Let $d$ be a square free integer. Show that $\mathbb{Z}[\sqrt{d}]$ is an integral domain.

Question

Let $d$ be a square free integer. Show that $\mathbb{Z}[\sqrt{d}]$ is an integral domain.

Here $1+0\sqrt{d}$ is identity of $\mathbb{Z}[\sqrt{d}]$. Also commutative property can be shown easily. But I can't show that $\mathbb{Z}[\sqrt{d}]$ has no zero divisor with $d$ is square free. Could you please help?

Which definition of this ring are you using? – Bill Dubuque May 01 '20 at 20:27 — Bill Dubuque, May 01 '20 at 20:27
Hint: Your ring is a subring of $\mathbb{C}$. – quasi May 01 '20 at 20:29 — quasi, May 01 '20 at 20:29

score 1 · Answer 1 · answered May 01 '20 at 20:32

For $a,b \in \Bbb Z$, note that if $a + b\sqrt{d}$ is a zero divisor, then the same must hold for the product $$ (a + b\sqrt{d})(a - b \sqrt{d}) = a^2 - db^2. $$ However, $0$ is the only zero divisor of $\Bbb Z \subset \Bbb Z[\sqrt{d}]$. So, we have $a^2 = db^2$. Because $d$ is square-free, this only occurs when $a=b=0$.

Let $d$ be a square free integer. Show that $\mathbb{Z}[\sqrt{d}]$ is an integral domain.

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