and fix $\sigma\in\operatorname{X}(K)$. Let $z\in\mathbb{C}$ such that
$|\sigma(Y)(z)|\geq r$ and write $s=|\sigma(Y)(z)|=|z-\sigma(\alpha)|$. Then

$\left|\frac{\sigma(e_{n})(z)}{\sigma(Y^{n})(z)}\right|\leq
s^{-n}\sum_{i=0}^{n-b-1}|\sigma(e_{n,i})|\cdot s^{i}\leq
s^{-n}\sum_{i=0}^{n-b-1}cs^{i}\leq\frac{cs^{-b}}{s-1}\leq\frac{cr^{-b}}{r-1}\leq\frac{1}{2}.$

It follows that

$\left|\frac{\sigma(g_{n})(z)}{\sigma(Y^{n})(z)}\right|=\left|1-\frac{\sigma(e_{n})(z)}{\sigma(Y^{n})(z)}\right|\geq\frac{1}{2}\quad\text{and}\quad|\sigma(g_{n})(z)|\geq\frac{|\sigma(Y^{n})(z)|}{2}\geq\frac{r^{n}}{2}\geq
r.$

Thus, if $|\sigma(g_{n})(z)|<r$, then $|z-\sigma(\alpha)|<r$. Taking the
maximum over all $\sigma\in\operatorname{X}(K)$ proves the theorem. ∎

######  Theorem 13.7 (Szegő).

Let $\alpha\in\overline{\mathbb{Q}}$ and $r\in\mathbb{R}_{>1}$. Then there
exist infinitely many $\beta\in\overline{\mathbb{Z}}$ such that
$|\alpha-\beta|_{\infty}<r$.

###### Proof.

Consider $K=\mathbb{Q}(\alpha)$ and let $R\subseteq K$ be some order of $K$.
By Theorem 13.6 there exists some monic non-constant $g\in R[X]$ such that for
all $z\in K_{\mathbb{C}}$ satisfying $\lvert g(z)\rvert_{K}<r$ we have $\lvert
z-\alpha\rvert_{K}<r$. Now $g^{n}-1\in R[X]$ is monic and non-constant for all
$n\in\mathbb{Z}_{\geq 1}$, so
$S=\\{\beta\in\overline{\mathbb{Z}}\,|\,(\exists\,n\in\mathbb{Z}_{\geq 1})\
g(\beta)^{n}=1\\}$ is infinite. Let $\beta\in S$ and $L=K(\beta)$. It now
suffices to show that $|\alpha-\beta|_{\infty}\leq r^{2}$.

Write
$\textup{P}=\prod_{\sigma\in\operatorname{X}(K)}\operatorname{X}_{\sigma}(L)$.
Let $\rho=(\rho_{\sigma})_{\sigma}\in\textup{P}$ and consider
$\beta_{\rho}=(\rho_{\sigma}(\beta))_{\sigma}\in\mathbb{C}^{\operatorname{X}(K)}=K_{\mathbb{C}}$.
Then

$\lvert g(\beta_{\rho})\rvert_{K}^{n}=\lvert
g^{n}(\beta_{\rho})\rvert_{K}=\max_{\sigma\in\operatorname{X}(K)}|\rho_{\sigma}(g^{n}(\beta))|=\max_{\sigma\in\operatorname{X}(K)}|\rho_{\sigma}(1)|=1,$

so $\lvert g(\beta_{\rho})\rvert_{K}=1<r$. Thus by definition of $g$ we have
$\lvert\alpha-\beta_{\rho}\rvert_{K}<r$. As this holds for all
$\rho\in\textup{P}$ we have
$|\alpha-\beta|_{\infty}=\max_{\rho\in\textup{P}}\,\lvert\alpha-\beta_{\rho}\rvert_{K}<r$,
as was to be shown. ∎

Theorem 1.2. If $\gamma\in\overline{\mathbb{Z}}$ satisfies $4<q(\gamma)$, then
$\gamma$ has infinitely many decompositions in $\overline{\mathbb{Z}}$.

###### Proof.

Let $\alpha=\gamma/2$. By Theorem 13.7 there are infinitely many
$\beta\in\overline{\mathbb{Z}}$ such that
$|\alpha-\beta|_{\infty}\leq\sqrt{q(\alpha)}$ as $q(\alpha)=q(\gamma)/4>1$. By
Lemma 13.2 each such $\beta$ satisfies $q(\alpha-\beta)\leq q(\alpha)$ and
thus $(\beta,\gamma-\beta)$ is a decomposition of $\gamma$ by Lemma 4.3. ∎

It follows from this theorem, as we will show later in the form of Proposition
15.4, that there are only finitely many indecomposables in
$\overline{\mathbb{Z}}$ of a given degree. Recall the definition of $r$ from
Definition 6.1.

######  Proposition 13.8.

The Hilbert lattice $\overline{\mathbb{Z}}$ has a covering radius
$2^{-1/4}\leq r(\overline{\mathbb{Z}})\leq 1$.

###### Proof.

By Theorem 1.6 we have
$2^{3/2}\leq\sup\\{q(\alpha)\,|\,\alpha\in\textup{indec}(\overline{\mathbb{Z}})\\}$
and consequently we get the lower bound
$2^{-1/4}\leq\sup\\{\|\alpha/2\|\,|\,\alpha\in\textup{indec}(\overline{\mathbb{Z}})\\}$.
For any $\alpha\in\textup{indec}(\overline{}\mathbb{Z})$ we have
$\|\alpha/2\|\leq\|\alpha/2-x\|$ for all $x\in\overline{\mathbb{Z}}$ by Lemma
4.3, and thus
$\alpha/2\in\overline{\operatorname{Vor}}(\overline{\mathbb{Z}})$ by Corollary
6.12. Therefore $2^{-1/4}\leq r(\overline{\mathbb{Z}})$ by Proposition 6.10.
For all $r>1$ and $\alpha\in\overline{\mathbb{Q}}$ there exist
$\beta\in\overline{\mathbb{Z}}$ such that $\|\alpha-\beta\|\leq r$ by Theorem
13.7. Taking the limit of $r$ down to $1$ and noting that
$\overline{\mathbb{Q}}=\mathbb{Q}\cdot\overline{\mathbb{Z}}$ is dense in the
Hilbert space of $\overline{\mathbb{Z}}$ proves the proposition. ∎

##  14\. Computing decompositions of algebraic integers of large square-norm

The proof of Theorem 13.7 and in turn Theorem 1.2 is constructive and rather
explicit as well. It is not too difficult to design an algorithm that given
$n\in\mathbb{Z}_{\geq 0}$, an order $R$ and an $\alpha\in R$ such that
$q(\alpha)>4$, computes $n$ distinct non-trivial decompositions of $\alpha$.
However, we will have to formalize how we encode the input and output.
Moreover, since $q$ maps to $\mathbb{R}$ we also have to deal with non-exact
arithmetic.

######  Convention 14.1.

We presuppose that we have decided on a way of encoding integers
$n\in\mathbb{Z}$ and pairs $(a,b)$ such that the length $\textup{len}(n)$ of
the encoding of $n$ grows at a rate $O(\log|n|)$ and
$\textup{len}((a,b))=O(\textup{len}(a)+\textup{len}(b))$. We then encode a
rational as a pair of integers, a numerator and denominator. We encode a
finite sequence of integers, a list, as nested pairs. An order is a
commutative ring which is free and finitely generated as $\mathbb{Z}$-module.
We encode an order as a non-negative integer $n$, its rank, together with
integers $(a_{ijk})_{1\leq i,j,k\leq n}$ such that $e_{i}\cdot
e_{j}=\sum_{k=1}^{n}a_{ijk}e_{k}$, where $e_{i}$ is the $i$-th standard basis
vector of $\mathbb{Z}^{n}$. We represent a number field $K$ with an order
$R\subseteq K$ such that $\mathbb{Q}R=K$. A homomorphism of modules out of a
free $\mathbb{Z}$ or $\mathbb{Q}$-module is encoded by a list of images of the
basis elements. Given an encoding for a commutative ring $R$ we encode a
polynomial in $R[X]$ as an integer $d$, its degree (with $\deg(0)=-1$), and a
list of its $d+1$ coefficients, i.e. a dense encoding.

######  Theorem 14.2.

There exist polynomial-time algorithms for the following problems: Take as
input a number field $K$ and

  1. (i)

an irreducible polynomial $f\in K[X]$, and compute the number field $K[X]/(f)$
and the natural inclusion $K\to K[X]/(f)$;

  2. (ii)

an element $\alpha\in K$, and compute the minimal polynomial of $\alpha$ over
$\mathbb{Q}$;

  3. (iii)

a non-zero polynomial $f\in K[X]$, and compute the monic irreducible factors
of $f$ over $K$.

There also exists an algorithm that takes as input a number field $K$ and

  1. (iv)

a polynomial $f\in K[X]$, and computes a splitting field of $f$ over $K$.

###### Proof.

Part (i) and (ii) are basic linear algebra. Part (iii) is a result due to A.
K. Lenstra [11]. Part (iv) is a repeated application of (iii) and (i). ∎

######  Definition 14.3.

An approximation of a real number $x\in\mathbb{R}$ is an algorithm that takes
as input a positive integer $n$ written in unary and computes a rational
number $x(n)$ such that $|x-x(n)|\leq 2^{-n}$. We define the analogous notion
for complex numbers as well by considering the real and imaginary parts
separately. A number for which there exists an approximation is called
computable. An algebraic real number $x$ is encoded as a minimal polynomial
$f$ over $\mathbb{Q}$ of $x$ together with an approximation of $x$.

Since there are only countably many algorithms and uncountably many real
numbers, not every real number is computable.

######  Lemma 14.4 (For example [1]).

There is an algorithm that takes as input a non-zero polynomial
$f\in\mathbb{Q}[X]$ and computes approximations for all complex roots of $f$
with multiplicities. ∎

As a direct consequence of Lemma 14.4 and Theorem 14.2.ii we have the
following proposition.

######  Proposition 14.5.

There is an algorithm that takes as input an algebraic number
$\alpha\in\overline{\mathbb{Q}}$ represented by its minimal polynomial and
computes an approximation of $q(\alpha)$, $N(\alpha)$ and $|\alpha|_{\infty}$.
∎

######  Proposition 14.6.

There is an algorithm that takes as input an algebraic real number $x$ and
algebraic numbers $\alpha,\beta\in\overline{\mathbb{Q}}$ and decides whether
$\langle\alpha,\beta\rangle=x$.

###### Proof.

Compute the normal closure $K$ of $\mathbb{Q}(\alpha,\beta)$ using Theorem
14.2.iv and compute the set A of conjugates of $\alpha$ in $K$ using Theorem
14.2.iii and similarly for $\beta$. Choose an embedding
$\rho\in\operatorname{X}(K)$. For all $\sigma\in\operatorname{X}(K)$ there
exist $\alpha_{\sigma}\in\textup{A}$ and $\beta_{\sigma}\in\textup{B}$ such
that $\sigma(\alpha)=\rho(\alpha_{\sigma})$ and
$\overline{\sigma(\beta)}=\rho(\beta_{\sigma})$ because $K$ is Galois, and we
may actually compute them since there are only finitely many candidates.
Factor the minimal polynomial $f$ of $x$ over $K$ using Theorem 14.2. Find a
$y$ among the roots of $f$ in $K$ such that $\rho(y)=x$. If it does not exist,
then $\langle\alpha,\beta\rangle\neq x$. Otherwise we may decide
$\tfrac{1}{[K:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(K)}\alpha_{\sigma}\beta_{\sigma}=y$
using exact arithmetic in $K$. ∎

######  Corollary 14.7.

There is an algorithm that takes as input algebraic integers $\alpha$, given
by a minimal polynomial over $\mathbb{Q}$, and $\beta$, given by a minimal
polynomial over $\mathbb{Q}(\alpha)$, and decides whether
$(\beta,\alpha-\beta)\in\textup{dec}(\alpha)$.

###### Proof.

Using Proposition 14.6 we can decide whether
$\langle\beta,\alpha-\beta\rangle=0$. If not, then after finitely many steps
approximating $\langle\beta,\alpha-\beta\rangle$ using Proposition 14.5 we may
decide whether $\langle\beta,\alpha-\beta\rangle\geq 0$. ∎

######  Lemma 14.8.

For every order $R$ in a number field $K$ and $\alpha\in K$ there exists a
rounding function $\lfloor\cdot\rceil:K[X-\alpha]\to R[X]$ as in Lemma 13.4,
which is computable and has a computable error constant.

###### Proof.

Note that the naive rounding function $\lfloor\cdot\rceil_{1}:K\to R$ with
respect to a basis $\alpha_{1},\dotsc,\alpha_{n}$ of $R$ is computable and has
error constant $\tfrac{1}{2}\sum_{i=1}^{n}|\alpha_{i}|_{\infty}$, which is
computable by Proposition 14.5. Finally note that the error function from
Lemma 13.4 is trivially computable given that $\lfloor\cdot\rceil_{1}$ is
computable, and it has the same computable error constant. ∎

######  Theorem 14.9.

There is an algorithm that takes as input an order $R$ in a number field $K$,
a rational number $r>1$ and an $\alpha\in K$, and computes a monic non-
constant polynomial $g\in R[X]$ such that for all $z\in K_{\mathbb{C}}$
satisfying $\lvert g(z)\rvert_{K}<r$ we have $\lvert z-\alpha\rvert_{K}<r$.

