tags: study notes number theory

Filling out details in Baker's Comprehensive Course

• Filling out details in Baker's Comprehensive Course
  • Theorem 11.2
    • "Then, by symmetry, we have …"
  • Theorem 12.2
    • Why λ_1 ⨉ ··· ⨉ λ_n = |det(A)| is the right condition
  • Theorem 12.3 (Dirichlet's Unit Theorem)
    • Properly applying Minkovski's theorem
    • Covolume of ring of integers = discriminant of number field
    • Kernel of the Log map
    • How and where "r = s + t - 1" is used in the proof
    • The conclusion of the proof
    • Wrapping up

Theorem 11.2

"Then, by symmetry, we have …"

We give two proofs of Baker's claim that for

$f\in {\mathcal{O}}_K[X]$

$$F=\prod _{\sigma :K\hookrightarrow \mathbb{C}}{f}^{\sigma }\in \mathbb{Z}[X]$$ Baker claims that this follows "by symmetry". One proof was provided to me in a comment by user @reuns and relies on the Primitive Element Theorem and the Symmetric Function Theorem; we give a second more elementary proof based on Lagrange interpolation of

$F$, but it is less in the "by symmetry" spirit.

Both proofs can be found on math.stackexchange.

Theorem 12.2

Why λ_1 ⨉ ··· ⨉ λ_n = |det(A)| is the right condition

We denote by

$\mathcal{L}(\cdot )$ the Lebesgue measure on

${\mathbb{R}}^n$. Consider the set

$$S=\{\mathbf{x}\in {\mathbb{R}}^n|A\mathbf{x}\in \prod _{i=1}^n[-{\lambda }_i,+{\lambda }_i]\}$$ Then

$A(S)=[-{\lambda }_1,+{\lambda }_1]\times \cdots \times [-{\lambda }_n,+{\lambda }_n]$ and thus

$$|det(A)|\cdot \mathcal{L}(S)=2^n{\lambda }_1\cdots {\lambda }_n$$ To apply the refinement of Minkovski's theorem with the lattice

${\mathbb{Z}}^n$ and the symmetric convex set

$S$, we need that

$$\frac{\mathcal{L}(S)}{2^n}\ge 1$$ that is

$${\lambda }_1\cdots {\lambda }_n\ge |det(A)|.$$

Theorem 12.3 (Dirichlet's Unit Theorem)

Properly applying Minkovski's theorem

Minor discrepancies

Let

$n={\dim }_{\mathbb{Q}}K$. We recall (corollary of the primitive element theorem) that there are precisely

$n$ ring (field) homomorphisms

${\sigma }_i:K\hookrightarrow \mathbb{C}$. We assume that the

${\sigma }_1,\dots ,{\sigma }_s$ are real and that the remaining

$2t$ are complex with

${\sigma }_{s+t+j}=\overline{{\sigma }_{s+j}}$ for

$j=1,\dots ,t$.

Applying Minkovski's theorem requires

• an
  $n$-dimensional real vector space
  $V$,
• $n$ linear
  $\mathbb{R}$-linear forms
  $L_1,\dots ,L_n:V\to \mathbb{R}$,
• which have to be linearly independent.

This is not quite the setup we have here. For several reasons:

• $K$ is a rational vector space,
• the
  ${\sigma }_i$ are
  $\mathbb{Q}$-linear,
• and they are are complex valued.

Fixing these discrepancies is easy enough. I highlighted them merely to pin point precisely what made me uneasy.

Linear independence of the field embeddings

The most important point, and which is quite independent from the other points, is the linear independence of the

${\sigma }_i\in {\mathrm{Lin}}_{\mathbb{Q}}(K,\mathbb{C})$. This follows from the fact that

$⟨\alpha \mid \beta {⟩}_K:={\mathrm{Tr}}_{K/\mathbb{Q}}(\alpha \beta )$ defines a (symmetric) nondegenerate bilinear form on

$K$. Indeed,

${\mathrm{Tr}}_{K/\mathbb{Q}}(1)=n$ so that

$\mathrm{\forall }\alpha \in K\setminus 0$,

$⟨\alpha \mid {\alpha }^{-1}{⟩}_K=n\ne 0$.

Lemma. The bilinear pairing

$⟨\alpha \mid \beta {⟩}_K:={\mathrm{Tr}}_{K/\mathbb{Q}}(\alpha \beta )$ is nondegenerate.

