= Solution
The <polynomial discriminant> satisfies
$$
\operatorname{disc}(f)=(-1)^{n(n-1)/2}\operatorname{Res}(f,f'),\qquad n=\deg f.
$$
A polynomial has no repeated roots over the algebraic closure exactly when it is coprime to its derivative. Consequently the assumed squarefree reduction has nonzero <resultant> with its derivative, so $p\nmid\operatorname{disc}(f)$. On the other hand the power basis is a basis of the <order in a number field> $\mathbb Z[\alpha]$, and the <discriminant-index formula for an integral lattice> gives
$$
\operatorname{disc}(f)=\operatorname{disc}(1,\alpha,\ldots,\alpha^{n-1})=[\mathcal O_L:\mathbb Z[\alpha]]^2d_L.
$$
Here the <field discriminant> $d_L$ is an integer. Therefore
$$
\boxed{p\nmid[\mathcal O_L:\mathbb Z[\alpha]].}
$$
This is <squarefree reduction excludes a prime from the power-basis index>; its argument establishes the index condition before applying the <Kummer-Dedekind theorem>.
Back to article page
