Exercise 4.3.6

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}}

Suppose that $\alpha$ and $\beta$ are algebraic over $F$ with minimal polynomials $f$ and $g$ respectively. Prove the {\bf Reciprocity theorem}: $f$ is irreducible over $F(\beta)$ if and only if $g$ is irreducible over $F(\alpha)$.

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}
Write $d_1 =[F(\alpha) : F], \delta_1= [F(\alpha,\beta):F(\alpha)], d_2 = [F(\beta):F],\delta_2 = [F(\alpha,\beta):F(\beta)]$.

The tower Theorem  gives the two relations
\begin{align}
[F(\alpha,\beta) :F] = \delta_1 d_1 = \delta_2 d_2. \label{eq:4.3.6}
\end{align}
Suppose that $f$ is irreducible over $F(\beta)$ (this makes sense because $f \in F[x]$ has a fortiori its coefficients in $F(\beta)$).

Then $f$ is the minimal polynomial of  $\alpha$ over $F(\beta)$, thus $$ \delta_2  = [F(\alpha,\beta),F(\beta)] = \deg(f) =d_1.$$

$\delta_2 = d_1$, combined with the relation \eqref{eq:4.3.6}, gives $\delta_1= d_2$.

Let $G$ be the minimal polynomial of  $\beta$ over $F(\alpha)$.

As $g \in F[x] \subset F(\alpha)[x]$, and $g(\beta)=0$, then $G \mid g$, and $\deg(g) = d_2 = \delta_1 = \deg(G)$, where $g$ and  $G$ are monic, thus $g=G$.

As $G$ is irreducible over $F(\alpha)$, $g$ is also irreducible over $F(\alpha)$.

We have proved:
\begin{center}
$f$ is irreducible over $F(\beta)$ $\Rightarrow$ $g$ is irreducible over $F(\alpha)$.
\end{center}
The proof of the converse is similar, by exchange of $\alpha,\beta$.

\begin{center}
$f$ is irreducible over $F(\beta)$ $\iff$ $g$ is irreducible over $F(\alpha)$.
\end{center}
\end{proof}

Exercise 4.3.6

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Exercise 4.3.6

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