Exercise 14.1.8

Question

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

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

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

ewcommand{\ssi} {si et seulement si }

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

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

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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}}

Let $f\in F[x]$ be irreducible of prime degree $p\geq 5$, where $F$ has characteristic 0, and let $\alpha
e\beta$ be roots of f in some splitting field. If $F(\alpha,\beta)$ contains all other roots of $f$, then $f$ is solvable by radicals by Theorem 14.1.1. But suppose that there is some third root $\gamma$ such that $\gamma\in F(\alpha,\beta)$. Is this enough to force $f$ to be solvable by radicals?
\begin{enumerate}
\item[(a)] Use the classification of transitive subgroups of $S_5$ from Section 13.2 to show that the answer is ``yes'' when p=5.
\item[(b)] Use the polynomial $x^7-154\,x+99$ from Example 13.3.10 to show that the answer is ``no'' when p=7.
\end{enumerate}

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}\,}

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\begin{proof}
\begin{enumerate}
\item[(a)]   
By hypothesis, $\deg(f) = p = 5$, and $\alpha 
e \beta$ are roots of $f$ in some splitting field.

Since $\alpha$ is a root of $f$, which is irreducible over $F$,
$$[F(\alpha) : F] = \deg(f) = p = 5.$$
Then $\beta$ is a root of $\frac{f(x)}{x-\alpha} \in F(\alpha)[x]$, so that the minimal polynomial of $\beta$ over $F(\alpha)$ has degree $d\leq p-1$. Thus
$$[F(\alpha,\beta) : F(\alpha] \leq p-1 = 4.$$
By the Tower Theorem,
$$[F(\alpha,\beta):F]  =   [F(\alpha,\beta):F(\alpha)] \ [F(\alpha):F] \leq p(p-1) = 20.$$
Now, suppose that there is some third root $\gamma$ such that $\gamma\in F(\alpha,\beta)$. Then $F(\alpha,\beta,\gamma) = F(\alpha,\beta)$. Let $\delta, \varepsilon$ be the remaining roots of $f$. Since the characteristic is $0$, the irreducible polynomial $f$ is separable.Then $\delta$ is a root of $\frac{f(x)}{(x-\alpha)(x-\beta)(x-\gamma)} \in F(\alpha,\beta,\gamma)[x]$, so that
$$[F(\alpha,\beta,\gamma,\delta):F(\alpha,\beta,\gamma)] \leq 2.$$
Since $F(\alpha,\beta,\gamma) = F(\alpha,\beta)$, the tower theorem gives
$$[F(\alpha,\beta,\gamma, \delta):F] \leq 40.$$
Moreover $\alpha + \beta +\gamma + \delta + \varepsilon = \sigma_1(\alpha,\beta,\gamma, \delta,\varepsilon) \in F$, thus $F(\alpha,\beta,\gamma, \delta,\varepsilon) = F(\alpha,\beta,\gamma, \delta)$. Write $L = F(\alpha,\beta,\gamma, \delta,\varepsilon)$ the splitting field of $f$ over $F$. We have proved
$$[L:F]\leq 40.$$
The classification of transitive subgroups of $S_5$ from Section 13.2 shows that any transitive subgroup of $S_5$ with cardinality $\le 40$ is a subgroup of $\mathrm{AGL}(1,\F_5)$, thus is solvable. So $\Gal(L/F)$ is a solvable group, where $F$ has characteristic $0$, therefore $f$ is solvable (Theorem 8.5.3).

To conclude, the answer is ``yes''  when $p = \deg(f) = 5$.
\item[(b)] To prove that the answer is ``no''  when $p = \deg(f) = 7$, we use the counterexample $f = x^7-154\, x+99$ from Example 13.3.10.

The polynomial $f$ is not solvable, since its Galois group is $\mathrm{GL}(3,\F_2)$, which is simple (Section 14.3) and not commutative, thus non solvable.

We prove that there are roots $\alpha,\beta, \gamma$ of $f$ such that $\gamma \in F(\alpha,\beta)$.

As in Example 13.3.10, consider the resolvant
$$\Theta_f(y) = \prod_{1\leq i < j < k \leq 7} \left(y - (\alpha_i + \alpha_j + \alpha_k)ight) \in \Q[y].$$
Then the factorization of $\Theta_f(y)$ over $\Q$ is 
$$\Theta_f(y) = g(y) h(y),$$
where the polynomials $g,h$, given in Example 13.3.10, are irreducible factors of degrees $7$ and $28$.

