We can also write it as follows:
$[\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}[\sqrt{2}]][\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]$
$[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]=\deg \text{Irr}(x^2-2)$
We have that $x^4-2x^2-1=(x^4-2x^2+1)-2=(x^2-1)^2-2=(x^2-1+\sqrt{2})(x^2-1-\sqrt{2})$.
We have that $\rho$ is the root of $x^4-2x^2-1$. Does this mean that $\rho$ is a root of one of the factors $(x^2-1+\sqrt{2})$ or $(x^2-1-\sqrt{2})$ ?
If this is true then we have to show that this factor is irreducible over $\mathbb{Q}[\sqrt{2}]$, right?
Suppose that it is reducible then it must be the product of two polynomials of first degree, $(x-c_1), (x-c_2)$,
where $c_1, c_2\in \mathbb{Q}[\sqrt{2}]$.
Is this correct so far?
Yes of course. If $\rho$ is a root of $f(x)g(x)$ then $\rho$ is a root of $f(x)$ or $g(x)$.
This is correct so far. To show that $f(x):=x^2-1\pm\sqrt{2}$ is irreducible over $\mathbf Q(\sqrt{2})$, one just needs to show that $f(x)$ does not have a root in $\mathbf Q(\sqrt{2})$. I haven't tried doing this but this seems doable.
Suppose that $x^2-1\pm\sqrt{2}$ is reducible over $\mathbf Q(\sqrt{2})$, then there are $c_1,c_2\in \mathbb{Q}(\sqrt{2})$, such that $x^2-1\pm\sqrt{2}=(x-c_1)(x-c_2)$.
Since $c_1, c_2\in \mathbb{Q}(\sqrt{2})$ we have that $c_1=a_1+b_1\sqrt{2}$ and $c_2=a_2+b_2\sqrt{2}$, for $a_1,a_2,b_1,b_2\in\mathbb{Q}$.
$$x^2-1\pm\sqrt{2}=(x-(a_1+b_1\sqrt{2}))(x-(a_2+b_2\sqrt{2}))\\ =x^2-(a_1+a_2+(b_1+b_2)\sqrt{2})x+(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})\\ =x^2-(a_1+a_2+(b_1+b_2)\sqrt{2})x+(a_1a_2+2b_1b_2+(a_1b_2+a_2b_1)\sqrt{2})$$
So, the following must hold:
$$a_1+a_2+(b_1+b_2)\sqrt{2}=0 \ \ \land \ \ a_1a_2+2b_1b_2+(a_1b_2+a_2b_1)\sqrt{2}=-1\pm \sqrt{2} \\ \Rightarrow \left ( a_1+a_2=0 \land b_1+b_2=0 \right ) \ \ \land \ \ \left (a_1a_2+2b_1b_2=-1 \land a_1b_2+a_2b_1=\pm 1\right ) \\ \Rightarrow a_1=-a_2 \land b_1=-b_2 \land \left (-a_2^2-2b_2^2=-1 \land -2a_2b_2=\pm 1\right ) \\ \Rightarrow a_1=-a_2 \land b_1=-b_2 \land \left (a_2^2+2b_2^2=1 \land a_2b_2=\mp \frac{1}{2}\right )$$
Is this correct so far?
How could we continue to get a contradiction?
This seems correct (thought I haven't checked the calculations carefully). What we want to show, in particular, is that $\sqrt{1+\sqrt{2}}$ is not in $\mathbf Q(\sqrt{2})$.
A simpler problem is to decide whether or not $\sqrt[4]{2}$ is in $\mathbf Q(\sqrt{2})$. The answer is no. This is because by Eisenstien $x^4-2$ is irreducible over $\mathbf Q$ and thus the minimal polynomial of $\sqrt[4]{2}$ over $\mathbf Q$ has degree $4$. If $\sqrt[4]{2}$ were in $\mathbf Q(\sqrt{2})$, then the degree of the minimal polynomial of $\sqrt[4]{2}$ over $\mathbf Q$ would be at most $2$.
I will get back to you on what to say about $\sqrt{1+\sqrt{2}}$. Give me some time (and remind me in a few days if you do not hear from me!).
Could we maybe do it as follows?
We have that $\rho$ is a root of $x^4-2x^2-1$ and that \begin{align*}x^4-2x^2-1&=(x^4-2x^2+1)-2\\ & =(x^2-1)^2-2\\ &=(x^2-1+\sqrt{2})(x^2-1-\sqrt{2}).\end{align*}
So, $\rho$ is also a root of oe of the factors $(x^2-1+\sqrt{2})$ ή $(x^2-1-\sqrt{2})$.
We have that $\rho^2-1\pm \sqrt{2}=0 \Rightarrow \sqrt{2}=\pm(1-\rho^2)\in \mathbb{Q}[\rho]$.
Therefore, $\mathbb{Q}[\rho, \sqrt{2}]=\mathbb{Q}[\rho]$.
So, $ [\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}]=[\mathbb{Q}[\rho]:\mathbb{Q}]$.
We have that $\rho$ is algebraic over $\mathbb{Q}$. So, we we he that the is an unique monic irreducible polynomal in $\mathbb{Q}[x]$ $p\in \mathbb{Q}[x]$ such that $p(\rho)=0$. This irreducible polynomial is the minimal polynomial.
Since irreducible monic polynomials ith common roots are equal, we have that $f=p$, so $f$ is the minimal polynomial of $\rho$ over $\mathbb{Q}$.
From a theorem that I found in my notes, we have that basis of the extension is $1, \rho, \rho^2, \rho^3$ and $[\mathbb{Q}[\rho]:\mathbb{Q}]=4$.
If we had $\sqrt{3}$ instead of $\sqrt{2}$, wouldwe do the same? Would we have to show that $\sqrt{3}\notin \mathbb{Q}[\rho]$ ? But how?