= Ring of integers of Q of pth root p
{title2=$\mathcal O_{\mathbb Q(\sqrt[p]{p})}=\mathbb Z[\sqrt[p]{p}]$}
For a <prime number> $p$ and $\alpha=\sqrt[p]{p}$, the polynomial $X^p-p$ is <Eisenstein> at $p$, and
$$
\mathcal O_{\mathbb Q(\alpha)}=\mathbb Z[\alpha].
$$
The <discriminant-index formula for an integral lattice> reduces the proof to the prime $p$. At the unique prime ideal above $p$, one has $v(\alpha)=1$ and $v(p)=p$, which rules out an integral element $g(\alpha)/p$ with $g\notin p\mathbb Z[X]$.
Back to article page
