# Ring of integers of $\mathbb{Q}(i,\sqrt{5})$

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?

• 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 – The Short One Mar 23 '19 at 21:41

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 $$5$$.
• 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 $\alpha$. – franz lemmermeyer Mar 24 '19 at 16:54
• 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! – Tomiri Mar 24 '19 at 17:49