Exercise 8.6.6

Question

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Let $k$ be a field and let $M = k(t)$, where $t$ is a variable. The goal of this exercise is to prove that if $n>1$, then there is no element $\beta \in M$ such that $\beta^n - \beta + t = 0$.
\be
\item[(a)] Write $\beta = A/B$, where $A,B \in k[t]$ are relatively prime polynomials. Prove that $\beta^n - \beta + t = 0$ implies that $B \mid A$ and hence $B$ is constant.
\item[(b)] Show that $A^n-A+t
e 0$ for all polynomials $A\in k[t]$.
\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}}

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\begin{proof}
\item[(a)] We assume that
$$\beta^n-\beta + t = 0, \qquad \beta = \frac{A}{B}, $$
where $A,B \in k[t]$ are relatively prime polynomials.
Then
$$\frac{A^n}{B^n} - \frac{A}{B} + t = 0,$$
$$A^n - A B^{n-1} + tB^n = 0.$$
Hence $B \mid A^n$, and $B \wedge A = 1$, so $B \wedge A^n = 1$, therefore $B \mid 1$, so $B$ is a constant, and $\beta = A/B$ is a polynomial.

\item[(b)] By part (a), if $\beta \in M$ satisfies $\beta^n-\beta + t= 0$, then $\beta \in k[t]$. Write $\beta = A \in k[t]$, then
$$A^n - A +t = 0.$$
Then $A \mid t$. As any polynomial of degree 1, $t$ is irreducible in $k[t]$, so 
\begin{center}
$A= \lambda $ or $A = \lambda t$, where $\lambda \in k^*$.
\end{center}
If $A = \lambda$ then $t \in k$, in contradiction with $\deg(t) = 1$.

If $A = \lambda t,\  \lambda \in k^*$, then $\lambda^n t^n + (1- \lambda) t = 0, \ n>1,$ shows that $t$ is algebraic over $k$, in contradiction with the definition of $t$ as a transcendental variable over $k$.

Conclusion: $x^n - x + t, \ n>1$ has no root in $M = k(t)$.
\end{proof}

Exercise 8.6.6

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Exercise 8.6.6

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