I know that the algebraic integers of $\mathbb Q[\sqrt{3}]$ is just $\mathbb Z[\sqrt{3}]$ and the algebraic integers of $\mathbb Q[\sqrt{5}]$ is just $\mathbb Z[{(1+\sqrt{5}})/2]$, and I was wondering how to deduce a general formula for the algebraic integers of $\mathbb Q[\sqrt{3},\sqrt{5}]$ or more generally the algebraic integers of $\mathbb Q[\sqrt{D},\sqrt{E}]$. Thank you.

  • 2
    $\begingroup$ Shouldn't that be $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$? $\endgroup$ – jgon Jan 9 '18 at 16:17
  • $\begingroup$ @jgon Why would that be? $\endgroup$ – P-S.D Jan 9 '18 at 16:22
  • $\begingroup$ Because 5 is 1 mod 4, so $\frac{1+\sqrt{5}}{2}$ is an algebraic integer. See: en.wikipedia.org/wiki/Quadratic_integer#Definition $\endgroup$ – jgon Jan 9 '18 at 16:24

I have seen that the ring of integers of $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is just $$ A = \mathbb{Z}\left[\sqrt{2},\sqrt{3},\frac{\sqrt{2}+\sqrt{6}}{2}\right] $$

with integral basis $\{1, \sqrt{2}, \sqrt{3}, \frac{\sqrt{2}+\sqrt{6}}{2}\}$. There are general formulas for the ring of integers in $K:=\mathbb{Q}[\sqrt{n},\sqrt{m}]$.

Proposition: Suppose $m \equiv 1\mod{4}$, $n \equiv k \equiv 2$ or $3 \mod{4}$. Then an integral basis for $\mathcal{O}_K$ is, with $k = mn/(m,n)^2$,

$$\left\{ 1, \frac{1 + \sqrt{m}}{2},\sqrt{n},\frac{\sqrt{n} + \sqrt{k}}{2} \right\}$$

Note that the ring of integers of $\mathbb Q[\sqrt{5}]$ is $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$, because $5\equiv 1\bmod 4$.

Reference with a proof:

On the ring of integers of a compositum of number fields


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