Exercise 13.4.8

Question

\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}}

Compute the Galois group of  $f=x^5-6x+3$ over $\mathbb{Q}$ using reduction modulo 11 and the method of Example 13.4.6.

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}

The discriminant calculation (cf. Ex.13.2.15):
$$\Delta(f) = 256a^5+3125b^4,$$
where $a=-6,b=3$. Then
\begin{align*}
\Delta(f) &= 2^8(-6)^5+5^53^4=3^4(5^5-3\cdot 2^{13})=3^4(3125-256\cdot96)=-3^4\cdot19\cdot1129.
\end{align*}
Reducing the polynomial modulo 11 gives:
$\bar f(x)=x^5+5x+3$.

The Sage instructions :
\begin{verbatim}
R.<x> = PolynomialRing(GF(11))
f = x^5 - 6*x + 3
f.factor()
\end{verbatim}
gives the irreducible factorization:
\begin{align*}
\bar f=(x^2 + 3x + 8)(x^3 + 8x^2 + x + 10).
\end{align*}
Since $\Delta(f)<0$, the Galois group $G_f\subset S_5$ of $f$ over $\mathbb{Q}$ is not a subgroup of $A_5$. The classification of transitive subgroups of $S_5$ given in (13.16) shows that $G_f =S_5$ or $G_f \simeq \mathrm{AGL}(1,\F_5)$.

Since $11
mid \Delta(f)$, by Theorem 13.4.5, $G_f$ contains the disjoint product of a 3-cycle and a 2-cycle, which has order $6$, thus the order of the Galois group is divisible by 3 and 2.

The order of  $\mathrm{AGL}(1,\F_5)$ is $20$. Hence the Galois group $G_f$ of $f$ over $\mathbb{Q}$ is  $G_f = S_5$.

\end{proof}

Exercise 13.4.8

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Exercise 13.4.8

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