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I'm trying to find units of $\mathbb Z[\sqrt{-5}]$.

So let $a,b\in \mathbb Z[\sqrt{-5}]$ s.t. $ab=1$. if $a=a_1+\sqrt{-5}a_2$ and $b=b_1+\sqrt{-5}b_2$, then we get $$\begin{cases}a_1b_1-5a_2b_2=1\\ a_1b_2+b_1a_2=0\end{cases}.$$ First equation give us $a_1,b_1\in \mathbb Z/5\mathbb Z$ and thus $$(a_1,b_1)\in \{(1,1),(2,3),(4,4)\}, $$ but how can I continue ? I was wondering to solve the second equation in $\mathbb Z/5\mathbb Z$, but I get $a_1b_2=4b_1a_2$, but it's unfortunately not conclusive.

user386627
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  • @VikrantDesai: Since I propose an other way to solve it, it's not duplicate ! – user386627 Feb 27 '17 at 18:07
  • Then you can add some appropriate tags like 'proof verification' and 'solution verification'. – Error 404 Feb 27 '17 at 18:10
  • @VikrantDesai: Try to think 2 minutes ! How can it be a "proof verification" or a "solution verification" since there is no proof, and I didn't give any solution ! I just gave a way of thinking that is not conclusive. (but thanks anyway for your link that give me an other way to prove it). I nevertheless want to proof using my way of thinking if it's possible. – user386627 Feb 27 '17 at 18:14
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    I said "appropriate" tags "like" .... and ....! – Error 404 Feb 27 '17 at 18:18
  • Appologies for acting like harsh. Actually I was not. But anyways why did you introduced $\Bbb Z / 5 \Bbb Z$? – Error 404 Feb 27 '17 at 18:27
  • @VikrantDesai: No problem :-) I introduced $\mathbb Z/5\mathbb Z$ because I was thinking that it could be easier to solve it there :-) (but may be not) – user386627 Feb 27 '17 at 18:36

2 Answers2

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If you equip $\mathbb Z[\sqrt{-5}]$ with the field norm $N$ defined by $N(a+b\sqrt{-5}) = a^2+5b^2$, then I leave it to you to check that the norm is multiplicative: $N(\alpha\beta)=N(\alpha)N(\beta)$. Thus, if $\alpha\in\mathbb Z[\sqrt{-5}]$ is a unit, $N(\alpha) = 1$, since there exists $\beta\in \mathbb Z[\sqrt{-5}]$ such that $\alpha\beta=1$, and the norm is multiplicative, so $N(\alpha\beta) = N(1) = 1 = N(\alpha)N(\beta)$.

Thus, the only units to be had are when $\alpha = \pm 1$.

Alex Ortiz
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There is a method in commutative algebra to solve such problems. Let $N:\mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}_+ $ is given by $N(a+b\sqrt{-5})=a^2+5b^2 $, called norm function. Its an ordinary theorem that $\alpha$ is invertible iff $N(\alpha)=1$. In this case, $a+b\sqrt{-5} $ is invertible iff $a^2+5b^2=1$. Is it enough?