Exercise 13.1.11

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}}

In Example 13.1.7 we showed that if $f=x^4+ax^3+bx^2+ax+1 \in \Z[x]$ is irreducible over $\Q$, then its Galois group is $\Z/2\Z 	imes \Z/2\Z$ if and only if there is $c \in \Q$ such that $4a^2+c^2 = (b+2)^2$.
\be
\item[(a)] Show that $c\in \Z$, and use the irreducibility of $f$ to prove that $c
e 0$. Hence we may assume that $c>0$, so that $(2a,c,b+2)$ is a Pythagorean triple.
\item[(b)] Show that $3^2+4^2 = 5^2, 5^2+12^2 = 13^2, 7^2 + 24^2 = 25^2$, and $8^2 + 15^2 = 17^2$ give two examples of polynomials with $\Z/2\Z 	imes \Z/2\Z$ as Galois group (two of the triples give reducible polynomials).
\ee

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}
\be
\item[(a)]  $c \in \Q$ is such that $c^2 = n \in \Z$. Write $c =a/b, b>0, a\wedge b = 1$. Then $a^2 = n b^2$. If $b
e 1$, there is a prime $p$ such that $p\mid b$. But then $p\mid a^2$, thus $p\mid a$, in contradiction with $a\wedge b = 1$. So $c \in \Z$.

If $c = 0$, then $(b+2)^2 = 4a^2$, so $b+2 =   2\varepsilon a$, $ b = -2+ 2 \varepsilon a$ , where $\varepsilon = \pm 1$.

In Exercise 9, we saw that
$$f=  x^4+ax^3+bx^2+ax+1 = (x^2-z_1x+1)(x^2 - z_2x+ 1),$$
where $z_1,z_2$  are the roots of $ z^2+az+b-2.$
Here $b = -2+2\varepsilon a$, so $z_1,z_2$ are the roots of $$z^2 +az -4+2\varepsilon a = (z+a-2\varepsilon)(z+2\varepsilon),$$
so
$$z_1 = -a+2\varepsilon \in \Z,\qquad  z_2 = -2 \varepsilon \in \Z,$$
so $f$ is not irreducible over $\Q$, in contradiction with the hypothesis.
We have proved that $c
e 0$ if $f$ is irreducible, and so $(2a,c,b+2)$ is a Pythagorean triple.

\item[(b)] 
$3^2+4^2 = 5^2$ gives $a=2, b =3$, and $f = x^4+2x^3+3x^2+2x+1 = (x^2+x+1)^2$ is not irreducible.

$5^2+12^2 = 13^2$ gives $a=6, b = 11$, and $f =  x^4+6x^3+11x^2+6x+1 = (x^2+3x+1)^2$ is not irreducible.

$7^2 + 24^2 = 25^2$ gives $a=12,b=23$, and $f= x^4 + 12x^3+23x^2+12x+1$ which is irreducible. So the Galois group of 
$$f= x^4 + 12x^3+23x^2+12x+1$$
is isomorphic to $\Z/2\Z 	imes \Z/2\Z$.

Verification with Sage:
\begin{verbatim}
R.<x> = QQ[]
f= x^4 + 12*x^3 + 23*x^2 + 12*x + 1
f.is_irreducible()
\end{verbatim}
\begin{center}
True
\end{center}
\begin{verbatim}
G = f.galois_group()
G.gens()
\end{verbatim}
$$ [(1,2)(3,4), (1,4)(2,3)]$$
\begin{verbatim}
G.structure_description()
\end{verbatim}
$$C2 	imes C2$$
\ee
\end{proof}

$8^2 + 15^2 = 17^2$ gives $a=4, b = 15$, and $f = x^4 + 4 x^3 + 15 x^2 + 4x+1$, which is irreducible.
The Galois group of 
$$f = x^4 + 4 x^3 + 15 x^2 + 4x+1$$
is isomorphic to $\Z/2\Z 	imes \Z/2\Z$.

Note: the polynomial associate to $7^2 + 24^2 = 25^2$ is 
\begin{align*}
f&= x^4 + 12x^3+23x^2+12x+1\
&= (x^2+6x+1)^2 - 15x^2\
&= (x^2+(6+\sqrt{15}) x +1) (x^2 + (6-\sqrt{15}) x +1)
\end{align*}
The discriminant of the first factor is $\Delta_1 = 47 + 12 \sqrt{15}$ and the discriminant of the second is $\Delta_2 = 47 - 12 \sqrt{15}$.
Since
$$\left(\sqrt{47 + 12 \sqrt{15}}ight)\left(\sqrt{47 -  12 \sqrt{15}}ight) = \sqrt{47^2 - 144 	imes 15} = \sqrt{49} = 7 \in \Q^*),$$
the splitting field of $f$ over $\Q$ is $\Q\left (\sqrt{47 + 12 \sqrt{15}}ight)$, which is a quadratic extension of a quadratic extension. The minimal polynomial of $a = \sqrt{47 + 12 \sqrt{15}}$ is $x^4-94x^2+49$, whose Galois group is also $\Z/2\Z 	imes \Z/2\Z$ (here $d=49$ is a square).

Exercise 13.1.11

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Exercise 13.1.11

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