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I read on Math.SE that $\mathbb{Q}(\sqrt{d})^\times \simeq (\mathbb{Q}^\times)^2$. I could certainly write the multiplication of two elements in $\mathbb{Q}(\sqrt{d})$ purely as an operation of some kind (which we also called "$\times$") defined over rationals $\mathbb{Q}$.

$$ (x + y \sqrt{d})\,(x + y\sqrt{d}) = \big( x_1 x_2 + d\, y_1 y_2\big) + \sqrt{d} \big( x_1 y_2 + x_2 y_1\big) $$

This formula doesn't tell us that the multiplicative group should behave in any nice way. The rank of $\mathbb{Q}$ is in a sense infinite since, for any set of primes $\{ p_1, p_2, \dots, p_k\}$ we have an injection:

$$ \langle p_1, \dots , p_k \rangle \simeq p_1^{\mathbb{Z}} \times p_2^{\mathbb{Z}} \times \dots \times p_k^{\mathbb{Z}} \simeq (\mathbb{Z},+)^k \hookrightarrow \mathbb{Q}^\times $$

yet we can't say something like $\mathbb{Q}^\times$ is the product of copies of $\mathbb{Z}$ for each prime. I do not wish to consider profinite integers at this moment. Adeles could be necessary, but those involve completions like $\mathbb{Q}_p$ and $\mathbb{R}$.

This logic could also be applied to the extension field $\mathbb{Q}(\sqrt{d})$. Here is a group action of $\mathbb{Q}$ on $\mathbb{Q}(\sqrt{d})$: $$ \big[q \in \mathbb{Q}\big] \times \big[ x + \sqrt{d}\,y \big] \in \mathbb{Q}(\sqrt{d}) \to \frac{ x + \sqrt{d}\,y}{q} = \frac{x}{q} + \sqrt{d} \,\frac{y}{q} \in \mathbb{Q}(\sqrt{d}) $$ This field also has infinitely any primes, but (letting $d > 0$) but I don't think there is any obvious way to partition the primes over $\mathbb{Q}(\sqrt{d})$ between the primes over $\mathbb{Q}$ and whatever is left over. In fact, certain primes split, e.g. $$ 7 = (3 + \sqrt{2})(3 - \sqrt{2}) $$ So this has start to become a question of how primes split in number fields.

If I write it algebraically, suppose I have two rings, then maybe this could be plausible: $$ \big( A \times B )^\times \simeq A^\times \oplus B^\times $$ Except what I need is a relation between the sum of two groups on the left and the on the right: $$ \big( A[x]/(x^2 - d) \big)^\times \simeq A^\times \times A^\times $$

In a way I could always renormalize the "real" coefficient to $1$ and define elements of the quotient ring:

$$ \big[1 + x \,\sqrt{d}\,\big] \times [1 + y \,\sqrt{d}\, \big] \equiv \left[ 1 + \frac{x+y}{1 + xy\,d}\sqrt{d} \,\right] $$

Bizarrely I have obtained the sum of arctangents formula. Hopefully that denominator is not zero. This group under multiplication could be $(\mathbb{Q})^\times$ I don't know any help would be appreciated.

cactus314
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    Can you provide a link to the M.SE discussion where this was asserted? – Thomas Andrews Nov 29 '17 at 17:12
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    Since $\mathbb Q^{>0,\times}$ is a free abelian group with generators the positive prime integers, you have that $\mathbb Q^{\times}={\pm 1}\times F$ where $F$ is a free abelian group on a countable basis. So $(\mathbb Q^{\times})^2\cong {\pm 1}^2\times F^2\cong {\pm 1}^2\times F.$ That would mean that $\mathbb Q[\sqrt d]$ would need four square roots of $1$, which isn't possible in a field. So you probably mean $$\mathbb Q[\sqrt d]^{\times}/{\pm 1}\equiv ( \mathbb Q^{>0,\times})^2$$ or that $\mathbb Q[\sqrt d]^{\times}\equiv {\pm 1}\times F$ for $F$ a free abelian group – Thomas Andrews Nov 29 '17 at 17:28
  • Note $\mathcal{I}_K$ the group of fractional ideals (the free group generated by the prime ideals of the Dedekind domain $\mathcal{O}_K$) is very different when $\mathcal{O}_K$ is a PID, a UFD, or none of these. – reuns Nov 29 '17 at 18:20
  • It is not true if the field $Q(\sqrt{d})$ contains a root of unity, for example when $d=-1$ or $d=-3$ – i. m. soloveichik Jan 10 '18 at 12:13

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