# show it is irreducible over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(i)$

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?

• Do they have roots over any of the respective field? – Bernard Mar 15 '17 at 17:36
• yes , they have – rian asd Mar 15 '17 at 17:39
• Which roots did you find for these polynomials? – Bernard Mar 15 '17 at 17:52
• can I find the root for the polynomial by USING (b±√ (b^2-4ac))/2a and see if it is belong to R , Q(i) , and Q(√ 2) or not? – rian asd Mar 15 '17 at 17:59
• This formula is valid in any field, provided $b^2-4ac$ has a square root. B.t.w. don't use the surd notation here: it denotes the positive (real) square root of a positive number. – Bernard Mar 15 '17 at 18:05

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 ...