I am trying on this exercise and I have proved that $\sqrt{2}\in \Bbb Q(\alpha)$ where $\alpha$ is the root of $x^4-4x^2+2$ as follows:

Solve for roots of $x^4-4x^2+2$ we have the roots are $\alpha_1=\sqrt{2+\sqrt{2}}$, $\alpha_2=-\sqrt{2+\sqrt{2}}$, $\alpha_3=\sqrt{2-\sqrt{2}}$, $\alpha_4=-\sqrt{2+\sqrt{2}}$. And taking the square of each root and subtract $2$ gives $\sqrt{2}$.

But when I tried to find the minimal polynomial of $\alpha$ over $\Bbb Q(\sqrt{2})$, since I have not been told to find the minimal polynomial of which root among these four. So I think the answer is $x^2-2-\sqrt{2}$ OR $x^2-2+\sqrt{2}$.

Could someone please check if it is correct? Or if it is wrong, could someone give a solution?


  • 2
    $\begingroup$ It really says "the root" in the exercise? Because, as you calculated, there are four different roots, so speaking of 'the' root would be very weird. (although they are very similar) $\endgroup$ – Verdruss Jul 24 '17 at 14:50
  • $\begingroup$ Yes it does... I find it weird as well. So is that the fact that the answer of the question is we cannot choose between these two minimal polynomials we obtain? $\endgroup$ – non-abelian group of order 9 Jul 24 '17 at 20:40
  • $\begingroup$ I would say so. Both are as you said minimal polynomials of roots of the polynomial. Maybe you could give more reasons for why these are mininal polynomials but you can't choose one of them to be 'the' mininal polynomial. $\endgroup$ – Verdruss Jul 24 '17 at 22:18
  • $\begingroup$ I think the problem is that $\sqrt{2}$ and $-\sqrt{2}$ are algebraically indistinguishable. Both are roots of $x^2 - 2$, and if we are thinking of extensions of $\mathbb{Q}$ abstractly, i.e., without embedding them in $\mathbb{R}$ or $\mathbb{C}$, then there is no natural choice for which one is $\sqrt{2}$ and which is $-\sqrt{2}$. If you instead denote the roots of $x^2 - 2$ by $\beta_1, \beta_2$ then the minimal polynomials would be $x^2 - (2 + \beta_1)$ and $x^2 - (2 + \beta_2)$ (though I guess you could equally as well say they are $x^2 - (2 - \beta_1)$ and $x^2 - (2 - \beta_2)$). $\endgroup$ – André 3000 Jul 24 '17 at 22:31

Your Answer

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

Browse other questions tagged or ask your own question.