###### Proof.

Let $\lfloor\cdot\rceil:K[X-\alpha]\to R[X]$ be the rounding map from Lemma
14.8. Then

  1. (1)

Compute some rational error constant $c$ for $\lfloor\cdot\rceil$;

  2. (2)

Compute some $d\in\mathbb{Z}_{\geq 1}$ such that $d\alpha\in R$ by clearing
denominators;

  3. (3)

Compute some $b\in\mathbb{Z}_{\geq 1}$ such that $2cr^{-b}\leq r-1$;

  4. (4)

Compute an $n\in\mathbb{Z}\cdot d^{b}\cdot b!$ such that $n>b$ and
$r^{n-1}\geq 2$;

  5. (5)

Compute and return $\lfloor(X-\alpha)^{n}\rceil$.

The algorithm is correct by the proof of Theorem 13.6. ∎

######  Theorem 14.10.

There is an algorithm that takes as input an $n\in\mathbb{Z}_{\geq 0}$ and
element $\alpha\in\overline{\mathbb{Z}}$ given by its minimal polynomial, and
decides whether $q(\alpha)>4$ and if so computes $n$ non-trivial
$(\beta,\gamma)\in\textup{dec}(\alpha)$ each represented by the minimal
polynomial of $\beta$ over $\mathbb{Z}[\alpha]$.

###### Proof.

To decide $q(\alpha)>4$, first decide whether $q(\alpha)=4$ using Proposition
14.6. If not, approximating $q(\alpha)$ will yield either $q(\alpha)<4$ or
$q(\alpha)>4$ after finitely many steps. We then follow the proof of Theorem
13.7 and Theorem 1.2.

Since $q(\alpha)>4$ by assumption we may compute using Proposition 14.5 a
rational $r$ such that $1<r\leq\sqrt{q(\alpha/2)}$. Compute using Theorem
14.2.i the number field $K=\mathbb{Q}(\alpha)$ and its order
$R=\mathbb{Z}[\alpha]$. Compute using Theorem 14.9 a monic non-constant $g\in
R[X]$ such that for all $z\in K_{\mathbb{C}}$ satisfying $\lvert
g(z)\rvert_{K}<r$ we have $\lvert z-\alpha\rvert_{K}<r$. Compute the monic
irreducible factors of $g^{k}-1$ over $K$ using Theorem 14.2.iii for $k$
ranging from $1$ to infinity until you have found among all of them $n$
distinct integral polynomials that are not equal to $X$ or $X-\alpha$. By the
proof of Theorem 13.7 this process terminates and produces non-trivial
decompositions. ∎

##  15\. Bounds on indecomposable algebraic integers

In this section we will prove an effective upper bound on the total number of
indecomposable algebraic integers of a given degree. In particular, we will
show that this number is finite. We do this by constructing a complete list of
candidates for indecomposability among all algebraic integers of given degree.
We alo give a lower bound on the number of indecomposables.

######  Proposition 15.1.

Suppose $\alpha\in\textup{indec}(\overline{\mathbb{Z}})$ has minimal
polynomial $f=\sum_{k=0}^{n}f_{n-k}X^{k}\in\mathbb{Z}[X]$ of degree
$n\in\mathbb{Z}_{\geq 1}$. Then $|f_{k}|\leq\binom{n}{k}2^{k}$ for all $0\leq
k\leq n$.

###### Proof.

Let $\alpha_{1},\dotsc,\alpha_{n}\in\mathbb{C}^{\times}$ be the roots of $f$.
We have Maclaurin’s inequalities (Theorem 11.2 in [3])

$s_{1}\geq s_{2}^{1/2}\geq s_{3}^{1/3}\geq\dotsm\geq
s_{n}^{1/n},\quad\text{where
}s_{k}=\binom{n}{k}^{-1}\cdot\sum_{\begin{subarray}{c}I\subseteq\\{1,\dotsc,n\\}\\\
|I|=k\end{subarray}}\prod_{i\in I}|\alpha_{i}|.$

By Theorem 1.2 we have that $q(\alpha)\leq 4$. Then by Lemma 2.14 we have

$s_{1}=\frac{1}{n}\sum_{i}|\alpha_{i}|\leq\Big{(}\frac{1}{n}\sum_{i}|\alpha_{i}|^{2}\Big{)}^{1/2}=\sqrt{q(\alpha)}\leq
2.$

Then
$|f_{k}|\leq\binom{n}{k}s_{k}\leq\binom{n}{k}s_{1}^{k}\leq\binom{n}{k}2^{k}$
for all $k$, as was to be shown. ∎

######  Corollary 15.2.

Suppose $\alpha\in\textup{indec}(\overline{\mathbb{Z}})$ has degree at most
$m$. Then there exists a monic polynomial
$g=\sum_{k=0}^{m}g_{m-k}X^{k}\in\mathbb{Z}[X]$ of degree $m$ such that
$g(\alpha)=0$ and $|g_{k}|\leq\binom{m}{k}2^{k}$ for all $0\leq k\leq m$.

###### Proof.

Let $f$ as in Proposition 15.1 and $g=X^{m-n}\cdot f$. Then
$|g_{k}|=|f_{k}|\leq\binom{n}{k}2^{k}\leq\binom{m}{k}2^{k}$. ∎

######  Proposition 15.3.

Considered as functions of $n\in\mathbb{Z}_{\geq 1}$, the following hold:

(i) $\displaystyle\log(n!)$ $\displaystyle=n\log n-n+O(\log n);$ (ii)
$\displaystyle\log\Big{(}\prod_{k=1}^{n}k^{k}\Big{)}$
$\displaystyle=\tfrac{1}{2}n^{2}\log n-\tfrac{1}{4}n^{2}+O(n\log n);$ (iii)
$\displaystyle\log\Big{(}\prod_{k=0}^{n}(k!)\Big{)}$
$\displaystyle=\tfrac{1}{2}n^{2}\log n-\tfrac{3}{4}n^{2}+O(n\log n);$ (iv)
$\displaystyle\log\Big{(}\prod_{k=0}^{n}\binom{n}{k}\Big{)}$
$\displaystyle=\tfrac{1}{2}n^{2}+O(n\log n).$

###### Proof.

(i) This is Stirling’s approximation, which is classical.

(ii) Note that $f(x)=x\log x$ is an increasing function on $\mathbb{R}_{\geq
1}$. Hence

$\log\Big{(}\prod_{k=1}^{n}k^{k}\Big{)}=\sum_{k=1}^{n}f(k)\leq\int_{1}^{n+1}f(x)\mathop{}\\!\mathrm{d}x=\Big{[}\tfrac{1}{2}x^{2}\log(x)-\tfrac{1}{4}x^{2}\Big{]}_{x=1}^{n+1}=\tfrac{1}{2}n^{2}\log(n)-\tfrac{1}{4}n^{2}+O(n\log(n)).$

We analogously get the same estimate for a lower bound by considering
$\int_{1}^{n}f(x)\mathop{}\\!\mathrm{d}x$.

(iii) From (i) and (ii) we get

$\displaystyle\log\Big{(}\prod_{k=0}^{n}(k!)\Big{)}=\sum_{k=1}^{n}\Big{(}k\log(k)-k+O(\log(k))\Big{)}=\big{(}\tfrac{1}{2}n^{2}\log(n)-\tfrac{1}{4}n^{2}\big{)}-\tfrac{1}{2}n^{2}+O(n\log(n)).$

(iv) We first rewrite the binomials in terms of factorials and then apply (i)
and (iii), so that

$\displaystyle\log\Big{(}\prod_{k=0}^{n}\binom{n}{k}\Big{)}$
$\displaystyle=\log\bigg{(}\frac{(n!)^{n}}{\big{(}\prod_{k=0}^{n}(k!)\big{)}^{2}}\bigg{)}=n\log(n!)-2\log\Big{(}\prod_{k=0}^{n}(k!)\Big{)}$
$\displaystyle=(n^{2}\log(n)-n^{2})-2(\tfrac{1}{2}n^{2}\log
n-\tfrac{3}{4}n^{2})+O(n\log n)=\tfrac{1}{2}n^{2}+O(n\log n),$

as was to be shown. ∎

######  Proposition 15.4.

Let $n\in\mathbb{Z}_{\geq 1}$. There are at most

$n\prod_{k=1}^{n}\Big{(}2\binom{n}{k}2^{k}+1\Big{)}=\exp\Big{(}\tfrac{\log(2)+1}{2}\,n^{2}+O(n\log(n))\Big{)}$

indecomposable elements in $\overline{\mathbb{Z}}$ of degree up to $n$.

###### Proof.

By Corollary 15.2 every indecomposable of degree at most $n$ is the root of a
monic polynomial $f=\sum_{k=0}^{n}f_{n-k}X^{k}$ such that
$|f_{k}|\leq\binom{n}{k}2^{k}$ for all $0\leq k\leq n$. Hence every such
polynomial corresponds to at most $n$ indecomposables. For every $0<k\leq n$
there are $2\binom{n}{k}2^{k}+1$ choices for $f_{k}$, and $f_{0}=1$, proving
the first upper bound. We may bound $2\binom{n}{k}2^{k}+1\leq
3\binom{n}{k}2^{k}$, so that by Proposition 15.3.iv we get

$n\prod_{k=1}^{n}\Big{(}2\binom{n}{k}2^{k}+1\Big{)}\leq n\cdot 3^{n}\cdot
2^{\binom{n+1}{2}}\cdot\prod_{k=0}^{n}\binom{n}{k}=\exp\Big{(}\tfrac{\log(2)+1}{2}\,n^{2}+O(n\log(n))\Big{)},$

as was to be shown. ∎

For $f\in\mathbb{Q}[X]$ monic write $q(f)$ for the average of the square
length of the roots of $f$ in $\mathbb{C}$, such that for all
$\alpha\in\overline{\mathbb{Z}}$ with minimal polynomial
$f_{\alpha}\in\mathbb{Q}[X]$ we get $q(\alpha)=q(f_{\alpha})$. Note that
$f=(X+2)^{n}$, although it is not irreducible, has $q(f)=4$ and attains the
bounds of Proposition 15.1. However, that does not imply that Proposition 15.4
cannot be improved, as it is not clear that all combinations of coefficients
occur for polynomials $f$ with $q(f)\leq 4$. Some small degree numerical
results might suggest improvements can be made.

degree | $1$ | $2$ | $3$ | $4$
---|---|---|---|---
# monic $f\in\mathbb{Z}[X]$ satisfying the conclusion to Proposition 15.1 | 5 | 81 | 5525 | 1786785
# monic $f\in\mathbb{Z}[X]$ satisfying $q(f)\leq 4$ | 5 | 49 | 989 | 48422
# $\alpha\in\overline{\mathbb{Z}}$ satisfying $q(\alpha)\leq 4$ | 5 | 39 | 739 | 40354

We also have the following lower bound.

######  Proposition 15.5.

Let $n\in\mathbb{Z}_{\geq 1}$. There are at least

$\exp\Big{(}\frac{\log 2}{4}n^{2}+O(n\log n)\Big{)}$

indecomposable algebraic integers of degree $n$.

###### Proof.

Let $n\in\mathbb{Z}_{\geq 1}$ and recall the definition of $\mathbb{Z}[X]_{n}$
from Definition 12.6. Consider the set

$S_{n}=\left\\{f=\sum_{k=0}^{n-1}f_{k}X^{k}\in
2X\mathbb{Z}[X]_{n-1}+2\,\middle|\,(\forall k)\
|f_{k}|(\sqrt{2})^{k}n\leq(\sqrt{2})^{n-1}\right\\}.$

For $f\in S_{n}$ consider $g=X^{n}-f$ and note that $g$ is irreducible by
Eisenstein’s criterion. Consider the ball $D\subseteq\mathbb{C}$ of radius
$r=(\sqrt{2})^{1-1/n}<\sqrt{2}$ around $0$. For all $z$ on the boundary of $D$
we have

$|f(z)|\leq\sum_{k=0}^{n-1}|f_{k}||z|^{k}\leq\sum_{k=0}^{n-1}|f_{k}|(\sqrt{2})^{k}\stackrel{{\scriptstyle\textup{(i)}}}{{\leq}}\sum_{k=0}^{n-1}\frac{(\sqrt{2})^{n-1}}{n}=(\sqrt{2})^{n-1}=|z|^{n},$

where (i) is strict for $n$ sufficiently large due to
$|f_{0}|n=2n<(\sqrt{2})^{n-1}$. Hence by Rouché’s Theorem the polynomials
$X^{n}$ and $g$ have the same number of roots in $D$. It follows that all
roots of $g$ in $\mathbb{C}$ have length less than $\sqrt{2}$. Thus
$q(\alpha)<2$ for all roots $\alpha\in\overline{\mathbb{Z}}$ of $g$, so
$\alpha$ is indecomposable by Proposition 8.1.

