 Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3

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.