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Find the Galois group of $x^3-2x-1$ over $\mathbb{Q}$ and over $\mathbb{Q}(\sqrt{5})$

$x^3-2x-1$ has $-1$ as root, so $(x^3-2x-1)/(x+1) = x^2-x-1$ which has $x = \frac{1}{2}\pm \frac{\sqrt{5}}{2}$ as roots. I must find the splitting fields of these roots over $\mathbb{Q}$ and over $\mathbb{Q}(\sqrt{5})$.

$\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)$ is the splitting field of the polynomial over $\mathbb{Q}$. We have $3$ roots and a polynomial of degree $3$ that contains these roots, so $\left[\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right):\mathbb{Q}\right] = 3$ and $\left|Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}\right)\right| = 3$, so $Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}\right) = \{p_1, p_2, p_3\}$, where $p_1$ is the identity automorphism, but what are $p_2$ and $p_3$?

Now, for $Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}(\sqrt{5})\right)$, there is only one root and its polynomial is $x+1$, so the degree is one and $\left|Gal\left(\mathbb{Q}\left(-1,\frac{1}{2}\pm\frac{\sqrt{5}}{2}\right)/\mathbb{Q}(\sqrt{5})\right)\right| = 1$, so the Galois group is the trivial one. Is this right?

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    $\begingroup$ The field exrtension is not of degree 3, since $-1$ is already in the base field. But you're right for the second part. $\endgroup$ – Gerry Myerson Oct 1 '17 at 5:14
  • $\begingroup$ @GerryMyerson so for the first case the degree is $2$, right? $\endgroup$ – Guerlando OCs Oct 1 '17 at 6:14
  • $\begingroup$ Yes${}{}{}{}{}$. $\endgroup$ – Gerry Myerson Oct 1 '17 at 11:46
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As you said, $p = x^3-2x-1 \in \mathbb{Q}[x]$ have roots $-1, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}$. So, the splitting field of $p$ over $\mathbb{Q}$ is $E = \mathbb{Q}(-1, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2})$ . But see: $E = \mathbb{Q}(\sqrt{5})$ (why?) and by Einsentein's Criterion $q = \min(\sqrt{5}, \mathbb{Q}) =x^2 - 5 \in \mathbb{Q}[x]$, then $[E:\mathbb{Q}] = \deg q = 2$. $\mathbb{Q}$ have characteristic zero, so it's separable and we have $|Gal(E/\mathbb{Q})| = [E:\mathbb{Q}] = 2$. $$Gal(\mathbb{Q}(\sqrt{5})/\mathbb{Q}) \simeq \mathbb{Z}_2$$

Let $\sigma \in Gal(E/\mathbb{Q})$. Note that $\{1, \sqrt{5}\}$ is a $\mathbb{Q}-$basis of the $\mathbb{Q}-$vector space $E$ and $Gal(E/\mathbb{Q}) \subset GL_{\mathbb{Q}}(E)$. Therefore, we just need to know $\sigma(1), \sigma(\sqrt{5})$. Of course $\sigma(1) = 1$. We also know that $\sigma(\sqrt{5})$ is a root of $q$, then $\sigma(\sqrt{5}) \in \{\sqrt{5}, -\sqrt{5}\}$.

If $\sigma(\sqrt{5}) = \sqrt{5}$, then $\sigma = id_E$. If $\sigma(\sqrt{5}) = - \sqrt{5}$, then $$\sigma: E \to E: \,a+b \sqrt{5} \mapsto a - b \sqrt{5}$$

For your last question: of course $Gal(E/E) = \{id_E\}$.

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  • $\begingroup$ Compare with $Gal(\mathbb{C}/\mathbb{R})$. $\endgroup$ – Thadeu Henrique Costa Oct 3 '17 at 0:12

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