# Proving unit of quartic number field is fundamental

Let $K = \mathbb Q(\alpha)$, for $\alpha$ a root of $a^4 + 4 \alpha^2 + 2 = 0$. I want to prove the group of units $\mathcal O_K^*$ equals $\langle -1, \alpha^2 + 1\rangle$.

I've found the ring of integers $\mathcal O_K$ is $\mathbb Z[\alpha]$, and that the number field has a trivial class group. Dirichlet's Unit Theorem tells me that since the number of pairs of complex embeddings is 2, I have that the group of units is of rank 1, i.e. $\mathcal O_K^* = \langle -1, \varepsilon \rangle$.

I suspect that the torsion group $\mu(K)$ equals $\langle -1 \rangle$ because we can embed $\alpha \mapsto \sqrt{-2 + \sqrt{2}} \in \mathbb C$, which has absolute value greater than 1, but I'm not sure whether that reasoning is correct.

One result that might be helpful is that I can prove that for any unit $\nu$, I have that $\nu^2$ is of the form $a + b \alpha^2$ for integers $a, b \in \mathbb Z$, but I tried exploring that and it didn't really lead anywhere.

First of all: your reasoning about the torsion is not correct: I can embed $$\mathbb{Q}(\sqrt{-3})$$ into $$\mathbb{C}$$ in the obvious way but it has torsion subgroup $$\mathbb{Z}/6\mathbb{Z}$$ altough $$\sqrt{-3}\in\mathbb{C}$$ has absolute value greater than $$1$$.
It is still true however that $$\mu(K)=\langle -1\rangle$$. Note that if $$\zeta_n\in K$$ for some integer $$n>1$$ then $$\mathbb{Q}(\zeta_n)\subset K$$, so by determining the subfields of $$K$$ you can determine the torsion. You can check that $$K/\mathbb{Q}$$ is Galois with cyclic group, so there is a unique quadratic subfield, which is $$\mathbb{Q}(\sqrt{2})$$. This is not cyclotomic, and $$K$$ itself also isn't ($$K$$ is not $$\mathbb{Q}(\zeta_5)$$ as $$5$$ is unramified in $$K$$, and not $$\mathbb{Q}(\zeta_8)$$ as it is not cyclic). Thus the only cyclotomic numberfield inside $$K$$ is $$\mathbb{Q}=\mathbb{Q}(\zeta_2)$$, hence $$\mu(K)=\langle -1\rangle$$.
Now some notation: let $$E=\mathbb{Q}(\sqrt{2})$$ be the quadratic subfield. You want to prove that $$\mathcal{O}_K^{*}=\mathcal{O}_E^{*}$$. You claim that $$\mathcal{O}_K^{*2}\subset\mathcal{O}_E^{*}$$, and I'd like to know how you proved that, because with that I can give a proof:
We have $$\mathcal{O}_K^{*2}\subset\mathcal{O}_E^{*}\subset\mathcal{O}_K^{*}$$ where $$[\mathcal{O}_K^{*}:\mathcal{O}_K^{*2}]=4$$. Thus $$\mathcal{O}_K^{*}=\mathcal{O}_E^{*}$$ follows by proving that $$\mathcal{O}_E^{*}/\mathcal{O}_K^{*2}$$ is a group of order $$4$$. Now $$-1\in\mathcal{O}_E^{*}$$ is not a square in $$\mathcal{O}_K^{*}$$ as we have seen that $$\mathbb{Q}(\zeta_4)\not\subset K$$, so $$-1$$ is a non-trivial element of $$\mathcal{O}_E^{*}/\mathcal{O}_K^{*2}$$. To see that $$\varepsilon=1+\alpha^2$$ is another non-trivial element, we need to show that $$\varepsilon$$ and $$-\varepsilon$$ are not a square in $$\mathcal{O}_K^{*}$$.
This is most easily done by finding suitable primes $$\mathfrak{p}$$ in $$\mathcal{O}_K$$ (one for $$\varepsilon$$ and one for $$-\varepsilon$$) such that $$\pm\varepsilon$$ is not a square in $$\mathcal{O}_K/\mathfrak{p}$$. Since $$\mathcal{O}_K=\mathbb{Z}[\alpha]$$ we use Kummer-Dedekinds factorisation theorem. After trying some primes we see that $$\mathfrak{p}=(17,\alpha-2)$$ is a prime of norm $$17$$ which settles both $$\varepsilon$$ and $$-\varepsilon$$: under the quotient map $$\mathcal{O}_K\to\mathcal{O}_K/\mathfrak{p}=\mathbb{F}_{17}$$ we see that $$\varepsilon=1+\alpha^2$$ maps to $$5\notin\mathbb{F}_{17}^{*2}$$. Note that as $$17\equiv 1\bmod 4$$ we have $$-1\in\mathbb{F}_{17}^{*2}$$, so also $$-\varepsilon$$ gets mapped to a non-square, which completes the argument.