I am reading 'Prime numbers and Irreducible polynomials' by Prof. M. Ram Murty, where he gives a proof of Cohn's Irreducibility theorem. I am posting the proof first and then will ask my questions.
Theorem: Let $b>2$ and let $p$ be a prime expanded as
$$
p=a_nb^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0 \ \ \text{where} \ \ \ 0 \le a_i \le b-1.
$$
Then the polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ is irreducible in $\mathbb{Q}[X]$.
In order to prove this theorem a lemma is used, which I state below. The proof of which is given here.
Lemma:
Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ belong to $\mathbb{Z}[X]$. Suppose that $a_n\ge1, \ a_{n-1}\ge 0 \ $ and $|a_i|\le H$ for $i=0,1,…,n-2,$ where $H$ is some positive constant. Then any complex zero $\alpha$ of $f(x)$ either has $\mathfrak{R}(\alpha)\le 0$ or satisfies
$$
|\alpha| < \frac{1+\sqrt{1+4H}}{2}
$$
Proof:
By Gauss lemma it suffices to consider reducibility over $\mathbb{Z}[X]$.
If $f(x)=g(x)h(x)$ with $g(x)$ and $h(x)$ noncostant polynomials in $\mathbb{Z}[X]$, then
$$
f(b)=p \implies \ \ \text{either} \ \ g(b)=\pm1 \ \ \text{or} \ \
h(b)=\pm1.
$$
WLOG assume $g(b)=\pm1$. Now $g$ is of the form
$$
g(x)=c\prod_i(x-\alpha_i)
$$
where $\alpha_i$ range over a certain subset of the zeros of $f$ and $c$ is the leading coefficient of $g(x)$.
By the lemma, every zero $\alpha$ of $f$ either has nonpositive real part or has an absolute value less than
$$
\frac{1+\sqrt{1+4(b-1)}}{2}
$$
In the former case, we simply have $|b-\alpha|\ge b$.
In the latter case, the fact that $b$ is atleast 3 gives
$$
|\alpha| < \frac{1+\sqrt{1+4(b-1)}}{2} \le b-1
$$
In particular, $|b-\alpha_i|>1$ for each $i$, from which we deduce that $g(b)>1$. Contradiction!
I have the following questions:
1. The author states that the proof breaks down for $b=2$. I cannot find how. Can someone please tell me how?
2. In this inequality below, used in the last part of the proof:
$$\frac{1+\sqrt{1+4(b-1)}}{2} \le b-1$$
does equality ever holds?
Edit:
3. How do we get the inequality
$$
\frac{1+\sqrt{1+4(b-1)}}{2} \le b-1 ?
$$
My thoughts for it is as follows:
If we consider the polynomial $x^2-x-(b-1)$. It has 2 roots
$$
x_1=\frac{1+\sqrt{1+4(b-1)}}{2} \ \ \text{and} \ \ x_2=\frac{1-\sqrt{1+4(b-1)}}{2}
$$
with $x_1x_2=-(b-1)$. Now note that since $b \ge 3$ we have both $|x_1|\ge 1$ and $|x_2|\ge 1$. Hence
$$
x_1=|x_1| \le |x_1x_2|=b-1
$$
Are the above line of arguments correct?
Best Answer
Your proof for (3) is good.
But it might be more illuminative to isolate the square root: $$\sqrt{4b-3} = \sqrt{1 + 4(b-1)} \le 2b - 3$$ If $b \ge \frac 32$, then both sides are non-negative, Since $f(x) =x^2$ is a strictly increasing function on the non-negative numbers, we can square both sides without changing the inequality: $$4b-2 \le (2b-3)^2\\0 \le 4b^2 - 16b + 12\\0\le 4(b-1)(b-3)$$ Which for $b\ge \frac 32$ is true exactly when $b \ge 3$.
Since every step taken is reversible, this tells us
Below $\frac 32$, a slightly different calculation is needed.