# Difference between revisions of "Cohn's criterion"

m |
m |
||

Line 30: | Line 30: | ||

− | {{ | + | {{stub}} |

## Revision as of 16:20, 14 August 2018

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 .

However, if Re , we have from our first lemma, that , so Re . Thus we have proved the lemma.

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.

*This article is a stub. Help us out by expanding it.*