Is it Integral domain?

Question

Is $ \mathbb{Z}[\sqrt {D}]$ an integral domain ?

My solution:

$$ \mathbb{Z}[\sqrt {D}]=\{ a+b\sqrt {D} , a,b \in \mathbb{Z}\}$$

No ,

Suppose $\mathbb {Z}[\sqrt {D}]$ is integral domain then Let $a, b \in \mathbb {Z}[\sqrt {D}]$ Where$ a=x +y\sqrt {D}$

$||a||$ $||b||=0$
$ \implies ||a|| =0 or ||b||=0$

If $||a|| =0 $then $x^2-y^2D=0$ So $x= y\sqrt {D}$

then it is contradiction when D is not perfect square , so it isn't Integra domain.

If D is prefect square then it is integral domain

Can anyone tell me if it is true or not? if not , any hint for solve ? Thanks

Really? Then you're probably not taking this stuff from algebraic number theory, right? And even more interesting: what then do you mean by $;\left|a\right|;$ ?? — DonAntonio, Mar 28 '18 at 12:36

score 3 · Answer 1 · answered Mar 28 '18 at 12:33

3

Hint:

Observe that no matter what $\;D\in\Bbb Q\;$ is, we always have $\;\Bbb Z(\sqrt D)\subset \Bbb C\;$ ...

answered Mar 28 '18 at 12:33

DonAntonio

But $\mathbb {Z} (\sqrt {D}} is not complex number – user508518 Mar 28 '18 at 12:39
@user508518 It is not even a number but a ring. Yet any element in that ring is , for sure, a complex number. – DonAntonio Mar 28 '18 at 12:54
So is it integral domain? – user508518 Mar 28 '18 at 12:56
@user508518 If it always contained in a field, what do you think? – DonAntonio Mar 28 '18 at 12:57
It is integral domain because field – user508518 Mar 28 '18 at 13:10
@user508518 No, this ring is not a field. Take $D=1$ for example. – Dietrich Burde Mar 28 '18 at 14:12

1 Answers1