Here is the problem :
Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ be a polynomial with integer coefficients such that $|a_0|$ is prime and $$|a_0|>|a_1| + |a_2| + \cdots + |a_n|.$$
Show that $f(x)$ is irreducible.
The solution is :
Let $\alpha$ be any complex zero of $f$. Suppose that $|\alpha| \le 1$, then
$$ |a_0|=|a_1\alpha + \cdots + a_n{\alpha}^n| \le |a_1| + \cdots + |a_n|,$$
a contradiction, therefore all the zeros of $f$ satisfies $|\alpha|>1$.
Now, suppose that $f(x)=g(x)h(x)$, where $g$ and $h$ are nonconstant integer polynomials. Then $a_0 = f(0) = g(0)h(0)$. Since $|a_0|$ is prime, one of $|g(0)|,|h(0)|$ equals 1. Say $|g(0)|=1$, and let $b$ be the leading coefficient of $g$. If ${\alpha}_1, \cdots, {\alpha}_k$ are the roots of $g$, then $|{\alpha}_1{\alpha}_2\cdots{\alpha}_k| = 1/|b| \le 1$. However, ${\alpha}_1, \cdots, {\alpha}_k$ are also the zeros of $f$, and so each has a magnitude greater than 1. Contradiction. Tberefore $f$ is irreducible.
Now I am left with one part that I don't understand, $1/|b| \le 1$. How can we know that $|b| \ge 1$? Any help is surely appreciated, thanks!
Best Answer
$b$ is an integer since it is a coefficient of $g$, so $|b|\geq 1$.