 Solution

Now let $n=m=p$ be prime and put $\alpha =\sqrt[p]{p}$ and $A=\mathbb{Z}[\alpha ]$. The Eisenstein criterion applied at $p$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the discriminant of elements of a number field of its power basis iswhere $f(X)=X^p-p$. By the discriminant-index formula for an integral lattice, every prime dividing $[{\mathcal{O}}_L:A]$ must therefore be $p$.

Suppose $A\ne {\mathcal{O}}_L$. The finite abelian group ${\mathcal{O}}_L/A$ is then a nontrivial finite p-group, so it contains an element of order $p$. Consequently there is an $x\in {\mathcal{O}}_L\setminus A$ such thatwith some $c_k$ not divisible by $p$.

The polynomial $X^p-p$ is Eisenstein, so total ramification from an Eisenstein polynomial gives a unique prime ideal $P$ above $p$ with normalized discrete valuationLet $k_0$ be the least index for which $p\nmid c_{k_0}$. The term $c_{k_0}{\alpha }^{k_0}$ has valuation $k_0<p$; every earlier term has valuation at least $p$, and every later term with coefficient prime to $p$ has a distinct, larger valuation. The non-Archimedean valuation therefore givesThis contradicts $x\in {\mathcal{O}}_L$, since an algebraic integer has nonnegative valuation at every prime ideal. Thus $A={\mathcal{O}}_L$, and