Take three roots $\alpha,\beta,\gamma$ of $f$ such that $y-(\alpha+\beta+ \gamma)$ is any linear factor of $g$, so that the minimal polynomial of $\alpha+\beta+ \gamma$ is $g$, with $\deg(g) = 7$, thus
$$[\Q(\alpha+\beta+\gamma) : \Q] = 7.$$
Now we prove that $\gamma \in F(\alpha,\beta)$. Consider the chain of extensions
$$\Q\subset \Q(\alpha) \subset \Q(\alpha,\beta) \subset \Q(\alpha,\beta,\gamma) \subset L,$$
where $L$ is the splitting field of $f$ over $\Q$.

The minimal polynomial of $\alpha$ over $\Q$ is $f$, thus $[\Q(\alpha) : \Q] = 7$, and 
$$[L:\Q] = |\Gal(L/\Q)| = |\mathrm{GL}(3,\F_2)| = 168 = 2^3 	imes 3 	imes 7.$$
By the Tower Theorem,
$$[L : \Q(\alpha)] = \frac{ [L : \Q]}{[ \Q(\alpha) : \Q]} = 2^3 	imes 3$$
is not divisible by 7.

Since $\gamma$ is a root of $f$, the minimal polynomial of $\gamma$ over $f$ divides $f$. Thus
$$[\Q(\alpha,\beta,\gamma) : \Q(\alpha,\beta)] = 1	ext{ or } 7.$$
If $[\Q(\alpha,\beta,\gamma) : \Q(\alpha,\beta)]  = 7$, by the Tower Theorem, $7$ divides $[L : \Q(\alpha)] = 2^3	imes 3$. This contradiction proves that 
$$[\Q(\alpha,\beta,\gamma) : \Q(\alpha,\beta)] = 1,$$
therefore $\gamma \in \Q(\alpha,\beta).$

In this example, there exist roots $\alpha 
e \beta$ of $f$, and some third root $\gamma$ such that $\gamma \in F(\alpha,\beta)$, but $f$ is not solvable.

This shows that the answer is ``no'' when $p =\deg(f) = 7$.
\end{enumerate}

\end{proof}
\qquad

Note: In the proof of the Proposition 13.3.9, we saw that $G_f$ must be conjugate to $\mathrm{GL}(3,\F_2)$. This means that there is some numbering of the roots
$$
\left\{
\begin{array}{ccc}
\F_2^3\setminus\{(0,0,0\} & 	o &\{\alpha \in L \, |\, f(\alpha) = 0\}\
(
u_1,
u_2,
u_3) &	o &\alpha_{
u_1,
u_2,
u_3}
\end{array}
ight.
$$
which verify that, for all $\sigma \in\Gal(L/F)$, there is some $g \in \mathrm{GL}(3,\F_2)$ such that
$$\sigma(\alpha_{
u_1,
u_2,
u_3}) = \alpha_{g \cdot (
u_1,
u_2,
u_3)}.$$

In this correspondance, the roots of $f$ are seen as nonzero vectors in $\F_2^3$, and the seven roots of $g$ correspond to the seven (unordered)  triples of  linearly dependent  nonzero vectors in $\F_2^3$. So the roots $\alpha,\beta, \gamma$ where chosen in the preceding proof such that the corresponding vectors $u,v,w$ verify $w= u+v$ (but not $\gamma = \alpha + \beta$) .

This is what we understand in the hint of D.A. Cox ``Regard the roots as the nonzero vectors of $\F_2^3$ and pick roots $\alpha,\beta, \gamma$ such that $\gamma = \alpha+\beta$''.

This last equality is not true in $L$, but true for the corresponding vectors in $\F_2^3$.

Moreover, let $\alpha
e \beta$ be {\it any} pair of roots. The corresponding vectors $u,v$ are such that $u,v,u+v = -u-v$ is not a base, so that the root $\gamma$ corresponding to $u+v$ is such that $ y - (\alpha+\beta+\gamma)$ is a factor of $g$, and the preceding proof shows that $\gamma \in \Q(\alpha,\beta)$. For each pair $\alpha
e \beta$ of roots of $f = x^7-154\, x+99$, there exists a third root $\gamma 
ot \in \{\alpha,\beta\}$ such that $\gamma \in F(\alpha,\beta)$.

\subsection{IMPRIMITIVE POLYNOMIALS OF PRIME-SQUARED DEGREE}

Exercise 14.1.8

Answers

0.1 IMPRIMITIVE POLYNOMIALS OF PRIME-SQUARED DEGREE

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Exercise 14.1.8

Answers

0.1 IMPRIMITIVE POLYNOMIALS OF PRIME-SQUARED DEGREE

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