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While working on a project, I came across the problem of finding the unit group of $\mathbb{Q}(\sqrt{2},\sqrt{-3})$. Dirichlet's Unit Theorem stated that there should be $2$ multiplicatively independent elements among the unit group. However, aside from the generator $1+\sqrt{2}$ and $\frac{1+\sqrt{-3}}{2}$, which is not considered a generator since $\left(\frac{1+\sqrt{-3}}{2}\right)^6=1$. I cannot find any other generator. Can someone offer help or an explanation? Thanks.

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$K=\Bbb{Q}(\sqrt2,\zeta_3),F=\Bbb{Q}(\sqrt2)$

$\qquad $ $f(w)=|w|^2$ is an homomorphism $O_K^\times \to O_F^\times$.

Let $\sigma\in Gal(K/\Bbb{Q}), (\sqrt2,\zeta_3)\to (-\sqrt2,\zeta_3)$. Since our extension is abelian $f\circ \sigma=\sigma \circ f$, if $f(w)=1$ then $f(\sigma(w))=1$ so all the $\Bbb{Q}$-conjugates of $w$ are on the unit circle. This implies that the coefficients of the characteristic polynomial $\prod_{g\in Gal(K/\Bbb{Q})} (x-g(w^n))\in \Bbb{Z}[x]$ of $w^n$ are bounded by those of $(x+1)^4$ independently of $n$, thus for some $m>n>0$, $w^n$ and $w^m$ have the same minimal polynomial and hence $w$ is a root of unity.

Also from that $K\subset \Bbb{Q}(\zeta_{12})$ and $[\Bbb{Q}(\zeta_l):\Bbb{Q}]=\varphi(l)$ we get that the roots of unity of $K$ are generated by $\zeta_6$, ie. $\ker(f)=\langle \zeta_6\rangle$.

Thus $$[O_K^\times / \langle \zeta_6 \rangle:O_F^\times/ \langle -1\rangle] = [f(O_K^\times):f(O_F^\times)]=1$$ which implies that $$O_K^\times = \langle \zeta_6\rangle O_F^\times=\langle \zeta_6\rangle (1+\sqrt2)^\Bbb{Z}$$ ($O_F^\times=\pm (1+\sqrt2)^\Bbb{Z}$ because $1+\sqrt2$ is the least unit $>1$, we know it from testing the roots of $x^2+ax\pm 1$ for $|a|\le 4$)

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Dirichlet's Unit Theorem states that the rank of the group of units $\mathcal O_K^\times$ is $r_1+r_2-1$ if $r_1$ is the number of real embeddings and $r_2$ is the number of (conjugate pairs of) complex embeddings of $K$. For $K=\mathbb Q(\sqrt2,\sqrt{-3})$, there are no real embeddings, since there are no real square roots of $-3$, and so $r_1=0$. As $r_1+2r_2=[K:\mathbb Q]=4$, we should have $r_2=2$, and so the unit group should be rank $1$ -- i.e. it should consist of the roots of unity in $K$ along with a single non-torsion generator (which you have found).

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    $\begingroup$ I think it needs a discussion why $1+\sqrt2$, a fundamental unit of $O_{\Bbb{Q}(\sqrt2)}$, stays a fundamental unit of $O_K$. $\endgroup$
    – reuns
    Dec 19, 2020 at 5:03
  • $\begingroup$ @reuns I agree that that's useful, and I think your answer does a great job of providing an explanation of how to compute the group of units and is thus more complete. My answer is only intended to provide a direct response to the "$2$ multiplicatively independent elements" portion, and I think if you combine both my answer and your answer it becomes a very thorough solution to the question at hand. $\endgroup$ Dec 19, 2020 at 6:14

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