We conclude that for $n$ sufficiently large there are at least
$n\cdot\\#S_{n}$ indecomposable algebraic integers of degree $n$, so it
remains to prove a lower bound on $\\#S_{n}$. Note that the coefficients of
$f\in S_{n}$ satisfy independent inequalities, so we may simply give a lower
bound per coefficient. Let $B=n-3\log_{2}(n)-2$, which is positive for $n$
sufficiently large. For $k>B$ we consider only $f_{k}=0$ and get a lower bound
of $1$ for this coefficient. For $0<k\leq B$ we have

$2\left\lfloor\frac{(\sqrt{2})^{n-k-1}}{2n}\right\rfloor+1\geq
2\Big{(}\frac{(\sqrt{2})^{n-k-1}}{2n}-1\Big{)}+1=\frac{(\sqrt{2})^{n-k-1}}{n}-1=\textup{(ii)}$

choices for $f_{k}$. Then for $n$ sufficiently large we have

$\frac{n}{(\sqrt{2})^{n-k-2}}\leq\frac{n}{n^{3/2}}\leq\sqrt{2}-1,\quad\text{so
that}\quad\textup{(ii)}\geq\frac{(\sqrt{2})^{n-k-2}}{n}.$

Hence $S_{n}$ contains, for $n$ sufficiently large, at least

$\prod_{k=1}^{B}\frac{(\sqrt{2})^{n-k-2}}{n}=\exp\Big{(}\frac{\log
2}{2}\sum_{k=1}^{B}(n-k-2)-B\log n\Big{)}=\exp\Big{(}\frac{\log
2}{4}n^{2}+O(n\log n)\Big{)}$

elements, from which the proposition follows. ∎

######  Corollary 15.6.

Let $n\in\mathbb{Z}_{\geq 1}$. There are at least

$\exp\Big{(}\frac{\log 2}{4}n^{2}+O(n\log n)\Big{)}$

indecomposable algebraic integers of degree up to $n$. ∎

From the upper and lower bound we may now conclude the following.

Theorem 1.3. There are least $\exp(\frac{1}{4}(\log 2)n^{2}+O(n\log n))$ and
at most $\exp(\frac{1}{2}(1+\log 2)n^{2}+O(n\log n))$ indecomposable algebraic
integers of degree up to $n$. ∎

##  16\. Fekete capacity theory

In this section we present a proof of a special case of Fekete’s theorem using
Minkowski’s convex body theorem. Fekete’s theorem can be thought of as a
partial converse to Theorem 13.7 of Szegő. Although this does not give us a
converse to Theorem 1.2, using similar techniques as in this section we will
later prove Theorem 1.4 in Section 19. The goal of this section is to showcase
the proof technique we will use to prove Theorem 1.4 so that we may later
improve clarity by brevity. Recall for $\alpha\in\overline{\mathbb{Q}}$ the
definition of the norm
$|\alpha|_{\infty}=\max_{\sigma\in\operatorname{X}(\overline{\mathbb{Q}})}|\sigma(\alpha)|$
from Section 13.

######  Theorem 16.1 (Fekete).

For each $\alpha\in\overline{\mathbb{Q}}$ and $0<r<1$ there exist only
finitely many $\beta\in\overline{\mathbb{Z}}$ such that
$|\beta-\alpha|_{\infty}\leq r$.

Just like for Szegő’s theorem, it is possible to derive an algorithmic
counterpart to Fekete’s theorem.

Combining Theorem 16.1 and Theorem 13.7, the point $r=1$ is still a
singularity. For $\alpha\in\overline{\mathbb{Z}}$ and $r=1$ clearly all
$\beta\in\alpha+\mu_{\infty}$ satisfy $|\beta-\alpha|_{\infty}\leq r$.
However, when $\alpha\not\in\overline{\mathbb{Z}}$ we do not know what happens
in general. We start with a volume computation.

######  Definition 16.2.

Let $A$ be an $\mathbb{R}$-algebra equipped with a real inner product. We
equip $A[Y]$ with an inner product

$\Big{\langle}\sum_{k=0}^{\infty}f_{k}Y^{k},\sum_{k=0}^{\infty}g_{k}Y^{k}\Big{\rangle}=\sum_{k=0}^{\infty}\langle
f_{k},g_{k}\rangle,$

which is the ‘standard’ inner product when we naturally identify $A[Y]$ with
$A^{(\mathbb{Z}_{\geq 0})}$. For $n\in\mathbb{Z}_{\geq 0}$ we equip
$A[Y]_{n}$, as defined in Definition 12.6, with the restriction of this inner
product.

######  Remark 16.3.

Obviously $\mathbb{R}$ is an $\mathbb{R}$-algebra with a real inner product.
We identify $\mathbb{C}$ with $\mathbb{R}^{2}$ by choosing $\mathbb{R}$-basis
$\\{1,\textup{i}\\}$, and equip $\mathbb{C}$ with the inner product induced by
the natural inner product on $\mathbb{R}^{2}$. Note that this inner product in
turn induces the standard norm of $\mathbb{C}$. For a number field $K$ we
remark that $K_{\mathbb{R}}$ has a real inner product as in Definition 12.3.

######  Lemma 16.4.

Let $A$ be an $\mathbb{R}$-algebra of dimension $d<\infty$. For all
$a,b\in\mathbb{R}$, $c\in A$ and $n\in\mathbb{Z}_{\geq 0}$ we have an
$\mathbb{R}$-linear transformation $\phi$ on $A[X]_{n}$ given by $f\mapsto
bf(a(X-c))$ with $\det\phi=(a^{n(n-1)/2}\cdot b^{n})^{d}$.

###### Proof.

Note that $\phi$ is trivially an $\mathbb{R}$-linear transformation. Choose an
$\mathbb{R}$-basis $\\{e_{1},\dotsc,e_{d}\\}$ for $A$. Writing $\phi$ as a
matrix with respect to the basis $\\{e_{i}X^{j}\,|\,1\leq i\leq d,\,0\leq
j<n\\}$ for $A[X]_{n}$ we note that $\phi$ is a lower triangular matrix with
diagonal entries $b,ba,ba^{2},\dotsc,ba^{n-1}$, each occurring with
multiplicity $d$. The determinant of $\phi$ is then simply the product of the
diagonal. ∎

######  Lemma 16.5.

Let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$ and let
$r\in\mathbb{R}_{>0}$. For $n\in\mathbb{Z}_{\geq 0}$ consider

$S_{n}(r)=\\{f\in\mathbb{F}[Y]_{n}\,|\,(\forall\,z\in\mathbb{C})\ |z|\leq
r\Rightarrow|f(z)|\leq r\\}.$

Then as function of $n$ we have
$\log\operatorname{vol}(S_{n}(r))\geq-\tfrac{1}{2}n^{2}\cdot[\mathbb{F}:\mathbb{R}]\cdot\log
r+O(n\log n)$.

###### Proof.

Write $S_{n}=S_{n}(1)$. By applying the transformation $f\mapsto rf(r^{-1}Y)$
to $\mathbb{F}[Y]_{n}$ we bijectively map $S_{n}$ to $S_{n}(r)$. From Lemma
16.4 it follows that
$\log\operatorname{vol}(S_{n}(r))=-\tfrac{1}{2}n^{2}\cdot[\mathbb{F}:\mathbb{R}]\log(r)+\log\operatorname{vol}(S_{n})+O(n\log
n)$. It remains to prove $\log\operatorname{vol}S_{n}\geq O(n\log n)$.

First suppose $\mathbb{F}=\mathbb{R}$. Consider the set

$T_{n}=\Big{\\{}\sum_{k=0}^{n-1}f_{k}Y^{k}\in\mathbb{R}[Y]_{n}\,\Big{|}\,\sum_{k=0}^{n-1}|f_{k}|\leq
1\Big{\\}}.$

Note that for all $f\in T_{n}$ and $z\in\mathbb{C}$ such that $|z|\leq 1$ we
have $|f(z)|\leq\sum_{k=0}^{n-1}|f_{k}|\leq 1$, so $f\in S_{n}$. Hence
$T_{n}\subseteq S_{n}$ and
$\operatorname{vol}(T_{n})\leq\operatorname{vol}(S_{n})$. With Proposition
15.3 we compute $\log\operatorname{vol}(T_{n})=\log(2^{n}/n!)=O(n\log n)$,
from which the lemma follows for $\mathbb{F}=\mathbb{R}$.

For $\mathbb{F}=\mathbb{C}$, note that we have an isometry
$\mathbb{R}[X]_{n}^{2}\to\mathbb{C}[X]_{n}$ given by $(f,g)\mapsto
f+\textup{i}\cdot g$. For $f,g\in\tfrac{1}{2}T_{n}$ and $z\in\mathbb{C}$ such
that $|z|\leq 1$ we have $|f(z)+\textup{i}\cdot
g(z)|\leq|f(z)|+|g(z)|\leq\tfrac{1}{2}+\tfrac{1}{2}=1$, so $f+\textup{i}\cdot
g\in S_{n}$. Hence
$\log\operatorname{vol}(S_{n})\geq\log(\operatorname{vol}(\tfrac{1}{2}T_{n})^{2})=2\log\operatorname{vol}(T_{n})-2n\log
2=O(n\log n)$, from which the lemma follows for $\mathbb{F}=\mathbb{C}$. ∎

######  Theorem 16.6.

Let $R$ be an order of a number field $K$ and let $0<r<1$. For all $\alpha\in
K$ there exists a non-zero $g\in R[X]$ such that for all $z\in
K_{\mathbb{R}}$, if $\lvert z-\alpha\rvert_{K}\leq r$ then $\lvert
g(z)\rvert_{K}\leq r$.

###### Proof.

Write $d=[K:\mathbb{Q}]$. Let $n\in\mathbb{Z}_{\geq 0}$ and consider the
lattice $\Lambda_{n}=R[X]_{n}$ in the inner product space
$K_{\mathbb{R}}[Y]_{n}$, where $Y=X-\alpha$. Note that
$\dim_{\mathbb{R}}K_{\mathbb{R}}[Y]_{n}=dn$ and that $\Lambda_{n}$ is a full-
rank lattice in $K_{\mathbb{R}}[Y]_{n}$ with
$\det(\Lambda_{n})=|\Delta(R)|^{n/2}$ by Lemma 16.4 and Theorem 12.5. Consider

$S_{n}=\big{\\{}f\in
K_{\mathbb{R}}[Y]_{n}\,\big{|}\,(\forall\,\sigma\in\operatorname{X}(K))\,(\forall\,z\in\mathbb{C})\
|z|\leq r\Rightarrow|\sigma(f)(z)|\leq r\big{\\}}$

and note that it is both symmetric and convex. Moreover, it follows from Lemma
16.5 that $\log\operatorname{vol}(S_{n})\geq-\tfrac{1}{2}n^{2}d\log r+O(n\log
n)$. Hence

$\log\Big{(}\frac{\operatorname{vol}(S_{n})}{2^{dn}\cdot\det(\Lambda_{n})}\Big{)}\geq-\tfrac{1}{2}n^{2}d\log
r+O(n\log n).$

Because $-\tfrac{1}{2}d\log r>0$ there exists some $n$ sufficiently large such
that $\operatorname{vol}(S_{n})>2^{dn}\det(\Lambda_{n})$. By Theorem 12.8
there then exists some non-zero $g\in\Lambda_{n}\cap S_{n}$ which as
polynomial in $X$ satisfies the requirements. ∎

###### Proof of Theorem 16.1.

Let $K=\mathbb{Q}(\alpha)$ and let $R\subseteq K$ be some order of $K$. Then
by Theorem 16.6 there exists some non-zero $g\in R[X]$ such that for all $z\in
K_{\mathbb{R}}$, if $\lvert z-\alpha\rvert_{K}\leq r$ then $\lvert
g(z)\rvert_{K}\leq r$. Suppose $\beta\in\overline{\mathbb{Z}}$ satisfies
$|\beta-\alpha|_{\infty}\leq r$ and let $L=K(\beta)$. Let
$(\rho_{\sigma})_{\sigma}\in\textup{P}=\prod_{\sigma\in\operatorname{X}(K)}\operatorname{X}_{\sigma}(L)$
such that $\overline{\rho_{\sigma}}=\rho_{\overline{\sigma}}$ for all
$\sigma\in\operatorname{X}(K)$, and consider
$z=(\rho_{\sigma}(\beta))_{\sigma}\in K_{\mathbb{R}}$. Then $\lvert
z-\alpha\rvert_{K}\leq|\beta-\alpha|_{\infty}\leq r$, so $\lvert
g(z)\rvert_{K}\leq r$. Consequently we have $|\rho(g(\beta))|\leq r$ for all
$\rho\in\operatorname{X}(L)$. Hence

$|N_{L/\mathbb{Q}}(g(\beta))|=\prod_{\rho\in\operatorname{X}(L)}|\rho(g(\beta))|\leq
r^{[L:\mathbb{Q}]}<1.$

As $g(\beta)\in\overline{\mathbb{Z}}$, we must then have $g(\beta)=0$. As
$\beta$ must be a root of $g$ and $g$ is non-zero, there can only be finitely
many $\beta$. ∎

##  17\. Reduction to exponentially bounded polynomials

We now prepare to prove Theorem 1.4. If there are only finitely many
decompositions of an algebraic integer $\alpha$, then certainly there exists a
non-zero polynomial $f\in\mathbb{Z}[X]$ such that $f(\beta)=0$ for all
decompositions $(\beta,\alpha-\beta)$ of $\alpha$. The goal is to exhibit such
a polynomial when $q(\alpha)$ is small using a lattice argument, similarly to
the proof of Theorem 16.1. In this section we derive an analytic sufficient
condition for a polynomial $f$ to have this property.

######  Definition 17.1.

Let $K$ be a number field. We define
$\mathcal{S}(K)=\operatorname{X}(K)\times\mathbb{C}$, the coproduct (i.e.
disjoint union) of measurable spaces of $\\#\operatorname{X}(K)$ copies of
$\mathbb{C}$, where $\mathbb{C}$ has the standard Lebesgue measurable space
structure. We write $\mathcal{M}(K)$ for the set of probability measures $\mu$
on $\mathcal{S}(K)$, i.e. all $\mu$ such that $\mu(\mathcal{S}(K))=1$.

