= Integral basis of a nonexceptional pure cubic field
{title2=$\mathcal O_{\mathbb Q(\sqrt[3]{m})}=\mathbb Z[\sqrt[3]{m}]$}
If $m>1$ is a <square-free integer>, $3\nmid m$, and $m\not\equiv\pm1\pmod9$, then $1,\sqrt[3]{m},\sqrt[3]{m^2}$ is an <integral basis> of the <pure cubic number field>, with <field discriminant> $-27m^2$. The <discriminant-index formula for an integral lattice> restricts a possible index to primes dividing $3m$. At primes dividing $m$ the <different exponent> is $2$ by <tame ramification>. A <shifted Eisenstein polynomial> gives total wild ramification at $3$, so its <different exponent> is at least $3$. These exponents exhaust the polynomial <discriminant> and force index one.
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