 Integral basis of a nonexceptional pure cubic field

If $m>1$ is a square-free integer, $3\nmid m$, and $m\not\equiv \pm 1(\mathrm{mod}9)$, 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.