I need some help with a lemma I need to prove. First I will provide some background with previous lemmas that I already have been able to prove. Maybe these lemmas are needed to proof the last lemma
Given $f(x)\in \mathbb{Z}[X]$ and $\Omega =\{f(z)|z\in \mathbb{Z}\}$, if $\Omega$ contains infinitely many prime numbers, then $f$ is irreducible in $\mathbb{Z}[X]$.
Let $f \in \mathbb{Z}[x]$. If $\alpha$ is a complex root of $f$, then so is the conjugate $\bar{\alpha}$.
Let $f(x) = a_mx^m + ... + a_1x + a_0$ be a polynomial with coefficients $a_i$ in $\mathbb{Z}$ for $i = 0,...,m$. Suppose that $\alpha_1, ..., \alpha_m$ are roots of f and $\Re e(\alpha_i) < b$ for all $i$ where $b$ is a fixed number. Then the polynomial f(x + b) has no missing coefficients and all the coefficients have the same sign.
So I have already been able to prove these previous lemmas. I am however struggling to prove the following lemma:
Let $f(x) \in \mathbb{Z}[x]$ be a polynomial with roots $\alpha_1, \alpha_2, ..., \alpha_n$. Suppose that there is an integer $b$ for which $f(b)$ is a prime, $f(b-1) \neq 0$ and $\Re e(\alpha_i) < b - \frac{1}{2}$ for all $1 \leq i \leq n$. Prove that $f(x)$ is irreducible in $\mathbb{Z}[x]$.
I am pretty sure there is a connection with at least some of the previous lemmas, but I have not been able to figure out a proof. Can someone help me with this please?
For the sake of contradiction assume $f(x)=g(x)h(x)$ is a non-trivial factorization. Then from $f(b)$ being a prime we can assume (without loss of generality) that $|g(b)|=1$. Now let $g(x)=a\prod (x-\beta_i)$ where $\beta_i$ are its roots (so in particular they are also roots of $f$). Since $f(b-1)=g(b-1)h(b-1)\neq 0$ we have $g(b-1)\neq 0$. The condition $\Re e(\beta_i) < b - \frac{1}{2}$ implies $\beta_i$ is closer to $b-1$ then it is to $b$, or in other words $|b-1-\beta_i|<|b-\beta_i|$. Hence $$ |g(b-1)|=|a|\prod |b-1-\beta_i|<|a|\prod |b-\beta_i|=|g(b)|=1. $$ So $|g(b-1)|<1$ and yet $|g(b-1)|\neq 0$, impossible as it is a non-negative integer.