######  Definition 17.2.

Let $K$ be a number field and $r\in\mathbb{R}_{>0}$. For $f\in
K_{\mathbb{R}}[Y]$ we say $f$ is exponentially bounded at radius $r$ if for
all $\mu\in\mathcal{M}(K)$ satisfying
$\int|z|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,z)<r^{2}$ it holds that
$\int\log|\sigma(f)(z)|\mathop{}\\!\mathrm{d}\mu(\sigma,z)<0$.

######  Proposition 17.3.

Let $\alpha\in\overline{\mathbb{Z}}$, $K=\mathbb{Q}(\alpha)$ and
$r>\sqrt{q(\alpha/2)}$. If $f\in\mathcal{O}_{K}[X]$ is exponentially bounded
at radius $r\in\mathbb{R}_{>0}$ as polynomial in the variable $Y=X-\alpha/2$,
then for all $(\beta,\gamma)\in\textup{dec}(\alpha)$ we have $f(\beta)=0$.

###### Proof.

Suppose $(\beta,\gamma)\in\textup{dec}(\alpha)$. Then $q(\beta-\alpha/2)\leq
q(\alpha/2)<r^{2}$ by Lemma 4.3. Let $L=K(\beta)$ and

$B=\\{(\sigma,\rho(\beta-\alpha/2))\in\mathcal{S}(K)\,|\,\rho\in\operatorname{X}_{\sigma}(L)\\},$

which has $\\#B=[L:\mathbb{Q}]$ and
$\\#(B\cap(\\{\sigma\\}\times\mathbb{C}))=[L:K]$ for all
$\sigma\in\operatorname{X}(K)$. Let $\mu\in\mathcal{M}(K)$ be the uniform
probability measure on $B$ and write $f_{Y}$ for $f$ as a polynomial in the
variable $Y$. Because
$\int|x|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,x)=q(\beta-\alpha/2)<r^{2}$ and
$f_{Y}$ is exponentially bounded at radius $r$ we get

$\displaystyle\log\big{(}N(f(\beta))^{[L:\mathbb{Q}]}\big{)}$
$\displaystyle=\log\prod_{\rho\in\operatorname{X}(L)}|\rho(f(\beta))|=\sum_{\rho\in\operatorname{X}(L)}\log|\rho(f_{Y}(\beta-\alpha/2))|$
$\displaystyle=[L:\mathbb{Q}]\cdot\int\log|\sigma(f_{Y})(x)|\mathop{}\\!\mathrm{d}\mu(\sigma,x)<0.$

We conclude that $N(f(\beta))<1$. Since $f(\beta)$ is integral we have
$f(\beta)=0$, as was to be shown. ∎

######  Example 17.4.

The set of polynomials of $K_{\mathbb{R}}$ exponentially bounded at radius $r$
is closed under multiplication and symmetric. However, we will show that it is
not convex.

Let $r=1$ and $K=\mathbb{Q}$. For all $c\in(-1,1)$ the constant polynomial $c$
is trivially exponentially bounded at any positive radius, in particular at
radius $1$. Also the polynomial $Y^{2}$ is exponentially bounded: For any
$\mu\in\mathcal{M}(K)$ such that
$\int|z|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,z)<1$ we have

$\int\log|z^{2}|\mathop{}\\!\mathrm{d}\mu(\sigma,z)\leq\log\int|z|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,z)<\log
1=0.$

Here the first inequality is Jensen’s inequality for integrals. When $\mu$ has
finite support, this comes down to Lemma 7.5. For $c\in(-1,1)$ and
$k\in\mathbb{Z}_{\geq 0}$ the product $cY^{2k}$ of exponentially bounded
polynomials at radius $1$ is exponentially bounded at radius $1$. We claim
that $\tfrac{1}{4}(1+Y^{2k})$ for $k$ sufficiently large, which is a convex
combination of $\tfrac{1}{2}$ and $\tfrac{1}{2}Y^{2k}$, is not exponentially
bounded at radius $1$. Taking $\mu\in\mathcal{M}(\mathbb{Q})$ with weight
$\tfrac{1}{5}$ at $2$ and remaining weight at $0$ we have
$\int|z|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,z)=\tfrac{4}{5}<1$, yet
$\int\log|\tfrac{1}{4}(1+Y^{2k})|\mathop{}\\!\mathrm{d}\mu(\sigma,z)=\tfrac{1}{5}\log(1+2^{2k})-\log
4\to\infty$ as $k\to\infty$. We conclude that the set of exponentially bounded
polynomials at radius $1$ is not convex. A similar argument works for all
radii and number fields.

######  Lemma 17.5.

Let $D\subseteq\mathbb{C}$ be a convex subset and let $f:D\to\mathbb{C}$ be
analytic. Then for distinct $x,y\in D$ we have

$\bigg{|}\frac{f(x)-f(y)}{x-y}\bigg{|}\leq\sup_{z\in D}|f^{\prime}(z)|.$

###### Proof.

Let $\gamma:[0,1]\to D$ be the parametrization of the straight line connecting
$x$ and $y$, which is well-defined since $D$ is convex. First note that

$\displaystyle\int_{0}^{1}f^{\prime}(\gamma(t))\mathop{}\\!\mathrm{d}t=\int_{0}^{1}f^{\prime}(tx+(1-t)y)\mathop{}\\!\mathrm{d}t=\frac{1}{x-y}\Big{[}f(tx+(1-t)y)\Big{]}_{t=0}^{1}=\frac{f(x)-f(y)}{x-y}.$

Then

$\displaystyle\bigg{|}\frac{f(x)-f(y)}{x-y}\bigg{|}=\bigg{|}\int_{0}^{1}f^{\prime}(\gamma(t))\mathop{}\\!\mathrm{d}t\bigg{|}\leq\int_{0}^{1}\big{|}f^{\prime}(\gamma(t))\big{|}\mathop{}\\!\mathrm{d}t\leq\int_{0}^{1}\big{(}\sup_{z\in
D}|f^{\prime}(z)|\big{)}\mathop{}\\!\mathrm{d}t=\sup_{z\in D}|f^{\prime}(z)|,$

as was to be shown. ∎

We will now translate the measure theoretic property of Definition 17.2 to an
analytic one. Our results in the coming sections only depend on the ‘only if’
part of the following equivalence.

######  Theorem 17.6.

Let $K$ be a number field, $0<r<1$ and $f\in K_{\mathbb{R}}[Y]$. Then $f$ is
exponentially bounded at radius $r$ if and only if there exist
$a\in\mathbb{R}_{>0}$ such that for all $\sigma\in\operatorname{X}(K)$ and
$z\in\mathbb{C}$ we have $|\sigma(f)(z)|\leq\exp(a(|z|^{2}-r^{2}))$.

###### Proof.

($\Leftarrow$) Suppose such $a$ exists. Let $\mu\in\mathcal{M}(K)$ such that
$\int|z|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,z)<r^{2}$. Then

$\int\log|\sigma(f)(z)|\mathop{}\\!\mathrm{d}\mu(\sigma,z)\leq\int
a(|z|^{2}-r^{2})\mathop{}\\!\mathrm{d}\mu(\sigma,z)<ar^{2}-ar^{2}=0,$

so $f$ is exponentially bounded at radius $r$.

($\Rightarrow$) Let $D_{0}=\\{z\in\mathbb{C}\,|\,|z|<r\\}$ and
$D_{\infty}=\\{z\in\mathbb{C}\,|\,|z|>r\\}$. For $c\in\\{0,\infty\\}$ let

$A_{c}=\big{\\{}a\in\mathbb{R}\,\big{|}\,(\forall\,\rho\in\operatorname{X}(K))\,(\forall\,z\in
D_{c})\ |\rho(f)(z)|\leq\exp(a(|z|^{2}-r^{2}))\big{\\}}.$

Firstly, we show that $A_{0}$ is non-empty. Let $\rho\in\operatorname{X}(K)$
and $z_{0}\in D_{0}$, and let $\mu\in\mathcal{M}(K)$ be the measure with
weight $1$ at $(\rho,z_{0})$. Then
$\int|x|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,x)=|z_{0}|^{2}<r^{2}$, so by
exponential boundedness

$|\rho(f)(z_{0})|=\exp\Big{(}\int\log|\sigma(f)(x)|\mathop{}\\!\mathrm{d}\mu(\sigma,x)\Big{)}<1.$

It follows that $0\in A_{0}$, and even $(-\infty,0]\subseteq A_{0}$. This
argument also shows that $\rho(f)$ is bounded by $1$ on the boundary of
$D_{0}$, the circle of radius $r$.

Secondly, we show that $A_{\infty}$ is non-empty. Since $\exp(|z|^{2}-r^{2})$
grows faster than any polynomial, there exists some $b>r$ such that
$|\rho(f)(z)|\leq\exp(|z|^{2}-r^{2})$ for all $|z|\geq b$. Write
$B=\\{x\in\mathbb{C}\,|\,|x|\leq b\\}$. Let $\rho\in\operatorname{X}(K)$ and
$z\in B\cap D_{\infty}$, and write $g=\rho(f)$ and $\theta=z/|z|$. As remarked
at the end of the previous paragraph we have $|g(r\theta)|\leq 1$, so that

$\displaystyle\log|g(z)|$
$\displaystyle\leq\log(1+|g(z)-g(r\theta)|)\leq|g(z)-g(r\theta)|=\frac{|z|^{2}-r^{2}}{|z|+r}\cdot\bigg{|}\frac{g(z)-g(r\theta)}{z-r\theta}\bigg{|}$
$\displaystyle\stackrel{{\scriptstyle*}}{{\leq}}(|z|^{2}-r^{2})\cdot\frac{\sup_{x\in
B}|g^{\prime}(x)|}{2r}\leq a(|z|^{2}-r^{2}),$

where $*$ follows from Lemma 17.5 and $a$ is the maximum of $1$ and all
$(2r)^{-1}\sup_{x\in B}|\rho(f)^{\prime}(x)|$ for
$\rho\in\operatorname{X}(K)$. Thus $a\in A_{\infty}$.

Thirdly, we show that $A_{0}\cap A_{\infty}$ is non-empty. Suppose for the
sake of contradiction that $A_{0}\cap A_{\infty}$ is empty. Clearly $A_{0}$
and $A_{\infty}$ are closed. Hence there exist reals that are neither in
$A_{0}$ nor $A_{\infty}$, and let $a$ be such a real number. It follows that
$a>0$. In turn, there exist $z_{0}\in D_{0}$ and $z_{\infty}\in D_{\infty}$
with $\rho_{0},\rho_{\infty}\in\operatorname{X}(K)$ such that
$|\rho_{0}(f)(z_{0})|>\exp(a(|z_{0}|^{2}-r^{2}))$ and
$|\rho_{\infty}(f)(z_{\infty})|>\exp(a(|z_{\infty}|^{2}-r^{2}))$. Choose
$t\in(0,1)$ such that $(1-t)|z_{0}|^{2}+t|z_{\infty}|^{2}<r^{2}$ and let $\mu$
be the measure that assigns weight $1-t$ to $(\rho_{0},z_{0})$ and weight $t$
to $(\rho_{\infty},z_{\infty})$. Then
$\int|z|^{2}\mathop{}\\!\mathrm{d}\mu(\sigma,z)<r^{2}$ and thus

$\displaystyle 0$
$\displaystyle>\int\log|\sigma(f)(z)|\mathop{}\\!\mathrm{d}\mu(\sigma,z)=(1-t)\log|\rho_{0}(f)(z_{0})|+t\log|\rho_{\infty}(f)(z_{\infty})|.$

Taking the limit of $t$ up to $s\in\mathbb{R}$ such that
$(1-s)|z_{0}|^{2}+s|z_{\infty}|^{2}=r^{2}$ we get

$0\geq(1-s)\log|\rho_{0}(f)(z_{0})|+s\log|\rho_{\infty}(f)(z_{\infty})|>a((1-s)|z_{0}|^{2}+s|z_{\infty}|^{2}-r^{2})=0,$

a contradiction. Hence $A_{0}\cap A_{\infty}\neq\emptyset$, as was to be
shown.

Note that $D_{0}\cup D_{\infty}$ is dense in $\mathbb{C}$, so any positive
$a\in A_{0}\cap A_{\infty}$ gives the inequality we set out to prove. Suppose
$a\in A_{0}\cap A_{\infty}$ is such that $a\leq 0$. Thus
$|\rho(f)(z)|\leq\exp(a(|z|^{2}-r^{2}))\leq 1$ for all $z\in D_{\infty}$ and
$\rho\in\operatorname{X}(K)$, so $\rho(f)$ is a constant function. However, as
$|\rho(f)(z)|<1$ for $z\in D_{0}$ as shown before, this constant is strictly
less than $1$. Let $c\in(0,1)$ be a constant that bounds $\rho(f)$ for all
$\rho\in\operatorname{X}(K)$. Then $-r^{-2}\log c\in A_{0}\cap A_{\infty}$ is
positive. Hence $A_{0}\cap A_{\infty}$ always contains a positive element. ∎

##  18\. Volume computation

The next step is to compute the volume of a symmetric convex set of
exponentially bounded polynomials. As in Lemma 16.5 it suffices for the sake
of volume computation to consider the case where the radius is $1$ and the
base field is $\mathbb{R}$. In view of Theorem 17.6, we consider the unit-ball
of the following norm.

