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How shall I show that $\mathbb Z[\sqrt d]=\{a+b\sqrt d~~\big|a,b\in \mathbb Z\}$ is not a field?

It is an integral domain, so the thing it lacks maybe is every element does not have a multiplicative inverse. How to show this?

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WLOG, $\sqrt{d} \notin \mathbb{Z}$ (otheriwse $\mathbb{Z}[\sqrt{d}] = \mathbb{Z}$ is not a field). Suppose $\sqrt{d}(a + b\sqrt{d}) = 1$. Then $a\sqrt{d} = 1 - bd$ is an integer which is only possible if $a = 0$. So $bd = 1 \implies d = 1$ which is a contradiction.

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  • $\begingroup$ What for the case when $a,b\in\mathbb{Q}$? It still isn't a field? $\endgroup$ – Yelon Oct 23 '17 at 10:36
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Is $2$ invertible in this ring? That is, $1/2\in \mathbb Z[\sqrt d]$?

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