Let $\mathcal{O}$ be a number ring. Letting $r$ and $2s$ be the number of real and complex embeddings of $\mathcal{O}$, the number ring $\mathcal{O}$ is a lattice in $\mathcal{O} \otimes \mathbb{R} \cong \mathbb{R}^{r+2s}$, and thus is Zariski dense in this vector space.


Question: What is the Zariski closure of the group $\mathcal{O}^{\times}$ in $\mathcal{O} \otimes \mathbb{R} \cong \mathbb{R}^{r+2s}$? There is one obvious polynomial condition coming from the fact that the norm is $\pm 1$. Are there others?



Zariski closure of set of units in a number ring
Let $x\_1,\dots, x\_n$ be the embeddings of $\mathcal O$ into $\overline{\mathbb Q}$. Every monomial in $x\_1,\dots, x\_n$ restricts to a character of $\mathcal O^\times$. If some polynomial in $x\_1,\dots, x\_n$ vanishes on units, then we get a linear relation among characters of $\mathcal O^\times$, unless two of the monomials in $x\_1,\dots, x\_n$ appearing in the polynomial give the same character of $\mathcal O^\times$. Thus all relations are generated by the relations of the form $\prod\_{i=1}^n x\_i^{d\_i} - \prod\_{i=1}^n x\_i^{e\_i}$ where these two monomials give the same character or equivalently by $1- \prod\_{i=1}^n x\_i^{e\_i}$ for $e\_i \in \mathbb Z$ where this monomial gives the trivial character.


Each complex conjugation element of the Galois group defines a permutation of the set $\{1,\dots,n\}$ of embeddings. If, for all units, we have $$ 1= \prod\_{i=1}^n x\_i^{e\_i}$$ then we have $$ 1 = \prod\_{i=1}^n x\_i^{e\_i} \overline{ \prod\_{i=1}^n x\_i^{e\_i} } = \prod\_{i=1}^n |x\_i|^{ 2e\_i } .$$ By Dirichlet's unit theorem, this can only happen if all the exponents on each real place are equal, and all the exponents on each complex place are double that. In other words, for each real place $i$ we must have $e\_i=n$ for some constant $n$, and for each complex place $i$ we must have $e\_i+ e\_{\sigma{i}}=2n$, so altogether we must have $e\_i + e\_{\sigma(i)}=2n$ for all $i$. 


This can only be no constant if the graph whose edges are $(i, \sigma(i))$ for all complex conjugations $\sigma$ has a bipartite connected component. In this case, because the graph is Galois-invariant, every component is bipartite, and because each connected graph is bipartite in at most one way, the graph of this set of functions is the number of components plus one.


We have to convert this information back into Galois theory, and then number theory. Because each vertex of the graph defines a part of its connected component, the stabilizer of a part contains the vertex stabilizer, hence corresponds to a subfield. Because each complex conjugation sends one part to the other part of the same graph, all complex conjugations in the Galois group of this field are equal, so it is a CM field. Moreover, the number of complex places of this field is the number of components (orbit-stabilizer theorem).


So we can see that the number of relations is at most the number of complex places of the maximum CM subfield, plus one. Is this exactly equal to the number of relations? Yes, because in such a CM field $K$ with totally real subfield $L$, a finite index subgroup of the unit group is contained in $L$, and so we have the one relation from $L$ plus $[L: \mathbb Q]$. Then we can pull these relations back to any extension of $K$ by the norm map. 


Thus, for $\mathcal O$ the ring of integers of a number field $F$, $K$ the maximal CM subfield, $L$ its maximal totally real subfield, the Zariski closure of the units is the inverse image under $N\_{F}^K$ of the Zariski closure of the units of $K$, which is a finite union of cosets of the norm one torus of $L$.


