Conductor of character What is the relation between conductor of a multiplicative character of a local field and the conductor of square of the character?  
 A: For a finite extension $K$ of $\mathbf{Q}_2$ of ramification index $e$, define the function $\lambda$ by $\lambda(n)=\mathrm{min}(2n, n+e)$ for every integer $n>0$.
Note that squaring $(\ )^2$ takes $U_n$ to $U_{\lambda(n)}$ and $U_{n+1}$ to $U_{\lambda(n)+1}$ for all $n>0$, and the resulting map
$$
U_n/U_{n+1}\to U_{\lambda(n)}/U_{\lambda(n+1)}
$$ 
is an isomorphism in all cases except for $n=e$, when you have an exact sequence
$$
1\to\mu_2\to U_e/U_{e+1}\to U_{2e}/U_{2e+1}\to\bar U_{2e}\to1
$$ 
where $\mu_2$ is the subgroup consisting of $1$ and $-1$, and $\bar U_{2e}$ is a group of order $2$ (in fact the image of $U_{2e}$ in $K^\times/K^{\times 2}$).  See for example Hasse's Number Theory, Chapter 15, or the presentation of the same material in Part III of 0711.3878.
From this remark you should be able to compute the conductor of $\chi^2$ in terms of the conductor of $\chi$, for every character $\chi:K^\times\to\mathbf{C}^\times$.
A: In case of odd residue characteristic, conductor of  the character and its square are same. In case when residue characteristic is 2, my observation is that conductor of square is related to valuation of 2(it falls down by v(2) aparently). can we expect an explicit formula in this case?
