# On Dirichlet’s Unit Theorem and the Class Group

So today, I’m going to write about something I’ve learnt that I think is really cool. Unsurprisingly, it’s related to the Class Group; I will get back to my set of posts on Baker’s Theorem and the Class Number Problem, think of this as a nice excursion. I might go quite fast over the basics of the Class Group, I refer to previous posts (and Wikipedia) to get the necessary background. Throughout, $K$ is a number field with ring of integers $\mathcal{O}_{K}$. I refer heavily to Neukirch’s “Algebraic Number Theory” in this post, a current favourite text of mine.

There are two major things that I think should be in a first course on Algebraic Number Theory; the Class Group (specifically, the finiteness of the Class Group), and Dirichlet’s Unit Theorem, both of which I state below for reference.

Theorem 1 Let $K$ be an algebraic number field. The Class Group of $K$ is finite.
Theorem 2: Dirichlet’s Unit Theorem Let $K$ be an algebraic number field with ring of integers $\mathcal{O}_{K}$. Let $r$ be the number of pairs of complex embeddings of $K$, and $s$ the number of real embeddings. The group of units $\mathcal{O}_{K}^{*}$ is the direct product of the finite cyclic group $\mu(K)$ (the group of roots of unity contained in $K$) and a free abelian group of rank $r+s-1$.

My aim for this post is to connect these two seemingly unrelated results, ending with a statement equivalent to both these theorems. I assume familiarity with a classic proof of both, especially of Dirichlet’s Unit Theorem. To get there however, we’ll have to generalise our ideas of fractional ideals.

As a reminder, a fractional ideal is a finitely generated $\mathcal{O}_{K}$-submodule $\mathfrak{a} \neq 0$ of $K$. I’ve discussed these in previous posts, so won’t spend too much time on them here. When we consider the set of places on $K$ (that is, the absolute values on $K$ up to equivalence), in a way these fractional ideals only take into account the finite places (after finishing my set of posts on Baker’s Theorem, I’ll write about places in some depth). Finite places correspond to prime ideals, hence we can see a link. We can extend this notion of fractional ideals then, by taking into consideration the infinite places/primes.

Define a replete ideal of $K$ to be an element of the group $J(\mathcal{\bar O}) := J(\mathcal{O}) \times \prod_{\mathfrak{p} \vert \infty}\mathbb{R}_{+}^{*}$, where $J(\mathcal{O})$ is the group of fractional ideals of $K$ and $\mathbb{R}_{+}^{*}$ denotes the multiplicative group of positive real integers.

I will unify this idea with the standard notation and put, for any infinite prime $\mathfrak{p}$ and any real number $\upsilon$, $\mathfrak{p}^{\upsilon}:= e^{\upsilon}$. That is, I follow the convention where we treat $e$ as an infinite prime number.

Further, given a set of real numbers $\upsilon_{\mathfrak{p}}$ with $\mathfrak{p} \vert \infty$ (that is, $\mathfrak{p}$ is an infinite prime), I choose to let $\prod_{\mathfrak{p} \vert \infty} \mathfrak{p}^{\upsilon_{\mathfrak{p}}}$ denote the vector $(\dots , e^{\upsilon_{\mathfrak{p}}}, ...)$ rather than the product of the quantities. This allows every replete ideal $\mathfrak{a} \in J(\mathcal{\overline O})$ to admit a unique product representation $\mathfrak{a} = \prod_{\mathfrak{p} \nmid \infty} \mathfrak{p}^{\upsilon_{\mathfrak{p}}} \times \prod_{\mathfrak{p} \vert \infty} \mathfrak{p}^{\upsilon_{\mathfrak{p}}}=\prod_{\mathfrak{p}}\mathfrak{p}^{\upsilon_{\mathfrak{p}}}$. We note that $\upsilon_{\mathfrak{p}} \in \mathbb{Z}$ when $\mathfrak{p}$ is a finite prime, and is a real number when infinite.

