Diagonalization of a cyclic shift over a field ($S\text{ diagonalizable over }K⟺X^n-1\text{ splits into distinct linear factors over }K$)

Let $S$ send successive standard basis vectors of $K^n$ to the next one, wrapping the last to the first. Its minimal polynomial is $X^n-1$: this polynomial annihilates it, and the cyclic vectors $e_1,S{e}_1,\dots ,S^{n-1}{e}_1$ form a basis, excluding any annihilator of smaller degree. The diagonalizability criterion using the minimal polynomial gives the displayed equivalence. A splitting polynomial is square-free exactly when the characteristic of a field does not divide $n$, as follows by taking its derivative $n{X}^{n-1}$ and checking common roots. Splitting over the specified field is a separate requirement. For $n=3$, the shift is diagonalizable over the complex numbers and ${\mathbb{F}}_7$, but not over ${\mathbb{F}}_2$ or ${\mathbb{F}}_3$.