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How do I find all divisors of $6$ in $\mathbb{Z}[\sqrt{-5}]$?

If I set $$ 6=(a+b\sqrt{-5})(c+d\sqrt{-5}), $$ where $a,b,c,d\in\mathbb{Z}$, I get $$\tag{*} \begin{cases} 6=ac-5bd,\\ 0=ad+bc. \end{cases} $$ (*) is a system of equations in $\mathbb{Z}$, is there a systematic way to solve it? (With systematic I mean without need of guessing.)

I know the following divisors of $6$: $$ 2,3,(1-\sqrt{-5}),(1+\sqrt{-5}) $$ are they all?

The question if these divisors are irreducible has been answered more times: here and here. But I have not found how to find out all divisors (actually here the question has been asked, but non answered).

user26857
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Mohamed Ali
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1 Answers1

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Starting from your system of equations $$\tag{*} \begin{cases} 6=ac-5bd,\\ 0=ad+bc, \end{cases}$$ the second shows that $ad=-bc$. If $ac=0$ then $6=-5bd$, a contradiction, so $ac\neq0$ and hence $$d=-\frac{bc}{a}\qquad\text{ and }\qquad b=-\frac{ad}{c}.$$ Let $g:=\gcd(a,c)$ and let $a',c'\in\Bbb{Z}$ be such that $a=a'g$ and $c=c'g$. The above shows that $a'\mid b$ and $c'\mid d$. Let $e,f\in\Bbb{Z}$ be such that $b=a'e$ and $d=c'f$. Then $$0=ad+bc=a'gc'f+a'ec'g=a'c'g(e+f),$$ which shows that $f=-e$. It follows that $$6=ac-5bd=a'gc'g-5a'ec'f=a'c'(g^2+5e^2),$$ This clearly implies $e\in\{-1,0,1\}$, and the corresponding values of $a$, $b$, $c$ and $d$ are easily found:

For $e=0$ we have $b=d=0$ and $6=ac$, corresponding to factorizations of $6$ in $\Bbb{Z}$.

For $e=\pm1$ we get $g=1$ and $a'c'=1$, so $a'=c'=\pm1$, corresponding to the factorizations \begin{eqnarray*} 6&=&(1+\sqrt{-5})(1-\sqrt{-5}),\\ 6&=&(1-\sqrt{-5})(1+\sqrt{-5}),\\ 6&=&(-1-\sqrt{-5})(-1+\sqrt{-5}),\\ 6&=&(-1+\sqrt{-5})(-1-\sqrt{-5}), \end{eqnarray*} which are all the same factorization, up to rearrangement of the factors and multiplication by units.

Servaes
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  • (1) Only unclear step: You say that $a$ is squarefree because from $a\mid \gcd(6,5b^2)$ follows that $a\mid 6$ and $6$ is squarefree? Why you drop also the $5$? (2) From $a\mid \gcd(6,5b^2)$ we could also already conclude that either (i) $b=0$ and $a\mid 6$, thus $a\in{\pm 1,\pm 2,\pm 3,\pm 6}$ or (ii) $b\ne 0$ and $a\in{\pm 1}$? – Mohamed Ali Mar 02 '19 at 13:23
  • The $5$ is 'dropped' because $\gcd(6,5b^2)=\gcd(6,b^2)$, because $\gcd(6,5)=1$. And indeed, because $a\mid 6$ it is squarefree, so if $a\mid b^2$ then $a\mid b$. I do not see how you conclude that either $b=0$ and $a\mid6$ or $b\neq0$ and $a=\pm1$; why not $a=b=6$, for example? – Servaes Mar 02 '19 at 13:38
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    How did you conclude that $a | 5b^2$, and not $a | 5b^2c$? – Gunnar Sveinsson Mar 02 '19 at 15:36
  • @GunnarSveinsson I did so wrongly, thanks for spotting! I have corrected my argument, I would dare say it's actually a bit simpler now. – Servaes Mar 02 '19 at 15:52