Important takeaways from here so we don’t get bogged down in theory; all fractional ideals are replete ideals and we can “factorise” replete ideals into a product over finite places times a product over infinite places. To $a \in K^{*}$ we associate the replete principal ideal $[a] =\prod_{\mathfrak{p}} \mathfrak{p}^{\upsilon_{\mathfrak{p}}(a)} = (a) \times \prod_{\mathfrak{p} \mid \infty} \mathfrak{p}^{\upsilon_{\mathfrak{p}}(a)}$, where $(a)$ is the principal fractional ideal.

From here, think back to the definition of the class group, we can see what we’re going to do. The replete principal ideals $P(\mathcal{\overline O})$ form a normal subgroup of the replete ideals $J(\mathcal{\overline O})$, so we define the replete ideal class group (or replete Picard group) to be $Pic(\mathcal{ \overline O}) = J(\mathcal{\overline O}) /P(\mathcal{\overline O})$.

These replete ideals behave essentially as we would want them to, we can define the absolute norm of a replete ideal which is multiplicative, we have some nice results if we consider extensions… I’m going to briefly discuss norms as we’ll shortly need them. Let $f_{\mathfrak{p}}$ be the residue class degree of $\mathfrak{p}$. We set $\mathfrak{N} (\mathfrak{p}) = p^{f_{\mathfrak{p}}}$ if $\mathfrak{p}$ is finite (where $\mathfrak{p}$ lies above $p$), and $\mathfrak{N}(\mathfrak{p}) = e^{f_{\mathfrak{p}}}$ if $\mathfrak{p}$ is infinite. This norm is multiplicative, so we can find the norm of any replete ideal $\mathfrak{a}$. The extension stuff is less relevant here, so I’ll leave it for another day.

We’re now going to consider divisors. I’ll define these straight away for replete divisors, but the analogous definition holds for fractional ideals as one would expect.

A replete (or Arakelov) divisor of $K$ is a formal sum $D=\sum_{\mathfrak{p}} \upsilon_{\mathfrak{p}} \mathfrak{p}$, where $\upsilon_{\mathfrak{p}}$ is an integer for the finite primes, a real number for the infinite primes and 0 for almost all $\mathfrak{p}$. These divisors form a group, which we’ll denote by $Div(\mathcal{\overline O})$. I’ll leave it as a comment that this is a locally compact topological group, if you want details comment.

We not consider the map $div : K^{*} \rightarrow Div(\mathcal{\overline O})$, given by $div(f) = \sum_{\mathfrak{p}} \upsilon_{\mathfrak{p}}(f) \mathfrak{p}$. We call elements of this form $div(f)$ replete principal divisors.

We’re now in a position to start to see how this relates to Dirichlet’s Unit Theorem. Consider the map $Div(\mathcal{\overline O}) \rightarrow \prod_{\mathfrak{p} \mid \infty} \mathbb{R}$, given by $\sum_{\mathfrak{p}} \upsilon_{\mathfrak{p}} \mathfrak{p} \mapsto (\upsilon_{\mathfrak{p}} f_{\mathfrak{p}})_{\mathfrak{p} \mid \infty}$, where for infinite primes, $f_{\mathfrak{p}} = [K_{\mathfrak{p}}:\mathbb{R}]$, with $K_{\mathfrak{p}}$ being the closure of $K$ with respect to $\mathfrak{p}$.

This is where I’m going to need familiarity with the “normal “, “standard” proof of Dirichlet’s Unit Theorem. The composite of the mapping above with the mapping $div: K^{*} \rightarrow Div(\mathcal{\overline O})$ is equal, up to sign, to the logarithm map of Minkowski theory, $\lambda : K^{*} \rightarrow \prod_{\mathfrak{p} \mid \infty} \mathbb{R}, \, \lambda (f) = (\dots , \log \lvert f \rvert_{\mathfrak{p}}, \dots)$. This maps the unit group $\mathcal{O}^{*}$ onto a complete lattice $\Gamma = \lambda (\mathcal{O}^{*})$ in the trace-zero space $H= \{ (x_{\mathfrak{p}}) \in \prod_{\mathfrak{p} \vert \infty} \mathbb{R} : \sum_{\mathfrak{p} \mid \infty} x_{\mathfrak{p}} = 0 \}$.

