Exercise 4.2.8

Question

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

ewcommand{\be} {\begin{enumerate}}

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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}}

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 $a\in \Z$ be a product of distinct prime numbers. Prove that $x^n-a$ is irreducible over $\Q$ for any $n\geq 1$. What does this imply about $\sqrt[n]{a}$ when $n\geq 2$.

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 $a= p_1\cdots p_r$ a product of distinct prime numbers.

We show that $f = x^n -a$ is irreducible over $\Q$. Suppose on the contrary that $f = x^n -a$ is reducible. By Gauss Lemma $f$ has a monic factor $g \in \mathbb{Z}[x], 1\leq \deg(g) <n$.

The decomposition of $f$ in $\C[x]$ is
 $$f =\prod_{\zeta \in \mathbb{U}_n} ( x - \zeta \sqrt[n]{a}).$$ 
 $\mathbb{C}[x]$ being a unique factorization domain, $$g = \prod_{\zeta \in A} ( x - \zeta \sqrt[n]{a}),\ \emptyset 
eq A \varsubsetneq \mathbb{U}_n,$$
where $\vert A \vert =s$ satisfies $1\leq s <n$.

As $g \in \mathbb{Z}[x]$, the constant term is an integer $N$, given by
$$N = \xi \sqrt[n] {a}^s,$$ where $\xi = \prod_{\zeta \in A} \zeta \in \mathbb{U}_n$ is a $n$-th root of unity.

Moreover $\xi = N/\sqrt[n] {a}^s \in \mathbb{R}$, thus $\xi = \pm1$, and $\sqrt[n] {a}^s = \pm N =M\in \mathbb{Z}$.

But then $p_1^s\cdots p_r^s = M^n$.

The unicity of the decomposition in prime factors shows that the $p_i$ are the only prime divisors of $M$: $M = p_1^{k_1}\cdots p_r^{k_r}$, and $s = nk_i, i=1,\ldots,r$.

Thus $n\mid s$, in contradiction with $1\leq s <n$.

Conclusion: $x^n -a$ is irreducible over $\mathbb{Q}$, if $a = p_1\cdots p_r$ is a product of distinct prime numbers.

\bigskip

The easy part of Proposition 4.2.6 shows that $x^n-a, n\geq 2$ has no root in  $\mathbb{Q}$, in other words $\sqrt[n]{a}$ is irrational, for every $a$ being a product of distinct prime numbers.
\end{proof}

Exercise 4.2.8

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Exercise 4.2.8

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