# Difference between revisions of "Cohn's criterion"

m (→Proof) |
(→Proof) |
||

Line 25: | Line 25: | ||

For <math>|z|\geq \frac{3}{2}</math>, then <math>|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}>1</math>. Therefore, <math>z</math> is not a root of <math>f(x)</math>. | For <math>|z|\geq \frac{3}{2}</math>, then <math>|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}>1</math>. Therefore, <math>z</math> is not a root of <math>f(x)</math>. | ||

− | |||

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

[[Category:Algebra]] | [[Category:Algebra]] |

## Revision as of 08:07, 4 March 2021

Let be a prime number, and an integer. If is the base- representation of , and , then is irreducible.

## Proof

The following proof is due to M. Ram Murty.

We start off with a lemma. Let . Suppose , , and . Then, any complex root of , , has a non positive real part or satisfies .

Proof: If and Re , note that: This means if , so .

If , this implies if and . Let . Since , one of and is 1. WLOG, assume . Let be the roots of . This means that . Therefore, is irreducible.

If , we will need to prove another lemma:

All of the zeroes of satisfy Re .

Proof: If , then the two polynomials are and , both of which satisfy our constraint. For , we get the polynomials , , , and , all of which satisfy the constraint. If ,

If Re , we have Re , and then For , then . Therefore, is not a root of .

To finish the proof, let . Since , one of and is 1. WLOG, assume . By our lemma, . Thus, if are the roots of , then . This is a contradiction, so is irreducible.