Why is $\mathbb{Z}[\sqrt{-5}]$ an integral domain?

Question

I could use some help with this.

I know that $\mathbb{Z}[\sqrt{-5}]=\{a+b\sqrt{-5\} }|a,b\in\mathbb{Z}\}$. I then put $$0=(a+b\sqrt{-5})(c+d\sqrt{-5})=ac-5bd+(ad+bc)\sqrt{-5}$$ which leaves me with $$0=ac-5bd$$ and $$0=(ad+bc)\sqrt{-5}$$

now how do I proceed from here? I know that I have to prove $a+b\sqrt{-5}=0$ for $c+d\sqrt{-5}\neq0$.

Please, do not radically change the subject of the question through editing after it has received an answer (you originally where asking why this ring was a domain but not an UFD). $\Bbb Z[\sqrt{-5}]$ is a domain because it is a subring of (say) $\Bbb C$, which is a field. — , Apr 14 '16 at 16:40
@G.Sassatelli I didn't 'radically change' anything. I cut the part of the question that was a duplicate, which I didn't know before. What remains as a question is still unanswered. How do I know that it is an integral domain? How does this follow from $\mathbb{C}$ being a field. — azureai, Apr 14 '16 at 17:03
The, it is a subring of $\Bbb Q[\sqrt{-5}]$, which is a field. — , Apr 14 '16 at 17:08
Because subrings of integral domains are integral domains, and fields are integral domains. — , Apr 14 '16 at 17:09
@G.Sassatelli Thank you. And is there a way to calculate directly from $a*b=0$ with $a,b$ being complex numbers and $b\neq 0$ that $a=0$? — azureai, Apr 14 '16 at 17:14

score 1 · Answer 1 · answered Apr 14 '16 at 13:19

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For the first part : numbers in $\mathbb{Z}[\sqrt{-5}]$ are in particular complex numbers. What happens when a product of two complex numbers is zero ?

For the second part : to show that your elements irreducible, try to see that no element has its norm equal to $2$ or $3$ (the norm of $a+b\sqrt{-5}$ is $a^2+5b^2$).

answered Apr 14 '16 at 13:19

Captain Lama

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Regarding the first part: Both real and imaginary part of the product are zero? – azureai Apr 14 '16 at 13:28
Well, yes, but that's a weird way to put it (you don't have to make any reference to real or imaginary parts). If $xy=0$ then $x=0$ and $y=0$ : $\mathbb{C}$ is integral. In general, you can see your first question as a special case of : a subring of an integral ring is integral. – Captain Lama Apr 14 '16 at 13:46
Ok but I want to prove this, not assume it. – azureai Apr 14 '16 at 13:53
But don't you already know that if $x,y\in \mathbb{C}$ then $xy=0$ implies $x=0$ and $y=0$ ? – Captain Lama Apr 14 '16 at 13:54
No. Can I prove it without using that $\mathbb{C}$ is a field? – azureai Apr 14 '16 at 13:58
Well, I guess you can, but is there such a thing as a course where you learn the notion of UFD but can't use the fact that complex numbers have a division ? – Captain Lama Apr 14 '16 at 14:10
How do I do it? – azureai Apr 14 '16 at 14:25

Why is $\mathbb{Z}[\sqrt{-5}]$ an integral domain?

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