Consider the ring $R=\mathbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}|a,b\in\mathbb{Z}\}$ with a square free $d\in\mathbb{Z}$.
What is meant by the notation $$R^{\times}=(\pm 1)\times\mathbb{Z},$$ where $R^\times$ is the group of units in $R$.
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1I suppose it is the product of the multiplicative group ${1,-1}$ with the additive group $\mathbf Z$. However, I don't see why it should be true. – Bernard Mar 03 '17 at 11:59
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What product do you mean? :) – Mar 03 '17 at 12:03
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The product of groups (the set of pairs of elements, endowed with the componentwise group law) – Bernard Mar 03 '17 at 12:04
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And in what way do we identify an element of the product with an element in $R$? – Mar 03 '17 at 12:06
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For example is $(1,3)$ in the product of the group. If we identify this with $1+3\sqrt{2}\in R$, then this is no unit in $R$. – Mar 03 '17 at 12:18
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I agree with, but I don't know the exact context, and as I said, I don't see why it should be true. Could you post a link to try to see what is meant? – Bernard Mar 03 '17 at 12:29
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1Is $d$ the same as $p$? Should the $=$ sign be a $\simeq$ sign? – ancient mathematician Mar 03 '17 at 12:31
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And see http://math.stackexchange.com/questions/280878/the-units-of-mathbb-z-sqrt2 also for a specific case. – ancient mathematician Mar 03 '17 at 12:34
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I don't have a context here but determine the group $R^\times$. Yes, $d=p$, sorry! In the next line we have $R^\times=(\pm 1)\times <\epsilon >$ with $\epsilon\in R$ the smallest unit $>1$. – Mar 03 '17 at 12:35
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@mathmarseille: Why do you assume that $\langle \epsilon \rangle = \Bbb Z$? And how do you compare elements in $R$ with $1$? – Alex M. Mar 03 '17 at 13:52
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I don't assume $
=\mathbb{Z}$. Where do you think I do this? I am just looking for $d>0$. Sorry, I should have said this. – Mar 03 '17 at 13:53
2 Answers
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This means that there exists a group isomorphism $R^\times \cong \{\pm 1 \} \times \mathbb{Z}$. In fact it will be given by "multiplication": There exists a unit $\varepsilon \in R$ such that every other unit is uniquely of the form $\pm \varepsilon^{n}$ for $n\in \mathbb{Z}$. Note that this is only true if $d>0$. So for finding such an isomorphism try to prove that the smallest unit $\varepsilon > 1$ in $\mathbb{Z}[\sqrt{d}]$ satisfies the following: Every unit in $R$ may be uniquely written as $\pm \varepsilon^{n}$ for $n\in \mathbb{Z}$. Here is a $\textbf{hint}$: For a unit $u$, look at $\pm u, \pm u^{-1}$, to reduce to the case of showing that a unit $x+y\sqrt{d}$ with $x,y \geq 0$ can be written as $\varepsilon^{n}$ for $n \geq 0$.
This is a special case of $\textit{Dirichlet's unit theorem}$ from Algebraic Number Theory:
For $K$ a number field, we have $\mathcal{O}^\ast \cong \mu(\mathcal{O}) \times \mathbb{Z}^{r_1+r_2-1}$, where $r_1$ denotes the number of real embeddings and $r_2$ denotes the number of pairs of complex embeddings, and $\mu(\mathcal{O})$ is the group of roots of unity in $\mathcal{O}$ for an order $\mathcal{O}$ of $K$).
Looking at the proof of this theorem tells you that the isomorphism is given by multiplication, i.e. if $\varepsilon_i$ denotes the image of $(1,e_i)$, where $e_i=(0,\ldots, 0, 1, 0,\ldots, 0)\in \mathbb{Z}^{r_1+r_2-1}$, then $(u,(n_1,\ldots, n_{r_1+r_2-1}))$ is identified with $u\cdot \varepsilon_1^{n_1}\ldots \varepsilon_{r_1+r_2-1}^{n_{r_1+r_2-1}}$.
In your particular case: $K=\mathbb{Q}(\sqrt{d})$, where $d>0$ is a square free integer and $\mathcal{O}=R$; we have $r_1=2$, $r_2=0$. But if a number field $K$ admits a real embedding, then $\mu(\mathcal{O})=\{\pm 1\}$, simply because any root of unity has to be mapped to a root of unity in $\mathbb{R}$.
user363120
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As suggested in the comments, the statement should probably be $R^\times \cong \{\pm 1\} \times \mathbb{Z}$. This is basically a special case of Dirichlet's unit theorem. Writing $R = \mathbb{Z}[\alpha]$, since $d > 0$, then $R$ has $2$ real embeddings, namely $\alpha \mapsto \pm \sqrt{d}$, where $\sqrt{d} \in \mathbb{R}$ is the usual positive square root. Then Dirichlet's unit theorem implies that $R^\times$ has rank $1$, hence is generated by a fundamental unit $\epsilon$. Thus $$ R^\times = \{\pm \epsilon^m : m \in \mathbb{Z}\} \cong \{\pm 1\} \times \mathbb{Z} \, . $$
Viktor Vaughn
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