######  Definition 18.1.

Let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$. We equip
$\mathbb{F}[Y]$ with the norm

$\|f\|_{\textup{e}}=\max_{z\in\mathbb{C}}\frac{|f(z)|}{\exp(|z|^{2})}.$

######  Lemma 18.2.

Consider the map $\phi:\mathbb{Z}_{\geq 0}\to\mathbb{R}_{\geq 0}$ given by

$\phi(n)=\begin{cases}\big{(}\frac{n}{2}\big{)}!&\textup{if $n$ is even}\\\
\big{(}\frac{n-1}{2}\big{)}!\cdot\sqrt{\frac{n+1}{2}}&\textup{if $n$ is
odd}\end{cases}.$

Then we have

$\log\Big{(}\prod_{k=0}^{n}\phi(k)\Big{)}=\tfrac{1}{4}n^{2}\log
n-(\tfrac{3}{8}+\tfrac{1}{4}\log 2)n^{2}+O(n\log n)$

and for all $x\in\mathbb{R}_{\geq 0}$ and $m\in\mathbb{Z}_{\geq 0}$ we have

$\frac{x^{2m+1}}{\phi(2m+1)}\leq\frac{1}{2}\bigg{(}\frac{x^{2m}}{\phi(2m)}+\frac{x^{2m+2}}{\phi(2m+1)}\bigg{)}.$

###### Proof.

Writing out the product we have

$\prod_{k=0}^{n}\phi(k)=\Bigg{(}\prod_{m=0}^{\lfloor
n/2\rfloor}\phi(2m)\Bigg{)}\Bigg{(}\prod_{m=0}^{\lfloor(n-1)/2\rfloor}\phi(2m+1)\Bigg{)}=\Bigg{(}\prod_{m=0}^{\lfloor
n/2\rfloor}m!\Bigg{)}\Bigg{(}\prod_{m=0}^{\lfloor(n-1)/2\rfloor}m!\Bigg{)}\Bigg{(}\prod_{m=0}^{\lfloor(n-1)/2\rfloor}\sqrt{m+1}\Bigg{)}.$

We then apply Proposition 15.3 to compute

$\displaystyle\smash{\log\Big{(}\prod_{k=0}^{n}\phi(k)\Big{)}}\vphantom{\scalebox{2.0}{1}}$
$\displaystyle=\lfloor\tfrac{n}{2}\rfloor^{2}\big{(}\tfrac{1}{2}\log\lfloor\tfrac{n}{2}\rfloor-\tfrac{3}{4}\big{)}+\lfloor\tfrac{n-1}{2}\rfloor^{2}\big{(}\tfrac{1}{2}\log\lfloor\tfrac{n-1}{2}\rfloor-\tfrac{3}{4}\big{)}+\lfloor\tfrac{n-1}{2}\rfloor\big{(}\log\lfloor\tfrac{n-1}{2}\rfloor-1\big{)}+O(n\log
n)$
$\displaystyle=\big{(}\tfrac{n}{2}\big{)}^{2}\big{(}\tfrac{1}{2}\log\tfrac{n}{2}-\tfrac{3}{4}\big{)}+\big{(}\tfrac{n}{2}\big{)}^{2}\big{(}\tfrac{1}{2}\log\tfrac{n}{2}-\tfrac{3}{4}\big{)}+0+O(n\log
n)$ $\displaystyle=\tfrac{1}{4}n^{2}\log n-(\tfrac{3}{8}+\tfrac{1}{4}\log
2)n^{2}+O(n\log n),$

proving the first part. For the second, let $m\in\mathbb{Z}_{\geq 0}$ and
$x\in\mathbb{R}_{\geq 0}$. Then

$\frac{1}{\phi(2m)}+\frac{x^{2}}{\phi(2m+2)}=\frac{1}{m!}\Big{(}\Big{(}1-\frac{x}{\sqrt{m+1}}\Big{)}^{2}+\frac{2x}{\sqrt{m+1}}\Big{)}\geq\frac{1}{m!}\frac{2x}{\sqrt{m+1}}=2\cdot\frac{x}{\phi(2m+1)},$

from which the second part follows. ∎

Recall the notation $R[X]_{n}$ from Definition 12.6, the subset of $R[X]$ of
polynomials of degree strictly less than $n$.

######  Proposition 18.3.

Write $S=\\{f\in\mathbb{R}[Y]\,|\,\|f\|_{\textup{e}}\leq 1\\}$. Then for
$n\in\mathbb{Z}_{\geq 0}$ we have

$\log\operatorname{vol}(S\cap\mathbb{R}[Y]_{n})\geq-\tfrac{1}{4}n^{2}\log
n+(\tfrac{3}{8}+\tfrac{1}{4}\log 2)n^{2}+O(n\log n).$

###### Proof.

Consider $\phi$ as in Lemma 18.2 and define

$T=\Big{\\{}\sum_{i=0}^{\infty}f_{i}Y^{i}\in\mathbb{R}[Y]\,\Big{|}\,(\forall
i)\ |f_{i}|\leq\frac{1}{2\phi(i)}\Big{\\}}.$

Then for all $f\in T$ and $z\in\mathbb{C}$ we have using Lemma 18.2 that

$\displaystyle|f(z)|$
$\displaystyle\leq\sum_{i=0}^{\infty}|f_{i}|\cdot|z|^{i}\leq\sum_{i=0}^{\infty}\frac{|z|^{i}}{2\phi(i)}=\frac{1}{2}\Bigg{[}\sum_{k=0}^{\infty}\frac{|z|^{2k}}{k!}+\sum_{k=0}^{\infty}\frac{|z|^{2k+1}}{\phi(2k+1)}\Bigg{]}$
$\displaystyle\leq\frac{1}{2}\Bigg{[}\sum_{k=0}^{\infty}\frac{|z|^{2k}}{k!}+\sum_{k=0}^{\infty}\frac{1}{2}\bigg{(}\frac{|z|^{2k}}{k!}+\frac{|z|^{2k+2}}{(k+1)!}\bigg{)}\Bigg{]}\leq\sum_{k=0}^{\infty}\frac{|z|^{2k}}{k!}=\exp|z^{2}|,$

so $f\in S$ and $T\subseteq S$. Then by Lemma 18.2 we have

$\log\operatorname{vol}(T\cap\mathbb{R}[Y]_{n})=-n\log
2-\log\Big{(}\prod_{k=0}^{n-1}\phi(k)\Big{)}=-\tfrac{1}{4}n^{2}\log
n+(\tfrac{3}{8}+\tfrac{1}{4}\log 2)n^{2}+O(n\log n),$

from which the proposition follows. ∎

Proposition 18.3 is sufficient for our purposes. It may interest a reader
attempting to improve our results that the lower bound of Proposition 18.3 is
actually an equality, which we will show in the remainder of this section.

######  Theorem 18.4 (Brunn–Minkowski inequality, Theorem 2.1 in [6]).

Let $n\in\mathbb{Z}_{\geq 1}$ and let $A,B\subseteq\mathbb{R}^{n}$ be bounded
non-empty measurable sets. Then for all $t\in[0,1]$ such that

$(1-t)A+tB=\\{(1-t)a+tb\,|\,a\in A,\,b\in B\\}$

is measurable we have the inequality

$\textup{vol}((1-t)A+tB)^{1/n}\geq(1-t)\textup{vol}(A)^{1/n}+t\textup{vol}(B)^{1/n}.\qed$

We will only apply this theorem to compact subsets of $\mathbb{R}^{n}$, which
are indeed measurable and bounded. Moreover, for $A,B\subseteq\mathbb{R}^{n}$
compact and $t\in(0,1)$ also the set $(1-t)A+tB$ is compact, hence measurable.

######  Corollary 18.5.

Let $n\in\mathbb{Z}_{\geq 0}$, let $V\subseteq\mathbb{R}^{n}$ be a subspace
and let $S\subseteq\mathbb{R}^{n}$ be a symmetric convex body. Then the map
that sends $x\in\mathbb{R}^{n}$ to $\operatorname{vol}_{V}(V\cap(S-x))$ takes
a maximum at $0$.

###### Proof.

Let $x\in\mathbb{R}^{n}$ and write $m=\dim V$ and $H_{x}=V\cap(S-x)$. If $m=0$
the corollary holds trivially, so suppose $m>0$. Note that $H_{-x}=-H_{x}$
since $S$ and $V$ are symmetric. Because $S$ and $V$ are convex we have
$\tfrac{1}{2}H_{x}+\tfrac{1}{2}H_{-x}\subseteq H_{0}$. Hence by Theorem 18.4
we have

$\textup{vol}(H_{0})^{1/m}\geq\textup{vol}(\tfrac{1}{2}H_{x}+\tfrac{1}{2}H_{-x})^{1/m}\geq\tfrac{1}{2}\textup{vol}(H_{x})^{1/m}+\tfrac{1}{2}\textup{vol}(H_{-x})^{1/m}=\textup{vol}(H_{x})^{1/m},$

from which the corollary follows. ∎

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3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
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} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}}}$ from Definition
5.1.

######  Corollary 18.6.

Let $n\in\mathbb{Z}_{\geq 0}$, let $U,V\subseteq\mathbb{R}^{n}$ be subspaces
such that $U\mathop{\mathchoice{\displaystyle\raisebox{2.4pt}{
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}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}}}V=\mathbb{R}^{n}$
and write $\pi$ for the projection
$U\mathop{\mathchoice{\displaystyle\raisebox{2.4pt}{ \leavevmode\hbox
to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}}}V\to U$. If
$S\subseteq\mathbb{R}^{n}$ is a symmetric convex body, then
$\operatorname{vol}_{\mathbb{R}^{n}}(S)\leq\operatorname{vol}_{U}(\pi
S)\cdot\operatorname{vol}_{V}(S\cap V)$.

###### Proof.

By Corollary 18.5 we have

$\operatorname{vol}(S)=\int_{\pi
S}\operatorname{vol}_{V}(V\cap(S-x))\mathop{}\\!\mathrm{d}x\leq\int_{\pi
S}\operatorname{vol}_{V}(V\cap
S)\mathop{}\\!\mathrm{d}x=\operatorname{vol}_{U}(\pi
S)\cdot\operatorname{vol}_{V}(V\cap S).$

∎

######  Theorem 18.7.

Write $S=\\{f\in\mathbb{R}[Y]\,|\,\|f\|_{\textup{e}}\leq 1\\}$. Then for
$n\in\mathbb{Z}_{\geq 0}$ we have

$\log\operatorname{vol}(S\cap\mathbb{R}[Y]_{n})=-\tfrac{1}{4}n^{2}\log
n+(\tfrac{3}{8}+\tfrac{1}{4}\log 2)n^{2}+O(n\log n).$

###### Proof.

We already proved a lower bound in Proposition 18.3, so it remains to prove an
upper bound. We will inductively show that

$\operatorname{vol}(S\cap\mathbb{R}[Y]_{n})\leq
2^{n}\prod_{k=1}^{n-1}\Big{(}\frac{2\textup{e}}{k}\Big{)}^{k/2}.$

It then follows from Proposition 15.3 that

$\log\operatorname{vol}(S\cap\mathbb{R}[Y]_{n})\leq n\log
2+\frac{1}{2}\sum_{k=1}^{n-1}k\big{(}\log(2\textup{e})-\log
k)=-\tfrac{1}{4}n^{2}\log n+(\tfrac{3}{8}+\tfrac{1}{4}\log 2)n^{2}+O(n\log
n).$

For $n=0$ and $n=1$ the inequality certainly holds. Now suppose the inequality
holds for $n\geq 1$. Write
$\mathbb{R}[Y]_{n+1}=(\mathbb{R}Y^{n})\mathop{\mathchoice{\displaystyle\raisebox{2.4pt}{
\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}}}\mathbb{R}[Y]_{n}$
and let $\pi:\mathbb{R}[Y]_{n+1}\to\mathbb{R}Y^{n}$ be the projection map. By
Corollary 18.6 it suffices to show for $n>0$ that
$\operatorname{vol}(\pi(S\cap\mathbb{R}[Y]_{n+1}))\leq
2(\frac{n}{2\textup{e}})^{-n/2}$. We do this by proving

$\pi(S\cap\mathbb{R}[Y]_{n+1})\stackrel{{\scriptstyle\textup{(i)}}}{{\subseteq}}S\cap(\mathbb{R}Y^{n})\stackrel{{\scriptstyle\textup{(ii)}}}{{\subseteq}}[-1,+1]\Big{(}\frac{2\textup{e}}{n}\Big{)}^{n/2}Y^{n}.$

(i) Suppose $f\in\mathbb{R}[Y]_{n+1}$ and
$\|f\|_{\textup{e}}<\|\pi(f)\|_{\textup{e}}$. Since $\pi(f)$ is a monomial,
the function $z\mapsto|\pi(f)(z)|\exp(-|z|^{2})$ takes its maximum on a circle
of radius say $r$. Then for all $z$ on this circle we have
$|f(z)|\leq\|f\|_{\textup{e}}\exp(r^{2})<\|\pi(f)\|_{\textup{e}}\exp(r^{2})=|\pi(f)(z)|$.
Hence by Rouché’s theorem, the polynomial $f-\pi(f)$ has as many roots as
$\pi(f)$ in the disk $\\{z\in\mathbb{C}\,|\,|z|\leq r\\}$, counting
multiplicities. However, since $f-\pi(f)$ has degree at most $n-1$ and
$\pi(f)$ has $n$ such roots, this is a contradiction. Hence
$\|\pi(f)\|_{\textup{e}}\leq\|f\|_{\textup{e}}$, from which (i) follows.

