Find a number field whose unit group is isomorphic to $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}.$
I'm trying to use Dirichlet's Unit Theorem to solve this problem. It states that if $K$ is a number field of signature $(r,s)$ and $\mu_K$ is the set of roots of unity in $K$, then the unit group $\mathcal{O}_K^{\times}$ of the ring of integers is isomorphic to $\mu_K \times \mathbb{Z}^{r+s−1}$ as an abelian group. So I suppose I want $r+s-1=1,$ or $r+s=2$. This forces $(r,s)=(0,2)$ because if there's at least one real embedding then $\mu_K$ is just $\{\pm 1\}$ so not $\mathbb{Z}/4\mathbb{Z}$.
Therefore I need a number field of degree $4$ with four complex embeddings and whose set of roots of unity is isomorphic to $\mathbb{Z}/4\mathbb{Z}$. The cyclotomic field $\mathbb{Q}(\zeta_5)$ doesn't work because it has more than $4$ elements in its set of roots of unity (and $\mathbb{Q}(\zeta_4)$ doesn't work because it doesn't have degree $4$). I suppose it will have to be $\mathbb{Q}(\alpha)$ where the minimal polynomial of $\alpha$ has degree $4$ but I haven't been able to find an example. Hints would be appreciated.
Best Answer
You are almost there! Since $K$ has a torsion element of order $4$, it contains $\zeta_4$ and thus contains $\mathbf{Q}(\zeta_4)$. Dirichlet's unit theorem then says that $K$ must be degree $4$ and thus a quadratic extension of $\mathbf{Q}(\zeta_4)$.
Now suppose that $K$ is any quadratic extension of $\mathbf{Q}(\zeta_4) = \mathbf{Q}(\sqrt{-1})$. The signature of $K$ is $(0,2)$ and so $K$ has unit rank one. Also, $K$ has an element $\zeta_4$ of order $4$. The only thing that remains is to find a $K$ that doesn't have any extra torsion. But the torsion subgroup of a number field is always cyclic and generated by a $m$th root of unity, or a $4n$th root of unity in this case since we already have a $4$th root of unity. So you just have to ensure that
$$\mathbf{Q}(\zeta_{4n}) \not\subset K$$
for any $n > 1$. The degree of $\mathbf{Q}(\zeta_{4n})$ is (Euler's $\varphi$ function) $\varphi(4n)$. This is $> 4$ for $n \ge 4$. So the answer is:
$K$ can be any quadratic extension of $\mathbf{Q}(\zeta_4)$ which doesn't equal $\mathbf{Q}(\zeta_8)$ or $\mathbf{Q}(\zeta_{12})$.
Since $\mathbf{Q}(\zeta_8) = \mathbf{Q}(\sqrt{-1},\sqrt{2}) = \mathbf{Q}(\sqrt{-1},\sqrt{-2})$ and $\mathbf{Q}(\zeta_{12}) = \mathbf{Q}(\sqrt{-1},\sqrt{-3}) = \mathbf{Q}(\sqrt{-1},\sqrt{2})$, you can find many such $K$, for example $K = \mathbf{Q}(\sqrt{-1},\sqrt{d})$ for any squarefree $\pm d > 3$. These are not the only examples --- the others are precisely all the quadratic extensions $K/\mathbf{Q}(\sqrt{-1})$ which are non-Galois over $\mathbf{Q}$ such as $\mathbf{Q}(i,\sqrt{3 + 4 i})$.