# The norm of the ring $\mathbb{Z}\left[\frac{1+\sqrt{-19}}{2}\right]$.

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!

• Think again. Is the codomain not $\mathbb Z$? Commented Sep 13, 2020 at 11:26
• 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. Commented Sep 13, 2020 at 11:27
• @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. Commented Sep 13, 2020 at 11:52
• $N\left(\frac{a+b\sqrt{-19}}{2}\right) = a^2 + ab + 5 b^2 \; \; \;$ Commented Sep 13, 2020 at 13:41

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.
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 $$\Bbb{Z}[\alpha]^{\times}=\{\pm1\}$$.