(ii) Consider the map $g:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ given by
$x\mapsto x^{n}\exp(-x^{2})$. Then

$\frac{\mathop{}\\!\mathrm{d}g}{\mathop{}\\!\mathrm{d}x}=x^{n-1}(n-2x^{2})\exp(-x^{2})=0\
\Longleftrightarrow\ x=0\ \lor\ x=\sqrt{n/2}.$

Hence $g$ takes a maximum at $(n/2)^{1/2}$. We conclude that
$\|Y^{n}\|_{\textup{e}}=g((n/2)^{1/2})=(\tfrac{n}{2\textup{e}})^{n/2}$. Thus
$\max\\{c\in\mathbb{R}\,|\,cY^{n}\in S\\}=(\tfrac{2\textup{e}}{n})^{n/2}$, as
was to be shown.

The theorem now follows by induction. ∎

##  19\. Proof of the main theorem

We are now ready to give a proof of Theorem 1.4.

######  Proposition 19.1.

Let $R$ be an order of a number field $K$, let $\alpha\in K$ and
$0<r^{2}<\frac{1}{2}\exp(\frac{1}{2})$. Then there exists some non-zero $f\in
R[X]$ such that $f(X-\alpha)$ is exponentially bounded at radius $r$.

###### Proof.

Let $n\in\mathbb{Z}_{\geq 1}$ and $d=[K:\mathbb{Q}]$. Write $Y=X-\alpha$ and
consider the real vector space $K_{\mathbb{R}}[Y]_{n}$, which we equip with an
inner product as in Definition 16.2 with respect to the variable $Y$. By
Theorem 12.5 and Lemma 16.4 the lattice $R[X]_{n}$ in $K_{\mathbb{R}}[Y]_{n}$
is full rank and has determinant
$\det(R[Y]_{n})=\left|\det(R)\right|^{n}=|\Delta(R)|^{n/2}$. For
$b\in\mathbb{R}_{\geq 0}$ consider

$\displaystyle S_{n}$ $\displaystyle=\\{f\in
K_{\mathbb{R}}[Y]_{n}\,|\,(\forall\,\sigma\in\operatorname{X}(K))\,(\forall\,z\in\mathbb{C})\
|\sigma(f)(z)|\leq\exp(bn(|z|^{2}-r^{2}))\\}$ $\displaystyle=\\{f\in
K_{\mathbb{R}}[Y]_{n}\,|\,(\forall\,\sigma\in\operatorname{X}(K))\
\|\exp(bnr^{2})\cdot\sigma(f)((bn)^{-1/2}\cdot Y)\|_{\textup{e}}\leq 1\\}.$

We have a natural orthogonal decomposition
$K_{\mathbb{R}}\cong\mathbb{R}^{u}\times\mathbb{C}^{v}$ for some
$u,v\in\mathbb{Z}_{\geq 0}$ which in turn gives an orthogonal decomposition
$K_{\mathbb{R}}[Y]_{n}=(\mathbb{R}[Y]_{n})^{u}\times(\mathbb{C}[Y]_{n})^{v}$.
Note that $S_{n}$ is simply a product over $\sigma\in\operatorname{X}(K)$ of

$S_{n}(\sigma)=\\{f\in\mathbb{F}[Y]_{n}\,|\,\|\exp(bnr^{2})\cdot\sigma(f)((bn)^{-1/2}\cdot
Y)\|_{\textup{e}}\leq 1\\}$

where $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$ depending on whether
$\sigma(K)\subseteq\mathbb{R}$. Then using Lemma 16.4 and Proposition 18.3 we
compute

$\displaystyle\log\operatorname{vol}(S_{n})$ $\displaystyle\geq
d\Big{(}\big{(}-\tfrac{1}{4}n^{2}\log n+(\tfrac{3}{8}+\tfrac{1}{4}\log
2)n^{2}\big{)}+\big{(}\tfrac{1}{4}n^{2}\log(bn)-bn^{2}r^{2}\big{)}\Big{)}+O(n\log
n)$ $\displaystyle=dn^{2}\cdot\epsilon(b)+O(n\log n),$

with $\epsilon(b)=\tfrac{1}{4}\log(2b)+\tfrac{3}{8}-r^{2}b$. Choosing
$b=(2r)^{-2}$ we get $\epsilon(b)=\tfrac{1}{4}(\tfrac{1}{2}-\log(2r^{2}))>0$.
Hence

$\log\Big{(}\frac{\operatorname{vol}(S_{n})}{2^{dn}\cdot|\Delta(R)|^{n/2}}\Big{)}\geq
dn^{2}\cdot\epsilon(b)+O(n\log n)\to\infty\quad(\text{as }n\to\infty).$

Thus by Minkowski’s theorem there exists for $n$ sufficiently large some non-
zero $g\in S_{n}\cap R[X]$. Because $g\in S_{n}$, this polynomial is
exponentially bounded at radius $r$ by Theorem 17.6. ∎

Theorem 1.4. If $\gamma\in\overline{\mathbb{Z}}$ satisfies
$q(\gamma)<2\exp(\tfrac{1}{2})\approx 3.297$, then $\gamma$ has only finitely
many decompositions.

###### Proof.

Let $\alpha=\gamma/2$, let $K=\mathbb{Q}(\gamma)$ and let $R$ be some order in
$K$. Choose $r\in\mathbb{R}_{>0}$ such that
$q(\alpha)<r^{2}<\tfrac{1}{2}\exp(\tfrac{1}{2})$. Then by Proposition 19.1
there exists some non-zero polynomial $f\in R[X]$ which as polynomial in
$Y=X-\alpha$ is exponentially bounded at radius $r$. Hence by Proposition 17.3
all $(\beta,\gamma-\beta)\in\text{dec}(\gamma)$ satisfy $f(\beta)=0$. As $f$
has only finitely many roots, the theorem follows. ∎

##  20\. Remarks on the proof of the main theorem

In this section we briefly discuss the proof of Theorem 1.4 and make some
practical remarks for explicit computation.

The proof of Theorem 1.4 proceeds in the following steps:

  1. (1)

We determine a sufficient condition for a polynomial to have all
decompositions of $\alpha$ as roots.

  2. (2)

We translate this condition into an analytic one.

  3. (3)

We determine the volume of a symmetric convex set of polynomials satisfying
this condition.

  4. (4)

We apply Minkowski’s convex body theorem to find integral polynomials in this
set.

Theorem 17.6 suggests that step (2) can hardly be improved upon. By Theorem
18.7 we correctly computed the volume of our symmetric convex set in step (3).
However, in order to make it convex we fixed the constant $a$ that comes out
of Theorem 17.6. It is easy to verify that we indeed made an optimal choice of
$a$ in Proposition 19.1, although that does not guarantee we chose the best
convex subset. If the weakest link in the proof is step (4), we likely require
a completely different approach. It should be noted however that Minkowski’s
convex body theorem is powerful enough to prove the classical Theorem 16.6.

A piece of information we did not exploit in our proof of the main theorem is
the following symmetry.

######  Proposition 20.1.

For all $\alpha\in\overline{\mathbb{Z}}$ we have an involution
$x\mapsto\alpha-x$ on $\overline{\mathbb{Z}}$ which induces action on
$\textup{dec}(\alpha)$ given by $(\beta,\gamma)\mapsto(\gamma,\beta)$. ∎

######  Corollary 20.2.

Let $\alpha\in\overline{\mathbb{Z}}$ and let $\mathbb{Z}[\alpha]\subseteq R$
be an order of $\mathbb{Q}(\alpha)$. For all $f\in R[X]$ such that
$f(\beta)=0$ for all $(\beta,\gamma)\in\textup{dec}(\alpha)$, also
$g=f(\alpha-X)\in R[X]$ satisfies $g(\beta)=0$ for all
$(\beta,\gamma)\in\textup{dec}(\alpha)$. ∎

Corollary 20.2 turns the involution on $\overline{\mathbb{Z}}$ into an
$R$-algebra automorphism on $R[X]$. We can incorporate this automorphism in
our proof of Theorem 1.4.

######  Proposition 20.3.

Let $\alpha\in\overline{\mathbb{Z}}$, let $\mathbb{Z}[\alpha]\subseteq R$ be
an order of $K=\mathbb{Q}(\alpha)$ and let $r\in\mathbb{R}_{>0}$. For all
$f\in R[X]$ such that $f$ as a polynomial in $Y=X-\alpha/2$ is exponentially
bounded at radius $r$, so is $f(X)\cdot f(\alpha-X)\in K[Y^{2}]$.

###### Proof.

Note that the involution $X\mapsto\alpha-X$ is with respect to $Y$ given by
$Y\mapsto-Y$. Hence if $f$ as a polynomial in $Y$ is exponentially bounded at
radius $r$, then so is $f(\alpha-X)$. As noted in Example 17.4, the set of
polynomials exponentially bounded at $r$ is closed under multiplication. Hence
$g=f(X)\cdot f(\alpha-X)$ is exponentially bounded at radius $r$. Now $g$ is
invariant under $Y\mapsto-Y$, meaning all coefficients at odd degree monomials
in $Y$ are zero, i.e. $g\in K[Y^{2}]$. ∎

An interesting question to ask is how dissimilar $f$ and $f(\alpha-X)$ can be
for exponentially bounded $f$. Certainly both should have $\beta$ as root for
all $(\beta,\gamma)\in\textup{dec}(\alpha)$ by Corollary 20.2. In the context
of finding ‘small’ $f$ algorithmically it seems that often $f$ and
$f(\alpha-X)$ are the same (up to sign).

As a consequence of Proposition 20.3, in the proof of Theorem 1.4 we may look
at the lattice $R[X(\alpha-X)]$ in $K_{\mathbb{R}}[Y^{2}]$ instead of $R[X]$
in $K_{\mathbb{R}}[Y]$. The effect is two-fold. Firstly, it simplifies the
volume computation of Proposition 18.3, as we no longer require the ad-hoc
function $\phi$ from Lemma 18.2. Secondly, any integral polynomial in our
symmetric body can be found in a lower dimensional lattice in Proposition
19.1. This follows from the suggested changes to Proposition 18.3, but can
heuristically be seen as follows. If $f$ is a solution in the original lattice
$R[X]$, then $f(X)\cdot f(\alpha-X)$ exists in our new lattice
$R[X(\alpha-X)]$ at the same dimension. However, as discussed before, $f$
might already be an element of $R[X(\alpha-X)]$ anyway, so we could have
already found $f$ at half the dimension. Neither of these changes have an
effect on the quality of our theoretical results. However, when we want to
compute decompositions in practice, the latter is very useful.

##  21\. Computing decompositions of algebraic integers of small square-norm

From the proof of Theorem 1.4 one easily derives the following result.

######  Theorem 21.1.

There exists an algorithm that takes as input an element
$\alpha\in\overline{\mathbb{Z}}$, and decides whether
$q(\alpha)<2\sqrt{\textup{e}}$ and if so, computes all decompositions of
$\alpha$.

###### Proof.

Clearly $q(\alpha)\neq 2\sqrt{\textup{e}}$ as the latter is not algebraic.
However, both are computable, and after finitely many steps of approximation
we can decide whether $q(\alpha)<2\sqrt{\textup{e}}$. If so, we can even
compute a rational $r$ such that $q(\alpha)<r^{2}<2\sqrt{\textup{e}}$.

We have an explicit formula for a lower bound on the volume of the set $S$ as
defined in the proof of Proposition 18.3. Thus we can compute a sufficiently
large $n$ such that Minkowski’s theorem, as in the proof of Proposition 19.1,
guarantees the existence of a non-zero lattice point in $S$. We may then
simply enumerate all lattice points to eventually find a polynomial $f$ as in
Proposition 19.1. We determine using Theorem 14.2 the monic irreducible
factors of $f$ and decide using Corollary 14.7 for each whether it gives rise
to a decomposition of $\alpha$. ∎

##  22\. Working out an example

We will now work out an example proving an algebraic integer $\alpha$ is
indecomposable, proceeding along the lines of our lattice algorithm. Of
course, this result will be trivial when $q(\alpha)\leq 2$ as we have seen in
Proposition 8.1, so we will choose $\alpha$ such that $q(\alpha)>2$. On the
other hand, the algorithms from Section 21 terminate faster the smaller
$q(\alpha)$ is, so for this example we will consider $\alpha=\sqrt[3]{3}$ with
$q(\alpha)=3^{2/3}\approx 2.080$.

Setup. Let $\alpha=\sqrt[3]{3}$ and let $r^{2}=6/11$, so that
$q(\alpha/2)<r^{2}<\tfrac{1}{2}\exp(\tfrac{1}{2})$. Write
$K=\mathbb{Q}(\alpha)$ and $R=\mathbb{Z}[\alpha]$ and consider the ring
$R[X]$. Writing $Y=X-\alpha/2$, we are looking for a polynomial $f\in R[X]$
such that $f$ as polynomial in $Y$ is exponentially bounded at radius $r$.
However, writing $Z=X(\alpha-X)$ we may instead look for such a polynomial in
$R[Z]$, as follows from Proposition 20.3.

