# M. Ram Murty's proof of Cohn's Irreducibility criteria.

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?

• question 2 is a simple algebra problem. Replace the inequality with an equality and solve for $b$. You will find a single solution. Jun 30, 2021 at 22:45
• And for question 1, that same inequality fails. $\frac{1+\sqrt{5}}2 \not\le 1$. Jun 30, 2021 at 22:50
• @PaulSinclair : Thankyou for clearing my doubts. I have edited and added a question. I have also added my views on it. If possible, please see and tell if it is ok? Jul 1, 2021 at 0:12

## 1 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

• For $$b \in \left[\frac 32,1\right), \frac{1+\sqrt{1+4(b-1)}}{2} > b-1$$
• For $$b = 3, \frac{1+\sqrt{1+4(b-1)}}{2} = b-1$$
• For $$b > 3, \frac{1+\sqrt{1+4(b-1)}}{2} \le b-1$$

Below $$\frac 32$$, a slightly different calculation is needed.