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Let $P = X^3 + X^2 - 2X -1$ be irreducible on $\mathbb{Q}$.

After some calculations it is easy to show that the roots of $P$ are the real roots $r_k = 2 \cos \frac{2k\pi}{7}$ for $k=1,2,3$.

Consider the extension $\mathbb{Q}[r_1]$. Then as $$ r_2 = 2 \cos 2 \frac{2 \pi}{7} = 2 \left(2 \cos^2 \frac{2 \pi}{7} -1 \right) = 4 \left(\cos \frac{2 \pi}{7}\right)^2 - 2 = r_1^2 - 2. $$ Thus the root $r_2$ is contained in $\mathbb{Q}[r_1]$. Does this imply that $\mathbb{Q}[r_1]$ is exactly the splitting field of $f$ and in particular contains all the roots of $f$?

Mathmo123
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vitalmath
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    Vieta's Formula implies that the roots $r_1, r_2, r_3$ satisfy $r_1 + r_2 + r_3 = -1$; in particular if two of the roots of $P$ are in some field, so is the third. – Travis Willse Mar 07 '22 at 21:11
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    Thank you! Isn't there another explanation based only on Galois theory? – vitalmath Mar 10 '22 at 21:18
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    Yes, and thanks to irreducibility we can do so without first verifying that $r_2 \in \Bbb Q[r_1]$: Since $P$ is irreducible and has square discriminant ($\Delta(P) = 49 = 7^2$), $\operatorname{Gal}(P) \cong A_3$, so $[\Bbb Q[r_1] : \Bbb Q] \mid 3$. – Travis Willse Mar 10 '22 at 22:35
  • So what is if we have a polynomial $P$ that has one root in $\mathbb{Q}$ in two roots $a$ and $b$ that are not in $\mathbb{Q}$? This field extension does not contain all the roots, but the splitting field that is generated by $a$ contains also $b$? (I still have som issues understanding Galois Theory that's why I ask) Thanks! – vitalmath Mar 11 '22 at 07:29
  • If a polynomial of degree $\geq 2$ has a root in $\Bbb Q$, then it's not irreducible over $\Bbb Q$. If a quadratic polynomial $P(X) := a X^2 + b X + c$ is irreducible, then the Galois group is $C_2$ and the nontrivial element acts by conjugation with respect to $\sqrt\Delta$, where $\Delta := b^2 - 4 a c$ is the discriminant. Put another way, we can write the roots as $r \pm s \sqrt\Delta$ for some $r, s \in \Bbb Q$ and $s \neq 0$, and the nontrivial element is characterized by $\sqrt\Delta \mapsto -\sqrt\Delta$. In particular, if $\Delta < 0$, the nontrivial element is complex conjugation. – Travis Willse Mar 11 '22 at 07:42
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    You can directly prove that if $a$ is a root then so is $a^2-2$ and these can't be equal so that the third root is also a rational function of $a$. Thus $\mathbb{Q} (a) $ is the splitting field. – Paramanand Singh Mar 17 '22 at 15:11
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    See https://math.stackexchange.com/questions/1767252/expressing-the-roots-of-a-cubic-as-polynomials-in-one-root – lhf Mar 20 '22 at 17:06

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