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I want to find the unit group of $\mathbb{Z}\left[\frac{1+\sqrt{-19}}{2}\right]$. To do this I want to use the same technique as one uses for rings of the form $\mathbb{Z}[\sqrt{d}]$ for some integer $d>1$ thats not a square. Here one can consider the norm function $$N:\mathbb{Z}[\sqrt{d}]\rightarrow\mathbb{Z}, N(x+y\sqrt{d})=(x+y\sqrt{d})(x-y\sqrt{d})=x^{2}-dy^{2}.$$ And then one can consider the Pell equation $x^{2}-dy^{2}=\pm 1$, to say something about the units.

But how is the norm defined on $\mathbb{Z}\left[\frac{1+\sqrt{-19}}{2}\right]$?

My guess would be something like $N\left(\frac{a+b\sqrt{-19}}{2}\right) = \frac{a^{2}+19b^{2}}{4}$. But then the codomain won't equal $\mathbb{Z}$ anymore. So in this case does one take $\mathbb{Q}$ as codomain?

Any help would be appreciated!

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    $\begingroup$ Think again. Is the codomain not $\mathbb Z$? $\endgroup$
    – Trebor
    Commented Sep 13, 2020 at 11:26
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    $\begingroup$ Your formula is right – the norm is a field-theoretical construction, it doesn’t care (at first glance) about rings of integers. Note that in your formula, $a$ and $b$ must be both odd or both even for the number to be an algebraic integer, and in both cases the norm is an integer. $\endgroup$
    – Aphelli
    Commented Sep 13, 2020 at 11:27
  • $\begingroup$ @Mindlack Ah I see, thank you. I thought that for instance $(1+0\sqrt{-19})/2$ was also a possible element. But the elements are actually of the form $a+b(\frac{1+\sqrt{-19}}{2})$ from which your statement easily follows. $\endgroup$
    – Peter
    Commented Sep 13, 2020 at 11:52
  • $\begingroup$ $N\left(\frac{a+b\sqrt{-19}}{2}\right) = a^2 + ab + 5 b^2 \; \; \; $ $\endgroup$
    – Will Jagy
    Commented Sep 13, 2020 at 13:41

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As noted in the comments it may be deceptive to write elements of $\mathbb{Z}\left[\frac{1+\sqrt{-19}}{2}\right]$ in the form $$\frac{a+b\sqrt{-19}}{2},$$ with $a$ and $b$ integers, because such a number is an element of the ring if and only if $a\equiv b\pmod{2}$. Indeed this follows easily from the fact that, by definition, every element of the ring $\mathbb{Z}\left[\frac{1+\sqrt{-19}}{2}\right]$ is of the form $$u\cdot1+v\cdot\frac{1+\sqrt{-19}}{2},$$ for integers $u$ and $v$.

From here you can indeed proceed by considering the norm, which is given by $$N\left(u+v\frac{1+\sqrt{-19}}{2}\right)=\left(u+v\frac{1+\sqrt{-19}}{2}\right)\left(u+v\frac{1-\sqrt{-19}}{2}\right)=u^2+uv+5v^2.$$ In general a norm is given by taking the product of the elements obtained by applying ring automorphisms that fix some subring. Here this fixed subring is just the subring $\Bbb{Z}$, which is of course fixed by any (unital) ring automorphism.

For this particular ring, every ring automorphism is determined by where it maps $\tfrac{1+\sqrt{-19}}{2}$. Because this is a root of $X^2-X+5$, any automorphism must take it to another root of this polynomial. Hence there are precisely two ring automorphisms, yielding the product above.

It is also worth noting the similarity between the minimal polynomial and the norm.

As for the unit group, Dirichlet's unit theorem tells you almost everything you need to know, but it seems like overkill in this context. Instead, continuing your argument with the norm:

If $u+v\alpha\in\mathbb{Z}\left[\alpha\right]$ is a unit, where $\alpha:=\frac{1+\sqrt{-19}}{2}$, then $$N(u+v\alpha)=u^2+uv+5v^2=\pm1.$$ It follows that $$\pm2=2u^2+2uv+10v^2=u^2+(u+v)^2+9v^2,$$ and hence that $v=0$ and $u=\pm1$. So $\Bbb{Z}[\alpha]^{\times}=\{\pm1\}$.

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