---
tag: 246SBC2M2N
kind: statement
role: theorem
name: Cohn's irreducibility criterion
msc: [11R09, 11C08, "12E05"]
uses: [00X2A9G9W3, 0SN0MAEKYX]
about: [00X2A9G9W3]
sources:
  - url: https://en.wikipedia.org/wiki/Cohn%27s_irreducibility_criterion
same_as:
  wikidata: Q2269888
  wikipedia: Cohn's irreducibility criterion
---

## Statement

Let $p \ge 10$ be a [[prime number|0SN0MAEKYX]] with decimal expansion $p = a_m 10^m + \dots + a_1 10 + a_0$, where $m \ge 1$, $a_i \in \set{0, 1, \dots, 9}$ and $a_m \ne 0$.
Then the polynomial $f = a_m x^m + \dots + a_1 x + a_0$ is [[irreducible|00X2A9G9W3]] in $\Z[x]$.

## Remarks

Proof idea: since the coefficients are digits, every complex root $\alpha$ of $f$ satisfies $\abs{10 - \alpha} > 1$.
So every factor $g \in \Z[x]$ of positive degree has $\abs{g(10)} > 1$.
A factorisation $f = g h$ into factors of positive degree would then write $p = f(10) = g(10) h(10)$ as a product of two integers of absolute value greater than $1$.
A constant factor is also excluded: it would divide all digits and $p$, but the digits are less than $p$.
Example: $113$ is prime, so $x^2 + x + 3$ is irreducible; indeed its discriminant $1 - 12 = -11$ is negative.
The theorem also holds in every base $b \ge 2$ in place of $10$.
The converse is false: $x^2 + 5$ is irreducible, but its value $105$ at $10$ is not prime.

## Explanations

### L1 human

Write a prime in decimal and read its digits as the coefficients of a polynomial.
That polynomial cannot be factored over the integers.
For example, the prime $1013$ gives $x^3 + x + 3$, which is irreducible.

### L2 human

The criterion links primality of a value with irreducibility of the polynomial.
The reverse direction, that an irreducible polynomial with suitable conditions takes infinitely many prime values, is the Bunyakovsky conjecture, which is open in every degree at least $2$.
The proof is a good example of locating roots in the complex plane to prove an algebraic fact, like [[Perron's criterion|5BCRQYVYR8]].