Now if

${\theta }_{\bullet }=({\theta }_1,\dots ,{\theta }_n)$ is some

$\mathbb{Q}$-basis of

$K$, the matrix

$$S=({\sigma }_j({\theta }_i){)}_{1\le i,j\le n}$$ is invertible since

$$S\cdot S^t=\underset{{\theta }_{\bullet }}{\mathrm{Mat}}(⟨\cdot \mid \cdot {⟩}_K)\in {\mathrm{GL}}_n(\mathbb{Q}).$$ In conclusion,

Lemma. The field embeddings

${\sigma }_1,\dots ,{\sigma }_n:K\hookrightarrow \mathbb{C}$ are

$\mathbb{Q}$-linearly independent.

Realification

When applying Minkovski we will implicitely apply it to

$V=K{\otimes }_{\mathbb{Q}}\mathbb{R}$ and to the maps

$$\{\begin{array}{lclr}{L}_i& =& {\sigma }_i& (i=1,\dots ,s)\\ L_{s+2j-1}& =& \mathrm{Re}({\sigma }_{s+j})& (j=1,\dots ,t)\\ L_{s+2j}& =& \mathrm{Im}({\sigma }_{s+j})& (j=1,\dots ,t)\end{array}$$ (we use the same notation for the

$\mathbb{R}$-linear extensions of

$\mathbb{Q}$-linear maps). We also notice that simple line operations convert the matrix

$$(L_i({\theta }_j))\Rightarrow ({\sigma }_i({\theta }_j))$$ at the cost of multiplying the determinant by

$(-2i{)}^t$. Since the matrix on the right is invertible, we get that the matrix on the left is aswell. This is the "

$(a_{ij})$"-matrix in Theorem 12.2.

Covolume of ring of integers = discriminant of number field

In this subsection we investigate why when applying Minkovski's theorem in Dirichlet's Unit Theorem, one requires

${\lambda }_1\cdots {\lambda }_n=\sqrt{\mathrm{disc}(K)}$ rather than

${\lambda }_1\cdots {\lambda }_n=\mathrm{covol}({\mathcal{O}}_K)$. The answer is, unsurprisingly, that

$\mathrm{covol}({\mathcal{O}}_K)=\sqrt{\mathrm{disc}(K)}$ relative to a natural volume form associated to the nondegenerate bilinear form

$⟨\cdot \mid \cdot {⟩}_K$.

Proposition. Let

$Q$ be a nondegenerate quadratic form on

$V$, a finite dimensional real vector space. Then there exists a unique translation invariant measure

${\mathcal{L}}_Q$ on

$V$ such that for any

$Q$-orthonormal basis

$${\mathcal{L}}_Q(\sum _{i=1}^n[0,1]e_i)=1.$$

The is easy:

${\mathcal{L}}_Q$ is simply the Lebesgue measure on

$V$ identified with

${\mathbb{R}}^n$ by choosing a

$Q$-orthonormal basis. This is independent on the choice of

$Q$-orthonormal basis. Indeed, for any base change matrix

$P=\underset{\mathcal{B}}{\mathrm{Mat}}(\mathcal{C})$,

$$\underset{\mathcal{C}}{\mathrm{Mat}}(Q))={}^{t}P\underset{\mathcal{B}}{\mathrm{Mat}}(Q)P$$ Therefore, if

$\mathcal{B}$ and

$\mathcal{C}$ are both

$Q$-orthonormal bases,

$|det(P)|=1$, and

${\mathcal{L}}_Q$ is well-defined.

Proposition. Let

$\mathrm{\Lambda }$ be a (full rank, discrete) lattice in

$V$; let

${\theta }_{\bullet }=({\theta }_1,\dots ,{\theta }_n)$ be a

$\mathbb{Z}$-basis of

$\mathrm{\Lambda }$. Then setting

${\mathrm{covol}}_Q(\mathrm{\Lambda }):={\mathcal{L}}_Q(\sum _{i=1}^n[0,1]{\theta }_i)$, we have

$$|det(\underset{{\theta }_{\bullet }}{\mathrm{Mat}}(Q))|={\mathrm{covol}}_Q(\mathrm{\Lambda }{)}^2$$

Let

${\theta }_{\bullet }$ be as above and let

$e_{\bullet }=(e_1,\dots ,e_n)$ be a

$Q$-orhtonormal basis of

$V$. We have

$$\begin{array}{rcl}\underset{{\theta }_{\bullet }}{\mathrm{Mat}}(Q))& =& {}^{t}P\underset{e_{\bullet }}{\mathrm{Mat}}(Q)P\\ & =& {}^{t}P\mathrm{Diag}(\pm 1)P\end{array}$$ where

$P=\underset{e_{\bullet }}{\mathrm{Mat}}({\theta }_{\bullet })$ is the matrix of the vectors of

