Uniformizer criterion for lower ramification groups

If $L/K$ is a finite totally ramified Galois extension and ${\pi }_L$ is a uniformizer, thenfor every $i\ge 0$.

For $L={\mathbb{Q}}_3({\zeta }_3,\alpha )$ with ${\alpha }^3=2$, the shifted cubic for $\alpha +1$ and quadratic for ${\zeta }_3-1$ are Eisenstein. Their compositum is totally ramified of degree six. The element $\pi =({\zeta }_3-1)/(\alpha +1)$ has valuation one. Order-three automorphisms move it with valuation two and transpositions with valuation one. The lower breaks are zero and one, the upper breaks zero and one-half, and the different exponent from ramification groups is seven.

For ${\mathbb{Q}}_2({\zeta }_8)/{\mathbb{Q}}_2$, the shifted polynomial $(1+T{)}^4+1$ is Eisenstein and ${\zeta }_8-1$ is a uniformizer. The three nonidentity automorphisms have displacement valuations two, four and two. Thus the lower breaks are one and three, the upper breaks one and two, and the different exponent from ramification groups is eight.

Let $K$ be an equal-characteristic-p local field and let $c\in K$ have $v_K(c)=-m<0$, with $p\nmid m$. If ${\alpha }^p-\alpha =c$, the Artin–Schreier extension $L=K(\alpha )$ has degree $p$ and is totally ramified: a root in $K$ would force the pole order to be divisible by $p$, and $p{v}_L(\alpha )=-me(L/K)$ forces $e=p$. Choose $1\le s<p$ and an integer $r$ with $rp-sm=1$. For a base uniformizer $t$, the element ${\pi }_L=t^r{\alpha }^s$ has valuation one. Under ${\sigma }_a(\alpha )=\alpha +a$, $a\in {\mathbb{F}}_p^{\times }$, the leading term in ${\sigma }_a({\pi }_L)-{\pi }_L$ has valuation $rp-(s-1)m=m+1$. The uniformizer criterion for lower ramification groups therefore gives the single lower break $m$. The Herbrand function is the identity up to that break, so the upper break is also $m$.

For the splitting field $M/{\mathbb{Q}}_3$ of $T^3-3T+3$, the cubic is Eisenstein and its discriminant is $-135$. The quadratic subfield is ${\mathbb{Q}}_3(\sqrt{3})$, and the degree-six extension is totally ramified. If $\alpha$ is a root and $s=\sqrt{3}$, then $\pi ={\alpha }^2/s$ is a uniformizer. Transpositions have $v_M(\sigma \pi -\pi )=1$, while three-cycles have value $2$, giving the displayed ramification groups. Its different exponent from ramification groups is $7$.

For odd $p$, the splitting field $K={\mathbb{Q}}_p({\zeta }_p,\sqrt[p]{p})$ is totally ramified of degree $p(p-1)$. Withthe lower ramification groups are

To verify the filtration, write ${\alpha }^p=p$ and $\lambda ={\zeta }_p-1$. The Eisenstein polynomials for $\alpha$ and $\lambda$ give coprime degrees $p$ and $p-1$, so their compositum has degree $p(p-1)$. Its element $\varpi =\lambda /\alpha$ has $p$-adic valuation $1/[p(p-1)]$, forcing total ramification. For an automorphism ${\sigma }_{a,b}$ with $\alpha \mapsto {\zeta }_p^a\alpha$ and ${\zeta }_p\mapsto {\zeta }_p^b$,Its residue is $b$, so the normalized displacement valuation is one if $b\ne 1$. If $b=1$ and $a\ne 0$, that valuation is $1+v_K({\zeta }_p^{-a}-1)=p+1$. The uniformizer criterion for lower ramification groups now gives every group displayed above.

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.