Exercise 7.1.12 (Transitivity of the action over the roots)

Question

Show by example that Corollary 7.1.11 becomes false if you drop the word `irreducible'.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

\begin{proof} 
Consider the polynomial $f(t) = (t^2+1)(t^2 - 2) \in \Q[t]$, whose splitting field over $\Q$ is $\Q(i,\sqrt{2})$. Then there is no $	heta \in \Gal_{\Q}(f) = \Gal( \Q(i,\sqrt{2}) : \Q) $ such that $	heta(i) = \sqrt{2}$. Indeed, the minimal polynomial of $i$ over $\Q$ is $t^2 + 1$, so that $	heta(i) = \pm i$. Therefore the action of $ \Gal_{\Q}(f)  $ over the roots $\{i, -i, \sqrt{2},-\sqrt{2}\}$ is not transitive.

\end{proof}

Exercise 7.1.12 (Transitivity of the action over the roots)

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Exercise 7.1.12 (Transitivity of the action over the roots)

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