Cohn's irreducibility criterion

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.

Going deeper

For a student

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.