Find all intermediate fields of extension $\mathbb{Q}(\sqrt{2} , \sqrt{3}) : \mathbb{Q}$ without using Galois correspondence. I'm in the middle of some notes which claim it should be possible to show that all the intermediate fields of the extension $\mathbb{Q}(\sqrt{2} , \sqrt{3}) : \mathbb{Q}$ are - 
$\mathbb{Q}(\sqrt{2}, \sqrt{3}), \mathbb{Q}, \mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{3}), \mathbb{Q}(\sqrt{6})$.
But this is before covering the Galois correspondence so we can't just find all subgroups of $\operatorname {Gal}(\mathbb{Q}(\sqrt{2} , \sqrt{3}) : \mathbb{Q})$ and count the number of subgroups to tell us there can be no more intermediate fields.

For example, one can show that the only intermediate fields of $\mathbb{Q}(\sqrt{2}) : \mathbb{Q}$ are $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}$ as follows:
By the Tower law, for an intermediate field $K$ where $\mathbb{Q} \subseteq K \subseteq \mathbb{Q}(\sqrt{2})$,
$|\mathbb{Q}(\sqrt{2}) : \mathbb{Q}| = |\mathbb{Q}(\sqrt{2}) : K| \cdot |K : \mathbb{Q}| = \deg(x^2 - 2) = 2$ as $x^2 - 2$ is irreducible over $\mathbb{Q}$.
Thus either $|\mathbb{Q}(\sqrt{2}) : K| = 1$, in which case $K=\mathbb{Q}(\sqrt{2})$
or
$|K : \mathbb{Q}| = 1$, in which case $K=\mathbb{Q}$

Using a similar method, I can get as far as showing that, since $|\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}| = |\mathbb{Q}(\sqrt{2}, \sqrt{3}) : K | \cdot |K : \mathbb{Q}| = 4$ for an intermediate field $K$, 
either $|\mathbb{Q}(\sqrt{2}, \sqrt{3}) : K | = 1$, in which case $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$
or $|K : \mathbb{Q}| = 1$, in which case $K = \mathbb{Q}$
or $|\mathbb{Q}(\sqrt{2}, \sqrt{3}) : K | = |K : \mathbb{Q}| = 2$
which is where I get stuck...
Any pointers appreciated!
 A: First, I think one should minimize the number of such raw exercises one does... unless these are meant to provide exactly the lesson that one does not want to work this way, as motivation for practical Galois theory.
But, anyway, assuming you know that $1,\sqrt{2}, \sqrt{3}, \sqrt{6}$ are linearly independent over $\mathbb Q$: any quadratic subextension will be generated by $\alpha=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ with $\alpha^2\in \mathbb Q$. In the case $a=0$, squaring produces $2b^2+3c^2+6d^2+2bc\sqrt{6}+4bd\sqrt{3}+6cd\sqrt{2}$. Looking at cases, at most one of $b,c,d$ can be non-zero. This produces the expected $\sqrt{2}$, $\sqrt{3}$, $\sqrt{6}$ generators for sub-fields.
Now take $a$ non-zero. Divide through to have $a=1$ without loss of generality. Again square the expression, and use the fact that the coefficients of $\sqrt{2}$, $\sqrt{3}$, $\sqrt{6}$ must be $0$. This is $2b+6cd=0$, $2c+4bd=0$, $2d+2bc=0$. We claim that all $b,c,d$ must be $0$. For example, if $b\not=0$, then $2b+6cd=0$ implies that $c,d$ are also non-zero, and symmetrically. Thus, if any is non-zero, all are. Substitute $c=-2bd$ into $2b+6cd=0$ to obtain $2b+6(-2bd)d=0$, or $6d^2=1$. Impossible. Thus, either we get the expected as in the previous paragraph, or, if $a\not=0$, then $\alpha\in\mathbb Q$.
Real lesson: don't do this.
A: One can show that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\alpha)$ where $\alpha=\sqrt{2}+\sqrt{3}$. This is done by showing that $\sqrt{2},\sqrt{3}$ are in $\mathbb{Q}(\alpha)$ by first showing that $\sqrt{6}$ is. This is a bit tedious (I remember this from a problem set, ha). Further computation shows that the irreducible polynomial for $\alpha$ over $\mathbb{Q}$ is $f(x)=x^4-10x^2+1$, which is of degree 4, as expected.
We can find the roots of this polynomial and factor it into 4 linear factors. Of course, one of the roots is $\alpha$. Call the other roots $\beta,\delta,\gamma$. The motivation for all this is to see the possible ways in which $f$ factors into two quadratics -- this will give us the possible field extensions. In particular, we can look at $(x-\alpha)(x-\beta),(x-\alpha)(x-\delta),(x-\alpha)(x-\gamma)$ as the possible polynomials for the 'top-of-the-tower' extensions. The coefficients for these polynomials must be in the field at the 'bottom-of-the-tower', which will force us to pick the possible 'bottom-of-the-tower' fields $K$ appropriately.
The set of roots of $f(x)$ can be computed to be $R=\{\pm\alpha,\pm(\sqrt{3}-\sqrt{2})\}$. The possible polynomials $g(x)$ that we get by the multiplication outlined above are: $-5-2\sqrt{6}+x^2,1-2 \sqrt{3} x+x^2,$ and $-1-2 \sqrt{2} x+x^2$. Hence we see that if these are the irreducible polynomials for the top extension, then $K$ must be $\mathbb{Q}(\sqrt{6}),\mathbb{Q}(\sqrt{3}),$ or $\mathbb{Q}(\sqrt{2})$, respectively.
Hope that helped. This is a rather messy approach, however -- I guess that's why we learn Galois theory in the first place!
