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Prove that $Z[√5] = \{a + b√5 | a, b ∈ Z\}$ is an integral domain.

Prove that $Z[√3i] = \{a + b√3i | a, b ∈ Z\}$ is an integral domain.

I'm trying to understand how to show that these are true. By use of the definition of integral domain, I think that first I would have to show that each of these are rings under addition and multiplication. Then show multiplication is also commutative. Then show that the semigroup with multiplication has an identity element (unity). I have already done these things.

My lack of understanding is with zero divisors. $∀x,y∈D:x∘y=0_D⟹x=0_D or y=0_D$ This is the definition I have but I don't know how to apply to the two given problems above.

Any help is appreciated.

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    $\begingroup$ Try to use that a subring of an integral domain is an integral domain. Can you think of an integral domain that has these 2 rings as a subring? $\endgroup$
    – Ward Beullens
    Apr 12, 2016 at 14:12
  • $\begingroup$ @WardBeullens We haven't learned subrings yet but having looked up the definition, that still went over my head. $\endgroup$
    – TfwBear
    Apr 12, 2016 at 14:32
  • $\begingroup$ Ok, forget about the subrings. As you mentioned you have to show that $xy = 0$ implies $x=0$ or $y=0$. But this is immediately true because it is true for all real numbers (for the fist case) and all complex numbers (second case) $\endgroup$
    – Ward Beullens
    Apr 12, 2016 at 14:38
  • $\begingroup$ Ah ok. That makes sense. Thank you. $\endgroup$
    – TfwBear
    Apr 12, 2016 at 14:42
  • $\begingroup$ What I was hinting at earlier was that the first ring is a subring of the real numbers. This means that it is a ring that 'sits inside' of the real numbers and it has the same rules for addition and multiplication. Then you could use that a subring of an integral domain is itself an integral domain. $\endgroup$
    – Ward Beullens
    Apr 12, 2016 at 14:48

1 Answer 1

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To check zero divisors in $\Bbb Z[\sqrt{5}]$, proceed as

$$(a+\sqrt{5}b)\cdot (c+\sqrt{5}d)=0.$$ Applying norm to both sides $$(a^2+5b^2)\cdot(c^2+5d^2)=0$$ $$\Leftrightarrow a^2+5b^2=0 \text{ or } c^2+5d^2=0$$ $$\Leftrightarrow a=b=0 \text{ or } c=d=0.$$

This shows that one of $a+\sqrt{5}b$ or $c+√5d$ is zero, which leads to $\Bbb Z[\sqrt{5}]$ is free from nontrivial zero divisors, which means $\Bbb Z[\sqrt{5}]$ is an integral domain.

Note:Similar arguments can be made in case of other rings like these.

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  • $\begingroup$ Why negative to this ? $\endgroup$
    – user525763
    Oct 14, 2018 at 15:04

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