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How do you prove that a particular set is an integral domain? is it enough to prove that there are no zero divisors to say that it is not an integral domain?

For example: a + bsqrt(2) : a,b are integers. Would it be sufficient for me to say that since we can have [a + bsqrt(2)]*[0+0*sqrt(2)] = 0 our set cannot be an integral domain ?

Thank you,

user26857
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Samir
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    An integral domain must also be a commutative ring. Also, $0+0\sqrt{2} = 0$ so you cannot say $a+b\sqrt{2}$ is not a zero divisor. – John Douma Oct 08 '15 at 19:14
  • @John Douma: You're wrong: a zero-divisor $x$ in a commutative ring is an element for which there exists a non-zero element $y $ such that $xy=0$. Hence $a+b\sqrt 2$ is a non-zero divisor if one of $a,b\ne 0$. – Bernard Oct 08 '15 at 19:21
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    @Bernard I don't believe I said otherwise. – John Douma Oct 08 '15 at 19:24
  • Well, mayb I misubderstood but ‘you cannot say $a+b\sqrt2$ is not a zero divisor’ means is is (double negation!). – Bernard Oct 08 '15 at 19:42
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    @Bernard It looked to me like Samir was saying that since $(a+b\sqrt{2})*(0+0\sqrt{2})=0$, there is nothing left to prove. I should have said he cannot say it is or isn't a zero divisor based on that. – John Douma Oct 08 '15 at 20:15
  • Ok, so how do I prove that it is indeed an integral domain? – Samir Oct 08 '15 at 21:09
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    You have to show that it is a commutative ring and that, given $ab=0$, one of $a$ or $b$ must be $0$. – John Douma Oct 08 '15 at 21:29

2 Answers2

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The ring $\mathbb{Z}[\sqrt{2}]$ is a subring of $\mathbb{C}$, which is a field, and hence has no nontrivial zero divisors. Hence $\mathbb{Z}[\sqrt{2}]$ itself is an integral domain.

Dietrich Burde
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The defining property of an integral domain is that we have the zero product law - i.e. if $ab=0$ then one of $a$ or $b$ is $0$. A statement like $$(a+b\sqrt{2})\cdot (0+0\sqrt{2})=0$$ does not contradict the zero product law, since $0+0\sqrt{2}=0$ (read this as "$0+0\sqrt{2}$ is the additive identity"), so the left hand does have a zero factor.

Milo Brandt
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