How can I show that $x^2-\sqrt{2}x+1$ is irreducible over $\mathbb{Q}\left(\sqrt{2}\right)$ and $\mathbb{R}$ and $x^2-\sqrt{2} ix-1$ is irreducible over $\mathbb{Q}\left( i\right)$?
I think showing it is irreducible over $\mathbb{Q}\left(\sqrt{2}\right)$ is the same as showing irreducible over $\mathbb{R}$, is it right?
Can I find the root for the polynomial by USING $\frac {-b \pm \sqrt{b^2 - 4ac}} {2a}$ and see if it is belong to $\mathbb{R}$, $\mathbb{Q}(i)$ or $\mathbb{Q}(\sqrt{2})$ or none of the above?
please any hint with that
Solve the respective quadratic equations over $\;R\;$ . For the first one we get:
$$\Delta_1=2-4=-2$$
so it has no roots in $\;\Bbb R\;$ and thus also not in $\;\Bbb Q(\sqrt2)\subset\Bbb R\;$ . Add the little details left.
For the second one we have, over $\;\Bbb C\;$ :
$$\Delta_2=-2+4=2\implies x_{1,2}=\frac{\sqrt2\,i\pm\sqrt2}2=\pm\frac1{\sqrt2}+\frac1{\sqrt2}i=\frac1{\sqrt2}(\pm1+i)$$
so if $\;x^2-\sqrt2\,ix-1=\left(x-\left(\frac1{\sqrt2}+\frac1{\sqrt2}i\right)\right)\left(x+\left(\frac1{\sqrt2}+\frac1{\sqrt2}i\right)\right)\;$ reducible over $\;\Bbb Q(i)\;$ , then
$$\frac1{\sqrt2}\in\Bbb Q(i)\implies \sqrt2\in\Bbb Q(i)$$
which is absurd as then we can write
$$\sqrt2=a+bi\;,\;\;a,b\in\Bbb Q\implies a=\sqrt2\,,\,\,b=0$$
by comparing real and imaginary real parts ...