Difference between revisions of "Cohn's criterion"

Revision as of 07:29, 4 March 2021

Let $p$ be a prime number, and $b\geq 2$ an integer. If $\overline{p_np_{n-1}\cdots p_1p_0}$ is the base- $b$ representation of $p$ , and $0\leq p_i<b$ , 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 $b=2$ , we will need to prove another lemma:

All of the zeroes of $f(x)$ satisfy Re $z<\frac{3}{2}$ .

Proof: If $n=1$ , 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 $n\geq 3$ , $|\frac{f(z)}{z^n}|\geq |1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-(\frac{1}{|z|^3}+\cdots+\frac{1}{|z|^n})>|1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-\frac{1}{|z|^2(|z|-1)}$

If Re $z\geq 0$ , 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 $f(x)$ .

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 $|g(2)|=|2-\phi_1||2-\phi_2|\cdots|2-\phi_n|>|1-\phi_1||1-\phi_2|\cdots|1-\phi_n|=|g(1)| < 1$ . This is a contradiction, so $f(x)$ is irreducible.

@@ Line 12: / Line 12: @@
 This means <math>g(z)>0</math> if <math>|z|\geq \frac{1+\sqrt{1+4H}}{2}</math>, so <math>|\phi|<\frac{1+\sqrt{1+4H}}{2}</math>.
-If <math>b\geq 3</math>, this implies <math>|b-\phi|>1</math> if <math>b\geq 3</math> and <math>f(\phi)=0</math>. Let <math>f(x)=g(x)h(x)</math>. Since <math>f(b)=p</math>, one of <math>|g(b)|</math> and <math>h(b)</math> is 1. WLOG, assume <math>g(b)=1</math>. Let <math>\phi_1, phi_2,\cdots,\phi_r</math> be the roots of <math>g(x)</math>. This means that <math>|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1</math>. Therefore, <math>f(x)</math> is irreducible.
+If <math>b\geq 3</math>, this implies <math>|b-\phi|>1</math> if <math>b\geq 3</math> and <math>f(\phi)=0</math>. Let <math>f(x)=g(x)h(x)</math>. Since <math>f(b)=p</math>, one of <math>|g(b)|</math> and <math>h(b)</math> is 1. WLOG, assume <math>g(b)=1</math>. Let <math>\phi_1, \phi_2,\cdots,\phi_r</math> be the roots of <math>g(x)</math>. This means that <math>|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1</math>. Therefore, <math>f(x)</math> is irreducible.
 If <math>b=2</math>, we will need to prove another lemma:

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Difference between revisions of "Cohn's criterion"

Revision as of 07:29, 4 March 2021

Proof