From here, we could prove that the kernel of $div$ is the group $\mu (K)$ of roots of unity of $K$, and that the image $P(\mathcal{O})$ is a discrete subgroup of $Div(\mathcal{\overline O})$, but I’ll leave that for another day (it’s all about the exact sequence $1 \rightarrow \mu (K) \rightarrow \mathcal{O^{*}} \rightarrow \Gamma \rightarrow 0$, where the second to last arrow is $\lambda$).

We now define $CH^{1}(\mathcal{\overline O}) = Div(\mathcal{\overline O})/P (\mathcal{\overline O})$ to be the replete divisor (or Arakelov) class group of $K$. We note we can do the exact same thing with fractional ideals, coming up with the divisor (or Chow) class group $CH^{1}(\mathcal{O})$, where we use the exact same construction, but with fractional rather than replete ideals. We’re almost at the point we wanted to get to, hang in there!

As $P (\mathcal{\overline O})$ is discrete in $Div(\mathcal{\overline O})$, it is closed, so $CH^{1}(\mathcal{\overline O})$ is a locally compact Hausdorff topological group with respect to the quotient topology. We consider a degree map from $CH^{1}(\mathcal{\overline O})$ onto the reals, induced by the continuous homomorphism $deg: Div(\mathcal{\overline O}) \rightarrow \mathbb{R}$. We define the by sending the replete divisor $D = \sum_{\mathfrak{p}} \upsilon_{\mathfrak{p}} \mathfrak{p}$ to the real number $\sum_{\mathfrak{p}} \upsilon_{\mathfrak{p}} \log \mathfrak{N}(\mathfrak{p}) = \log (\prod_{\mathfrak{p}} \mathfrak{N}(\mathfrak{p})^{\upsilon_{\mathfrak{p}}})$. By the product formula for places, we find that for any replete principal divisor $div(f) \in P(\mathcal{\overline O}), \, deg(div(f))=0$. This shows that $deg$ is a well-defined continuous homomorphism.

We are now at the point where we can finish and connect all the things we’ve been talking about. The kernel $CH^{1}(\mathcal{\overline O})^{0}$ of the degree map is made up from the unit group $\mathcal{O^{*}}$ and the ideal class group $Cl(K) \cong CH^{1}(\mathcal{O})$, as can be seen from the following proposition.

Proposition Let $\Gamma = \lambda(\mathcal{O}^{*})$ denote the complete lattice of units in trace-zero space $H= \{ (x_{\mathfrak{p}}) \in \prod_{\mathfrak{p} \vert \infty} \mathbb{R} : \sum_{\mathfrak{p} \mid \infty} x_{\mathfrak{p}} = 0 \}$. There is an exact sequence $0 \rightarrow H/\Gamma \rightarrow CH^{1}(\mathcal{\overline O})^{0} \rightarrow CH^{1}(\mathcal{O}) \rightarrow 0.$

The proof of this isn’t too hard, and isn’t too instructive, so I omit it. From this point, the finiteness of the class number and Dirichlet’s Unit Theorem merge into (and are equivalent to) the fact that $CH^{1}(\mathcal{\overline O})^{0}$ is compact.

Theorem The group $CH^{1}(\mathcal{\overline O})^{0}$ is compact.

To see this we consider the exact sequence from the previous proposition. As $\Gamma$ is a complete lattice in $H$, $H/ \Gamma$ is a compact torus, of dimension $r+s-1$, with $r, \, s$ defined as above (this is where we get Dirichlet’s Unit Theorem from!). We obtain $CH^{1}(\mathcal{ \overline O})^{0}$ as the union of finitely many (indeed, $h_{K}$ many) compact cosets of $CH^{1}(\mathcal{\overline O})^{0}$, giving us that $CH^{1}(\mathcal{ \overline O})^{0}$ is itself compact, and that $CH^{1}(\mathcal{O}) \cong Cl_{K}$ is finite.