Let $K:=\Bbb Q(\sqrt[]a,\sqrt[]b)$ where $a,b$ coprime squarefree $\ne 1$ integers
Find an integral basis for the case $a ≡ 1\bmod 4$, $b ≡\ 2, 3\bmod 4$.
My attempt:
Denote $A:=\Bbb Q(\sqrt[]a)$, $B:=\Bbb Q(\sqrt[]b)$
We know that $\mathcal O_A=\Bbb Z[\frac{1+\sqrt[]a}{2}]$ and $\mathcal O_B=\Bbb Z[\sqrt[]b]$.
We know also that $\operatorname{disc}(\mathcal O_A)=a$ and $\operatorname{disc}(\mathcal O_B)=4b$,
so $\gcd(\operatorname{disc}(\mathcal O_A),\operatorname{disc}(\mathcal O_B))=1$ and $\mathcal O_{AB}=\mathcal O_K=\mathcal O_A \mathcal O_B$
Therefore since $\mathcal O_K$ is a ring, the family of integral elements $\{1,\frac{1+\sqrt[]a}{2},\sqrt[]b,\frac{1+\sqrt[]a}{2}\sqrt[]b\}$ generates $\mathcal O_K$ and is composed of integral elements since the product of integral elements is integral.
I showed that this family is linearly independent since $\{1,\sqrt[]a,\sqrt[]b,\sqrt[]{ab}\}$ is a basis of $\Bbb Z[\sqrt[]a,\sqrt[]b]$, so its elements are linearly independent.
Is that correct? is there a more elegant way to find a basis?
Thank you for your help.