${\theta }_{\bullet }$ relative to the basis

$e_{\bullet }$. Also, if we define

$g\in \mathrm{GL}(V)$ by

$g(e_i)={\theta }_i$, then

$$g(\sum _{i=1}^n[0,1]e_i)=\sum _{i=1}^n[0,1]{\theta }_i$$ and so

${\mathrm{covol}}_Q(\mathrm{\Lambda })=|det(g)|$. Since

$\underset{e_{\bullet }}{\mathrm{Mat}}(g)=P$ this imposes

$${\mathrm{covol}}_Q(\mathrm{\Lambda })=|det(P)|.$$

Corollary. If

$K$ is a number field,

${\mathrm{covol}}_Q({\mathcal{O}}_K)=\sqrt{|\mathrm{disc}(K)|}$.

The real vector space

$V=K{\otimes }_{\mathbb{Q}}\mathbb{R}$ inherits the nondegenerate bilinear form

$⟨\cdot \mid \cdot {⟩}_K$ from

$K$. The statement above is w.r.t. the associated volume form. Recall that the discriminant of

$K$ is defined as the determinant

$$\begin{array}{rcl}\mathrm{disc}(K)& =& det({\mathrm{Tr}}_{K/\mathbb{Q}}({\theta }_i{\theta }_j))\\ & =& det(\underset{{\theta }_{\bullet }}{\mathrm{Mat}}(Q))\\ & =& det(P{)}^2\end{array}$$ of any integral basis

${\theta }_{\bullet }=({\theta }_1,\dots ,{\theta }_n)$ of

${\mathcal{O}}_K$. That is,

$$\mathrm{disc}(K)={\mathrm{covol}}_Q(\mathrm{\Lambda }{)}^2.$$

Kernel of the Log map

By construction, the kernel of the Log map

$${\mathrm{Log}}_K:\{\begin{array}{ccl}{\mathcal{O}}_K\setminus 0& ⟶& {\mathbb{R}}^r\\ x& ⟼& (\ln |{\sigma }_1(x)|,\dots ,\ln |{\sigma }_s(x)|,\ln |{\sigma }_{s+1}(x)|,\dots ,\ln |{\sigma }_r(x)|)\end{array}$$ where

$r=s+t-1$. Its kernel is comprised of those algebraic integers of

$K$ all of whose "conjugates" lie on the unit circle.

Lemma. Let

$\xi \in {\mathcal{O}}_K$ have all its conjugates of modulus

$1$, then

$\xi$ is a root of unity, i.e. the kernel of

$K$'s Log map is the group of roots of unity in

$K$.

Clearly, every root of unity in

$K$ is an algebraic integer and lies in the kernel of the Log map:

$${\mu }_K\subset \ker ({\mathrm{Log}}_K)$$ (

${\mu }_K$ is the group of all roots of unity in

$K$). Conversely, if

$x\in {\mathcal{O}}_K$ has all its conjugates of modulus

$1$, then its minimal polynomial lies in a finite set of integer polynomials by Kummer's argument. Therefore

$\ker ({\mathrm{Log}}_K)$ is a finite subgroup of

$\mathcal{U}(K)$. If

$\xi$ is in the kernel, then so are all its powers, and thus

$\xi$ is a root of unity. If

$N=\mathrm{\#}\ker ({\mathrm{Log}}_K)$, then if follows that

$\ker ({\mathrm{Log}}_K)\subset {\mu }_N$ is a subgroup of the group of

$N$-th roots of unity. Equality follows from cardinality.

Note. The lemma is wrong if one changes "

$\xi \in {\mathcal{O}}_K$" to "

$\xi \in K$": the point

$\frac{3}{5}+\frac{4}{5}i$ and its (sole) conjugate

$\frac{3}{5}-\frac{4}{5}i$ have both modulus

$1$ yet aren't roots of unity: if they were they would have to be integers in

$\mathbb{Q}(i)=\mathbb{Q}(\sqrt{-1})$ which they are not since their coordinates aren't integers (note

$-1\equiv 3[4]$).

Note. The Log map clearly makes sense as a map

$K^{\times }\to {\mathbb{R}}^r$ where its defines a group homomorphism. For this note we only care about it along the subset

${\mathcal{O}}_K\setminus 0$.

