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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

blueplusgreen
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  • You also need to say that $p$ is monic. Then compare the answers from the duplicates, e.g., with this question. – Dietrich Burde Jan 15 '18 at 10:15
  • @DietrichBurde, alright, I agree. With this terms, can I conclude that the degree is $4$? – blueplusgreen Jan 15 '18 at 10:18
  • Yes, you can now argue that $p$ has the smallest degree with these properties (see the duplicate). – Dietrich Burde Jan 15 '18 at 10:24
  • The linked one? I'm not sure I understand the answers. The notation is unclear to me, too (i.e. $\mathbb Q [\sqrt 2]$) – blueplusgreen Jan 15 '18 at 10:28
  • What is unclear? Igor's answer is: First, show that $\sqrt{3}$ is not in the quadratic extension generated by $\sqrt{2}.$ That means that the degree of the extension is at least $4.$ But you have found a polynomial of degree $4,$ so it must be minimal. – Dietrich Burde Jan 15 '18 at 10:29
  • You have to show that $\sqrt{3}$ is not in $\mathbb{Q}(\sqrt{2})$. This is easy. Look around at MSE. – Dietrich Burde Jan 15 '18 at 10:30
  • If I understand correctly (after Googling): I need to show that there are no $a,b\in \mathbb Z$ such that $a + b\sqrt 2 = \sqrt 3$. right? – blueplusgreen Jan 15 '18 at 10:34
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    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. – Mathmo123 Jan 15 '18 at 12:38
  • @Mathmo123 But monic is relevant, because by definition it says "The coefficient of the highest-degree term in the polynomial is required to be $1$", see wikipedia? – Dietrich Burde Jan 15 '18 at 12:50
  • @DietrichBurde I guess to conclude that $p(x)$ is the minimal polynomial (with that convention), then monic is relevant. But monic is just a convention to ensure that the minimal polynomial is unique. To check the degree of an algebraic number, it is only the degree of the polynomial that matters. – Mathmo123 Jan 15 '18 at 13:47
  • @Mathmo123 Yes, sure. – Dietrich Burde Jan 15 '18 at 15:07

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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 $\Bbb Q$.

Lubin
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