# What is the unit group of $\mathbb{Q}(\sqrt{2},\sqrt{-3})$?

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.

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

$$\qquad$$ $$f(w)=|w|^2$$ 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 $$|a|\le 4$$)

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).

• I think it needs a discussion why $1+\sqrt2$, a fundamental unit of $O_{\Bbb{Q}(\sqrt2)}$, stays a fundamental unit of $O_K$. Dec 19, 2020 at 5:03
• @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. Dec 19, 2020 at 6:14