Show $1+\sqrt{5}$ is irreducible in $\mathbb{Z}[\sqrt{5}]$. Similar questions use the norm ( see e.g. How to show $1 + \sqrt{ 5 }$ is irreducible ), but I do not know yet about such a mapping, so I can't use it here.
I want to show now that if $$1+\sqrt{5}=(a+b\sqrt{5})(c+d\sqrt{5})$$ at least one of them is a unit i.e. has a multiplicative inverse in $\mathbb{Z}[\sqrt{5}]$. First I tried to characterise the units in $\mathbb{Z}[\sqrt{5}]$ by looking at $$1=(a+b\sqrt{5})(c+d\sqrt{5})$$ and trying to see which properties $a,b,c,d$ need to have to be a unit. But that did not get me very far to be honest. Rearranging the original equation got me nowhere good either.
How can I approach this without using the norm?
$\endgroup$ 21 Answer
$\begingroup$I'm assuming you meant $\mathbb Z[\sqrt 5]$, since $\mathbb Q[\sqrt 5]$ is a field. $$1+\sqrt{5}=(a+b\sqrt{5})(c+d\sqrt{5}) = (ac+5bd) + \sqrt 5(bc+da)$$ $$\implies ac+5bd =1\qquad bc+da=1$$ $$ \implies 1-\sqrt{5}= (ac+5bd) - \sqrt 5(bc+da) =(a-b\sqrt{5})(c-d\sqrt{5}) $$ $$ \implies (1+\sqrt{5})(1-\sqrt{5}) = (a+b\sqrt{5})(c+d\sqrt{5})(a-b\sqrt{5})(c-d\sqrt{5}) $$ $$ \implies -4 = (a^2-5b^2)(c^2-5d^2) $$ Now, if $a^2-5b^2 = \pm 1$, then $(a+b\sqrt{5})(a-b\sqrt{5}) = \pm 1$, so it is a unit. The same holds for $c^2-5d^2$. So If $1+\sqrt{5}$ is reducible, you have $a^2-5b^2=\pm 2$, $c^2-5d^2=\mp 2$ that is impossible modulus $4$.
$\endgroup$ 5
