Exercise 13.2.12

Question

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ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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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}\,}

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

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Let $f = x^5 - 6x+3 \in \Q[x]$. Compute $\Delta(f)$ and $	heta_f(y)$ and show that $	heta_f(y)$ is irreducible over $\Q$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

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}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
By the Sch"onemann-Eisenstein Criterion for $p=3$, we know that $f$ is irreducible over $\Q$.

The discriminant $f = x^5 +ax+b, a,b\in \Q$ is given (see Ex. 15) by
$$\Delta(f) = 256a^5+3125b^4,$$
so 
$$\Delta(f) = -256 \cdot 6^5 + 3125 \cdot 3^4 = -1737531.$$
If we apply on the resolvent $	heta_f(y)$ the evaluation $\sigma_1\mapsto 0, \sigma_2\mapsto 0, \sigma_3\mapsto 0, \sigma_4 \mapsto a, \sigma_5 \mapsto -b$, we obtain
$$	heta_f(y) =  {\left(y^{3} - 20 \, a y^{2}  + 240 \,a^{2} y + 320 \, a^{3} ight)}^{2} - 2^{10}(256 \, a^{5}  + 3125 \, b^{4}) y $$

With $a = -6, b = 3$, we obtain
\begin{align*}
	heta_f(y) &={\left(y^{3} + 120 \, y^{2} + 8640 \, y - 69120ight)}^{2} + 2^{10}\, 1737531 \, y\
&= y^{6} + 240 \, y^{5} + 31680 \, y^{4} + 1935360 \, y^{3} + 58060800 \,
y^{2} + 584838144 \, y + 4777574400
\end{align*}
The Sch"onemann-Eisenstein Criterion doesn't apply.

With Sage, we obtain
\begin{verbatim}
R.<y> = QQ[]
p=y^6 + 240*y^5 + 31680*y^4 + 1935360*y^3 + 58060800*y^2 + 584838144*y + 4777574400
p.is_irreducible()
\end{verbatim}
\begin{center}
True
\end{center}

$	heta_f(y)$ is irreducible over $\Q$. A fortiori, $	heta_f(y)$ has no root in $\Q$.

Since $\Delta(f)<0$ is not a square in $\Q$, the Galois group of $f$ is $S_5$.
\end{proof}

Exercise 13.2.12

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Exercise 13.2.12

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