Exercise 8.5.6

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}}

The proof of Theorem 8.5.9 used the Theorem of the Primitive Element to show that $\R$ has no extension of odd degree $>1$. Prove this without using primitive elements.

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}
We show that $\R$ has no extension $L$ of odd degree $d = [L:\R]>1$, knowing that every polynomial with an odd degree has a real root by the Intermediate Value Theorem.

As $[L : \R] >1$, there exists $\alpha \in L , \alpha 
ot \in \R$. Let $p$ the minimal polynomial of $\alpha$ over $\R$. By the Tower Theorem,
$$ [L : \R] = [L : \R(\alpha)][\R(\alpha) : \R],$$ 
hence $\deg(p) =  [\R(\alpha) : \R]$ divides the odd integer $d = [L:\R]$ , so $\deg(p)$ is odd. Therefore $p$ has a real root.
As $p$ is irreducible over $\R$, its degree is $\deg(p) = 1$, which implies $ [\R(\alpha) : \R]=1$, so $\alpha \in \R$, in contradiction with the definition of $\alpha$.

Conclusion: $\R$ has no extension of an odd degree greater than 1.
\end{proof}

Exercise 8.5.6

Answers

Comments

Exercise 8.5.6

Answers

Comments

Add answer