# Cohn's criterion

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Let $p$ be a prime number, and $b\geq 2$ an integer. If $\overline{p_np_{n-1}\cdotsp_1}$ (Error compiling LaTeX. ! Undefined control sequence.) is the base- $b$ representation of $p$, and $0\leq p_i, then $$f(x)=p_nx^n+p_{n-1}x^{n-1}+\cdots+P_1x+p_0$$ is irreducible.

## Proof

The following proof is due to M. Ram Murty.

We start off with a lemma. Let $g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]$. Suppose $a_n\geq 1$, $a-{n-1}\geq 0$, and $|a_i|\leq H$. Then, any complex root of $f(x)$, $\phi$, has a non positive real part or satisfies $|\phi|<\frac{1+\sqrt{1+4H}}{2}$.

Proof: If $|z|>1$ and Re $z>0$, note that: $$|\frac{g(z)}{z^n}|\geq |a_n+\frac{a_{n-1}}{z}|-H(\frac{1}{|z|^2}+\cdots+\frac{1}{|z|^n})>Re(a_n+\frac{a_{n-1}}{z})-\frac{H}{|z|^2-|z|}\geq 1-\frac{H}{|z|^2-|z|}=\frac{|z|^2-|z|-H}{|z|^2-|z|}$$ This means $g(z)>0$ if $|z|\geq \frac{1+\sqrt{1+4H}}{2}$, so $|\phi|<\frac{1+\sqrt{1+4H}}{2}$.

If $b\geq 3$, this implies $|b-\phi|>1$ if $b\geq 3$ and $f(\phi)=0$. Let $f(x)=g(x)h(x)$. Since $f(b)=p$, one of $|g(b)|$ and $h(b)$ is 1. WLOG, assume $g(b)=1$. Let $\phi_1, phi_2,\cdots,\phi_r$ be the roots of $g(x)$. This means that $|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1$. Therefore, $f(x) is irreducible. If$ (Error compiling LaTeX. ! Missing $inserted.)b=2$, we will need to prove another lemma:

All of the zeroes of$(Error compiling LaTeX. ! Missing$ inserted.)f(x) $satisfy Re$z>\frac{3}{2}$. Proof: If$ (Error compiling LaTeX. ! Missing $inserted.)n=1 $, then the two polynomials are$x $and$x\pm 1 $, both of which satisfy our constraint. For$n=2 $, we get the polynomials$x^2 $,$x^2\pm x $,$x^2\pm 1 $, and$x^2\pm x\pm 1 $, all of which satisfy the constraint. If$n\geq 3$, <cmath>|\frac{f(z)}{z^n}|\geq |1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-(\frac{1}{|z|^3}+\cdots+\frac{1}{|z|^n})>|1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-\frac{1}{|z|^2(|z|-1)}</cmath>

If Re$(Error compiling LaTeX. ! Missing$ inserted.)z\geq 0 $, we have Re$\frac{1}{z^2}\geq 0 $, and then |\frac{f(z)}{z^n}|>1-\frac{1}{|z|^2(|z|-1)} For$|z|\geq \frac{3}{2} $, then$|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}>1 $. Therefore,$z $is not a root of$f(x).

However, if Re $z<0$, we have from our first lemma, that $|z|<\frac{1+\sqrt{5}}{2}$, so Re $z<\frac{1+\sqrt{5}}{2\sqrt{2}}<\frac{3}{2}$. Thus we have proved the lemma.

To finish the proof, let $f(x)=g(x)h(x)$. Since $f(2)=p$, one of $g(2)$ and $h(2)$ is 1. WLOG, assume $g(2)=1$. By our lemma, $|z-2|>|z-1|$. Thus, if $\phi_1, \phi_2,\cdots,\phi_r$ are the roots of $g(x)$, then $|g(2)|=|2-\phi_1||2-\phi_2|\cdots|2-\phi_n|>|1-\phi_1||1-\phi_2|\cdots|1-\phi_n|=|g(1)|\leq 1$. This is a contradiction, so $f(x)$ is irreducible.