I was looking for an elementary (or involving introductory level abstract algebra/analysis) proof of Cohn's Irreduciblity Criterion:
If
$$ a_0, a_1, \dots, a_n \in \Bbb{Z} $$
and
$$ 0 \le a_i \le t$$
and
$$ a_0 + a_1t + a_2t^2 +\cdots + a_nt^n \in \{ \text{Primes} \} $$
then the polynomial
$$ a_0 + a_1 x + a_2 x^2 +\cdots+ a_nx^n $$
is irreducible.
I began to plot out a proof. Assume that we have a polynomial that does evaluate to a prime for some $t$ satisfying the inequalities above. Then it follows, that if this polynomial is factorable, it would be factored into
$$ (b_0 + b_1x + b_2x^2 + \cdots+ b_rx^r )(c_0 + c_1x + c_2x^2 + \cdots+ c_jx^j) $$
Whereas Without loss of generality we can assume
$$ c_0 + c_1t + c_2t^2 + \cdots+ c_jt^j = \pm P$$
(the prime in question is $P$)
$$ b_0 + b_1t + b_2t^2 + \cdots+ b_rt^r = \pm 1 $$
From here I am not clear how to proceed.
Another Idea:
We can try to do something with induction. Lets start with the base case of the polynomial $b_0 + b_1x +\cdots $ being linear
$$ b_0 + b_1 t = 1 $$
Tells us that
$$ t = \frac{1 – b_0}{b_1} $$
But since $b_0 \ge 0, b_1 \ge 0 , t \ge 0$ it follows that $b_0 = 0$ (but then the value P never could have been prime since it would be divisible by t)
So we conclude that the polynomial has no linear factors that way. On the flip side it could be that
$$ b_0 + b_1 t = -1$$
$$ t = \frac{-1-b_0}{b_1} $$
Theres no $b_0 > 0$ that can make this expression greater than 0. So this case is covered.
Thus we conclude there are NO linear factors.
But I have no idea how to generalize this technique in a way that knocks out other polynomials too. especially given that from degree $5$ onwards there isn't even an algebraic formula for me to work with, expressing $t$.
Best Answer
I suggest the following solution:
But first, we need the following lemma:
Lemma 1
For all polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x+a_0 \in \mathbb{C[x]}$, if $z$ is a solution of the polynomial then \begin{align} |z| \le 1+ \max\{\frac{a_i}{a_n}|i\in\{0,1,2,\cdots,n\}\} \end{align}
I will leave this as an exercise for whoever reading this.
Now back to the problem
Solution
If $P(x)$ is reducible over $\mathbb{Q[x]}$, then it is reducible over $\mathbb{Z[x]}$, according to Gauss lemma.
hence, $P(x)=H(x) \times Q(x)$
Thus $P(10)=H(10)Q(10) \in \mathbb{P}$
So either $|H(10)|=1$ or $|Q(10)|=1$. Assume it's $H(x)$.
Denote $z_1,z_2, \cdots, z_k$ be the solutions of $H(x)$. Then,
\begin{cases} |z_i| \le 1+\frac92 \\ H(10)=\alpha (10-x_1)(10-x_2)\cdots(10-x_k) \end{cases}
But then $|H(10)|\ge\alpha\times(\frac{11}{2})^k >1$, a contradiction.
Hence we have QED. $\square$.