Finding a polynomial. It is quite involved to systematically find short
vectors in a lattice. Instead we will employ a more ad-hoc approach, more
along the lines of Theorem 13.7. We guess that our polynomial $f$ will be
monic in $Z$ of some degree $n$. We start with $Z^{n}$ and then greedily
subtract $\mathbb{Z}[\alpha]$-multiples of lower degree powers of $Z$ such
that the resulting polynomial in $Y$ becomes ‘small’, i.e. has small
coefficients under every embedding $K\to\mathbb{C}$ with lower degree terms
weighing more heavily. Efficitively, we are applying a rounding function in
the sense of Lemma 13.4. Note that $Z=-Y^{2}+\alpha^{2}/4$. Similarly as in
the proof of Theorem 13.7, taking $n=4$ the $Y^{6}$ term becomes integral,
which is useful. Thus we will try $n=4$. We compute:

The remaining coefficients with respect to $Y$ look pretty small in every
embedding $K\to\mathbb{C}$, so we guess

$f(Y)=Y^{8}-\tfrac{1}{8}\alpha
Y^{4}-\tfrac{3}{8}Y^{2}+\tfrac{21}{256}\alpha^{2}=Z^{4}-\alpha^{2}Z^{3}+\alpha
Z^{2}\in R[Z].$

is going to be exponentially bounded at radius $r$ as polynomial in $Y$.

Proving exponentially boundedness. If we take $b:\mathbb{R}_{\geq
0}\to\mathbb{R}_{\geq 0}$ given by

$b(w)=w^{4}+\tfrac{1}{8}\cdot 3^{1/3}\cdot w^{2}+\tfrac{3}{8}\cdot
w+\tfrac{21}{256}\cdot 3^{2/3},$

then for all $\sigma\in\operatorname{X}(K)$ and $z\in\mathbb{C}$ we have
$|\sigma(f)(z)|\leq b(|z|^{2})$. To prove that $f$ is exponentially bounded at
radius $r$ it suffices to find $a\in\mathbb{R}$ such that
$b(w)\leq\exp(a(w-r^{2}))$ for all $w\in\mathbb{R}_{\geq 0}$. Because
$b(0)=\frac{21}{256}3^{2/3}$, we must have $a\leq-\log(b(0))/r^{2}\approx
3.4$. We will try $a=3$ for simplicity. Consider the function
$B(w)=b(w)\cdot\exp(-a(w-r^{2}))$, for which we want to show $B(w)\leq 1$ for
all $w$. Then

$0=B^{\prime}(w)=\exp(-a(w-r^{2}))(b^{\prime}(w)-ab(w))$

if and only if $ab(w)-b^{\prime}(w)=0$. Since the latter is simply a
polynomial equation we will find using standard techniques that it has no
positive real roots. We compute:

$\displaystyle ab(w)-b^{\prime}(w)$
$\displaystyle=3w^{4}-4w^{3}+\tfrac{3}{8}3^{1/3}w^{2}+(\tfrac{9}{8}-\tfrac{1}{4}3^{1/3})w+(\tfrac{63}{256}3^{2/3}-\tfrac{3}{8})$
$\displaystyle>3w^{4}-4w^{3}+\tfrac{3}{5}w^{2}+\tfrac{3}{4}w+\tfrac{1}{4}.$

For $1\leq w$ we get
$ab(w)-b^{\prime}(w)>3w^{4}-4w^{3}+\tfrac{3}{2}=w^{2}(3^{1/2}w-2\cdot
3^{-1/2})^{2}+(\tfrac{3}{2}-\tfrac{4}{3}w^{2})\geq 0$ and for $0<w\leq 1$ we
get
$ab(w)-b^{\prime}(w)>3w^{4}-4w^{3}+\tfrac{3}{2}w^{2}=3w^{2}(w^{2}-\tfrac{4}{3}w+\tfrac{1}{2})\geq
3w^{2}(w-2^{-1/2})^{2}\geq 0$. Hence $B$ has no local maxima besides possibly
at $0$, and because $B(w)\to 0$ as $w\to\infty$ we conclude that $B$ takes a
maximum at $0$. Therefore $b$ is bounded by $w\mapsto\exp(a(w-r^{2}))$ and
thus $f$ is exponentially bounded at radius $r$.

Finding decompositions. Writing $f$ as a polynomial in $X$ we get

$\displaystyle f$ $\displaystyle=X^{8}-4\alpha
X^{7}+7\alpha^{2}X^{6}-21X^{5}+13\alpha X^{4}-5\alpha^{2}X^{3}+3X^{2}$
$\displaystyle=X^{2}\cdot(\alpha-X)^{2}\cdot(X^{4}-2\alpha
X^{3}+2\alpha^{2}X^{2}-3X+\alpha).$

By Proposition 17.3 all decompositions of $\alpha$ can be found among the
roots of $f$. The factors $X$ and $\alpha-X$ correspond to the trivial
decompositions $(0,\alpha)$ and $(\alpha,0)$ of $\alpha$. The polynomial
$h=X^{4}-2\alpha X^{3}+2\alpha^{2}X^{2}-3X+\alpha$ is irreducible as it is
Eisenstein at the prime $(\alpha)$. Let $\beta\in\overline{\mathbb{Z}}$ be a
root of $h$. By Lemma 7.7 we have $q(\beta)\geq
N(\beta)^{2}=N(h(0))^{2}=3^{2/3}=q(\alpha)$. We can only have
$q(\beta)+q(\alpha-\beta)\leq q(\alpha)$ if $q(\alpha-\beta)=0$, i.e.
$\alpha=\beta$, which is impossible. Hence $\alpha$ is indecomposable by Lemma
4.3.

##  23\. Enumeration of indecomposable algebraic integers of degree 3

In this section we discuss our attempt to compute the indecomposable algebraic
integers of degree 3 and derive Theorem 1.5. These results are obtained by a
computer program written in Sage.

We will write $f_{\alpha}\in\mathbb{Z}[x]$ for the minimal polynomial of
$\alpha\in\overline{\mathbb{Z}}$. We save ourselves some work by only
considering $\alpha$ up to ‘trivial isometries’ of $\overline{\mathbb{Z}}$,
namely those of $\mu_{\infty}\rtimes\textup{Gal}(\overline{\mathbb{Q}})$ as in
Lemma 7.13. Using Proposition 15.1 we can compute a set of $5525$ polynomials
among which we can find all minimal polynomials of the indecomposable
algebraic integers of degree 3. Among those $5525$ polynomials $f$ only $700$
are in fact irreducible with $q(\alpha)\leq 4$ for all roots $\alpha$ of $f$.
We already eliminated the Galois action by considering minimal polynomials
instead of elements, and by choosing only one element of each
$\mu_{\infty}$-orbit $\\{f,-f(-x)\\}$ we eliminate the action of
$\mu_{\infty}$, and end up with ‘only’ $350$ polynomials to check. Of those
$350$, there are $27$ polynomials $f_{\alpha}$ such that $q(\alpha)<2$, so
that $\alpha$ is indecomposable by Proposition 8.1.

Small degree decompositions. For $95$ polynomials $f$, the roots $\alpha$ of
$f$ have a non-trivial decomposition in the ring of integers of
$\mathbb{Q}(\alpha)$. For $116$ of the remaining polynomials $f_{\alpha}$ we
can find a non-trivial decomposition $(\beta,\alpha-\beta)$ of $\alpha$ with
$\beta$ in the ring of integers of a degree 2 extension of
$\mathbb{Q}(\alpha)$. Of those $116$ there are $84$ for which the minimal
polynomial $g_{\beta}$ of $\beta$ over $\mathbb{Q}(\alpha)$ is of the form
$x^{2}-\alpha x\pm 1$, a polynomial we encountered in the proof of Lemma 11.2.
The remaining $32$ polynomials and corresponding decompositions can be found
in Table 1. We are now left with $112$ polynomials to check.

$q(\alpha)$ | $f_{\alpha}$ | $g_{\beta}$
---|---|---
$2.2844$ | $x^{3}-x^{2}+3x+1$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}+1)$
$2.5198$ | $x^{3}+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}$
$2.6178$ | $x^{3}+2x^{2}+4x+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}$
$2.7246$ | $x^{3}+2x+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}$
$2.7850$ | $x^{3}-2x^{2}+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}$
$2.8582$ | $x^{3}+2x^{2}-x+2$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}+\alpha)$
$2.8657$ | $x^{3}-2x^{2}+3x+2$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}-\alpha)$
$2.9073$ | $x^{3}+2x^{2}-2x+1$ | $x^{2}-\alpha x-\alpha$
$2.9379$ | $x^{3}+2x^{2}+x+4$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}+\alpha)$
$2.9380$ | $x^{3}-x^{2}-x+5$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}-1)$
$2.9516$ | $x^{3}+x^{2}+x+5$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}+1)$
$3.0000$ | $x^{3}+x^{2}-4x+1$ | $x^{2}-\left(\alpha+1\right)x+1$
$3.1102$ | $x^{3}+x^{2}-2x+4$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}+\alpha)$
$3.2714$ | $x^{3}+2x^{2}+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}$
$3.2849$ | $x^{3}-2x^{2}+2x+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}$
$3.2981$ | $x^{3}+2x^{2}-2x+2$ | $x^{2}-\left(\alpha-1\right)x-(\alpha-1)$
$q(\alpha)$ | $f_{\alpha}$ | $g_{\beta}$
---|---|---
$3.3321$ | $x^{3}+3x+5$ | $x^{2}-\alpha x+\frac{1}{3}(\alpha^{2}+\alpha+1)$
$3.3333$ | $x^{3}-2x^{2}-3x+1$ | $x^{2}-\left(\alpha-1\right)x+1$
$3.3378$ | $x^{3}+x^{2}-4x+3$ | $x^{2}-\left(\alpha-1\right)x-(\alpha-1)$
$3.3853$ | $x^{3}+4x+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}+1$
$3.4436$ | $x^{3}-2x^{2}-x+6$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}-\alpha)$
$3.5716$ | $x^{3}+2x^{2}+6x+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}+1$
$3.6432$ | $x^{3}+2x^{2}+6x+2$ | $x^{2}-\left(\alpha+1\right)x-1$
$3.6830$ | $x^{3}+x^{2}-4x+4$ | $x^{2}-\alpha x+\frac{1}{2}\alpha^{2}+\frac{1}{2}\alpha-1$
$3.6839$ | $x^{3}-x^{2}-3x+7$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}-1)$
$3.7159$ | $x^{3}+x^{2}+3x+7$ | $x^{2}-\alpha x+\frac{1}{2}(\alpha^{2}+1)$
$3.7362$ | $x^{3}+2x^{2}-3x+2$ | $x^{2}-\alpha x-\alpha$
$3.7962$ | $x^{3}+2x^{2}+6x+1$ | $x^{2}-\left(\alpha-1\right)x-\alpha-1$
$3.8854$ | $x^{3}-3x+7$ | $x^{2}-\alpha x+\frac{1}{3}(\alpha^{2}-\alpha+1)$
$3.9111$ | $x^{3}-2x^{2}+5x+2$ | $x^{2}-\left(\alpha-1\right)x-1$
$3.9938$ | $x^{3}+x^{2}-5x+4$ | $x^{2}-\left(\alpha+1\right)x+1$
$4.0000$ | $x^{3}-2x^{2}-4x+1$ | $x^{2}-\alpha x+\alpha$

Table 1.  A table of minimal polynomials $f_{\alpha}$ together with a minimal
polynomial $g_{\beta}$ of degree 2 over $\mathbb{Q}(\alpha)$ such that
$(\beta,\alpha-\beta)$ is a non-trivial decomposition of $\alpha$.

Large degree decompositions. To find decompositions in higher degree
extensions we implemented a lattice algorithm. Since we are interested in
finding only one decomposition instead of all of them, and since verifying
whether something is a decomposition is computationally easy, we can get away
with a lot of heuristics. For $(\beta,\alpha-\beta)\in\textup{dec}(\alpha)$ we
have, on average of squares over all embeddings of $\mathbb{Q}(\alpha,\beta)$
in $\mathbb{C}$, that $|\beta-\alpha/2|\leq\sqrt{q(\alpha/2)}=r$ by Lemma 4.3.
Hence if we write $g_{\beta}=\sum_{i}c_{i}(x-\alpha/2)^{i}$ we have that
$\sum_{i}|c_{i}|r^{i}$ should be small. It is useful for our lattice algorithm
to instead consider the 2-norm
$(\sum_{i}r^{i}\sum_{\sigma}|\sigma(c_{i})|^{2})^{1/2}$ and hope this does not
affect the quality of our results for the worse. We enumerate small
polynomials $\epsilon\in\mathbb{Q}(\alpha)[x]$ of degree less than
$d\in\mathbb{Z}_{>0}$ such that $(x-\alpha/2)^{d}-x^{d}+\epsilon$ is in the
lattice of integral polynomials, and thus $(x-\alpha/2)^{d}+\epsilon$ is
monic, integral and small. We then verify for each of those whether they
induce a decomposition of $\alpha$. The $41$ polynomials $f_{\alpha}$ for
which this method has found a non-trivial decomposition $(\beta,\alpha-\beta)$
of $\alpha$ with $g_{\beta}$ of degree greater than $2$ are listed in Table 2
together with the polynomial $g_{\beta}$ found. This leaves $71$ polynomials
to check and gives an upper bound of $6\cdot 98=588$ on the number of
indecomposable algebraic integers of degree 3.

