I already showed out that the four numbers: $\pm \sqrt 2 \pm \sqrt 3$ are the root of the polynomial $p(x) = x^4 -10x^2 +1$.

In addition, I showed that $p(x)$ is irreducible in $\mathbb Q [X]$

Are the two claims sufficient to conclude that the degree of those four numbers is $4$? If not, how do I show that?


  • $\begingroup$ You also need to say that $p$ is monic. Then compare the answers from the duplicates, e.g., with this question. $\endgroup$ – Dietrich Burde Jan 15 '18 at 10:15
  • $\begingroup$ @DietrichBurde, alright, I agree. With this terms, can I conclude that the degree is $4$? $\endgroup$ – blueplusgreen Jan 15 '18 at 10:18
  • $\begingroup$ Yes, you can now argue that $p$ has the smallest degree with these properties (see the duplicate). $\endgroup$ – Dietrich Burde Jan 15 '18 at 10:24
  • $\begingroup$ The linked one? I'm not sure I understand the answers. The notation is unclear to me, too (i.e. $\mathbb Q [\sqrt 2]$) $\endgroup$ – blueplusgreen Jan 15 '18 at 10:28
  • 1
    $\begingroup$ To answer your question, yes those two claims are sufficient. In general, the minimal polynomial of $\mathbb Q(\alpha)$ over $\mathbb Q$ is the degree of the minimal polynomial of $\alpha$ (being monic is not relevant). There are other, more sophisticated ways of reaching the same result. For that, you should see the links @DietrichBurde has referred you to. $\endgroup$ – Mathmo123 Jan 15 '18 at 12:38

Quite generally, if $k$ is a field and $f(X)\in k[X]$ is irreducible in that ring and has degree $n$, then any root $\rho$ of $f$ will be of degree $n$ over $k$, if by that you mean that $\bigl[k(\rho):k\bigr]=n$. You claim that you’ve proved both of the required properties, so yes, your numbers are of degree four over $\Bbb Q$.

| cite | improve this answer | |

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.