Exercise 4.4.6

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}}

Let $F$ be a field. Show that other than the elements of $F$ itself, no elements of $F(x)$ are algebraic over $F$.

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}
Let $f \in F(x), f
eq 0$. Then $f =p/q,\  p,q \in F[x], \ p \wedge q = 1,\  p
eq 0, q
eq 0$.

If $f$ is algebraic over $F$, let $P  = \sum_{i=0}^n a_i x^i \in F[x]$ be the minimal polynomial of $f$ over $F$, with $\deg(P) = n$. Then $a_n = 1 
eq 0$, and $a_0 
eq 0$ (if $a_0 = 0$,  $P/x$ has the root $f$ and so $P$ should not be the minimal polynomial). Then 
$$a_{n}\left ( \frac{p}{q}ight )^n+a_{n-1}\left ( \frac{p}{q}ight )^{n-1}+\cdots+a_0 = 0,$$
thus   $$a_n p^n + a_{n-1} p^{n-1}q+\cdots+a_0 q^n = 0.$$

This equality, with $a_0 
eq 0, a_n 
eq 0$, shows that $p \mid q^n$, with $p\wedge q= 1$, so $p \wedge q^n=1$ shows that $p \mid 1$. Similarly $q\mid 1$. Thus $\deg(p) =\deg(q) =0$, and so $f=p/q \in F$.

The only elements of $F(x)$ which are algebraic over $F$ are the elements of $F$.
\end{proof}

Exercise 4.4.6

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Exercise 4.4.6

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