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

  • $\begingroup$ Do they have roots over any of the respective field? $\endgroup$ – Bernard Mar 15 '17 at 17:36
  • $\begingroup$ yes , they have $\endgroup$ – rian asd Mar 15 '17 at 17:39
  • $\begingroup$ Which roots did you find for these polynomials? $\endgroup$ – Bernard Mar 15 '17 at 17:52
  • $\begingroup$ 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? $\endgroup$ – rian asd Mar 15 '17 at 17:59
  • $\begingroup$ 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. $\endgroup$ – Bernard Mar 15 '17 at 18:05

Solve the respective quadratic equations over $\;R\;$ . For the first one we get:


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 ...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.