# Difference between revisions of "Cohn's criterion"

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Let <math>p</math> be a prime number, and <math>b\geq 2</math> an integer. If <math>\overline{p_np_{n-1}\cdots p_1p_0}</math> is the base-<math>b</math> representation of <math>p</math>, and <math>0\leq p_i<b</math>, then | Let <math>p</math> be a prime number, and <math>b\geq 2</math> an integer. If <math>\overline{p_np_{n-1}\cdots p_1p_0}</math> is the base-<math>b</math> representation of <math>p</math>, and <math>0\leq p_i<b</math>, then | ||

− | <cmath>f(x)=p_nx^n+p_{n-1}x^{n-1}+\cdots+ | + | <cmath>f(x)=p_nx^n+p_{n-1}x^{n-1}+\cdots+p_1x+p_0</cmath> |

is irreducible. | is irreducible. | ||

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− | + | [[Category:Algebra]] |

## Revision as of 03:48, 12 February 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 .

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.