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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!

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    Hi, please use MathJax to write maths. Also, what have you tried ? – SacAndSac Apr 28 '22 at 09:57
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    A general technique you might want to take advantage when doing these kinds of things "by hand" is that trace and norm of an integral element are integers. This gives certain integers that are polynomials in $a,b,c,d$. Sometimes, massaging those polynomial expressions enough will reveal that $a,b,c,d$ themselves are integers. – Thorgott Apr 29 '22 at 18:34
  • Please be more attentive to how you typeset square roots: use braces around the entire expression meant to be inside the square root or the sqrt command is only applied to the first symbol after it, leading to 2-digit numbers like 11 or 66 where the square root only extends over the leftmost digit. I fixed this everywhere except your last $\sqrt 66$, which looks weird. Try to fix that yourself. – KCd May 01 '22 at 18:47

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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.

Steven Mai
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