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.


1 Answer 1


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.


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