Lattice of intermediate fields of $\mathbb{Q}(i,\sqrt{2},\sqrt{3}) / \mathbb{Q}$ I need to construct the lattice of intermediate fields of $\mathbb{Q}(i,\sqrt{2},\sqrt{3}) / \mathbb{Q}$.  I think I know how to do this by simply listing them, but it seems that the picture I get will be pretty huge, with a web of tangled lines.  Is there a nicer way?
 A: Note that if we let $\omega$ be a $24^{th}$ root of unity has degree $\phi(24)=\phi(3)\cdot\phi(8)=8$ over $\Bbb Q$. I claim the field $K=\Bbb Q(i,\sqrt{2},\sqrt{3})$ is exactly $L=\Bbb Q(\omega)$. To see this we note that

$$ \begin{cases}\zeta_8 =\exp\left({2\pi i\over 8}\right)=\cos\left({2\pi\over 8}\right)+i\sin\left({2\pi i\over 8}\right)={1\over\sqrt 2}+{i\over\sqrt{2}} \\ \zeta_3=\exp\left({2\pi i\over 3}\right)=\cos \left({2\pi \over 3}\right)+i\sin\left({2\pi\over 3}\right)=-{1\over 2}+i{\sqrt 3\over 2}\end{cases}$$

so that $L\subseteq K$ and by degree considerations they're equal. So the Galois group is just

$$G=\operatorname{Gal}\left(K/\Bbb Q\right)=\left(\Bbb Z/8\Bbb Z\right)^*\times\left(\Bbb Z/3\Bbb Z\right)^*$$

which is easier to work with than the abstract $\left(\Bbb Z/2\Bbb Z\right)^3$ since we know what congruences the exponents have to satisfy for the sub-group conditions. We know that
$$\begin{cases}\left(\Bbb Z/8\Bbb Z\right)^*=\langle -1, 5\rangle \\ \left(\Bbb Z/3\Bbb Z\right)^*=\langle -2\rangle\end{cases}$$
The fact that we know all the generators allows us to figure out the fixed fields either by one of two methods:  either we can use the period theory to look at sums of powers of $\omega$ as generators of sub-extensions (not hard) or we can use the three generators $i,\sqrt{2},\sqrt{3}$ and note that each generator in our chosen generating set for the group affects exactly one of those three generators for the field. For staying as close as possible to the original problem, I opt to illustrate the latter approach, and use the same name for the order $2$ generators as above.
The subfields are then determined in degree by how many generators are present: $0$ generates the trivial group which fixes the top field, $1$ will give an index-$4$ subgroup, hence a quartic sub-extension, $2$ will give an index-$2$ subgroup corresponding to a quadratic extension, and $G$ of course fixes $\Bbb Q$. We list them systematically and note that any omitted pairs of generators can be seen to be redundant in the list
$$\begin{cases}
\{1\} \leftrightarrow \Bbb Q(\omega) \\ 
\langle -1\rangle \longleftrightarrow \Bbb Q(\sqrt 2,\sqrt 3) \\ 
\langle 5\rangle \longleftrightarrow \Bbb Q(i,\sqrt 3) =\Bbb Q(\zeta_{12})\\ 
\langle 2 \rangle \longleftrightarrow \Bbb Q(i, \sqrt 2)=\Bbb Q(\zeta_8) \\ 
\langle -2\rangle \longleftrightarrow \Bbb Q(i\sqrt 3, \sqrt 2) \\
\langle -5\rangle \longleftrightarrow \Bbb Q(i\sqrt 2, \sqrt 3) \\
\langle 10\rangle \longleftrightarrow \Bbb Q(i, \sqrt 6) \\
\langle -10\rangle \longleftrightarrow \Bbb Q(i\sqrt 2+i\sqrt 3+\sqrt 6) \\
\langle -1,5\rangle \longleftrightarrow \Bbb Q(\sqrt 3) \\ 
\langle -1, 2\rangle \longleftrightarrow \Bbb Q(\sqrt 2) \\ 
\langle -1, 10\rangle \longleftrightarrow \Bbb Q(\sqrt 6) \\
\langle 5,2\rangle \longleftrightarrow \Bbb Q(i) =\Bbb Q(\zeta_4)\\ 
\langle 5,-2\rangle \longleftrightarrow \Bbb Q(i\sqrt 3)=\Bbb Q(\zeta_3) \\
\langle 2,-5\rangle \longleftrightarrow \Bbb Q(i\sqrt 2) \\
\langle -2,-5\rangle \longleftrightarrow \Bbb Q(i\sqrt 6) \\
G \longleftrightarrow \Bbb Q 
\end{cases}$$
The lattice, of course, is easily done from this list since containment can be listed from the canonical generators I've chosen.
Edit Heck, why not even just draw the sucker. Note that every single field is contained in $K$ so we'll omit the top of the diagram, and similarly we'll omit $\Bbb Q$ at the bottom. Since drawing the large diagram is cumbersome, I'll just list the quartic subfields and their subfields, and you can reproduce this on paper by just not double-listing things.
$$\begin{matrix} && \Bbb Q(\sqrt{2},i) & \\ 
&\huge\diagup & \huge| & \huge\diagdown \\ 
\Bbb Q(\sqrt{2}) & & \Bbb Q(i\sqrt{2})& & \Bbb Q(i)\end{matrix}\qquad\begin{matrix} && \Bbb Q(\sqrt{3},i) & \\ 
&\huge\diagup & \huge| & \huge\diagdown \\ 
\Bbb Q(\sqrt{3}) & & \Bbb Q(i\sqrt{3})& & \Bbb Q(i)\end{matrix}$$

$$\begin{matrix} && \Bbb Q(\sqrt{6},i) & \\ 
&\huge\diagup & \huge| & \huge\diagdown \\ 
\Bbb Q(\sqrt{6}) & & \Bbb Q(i\sqrt{6})& & \Bbb Q(i)\end{matrix} \qquad \begin{matrix} && \Bbb Q(i\sqrt{3},\sqrt 2) & \\ 
&\huge\diagup & \huge| & \huge\diagdown \\ 
\Bbb Q(i\sqrt{3}) & & \Bbb Q(i\sqrt{6})& & \Bbb Q(\sqrt 2)\end{matrix}$$

$$\begin{matrix} && \Bbb Q(\sqrt{2},\sqrt 3) & \\ 
&\huge\diagup & \huge| & \huge\diagdown \\ 
\Bbb Q(\sqrt{2}) & & \Bbb Q(\sqrt{6})& & \Bbb Q(\sqrt 3)\end{matrix} \qquad\begin{matrix} && \Bbb Q(i\sqrt{2},\sqrt 3) & \\ 
&\huge\diagup & \huge| & \huge\diagdown \\ 
\Bbb Q(i\sqrt{2}) & & \Bbb Q(i\sqrt{6})& & \Bbb Q(\sqrt 3)\end{matrix}$$

$$\begin{matrix} && \Bbb Q(i\sqrt{2}+i\sqrt 3+\sqrt 6) & \\ 
&\huge\diagup & \huge| & \huge\diagdown \\ 
\Bbb Q(i\sqrt{2}) & & \Bbb Q(i\sqrt{3})& & \Bbb Q(\sqrt 6)\end{matrix}$$
