show · lf.discriminant_exponent all knowls · up · search:

Let $L/K$ be a finite extension of $p$-adic fields, with rings of integers $\mathcal{O}_L$ and $\mathcal{O}_K$ and uniformizers $\pi_L$ and $\pi_K$. The discriminant of $L/K$ is the square of the determinant of the matrix \[ \left( \begin{array}{ccc} \sigma_1(\beta_1) & \cdots & \sigma_1(\beta_n) \\ \vdots & & \vdots \\ \sigma_n(\beta_1) & \cdots & \sigma_n(\beta_n) \\ \end{array} \right) \] where $\sigma_1,..., \sigma_n$ are the embeddings of $L$ into an algebraic closure $\overline{K}$, and $\{\beta_1, \ldots, \beta_n\}$ is a basis for $\mathcal{O}_L$ as a free $\mathcal{O}_K$-module.

The discriminant of $L/K$ is an element of $K^\times$ which is well-defined up to the square of a unit. Thus, it is of the form $\pi_K^c u$ where $u \in \mathcal{O}_K^\times$ is a unit. The value $c$ is the discriminant exponent for $L/K$. Together with the discriminant root field of $L/K$, it determines the discriminant of $L/K$ (up to the square of a unit).

The base discriminant exponent $c_0$ of $L/K$ is the discriminant exponent of $K/\Q_p$. The absolute discriminant exponent $c_{\mathrm{abs}}$ of $L/K$ is the discriminant exponent of $L/\Q_p$.

Authors:
Knowl status:
  • Review status: reviewed
  • Last edited by David Roe on 2024-11-12 00:43:40
Referred to by:
History: (expand/hide all) Differences (show/hide)