Let $f(x)=x^4-2x^2-1$ $\in$ $\mathbb{Q}[x]$ and $t$ a root of $f$.Find the degree of the extension $ \mathbb{Q} [\sqrt{2},t]/ \mathbb{Q} $.
We know from eisenstein cretirion that $f$ is irreducible over $\mathbb{Q}$.
Can someone help me to prove this without finding the roots of $f$?
Thank you in advance!
Note that the $f$ splits in $\Bbb Q(\sqrt2)$ into $(x^2-1 + \sqrt2)(x^2 - 1 - \sqrt2)$. No matter which of these two factors we choose $t$ to be a root of, the degree $[\Bbb Q(t, \sqrt 2):\Bbb Q(\sqrt2)]$ is clearly not larger than $2$.
If we choose $t$ to be a root of $x^2 - 1 + \sqrt2$, then $t$ is not a real number, so $[\Bbb Q(t, \sqrt 2):\Bbb Q(\sqrt2)]$ cannot be $1$ and therefore must be $2$.
If we choose $t$ to be a root of $x^2 - 1 - \sqrt2$, then $[\Bbb Q(t, \sqrt 2):\Bbb Q(\sqrt2)]$ is still $2$, but it's not as immediate to see. The most straight-forward way is to assume that the degree is $1$ (i.e. that $t \in \Bbb Q(\sqrt2)$, or in other words, $t = a + b\sqrt2$ for rational $a, b$) and derive a contradiction. We get $$ (a + b\sqrt2)^2 - 1 - \sqrt2 = 0\\ a^2 + 2b^2 - 1 + (2ab - 1)\sqrt2 = 0\\ a^2 + 2b^2 - 1 = 0 \quad \bigwedge \quad 2ab - 1 = 0\\ a^2 + 2b^2 -1 = 0\quad \bigwedge \quad a = \frac{1}{2b}\\ \left(\frac{1}{2b}\right)^2 + 2b^2 -1 = 0\\ \left(\frac{1}{2b}\right)^2 - 2 + 2b^2 = -1\\ \left(\frac{1}{2b} - 2b\right)^2 = -1 $$ which clearly isn't possible for rational $b$. Therefore we must have $[\Bbb Q(t, \sqrt 2):\Bbb Q(\sqrt2)] = 2$.
In any case, we get $[\Bbb Q(t, \sqrt 2), \Bbb Q] = 4$.