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Let $\mathbb{Z}[\sqrt{7}]=\{a+b\sqrt{7}\mid a,b\in\mathbb{Z}\}$. Let $\mathbb{Z}[\sqrt{7}]$ have the usual addition and multiplication, namely $$(a+b\sqrt{7})+(c+d\sqrt{7})=(a+c)+(b+d)\sqrt{7}$$ and $$(a+b\sqrt{7})(c+d\sqrt{7})=(ac+7bd)+(ad+bc)\sqrt{7}$$

Let $\mathbb{Z}[\sqrt{7}]^\times$ be the group of units of $\mathbb{Z}[\sqrt{7}]$. What familiar group is $\mathbb{Z}[\sqrt{7}]^\times$ isomorphic to?

I'm aware "familiar" is a vague term. Examples of "familiar" groups would be things like $D_{2n},S_n,A_n,\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C},Z_n,GL(n,F),SL(n,F)$

I know some vague facts about this, but nothing too concrete. I know that if $(a+b\sqrt{7})$ is invertible, there exists some $(c+d\sqrt{7})$ such that $(ac+7bd)+(ad+bc)\sqrt{7}=1$, so $ac+7bd=1$ and $ad+bc=0$. I'm also aware that there's a natural norm on $\mathbb{Z}[\sqrt{7}]$, namely $N(a+b\sqrt{7})=a^2-7b^2$. As the norm is multiplicative, we know that for $x\in\mathbb{Z}[\sqrt{7}]^\times$, $N(x)=\pm 1$.

I believe this problem can also be solved through number theory (such as this question), but I don't know any number theory currently, and the context in which this problem was given assumed no familiarity with number theory.

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  • $\begingroup$ Find the inverse of such an element in $\mathbb{C}$. What conditions do you need for this to also be in $\mathbb{Z}[\sqrt{7}]$? $\endgroup$
    – Matt B
    Commented May 20, 2016 at 21:19
  • $\begingroup$ @MattB in $\mathbb{C}$, we have that $(a+b\sqrt{7})^{-1}=\frac{1}{a+b\sqrt{7}}=\frac{a-b\sqrt{7}}{a^2-7b^2}$. For this division to be ok in $\mathbb{Z}[\sqrt{7}]$, we need $a^2-7b^2=\pm 1$, which was our condition from the norm $N(a+b\sqrt{7})=\pm 1$. My issue is that I feel like I could easily find some examples of units in $\mathbb{Z}[\sqrt{7}]$ by finding integer solutions to $a^2-7b^2=\pm 1$, but I don't see how this will tell me the group structure. I know that the group will be abelian, and I'm guessing it may be finitely generated (although that guess isn't as motivated as well). $\endgroup$
    – Mark Schultz-Wu
    Commented May 20, 2016 at 21:24
  • $\begingroup$ @MattB should be $\Bbb Z \oplus \Bbb Z/2\Bbb Z$ rather than $\Bbb Z^2$ by my reading $\endgroup$
    – Rolf Hoyer
    Commented May 20, 2016 at 21:33
  • $\begingroup$ @RolfHoyer you're right, I completely forgot to subtract the one. And I was ignoring the torsion for the moment, but yes you're also right there. $\endgroup$
    – Matt B
    Commented May 20, 2016 at 21:34
  • $\begingroup$ To expand on @RolfHoyer 's comment, we also have to take into account the elements of finite order, which will just be the roots of unity in our field. In this case, they're just $\pm 1$ which give $\mathbb{Z}/2\mathbb{Z}$. $\endgroup$
    – Matt B
    Commented May 20, 2016 at 21:37

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It is known the this group is isomorphic to the product of $\mathbf Z/2\mathbf Z$ with a monogenous group (infinite cyclic group, isomorphic with $\mathbf Z$.

The structure of all groups of units of algebraic number fields is describes in Dirichlet's unit theorem.

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    $\begingroup$ And by the method of using continued fractions to find solutions to Pell's equation, the cyclic part is generated in this case by $8 + 3\sqrt{7}$. $\endgroup$
    – Daniel Schepler
    Commented Jul 13, 2017 at 0:22

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