How and where "r = s + t - 1" is used in the proof

"r = s + t - 1" is enough

This is simply the observation that for any unit

$u\in {\mathcal{U}}_K$,

$${\displaystyle 1=|N_{K/\mathbb{Q}}(u)|=|\prod _{\sigma \hookrightarrow \mathbb{C}}\sigma (u)|=\prod _{i=1}^s|{\sigma }_i(u)|\cdot \prod _{j=1}^t|{\sigma }_{s+j}(u){|}^2}$$ so that

$${\displaystyle 0=\sum _{i=1}^s\ln |{\sigma }_i(u)|+2\sum _{j=1}^t\ln |{\sigma }_{s+j}(u)|}$$ In other words, the more natural "full Log map"

${\mathrm{Log}}_K^{\prime }(u)=(\ln |{\sigma }_i(u)|{)}_{1\le i\le s+t}$ has its image in the hyperplane

$\sum _{1\le i\le s}{x}_i+2\sum _{s<j\le s+t}{x}_j=0$. The "missing modulus"

$\ln |{\sigma }_{r+1}(u)|=\ln |{\sigma }_{s+t}(u)|$ is redundant information.

How the precise value "r = s + t - 1" is used

The beginning of the proof works without modification for "shorter Log maps" (i.e. by including fewer than

$r$ of the

${\sigma }_i$). The fact that

$r=s+t-1$ isn't used until one invokes the proposition below.

Let

${\eta }_1,\dots ,{\eta }_r\in {\mathcal{U}}_K$ be the family of units constructed in Dirichlet's Unit Theorem (DUT). By construction, the vectors

${\mathrm{Log}}_K({\eta }_i)$,

$i=1,\dots ,r$, are linearly independent in

${\mathbb{R}}^r$, and

$⟨{\eta }_1,\dots ,{\eta }_r⟩$^[1] is a free abelian group of rank

$r=s+t-1$ within

${\mathcal{U}}_K$.

Proposition. The quotient group

${\mathcal{U}}_K/⟨{\eta }_1,\dots ,{\eta }_r⟩$ is finite.

Proof. This is the proposition that is used in the proof of the DUT. Its proof is basically that of corollary 12.1. Let

$u\in {\mathcal{U}}_K$:

• $u$ is equivalent (up to an element in
  $⟨{\eta }_1,\dots ,{\eta }_r⟩$) to a unit
  $v$ such that for all
  $\sigma :K\hookrightarrow \mathbb{C}$,
  $|\sigma (v)|\le C$ for some absolute constant
  $C>0$;
• $v$'s minimal polynomial thus belongs to a finite set of integer polynomialls;
• thus
  $v$ belongs to the finite set of the
  $K$-roots of these polynomials,
• and thus
  ${\mathcal{U}}_K/⟨{\eta }_1,\dots ,{\eta }_r⟩$ is finite.

The unit

$v$ is constructed as in Baker's proof by taking a nearby vector

$\mathrm{Log}({\eta }_1^{m_1}\cdots {\eta }_r^{m_r})$ to

$\mathrm{Log}(u)$ in the (discrete, full rank) lattice

$\mathrm{Log}(⟨{\eta }_1,\dots ,{\eta }_r⟩)\subset {\mathbb{R}}^r$ and setting

$$v=u\cdot {\eta }_1^{-m_1}\cdots {\eta }_r^{-m_r}.$$ Here "nearby" means w.r.t. the

$\Vert \cdot {\Vert }_{\mathrm{\infty }}$-norm on

${\mathbb{R}}^r$. We are not necessarily interested in the closest lattice point, merely in the fact that there is a uniform error bound

$\delta >0$ on the coordinates

$|\ln |{\sigma }_i(v)||$. This is clear, for by construction, if

$v$ was constructed by

1. first: expressing
   $\mathrm{Log}(u)$ in the
   $\mathbb{R}$-basis
   $(\mathrm{Log}({\eta }_1),\dots ,\mathrm{Log}({\eta }_r))$ as
   $$\mathrm{Log}(u)=\sum _{i=1}^r{u}_i\mathrm{Log}({\eta }_i)$$
2. defining
   $m_i=\lfloor u_i\rfloor$
3. setting
   $v=u\cdot {\eta }_1^{-m_1}\cdots {\eta }_r^{-m_r}$,

then for

$i=1,\dots ,r$:

$$\begin{array}{rcl}|\ln |{\sigma }_i(v)||& \le & r\cdot \underset{j=1,\dots ,r}{max}|\ln |{\sigma }_i({\eta }_j)||\\ & \le & \underset{=\delta }{\underbrace{r\cdot \underset{i=1,\dots ,r}{max}\underset{j=1,\dots ,r}{max}|\ln |{\sigma }_i({\eta }_j)||}}\end{array}$$

The only noteworthy point is how one deduces the boundedness of the length of the "single missing conjugate"

