Integral Basis of $O_k$ Let $K=Q(\sqrt 6,\sqrt{11})$. Write $α ∈ O_K$ and its conjugates in terms of a $Q$-basis.
And show that an integral basis of $O_K$ is given by
${1,\sqrt 6,\sqrt {11},\frac{\sqrt 6+\sqrt{66}}2 }$, from first principles.
I'm really not sure how to do this, if anyone could help I'd really appreciate it!

You can consider the traces of $\alpha = a+b\sqrt{6}+c\sqrt{11}+d\sqrt{66}$ over the 3 quadratic subfields, noting that this gives an algebraic integer (in that subfield).The three subfields are $L_1 = \mathbb{Q}(\sqrt{6})$, $L_2=\mathbb{Q}(\sqrt{11})$ and $L_3=\mathbb{Q}(\sqrt{66})$ having integers $\mathbb{Z}[6]$, $\mathbb{Z}[11]$, $\mathbb{Z}[66]$.
$Tr_{K/L_1}(\alpha)=2a+2b\sqrt{6}$, so $a=A/2, b=B/2,$ where $A,B$ are integers. Similarly, we get $c=C/2,d=D/2$. Then from here, we can check which of the 16 possibilities are algebraic integers, for $A,B,C,D \in \{0,1\}$. To do this, we can compute the norm over the quadratic subfields. For example,
$N_{K/L_1}(\alpha)= \frac{1}{4} N_{K/L_1}((A+B\sqrt{6})+\sqrt{11}(C+D\sqrt{6})=\frac{1}{4}(A^2+2AB\sqrt{6}+6B^2-11(C^2+2CD\sqrt{6}+6D^2)) = \frac{A^2+6B^2-11C^2-66D^2}{4} + \sqrt{6}\frac{AB-11CD}{2}$
Thus $$A^2+6B^2-11C^2-66D^2=A^2+2B^2+C^2+2D^2=0 \mod 4$$, $$AB-11CD=AB+CD=0\mod 2$$
For, $N_{K/L_2}$ swap $B,C$ and $6,11$. So
$A^2+11C^2-6B^2-66D^2=0 \mod 4$, $$AC-6BD=0 =AC\mod 2$$.
Now do case work. $AC=0 \mod 2$, so $A=0$ or $C=0$.
If $A=0,C=0$, then we must have $B=D=1,0$.
If $A=0, C=1$, then contradiction since $A^2+2B^2+C^2+2D^2$ is odd.
Similarly for $A=1,C=0$.
Now we verify that the thing we found $\frac{\sqrt{6}+\sqrt{66}}{2}$ is an alg integer, as you've done. And then this gives an integral basis after adding the 3 other "non-halved" basis vectors.
Would love to know if there's a nicer way to do this.
