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The book I'm reading casually mention that $\mathbb{Z}[\sqrt{5}]$ isn't integrally closed without explaining why. Could someone show me why this is the case?

Bérénice
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Sam Spiro
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    Because the golden ratio $,\varphi = (1!+!\sqrt{5})/2,$ is a root of $,x^2-x-1\ $ so the Rational Root Test fails. – Bill Dubuque Aug 12 '16 at 20:48

1 Answers1

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$\frac{1+\sqrt{5}}{2}$ is integral over $\mathbb{Z}$ (hence over $\mathbb{Z}[\sqrt{5}]$) and is contained in the field of fractions $\mathbb{Q}(\sqrt{5})$ of $\mathbb{Z}[\sqrt{5}]$, but is not contained in $\mathbb{Z}[\sqrt{5}]$.

In general, the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[\sqrt{d}]$ if $d\equiv 2,3$ (mod $4$), and is $\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$ if $d\equiv 1$ (mod $4$).

carmichael561
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