$|{\sigma }_{r+1}(v)|=|{\sigma }_{s+t}(v)|$. We get it from the previously made observation:

$$0=\sum _{i=1}^s\ln |{\sigma }_i(v)|+2\sum _{j=1}^t\ln |{\sigma }_{s+j}(v)|\text{thus}|\ln |{\sigma }_{s+t}(v)||\le \underset{\le r\delta }{\underbrace{\sum _{i=1}^r|\ln |{\sigma }_i(v)||}}$$ (not bothering with the

$\frac{1}{2}$ factors one would expect on the first

$s$ terms allows us not to have to differentiate between the case

$t>0$ and

$t=0$.) This (trivial) computation is precisely where we use the fact that we used

$r=s+t-1$: had we chosen a value lower, we would not have been able to untangle the remaining moduli. Taking

$r=s+t-1$ allows us to have a single modulus to dominate.

The remaining points are easy consequences of the boundedness of the conjugates.

The conclusion of the proof

Once Baker knows that the quotient is finite, the proof is nearly done. We put

$N=\mathrm{\#}({\mathcal{U}}_K/⟨{\eta }_1,\dots ,{\eta }_r⟩)$. Then

${\mathcal{U}}_K^N\subset ⟨{\eta }_1,\dots ,{\eta }_r⟩$^[2], where we set

${\mathcal{U}}_K^N=\{{ϵ}^N\mid ϵ\in {\mathcal{U}}_K\}$. We note that

${\mathrm{Log}}_K({\mathcal{U}}_K^N)$ includes the full rank sugroup

$$⟨{\mathrm{Log}}_K({\eta }_1^N),\dots ,{\mathrm{Log}}_K({\eta }_r^N)⟩=N⟨{\mathrm{Log}}_K({\eta }_1),\dots ,{\mathrm{Log}}_K({\eta }_r)⟩$$ of

$⟨{\mathrm{Log}}_K({\eta }_1),\dots ,{\mathrm{Log}}_K({\eta }_r)⟩$, hence has full rank itself. Using Lemma 11.3, there exists a basis

$({\mathrm{Log}}_K({ϵ}_1^N),\dots ,{\mathrm{Log}}_K({ϵ}_r^N))$ of

${\mathrm{Log}}_K({\mathcal{U}}_K^N)$ such that its matrix w.r.t. the basis

$({\mathrm{Log}}_K({\eta }_1),\dots ,{\mathrm{Log}}_K({\eta }_r))$ is upper triangular. I don't believe the "upper triangular" property is important here, though.

If

$u\in {\mathcal{U}}_K$ is a unit, then there exist

$j_1,\dots ,j_r\in \mathbb{Z}$ such that

$${\mathrm{Log}}_K(u^N)=\sum _{k=1}^r{j}_k\cdot {\mathrm{Log}}_K({ϵ}_k^N)$$ therefore

$$u\cdot {ϵ}_1^{-j_r}\cdots {ϵ}_r^{-j_r}\equiv \text{some root of unity}$$ (indeed, raising the LHS to the

$N$-th power lands it in the kernel of the Log map; the kernel of the Log map is the set of roots of unity in

$K$; the LHS is thus a root of unity in

$K$) i.e.

$$u=\rho \cdot {ϵ}_1^{j_r}\cdots {ϵ}_r^{j_r}$$ for some

$N$-th root of unity

$\rho \in {\mathcal{U}}_K$.

Wrapping up

In conclusion, we get a split short exact sequence since its final term is free abelian:

$$0\to {\mu }_K\hookrightarrow {\mathcal{U}}_K\twoheadrightarrow \underset{{\displaystyle \simeq {\mathbb{Z}}^r}}{\underbrace{\mathrm{Log}({\mathcal{U}}_K)}}\to 0$$ and so

${\mathcal{U}}_K\simeq {\mu }_K\times {\mathbb{Z}}^r$.

---

1. We use here the notation

   $⟨g_1,\dots ,g_r⟩$ for
   $g_1,\dots ,g_r\in G$ to denote the subgroup of a group
   $G$ generated by the
   $g_i$. This should not be confused with the ideal
   $⟨{\eta }_1,\dots ,{\eta }_r⟩\subset {\mathcal{O}}_K$ generated by the
   ${\eta }_i$. Since ↩︎

2. In this subsection we write

   $⟨v_1,\dots ,v_n⟩$ for the subgroup of some group
   $\mathrm{\Lambda }$ generated by elements
   $v_1,\dots ,v_n\in \mathrm{\Lambda }$. Here
   $\mathrm{\Lambda }=\mathrm{Log}({\mathcal{U}}_K)\subset {\mathbb{R}}^r$ is actually a lattice. There should be no possible confusion with the analogous notation for ideals. ↩︎