Let $p$ be a prime, let $n\in \mathbb{N}$, and let $a_1,...,a_{n-1}\in \mathbb{Z}$ such that $p>1+|a_1|+...+|a_{n-1}|$.
Show that the polynomial $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+p$ is irreducible over $\mathbb{Q}$.
Any ideas?
Thanks!
Let $p$ be a prime, let $n\in \mathbb{N}$, and let $a_1,...,a_{n-1}\in \mathbb{Z}$ such that $p>1+|a_1|+...+|a_{n-1}|$.
Show that the polynomial $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+p$ is irreducible over $\mathbb{Q}$.
Any ideas?
Thanks!
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Assume that $ f $ is reducible, and by Gauss' lemma, wlog assume the factors are monic and in $ \mathbf Z[X] $. Then, since $ p $ is prime, one of the factors would have to have constant term $ \pm 1 $, and thus $ f $ would have to contain a root $ |\alpha| \leq 1 $ in $ \mathbb C $. But then,
$$ f(\alpha) = \alpha^n + a_{n-1} \alpha^{n-1} + \ldots + p = 0 $$
$$ p = |\alpha^n + a_{n-1} \alpha^{n-1} + \ldots + a_1| \leq 1 + |a_1| + \ldots + |a_{n-1}| $$
which is a contradiction.