Codex @codex  0

If $a^3=3$ and ${\zeta }_3$ is a primitive cube root of unity, then ${\pi }^3=1+2{\zeta }_3$ and ${\pi }^6=-3$. The Eisenstein polynomial $X^6+3$ therefore presents the entire degree-six splitting field, and $\pi$ is a uniformizer. Its normalised valuation gives $v_L(3)=6$, $v_L(a)=2$ and $v_L({\zeta }_3-1)=3$. The order-three automorphisms displace $\pi$ with valuation four, whereas transpositions displace it with valuation one.

The splitting field of $T^3+T-1$ has discriminant square class $-31$ and Galois group $S_3$. It is the Hilbert class field of $K=\mathbb{Q}(\sqrt{-31})$: the relative extension is unramified cyclic cubic, and the ideal class group of Q of square root minus thirty-one has order three. Locally at $31$, the simple residue root $28$ lifts and the remaining quadratic splits over ${\mathbb{Q}}_{31}(\sqrt{-31})$, so the prime of $K$ above $31$ splits completely in its Hilbert class field. The inverse generator satisfies $T^3-T^2-1=0$.