$q(\alpha)$ | $f_{\alpha}$ |  $g_{\beta}$
---|---|---
$2.9240$ | $x^{3}+5$ |  $x^{4}-2\alpha x^{3}+2\alpha^{2}x^{2}+5x-\alpha$
$3.0103$ | $x^{3}-x^{2}+5$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(-3\alpha^{2}-2\alpha+15)x^{3}+(3\alpha^{2}-7\alpha-6)x^{2}+(\alpha^{2}+\alpha+5)x-\alpha+1$
$3.2595$ | $x^{3}+2x^{2}-x+3$ |  $x^{4}-2\alpha x^{3}+(2\alpha^{2}+\alpha-1)x^{2}+(\alpha^{2}+3)x-\alpha+1$
$3.2624$ | $x^{3}-2x^{2}+3x+3$ |  $x^{4}-2\alpha x^{3}+(2\alpha^{2}-\alpha+1)x^{2}+(-\alpha^{2}+2\alpha+3)x-\alpha$
$3.3019$ | $x^{3}+6$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+18x^{3}-8\alpha x^{2}+2\alpha^{2}x+1$
$3.3378$ | $x^{3}+x+6$ |  $x^{4}-2\alpha x^{3}+(\frac{3}{2}\alpha^{2}-\frac{1}{2}\alpha)x^{2}+(\frac{1}{2}\alpha^{2}+\frac{1}{2}\alpha+3)x+1$
$3.3656$ | $x^{3}-2x^{2}+4x+2$ |  $x^{4}-2\alpha x^{3}+(2\alpha^{2}-\alpha+2)x^{2}+(-\alpha^{2}+2\alpha+2)x-\alpha$
$3.3709$ | $x^{3}-x^{2}+2x+5$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(-3\alpha^{2}+6\alpha+15)x^{3}+(-\alpha^{2}-10\alpha-5)x^{2}+3\alpha^{2}x+3$
$3.3863$ | $x^{3}+x^{2}-3x+4$ |  $x^{8}-4\alpha x^{7}+7\alpha^{2}x^{6}+(7\alpha^{2}-21\alpha+28)x^{5}+(18\alpha^{2}-30\alpha+17)x^{4}+(19\alpha^{2}-30\alpha+32)x^{3}+(13\alpha^{2}-24\alpha+19)x^{2}+(6\alpha^{2}-9\alpha+8)x+\alpha^{2}-2\alpha+2$
$3.3895$ | $x^{3}-x^{2}+6$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(-3\alpha^{2}-2\alpha+18)x^{3}+(3\alpha^{2}-8\alpha-7)x^{2}+(\alpha^{2}+\alpha+6)x-\alpha+1$
$3.4190$ | $x^{3}-2x+6$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}-1)x^{4}+(-4\alpha+18)x^{3}+(\alpha^{2}-8\alpha)x^{2}+2\alpha^{2}x+1$
$3.4338$ | $x^{3}-x^{2}+4x+3$ |  $x^{8}-4\alpha x^{7}+(7\alpha^{2}-1)x^{6}+(-7\alpha^{2}+31\alpha+21)x^{5}+(-17\alpha^{2}-30\alpha-13)x^{4}+(19\alpha^{2}-27\alpha-24)x^{3}+(4\alpha^{2}+26\alpha+13)x^{2}+(-5\alpha^{2}+3\alpha+3)x-2\alpha-1$
$3.4438$ | $x^{3}+x^{2}+6$ |  $x^{10}-5\alpha x^{9}+11\alpha^{2}x^{8}+(14\alpha^{2}+84)x^{7}+(11\alpha^{2}-69\alpha+68)x^{6}+(44\alpha^{2}-36\alpha+30)x^{5}+(29\alpha^{2}-5\alpha+99)x^{4}+(5\alpha^{2}-24\alpha+42)x^{3}+(3\alpha^{2}-6\alpha)x^{2}-1$
$3.4537$ | $x^{3}+2x+6$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(4\alpha+18)x^{3}+(-\alpha^{2}-8\alpha)x^{2}+2\alpha^{2}x+1$
$3.4767$ | $x^{3}+x^{2}-2x+5$ |  $x^{14}-7\alpha x^{13}+23\alpha^{2}x^{12}+(47\alpha^{2}-94\alpha+235)x^{11}+(200\alpha^{2}-467\alpha+333)x^{10}+(695\alpha^{2}-761\alpha+1040)x^{9}+(1143\alpha^{2}-1912\alpha+2739)x^{8}+(1840\alpha^{2}-3030\alpha+3435)x^{7}+(2256\alpha^{2}-3290\alpha+4258)x^{6}+(1960\alpha^{2}-3100\alpha+3990)x^{5}+(1343\alpha^{2}-2101\alpha+2601)x^{4}+(667\alpha^{2}-1022\alpha+1300)x^{3}+(226\alpha^{2}-354\alpha+447)x^{2}+(49\alpha^{2}-75\alpha+95)x+5\alpha^{2}-8\alpha+10$
$3.4858$ | $x^{3}+2x^{2}+2x+6$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(6\alpha^{2}+4\alpha+18)x^{3}+(4\alpha^{2}-3\alpha+16)x^{2}+(3\alpha^{2}-2\alpha+6)x+\alpha^{2}+2$
$3.5833$ | $x^{3}-2x^{2}+6$ |  $x^{8}-4\alpha x^{7}+(7\alpha^{2}+\alpha)x^{6}+(-17\alpha^{2}+42)x^{5}+(25\alpha^{2}-26\alpha-75)x^{4}+(-14\alpha^{2}+36\alpha+72)x^{3}+(-\alpha^{2}-22\alpha-29)x^{2}+(4\alpha^{2}+5\alpha)x-\alpha^{2}+2$
$3.6133$ | $x^{3}+x^{2}-x+6$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+\alpha)x^{4}+(\alpha^{2}-3\alpha+18)x^{3}+(\alpha^{2}-8\alpha-1)x^{2}+(2\alpha^{2}+\alpha)x+1$
$3.6361$ | $x^{3}+2x^{2}-x+4$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}-1)x^{4}+(6\alpha^{2}-\alpha+12)x^{3}+(5\alpha^{2}-8\alpha+11)x^{2}+(4\alpha^{2}-4\alpha+4)x+\alpha^{2}-\alpha+2$
$3.6370$ | $x^{3}-2x^{2}+3x+4$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(-6\alpha^{2}+7\alpha+12)x^{3}+(3\alpha^{2}-14\alpha-10)x^{2}+(2\alpha^{2}+5\alpha+4)x-\alpha^{2}+1$
$3.6521$ | $x^{3}+2x^{2}+5$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(6\alpha^{2}+15)x^{3}+(5\alpha^{2}-7\alpha+13)x^{2}+(4\alpha^{2}-3\alpha+5)x+\alpha^{2}+2$
$3.6593$ | $x^{3}+7$ |  $x^{3}-2\alpha x^{2}+\alpha^{2}x+1$
$3.6785$ | $x^{3}-x^{2}-x+7$ |  $x^{10}-5\alpha x^{9}+(11\alpha^{2}+1)x^{8}+(-14\alpha^{2}-18\alpha+98)x^{7}+(30\alpha^{2}-68\alpha-81)x^{6}+(16\alpha^{2}+27\alpha+140)x^{5}+(-13\alpha^{2}-70\alpha+18)x^{4}+(24\alpha^{2}+8\alpha-14)x^{3}+(-8\alpha^{2}-5\alpha+31)x^{2}+(2\alpha^{2}-2\alpha-7)x+1$
$3.6878$ | $x^{3}-x+7$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(-3\alpha+21)x^{3}+(\alpha^{2}-9\alpha)x^{2}+2\alpha^{2}x+1$
$3.6915$ | $x^{3}+x+7$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(3\alpha+21)x^{3}+(-\alpha^{2}-9\alpha)x^{2}+2\alpha^{2}x+1$
$3.6939$ | $x^{3}+x^{2}+x+7$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+\alpha+1)x^{4}+(\alpha^{2}+\alpha+21)x^{3}+(-9\alpha-1)x^{2}+(2\alpha^{2}+\alpha)x+1$
$3.7123$ | $x^{3}-2x^{2}+4x+3$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(-6\alpha^{2}+12\alpha+9)x^{3}+(-15\alpha-7)x^{2}+(4\alpha^{2}+\alpha)x-\alpha^{2}+2\alpha+1$
$3.7228$ | $x^{3}+2x^{2}+x+6$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(6\alpha^{2}+3\alpha+18)x^{3}+(4\alpha^{2}-5\alpha+16)x^{2}+(3\alpha^{2}-3\alpha+6)x+\alpha^{2}+2$
$3.7238$ | $x^{3}-x^{2}+2x+6$ |  $x^{8}-4\alpha x^{7}+7\alpha^{2}x^{6}+(-7\alpha^{2}+14\alpha+42)x^{5}+(-4\alpha^{2}-35\alpha-26)x^{4}+(15\alpha^{2}+8\alpha-6)x^{3}+(-6\alpha^{2}+6\alpha+20)x^{2}+(\alpha^{2}-4\alpha-6)x+\alpha+1$
$3.7345$ | $x^{3}+x^{2}-3x+5$ |  $x^{3}-2\alpha x^{2}+(\frac{3}{2}\alpha^{2}-\frac{1}{2})x+\frac{1}{2}\alpha^{2}-\alpha+\frac{3}{2}$
$3.7479$ | $x^{3}-x^{2}+7$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(-3\alpha^{2}-2\alpha+21)x^{3}+(3\alpha^{2}-9\alpha-9)x^{2}+(\alpha^{2}+2\alpha+7)x-\alpha$
$3.7664$ | $x^{3}-2x+7$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}-1)x^{4}+(-4\alpha+21)x^{3}+(\alpha^{2}-9\alpha)x^{2}+2\alpha^{2}x+1$
$3.7771$ | $x^{3}-2x^{2}+x+6$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(-6\alpha^{2}+3\alpha+18)x^{3}+(4\alpha^{2}-10\alpha-15)x^{2}+(4\alpha+6)x-\alpha-1$
$3.7949$ | $x^{3}+2x+7$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(4\alpha+21)x^{3}+(-\alpha^{2}-9\alpha)x^{2}+2\alpha^{2}x+1$
$3.7995$ | $x^{3}+x^{2}+7$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+\alpha)x^{4}+(\alpha^{2}+21)x^{3}+(-9\alpha-1)x^{2}+(2\alpha^{2}+\alpha)x+1$
$3.8253$ | $x^{3}+x^{2}-2x+6$ |  $x^{6}-3\alpha x^{5}+4\alpha^{2}x^{4}+(3\alpha^{2}-6\alpha+18)x^{3}+(4\alpha^{2}-10\alpha+9)x^{2}+(3\alpha^{2}-5\alpha+6)x+\alpha^{2}-\alpha+1$
$3.8560$ | $x^{3}+2x^{2}+2x+7$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(6\alpha^{2}+4\alpha+21)x^{3}+(4\alpha^{2}-4\alpha+19)x^{2}+(3\alpha^{2}-3\alpha+7)x+\alpha^{2}+2$
$3.8739$ | $x^{3}-x^{2}+x+7$ |  $x^{3}-2\alpha x^{2}+(\frac{3}{2}\alpha^{2}+\frac{1}{2})x-\frac{1}{2}\alpha^{2}+\frac{5}{2}$
$3.9466$ | $x^{3}-2x^{2}+7$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}-1)x^{4}+(-6\alpha^{2}+2\alpha+21)x^{3}+(4\alpha^{2}-10\alpha-19)x^{2}+(\alpha^{2}+5\alpha+7)x-\alpha^{2}-\alpha+1$
$3.9928$ | $x^{3}+2x^{2}-x+5$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}-1)x^{4}+(6\alpha^{2}-\alpha+15)x^{3}+(5\alpha^{2}-9\alpha+13)x^{2}+(4\alpha^{2}-4\alpha+5)x+\alpha^{2}-\alpha+1$
$3.9948$ | $x^{3}-x^{2}+3x+6$ |  $x^{6}-3\alpha x^{5}+(4\alpha^{2}+1)x^{4}+(-3\alpha^{2}+7\alpha+18)x^{3}+(-\alpha^{2}-12\alpha-9)x^{2}+(3\alpha^{2}+3\alpha)x-\alpha^{2}+\alpha+1$

Table 2.  A table of minimal polynomials $f_{\alpha}$ together with a minimal
polynomial $g_{\beta}$ of degree greater than 2 over $\mathbb{Q}(\alpha)$ such
that $(\beta,\alpha-\beta)$ is a non-trivial decomposition of $\alpha$.

Indecomposables. On the other hand, we want to prove that certain $\alpha$ are
indecomposable. To this end, we implemented a lattice algorithm similar to
that of Theorem 21.1. To hopefully speed up the algorithm we also apply the
dimension reducing symmetry trick discussed in Section 20. Writing $R$ for the
ring of integers of $\mathbb{Q}(\alpha)$ and $z=x(\alpha-x)$, we enumerate
short $g\in R[z]$ for which we verify whether $g$ as polynomial in
$y=x-\alpha/2$ is exponentially bounded. The $32$ polynomials $f_{\alpha}$ for
which we found such a $g$ proving indecomposability of $\alpha$ are listed in
Table 3. We present $g$ in factored form for compactness. This leaves $39$
polynomials undetermined and gives a lower bound of $6\cdot 59=354$ on the
number of indecomposable algebraic integers of degree $3$.

$q(\alpha)$ | $f_{\alpha}$ |  $g$
---|---|---
$2.0000$ | $x^{3}-2x^{2}-x+1$ |  $\left(\alpha^{2}-2\alpha\right)\cdot x\cdot(x-\alpha)\cdot(x^{2}-\alpha x+\alpha^{2}-\alpha-1)$
$2.0347$ | $x^{3}+x^{2}+3x+2$ |  $x^{2}\cdot(x-\alpha)^{2}\cdot(x^{2}-\alpha x-1)$
$2.0780$ | $x^{3}-2x^{2}+x+2$ |  $(x-1)\cdot(x-\alpha+1)\cdot x^{2}\cdot(x-\alpha)^{2}$