 Solution

The inverse different is the trace-dual latticeThe integral closure ${\mathcal{O}}_L$ is finite free over the complete discrete valuation ring ${\mathcal{O}}_K$. Choose an integral basis and use the nondegenerate trace pairing to form its dual basis over $K$. This exhibits ${\mathfrak{D}}^{-1}$ as a full, finite ${\mathcal{O}}_K$-lattice. It is stable under multiplication by ${\mathcal{O}}_L$, because $b{\mathcal{O}}_L\subseteq {\mathcal{O}}_L$ for $b\in {\mathcal{O}}_L$, and bounded denominators make it a fractional ideal of ${\mathcal{O}}_L$. Integral elements have integral field traces, so ${\mathcal{O}}_L\subseteq {\mathfrak{D}}^{-1}$. A nonzero fractional ideal of a discrete valuation ring is invertible; its inverse is consequently an integral ideal, the different ideal ${\mathfrak{D}}_{L/K}$.

Assume now $A={\mathcal{O}}_K[x]={\mathcal{O}}_L$, with monic separable minimal polynomial $g$ of degree $n$. Lagrange interpolation at its distinct roots givesIndeed, these are the leading coefficients in the interpolation formula for $X^j$. For any element of $A$, this trace is the coefficient of $X^{n-1}$ in its degree-less-than-$n$ representative modulo $g$. The resulting pairing on the basis $1,x,\dots ,x^{n-1}$ is integral and unimodular: reversing the order of one basis makes its matrix triangular with diagonal ones, since entries vanish when the exponent sum is less than $n-1$. Thus it identifies $A$ with its full ${\mathcal{O}}_K$-dual. Translating back to the trace pairing proves

For a totally ramified extension of degree $n$, any uniformizer ${\pi }_L$ is an Eisenstein generator of a totally ramified extension, and ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. To see the latter equality, use the common residue field to expand an integral element in powers of ${\pi }_L$ with digits from ${\mathcal{O}}_K$; reduce powers using its Eisenstein polynomial and take limits in the finite complete ${\mathcal{O}}_K$-module generated by $1,{\pi }_L,\dots ,{\pi }_L^{n-1}$. Write that polynomial as $X^n+{\sum }_{j<n}{a}_j{X}^j$, with $v_L(a_j)\ge n$ for $j<n$. When $p\nmid n$, the derivative's leading term has valuation $n-1$, while every other nonzero derivative term has valuation at least $n$. There can be no cancellation of the unique smallest term. Therefore

For the prime-power p-adic cyclotomic extension, put $\zeta ={\zeta }_{p^r}$ and $e=p^{r-1}(p-1)$. Modulo $p$, the shifted cyclotomic polynomial ${\Phi }_{p^r}(1+T)$ is $T^e$, while its constant term is $p$, not a multiple of $p^2$. It is therefore Eisenstein, so $\pi =\zeta -1$ is a uniformizer and ${\mathcal{O}}_{K_r}={\mathbb{Z}}_p[\zeta ]$. Fromwe obtain, by differentiating the numerator at $\zeta$,The denominator is a primitive-$p$ root of unity minus one and has $K_r$-valuation $p^{r-1}$. The different exponent is consequently $re-p^{r-1}$, givingThis includes $p=2,r=1$: the extension is trivial and its different exponent is zero.