= Solution
For the finite <separable field extension> define the <inverse different>, or <codifferent>, by
$$
\mathfrak D_{L/K}^{-1}
=\{x\in L:\operatorname{Tr}_{L/K}(x\mathcal O_L)
\subseteq\mathcal O_K\}.
$$
It is a <fractional ideal> of $\mathcal O_L$. The <different ideal> is its inverse <fractional ideal>.
Suppose the integer ring is generated by $\alpha$, with monic <minimal polynomial of an algebraic element> $g$ of degree $d$. We prove the derivative formula using the <coefficient pairing for a monogenic algebra>. If $\alpha_1,\ldots,\alpha_d$ are its distinct conjugate <polynomial roots>, <Lagrange interpolation> for a <polynomial> $h$ of degree less than $d$ gives
$$
h(T)=\sum_{i=1}^d h(\alpha_i)
\frac{g(T)}{(T-\alpha_i)g'(\alpha_i)}.
$$
Comparing <coefficients> of $T^{d-1}$ yields
$$
\operatorname{Tr}_{L/K}\!\left(\frac{h(\alpha)}{g'(\alpha)}\right)
=[T^{d-1}]h(T).
$$
For any integral <polynomial>, first reduce modulo $g$; the <coefficients> stay integral because $g$ is monic over $\mathcal O_K$.
It follows that $g'(\alpha)^{-1}\mathcal O_L$ lies in the <codifferent>. To prove equality, consider the <matrix>
$$
A_{ij}=\operatorname{Tr}_{L/K}
\left(\frac{\alpha^{i+j}}{g'(\alpha)}\right),
\qquad 0\le i,j\le d-1.
$$
All entries are integral. Entries with $i+j<d-1$ are zero, and those with $i+j=d-1$ are one. Reversing the column order makes the <matrix> triangular with diagonal ones; thus its <determinant> is a <unit> of $\mathcal O_K$.
Write an arbitrary $x\in L$ as $x=g'(\alpha)^{-1}\sum_j b_j\alpha^j$ with $b_j\in K$. Membership in the <codifferent> says $A(b_j)_j\in\mathcal O_K^d$. Since $A$ is invertible over $\mathcal O_K$, this forces every $b_j$ integral. Therefore
$$
\mathfrak D_{L/K}^{-1}=g'(\alpha)^{-1}\mathcal O_L,
\qquad
\boxed{\mathfrak D_{L/K}=(g'(\alpha)).}
$$
Separability ensures that $g'(\alpha)\ne0$; the <trace pairing> calculation also proves that no extra factor has been overlooked.
Back to article page
