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$2,1+\sqrt{-5}$ is an integral basis for the ideal generated by them in $\mathbb{Z}[\sqrt{-5}]$. Is there a quick way to see this? What if these two are replaced with another pair?

My method: Write elements of $\mathbb{Z}[\sqrt{-5}]$ as $(a,b)$, meaning $a+ b\sqrt{-5}$. It suffices to show that, for any $(a,b)$, $(c,d)$, we have $(a,b)(2,0) + (c,d)(1,1) = (e,f)$ where $e \equiv f \mod{2}$. Modulo 2, the left hand side is $(c,d)(1,1) \equiv (c+d,c+d)$. QED

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Is clear that, if $a.2+b(1+\sqrt{-5})=0$, with $a,b\in \mathbb{Z}$, then $a=b=0$. If $h\in I=\langle 2, 1+\sqrt{-5}\rangle$, then exists $x,y,z,w\in \mathbb{Z}$ such that $$ h=2.(x+y\sqrt{-5})+(1+\sqrt{-5})(z+w\sqrt{-5}), $$ since $\mathbb{Z}[\sqrt{-5}]=\{a+b\sqrt{-5};\, a,b\in \mathbb{Z}\}$. So, $$ h=2x+z-5w+(2y+z+w)\sqrt{-5}=2x+z-5w+(2y+z+w)(1+\sqrt{-5})-(2y+z+w) $$ i.e., $$ h=2(x-y)-6w+(2y+z+w)(1+\sqrt{-5})=[(x-y)-3w].2+(2y+z+w)(1+\sqrt{-5}), $$ then defining $a=(x-y)-3w$ and $b=2y+z+w$, have $a,b\in\mathbb{Z}$ and $h=a.2+b(1+\sqrt{-5})$. Showing that $2, 1+\sqrt{-5}$ generate $I$ and then $2, 1+\sqrt{-5}$ is integral basis for $I$.

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@user71815, Clerical error, the correct is $ \mathbb{Z} [\sqrt {-5}] $! Sorry for not being a quick argument! I can not guess what you had done before! – Edson Sampaio May 8 '14 at 18:46
Ok, I'll show you what I did, which is actually quicker than your way. Please see the edit. – user71815 May 8 '14 at 19:05

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