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I'm trying to find the ring of integers $A_L$ of $\mathbb{Q}(i,\sqrt{5})$. I know that the ring of integers of $\mathbb{Q}(i)$ is $\mathbb{Z}[i]$ and that the one of $\mathbb{Q}(\sqrt{5})$ is $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$. I would like to say that $A_L=\mathbb{Z}\left[\frac{1+\sqrt{5}}{2},i\right]$.

From Integral basis of the ring of integers of an extension, given integral bases of the rings of integers of subfields I understood it is possible to say $A_L=\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]\mathbb{Z}\left[i\right]$ using the fact the discriminants of the two integer basis are coprime (Here we use a result that is possible to find in Marcus' book).

Is there a way to find $A_L$ in a more direct way without using that result?

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    $\begingroup$ For what it's worth, multiplying units of the intermediate rings could give you the defining polynomial, e.g., $$\frac{-i}{2} - \frac{\sqrt 5}{2}$$ has polynomial $x^4 + 3x^2 + 1$, from which you can find its entry in the LMFDB: lmfdb.com/NumberField/4.0.400.1 $\endgroup$ – The Short One Mar 23 '19 at 21:41
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Write $\omega = \frac{1 + \sqrt{5}}2$. Then all elements $\alpha = a + bi + c\omega + di\omega$ where $a, b, c, d$ are integers. If the ring of integers is larger, there must be algebraic integers of the form $\alpha/2$ or $\alpha/5$ since $2$ and $5$ are the only prime divisors of the discriminant of the subring generated by $i$ and $\omega$. Now all you have to do is show that if $\alpha/2$ is an algebraic integer, then $a$, $b$, $c$, $d$ are divisible by $2$, and then do the same with respect to the prime $5$.

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    $\begingroup$ This must be in Marcus. If $K$ is any number field generated by an algebraic integer $\alpha$, and if $\alpha$ has discriminant $D$, then any algebraic integer in $K$ is a fraction with denominator $D$, whose numerator is a ${\mathbb Z}$-linear combination of powers of $\alpha$. $\endgroup$ – franz lemmermeyer Mar 24 '19 at 16:54
  • $\begingroup$ I'm sorry I deleted the question just two minutes after your comment because I didn't see it and I managed to solve the problem manually. I show it in our case: Let $ y= a+bα+c\beta+d\alpha\beta \in A_L$ with $ a,b,c,d \in \mathbb{Q} $. We know that $$ D(a,bα,c\beta,d\alpha\beta)= (abcd)^2 \cdot D(1,α,\beta,α\beta).$$ It follows that $ (abcd)^2 = D(a,bα,c\beta,d\alpha\beta) D(1,α,\beta,α\beta)^{-1}$ and $ D(1,α,\beta,α\beta)\in \mathbb{Z} $. So we have trivially the result. I hope this is right and useful! $\endgroup$ – Tomiri Mar 24 '19 at 17:49

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