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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?

Thanks

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $\Bbb Q$.

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