Capelli's theorem

Remarks

The conditions are necessary: if $a = b^p$ with $p \mid n$, then $y^p - b^p$ has the factor $y - b$, and $y = x^{n/p}$ gives a factor of $x^n - a$. If $a = -4c^4$ and $4 \mid n$, the identity $y^4 + 4c^4 = (y^2 + 2cy + 2c^2)(y^2 - 2cy + 2c^2)$ with $y = x^{n/4}$ gives a factorisation. The proof of sufficiency first treats $n$ prime with a norm argument, then prime powers, and combines them with the multiplicativity of degrees in towers of extensions. Example: $x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)$ over $\Q$, although $-4$ is not a square in $\Q$; this is the case $c = 1$ of clause 2. Example: $x^n - 2$ is irreducible over $\Q$ for every $n \ge 1$. For $n$ odd, clause 2 is empty and the test reduces to $p$-th powers for the primes $p \mid n$.

Going deeper

For a student

The theorem is the basic tool to study binomial equations $x^n = a$ and the radical extensions $K(\sqrt[n]{a})$. It gives the degree $[K(\sqrt[n]{a}) : K] = n$ when the two clauses hold. This is used in the theory of radical extensions and of Kummer extensions.