# Show that $f(x)$ is irreducible

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!

$$b$$ is an integer since it is a coefficient of $$g$$, so $$|b|\geq 1$$.