Valuations and different ideal in local fields

Question

We know that if we have a totally ramified finite extension of local fields $L/K$ then the different ideal is $D_{L/K} = ( h'(\pi))$ i.e. the ideal generated by te derivative of the minimal polinomial of $\pi$ (a uniformizer of L). My question is: since the uniformizer is not unique then if we consider another one uniformizer $\tilde{\pi}$ what happens to $D_{L/K}$ and its valuation is still the same? (i.e. the valuation of the minimal polynomial ?) Thank you for the answers!

Answer 1

Another uniformizer has the form $\tilde{\pi} = u \pi$ for some unit $u \in O_L^{\times}$. The minimal polynomial $\tilde h$ is just $\tilde h(X) = h(u^{-1}X)$, and in particular we get $$\tilde h'(\tilde{\pi}) = u^{-1} h'(u^{-1} \tilde{\pi}) = u^{-1} h(\pi).$$ Thus the ideal generated by $\tilde h'(\tilde{\pi})$ in $O_L$ is the same as the ideal generated by $h'({\pi})$.

Answer 2

Your question is a bit "fuzzy" because the formula $D_{L/K}=(h'(\pi))$ which you give is not the definition of the discriminant, but only a consequence in the case of a totally ramified extension .

1) If you admit that there is an intrinsic definition , then of course the formula does not depend on the choice of $\pi$. I recall here that $L/K$ is a finite extension of local fields, where a "local field" $K$ means that $K$ is the field of a complete discrete valuation ring $A_K$ whose residual field is perfect. The codifferent of $A_L$ over $A_K$ is the fractionary ideal consisting of the $y \in L$ s.t. $Tr(xy)\in A_K$ for all $x\in A_L$. The different ideal $D_{L/K}$ is defined as the inverse of the codifferent. It is a general property of local fields that $A_L$ is a monogeneous $A_K$-algebra, say $A_L=A_K[\alpha]$ for a certain $\alpha\in A_L$. From this one can derive that $D_{L/K}=(f'(\alpha))$, where $f$ is the minimal polynomial of $\alpha$. If $L/K$ is totally ramified, one can take for $\alpha$ a uniformizer $\pi$ of $L$.

2) If you start from the formula, the answer given by Watson is the most natural proof of the independence from the choice of $\pi$. If you suppose moreover that $L/K$ is Galois (not necessarily totally ramified), there is another natural proof which gives further information. For $s\in G:=Gal(L/K)$, let $i_G(s)=v_L(s(\alpha)-\alpha)$, and for any integer $i\ge -1$, define the ramification subgroup $G_i$ by $s\in G_i$ iff $i_G(s)\ge i+1$, iff $v_L(s(a)-a)\ge i+1$ for all $a\in A_L$. Since the minimal polynomial $f$ of $\alpha$ verifies $f(X)={\prod}_G (s(\alpha)-\alpha)$, one has $f'(\alpha)={\prod}_{s\neq1} (\alpha-s(\alpha))$, and so $v_L(D_{L/K})={\sum}_{s\neq1} i_G(s)$. Notice that the function $i_G$ is $1$ on $G_{i-1}$ minus $G_i$ and let $r_i=$ ord$G_i -1$. Then $v_L(D_{L/K})={\sum}_{s\neq1} i_G(s)=\sum_i i(r_{i-1}-r_i)$ independently from the choice of $\alpha$. For further details on the ramification subgroups, see Serre's "LocalFields", cap.IV, §1 .