Question related to trace function I was reading this document called Discriminants and Ramified Primes.
Suppose $O_K$ is the ring of algebraic integers of a number field $K$.
Let $\mathcal{P}$ be a prime ideal of $O_K$ above $p$ such that
$O_K/\mathcal{P}$ is a finite field of characteristic $p$.
Near the bottom of page 4 it states:
"We want to prove $\text{disc}_{\mathbb{Z}/p\mathbb{Z}} (O_K/\mathcal{P}) \not = \bar{0}$.
If this discriminant is  $\bar{0}$ then (because $O_K/\mathcal{P}$ is a field) the trace function $\operatorname{Tr} : O_K / \mathcal{P} \rightarrow \mathbb{Z}/p\mathbb{Z}$ is identically zero. "
I don't quite understand why the trace would be identically zero in this situation.
Could someone give me an explanation for this? Thank you very much!
 A: This has nothing to do with number theory, finite fields, or characteristic $p$, but is a general feature of discriminants of field extensions. 
Let $L/K$ be a finite extension of fields.  The discriminant of a basis $\{e_1,\dots,e_n\}$ of this extension is defined to be the determinant of the matrix $({\rm Tr}_{L/K}(e_ie_j))$. We want to show this determinant is $0$ if and only if the trace mapping ${\rm Tr}_{L/K} \colon L \rightarrow K$ is identically $0$. (Changing the basis for $L/K$ changes the determinant by a nonzero square factor in $K$, so saying that the determinant is $0$ is independent of the choice of basis.)
Obviously if the trace mapping is identically $0$ then the matrix is $O$, so the determinant is $0$. 
Now assume, conversely, that $\det({\rm Tr}_{L/K}(e_ie_j)) = 0$. A square matrix with entries in $K$ and determinant $0$ has a nontrivial kernel in $K^n$, so there is a nonzero (column) vector $v = (c_1,\dots,c_n)$ in $K^n$ such that $({\rm Tr}_{L/K}(e_ie_j))v = 0$.
Write this vector equation as a set of $n$ linear equations (and abbreviate ${\rm Tr}_{L/K}$ to ${\rm Tr}$):
\begin{eqnarray*}
{\rm Tr}(e_1e_1)c_1 + \cdots + {\rm Tr}(e_1e_n)c_n & = & 0 \\
{\rm Tr}(e_2e_1)c_1 + \cdots + {\rm Tr}(e_2e_n)c_n & = & 0 \\
 & \vdots & \\
{\rm Tr}(e_ne_1)c_1 + \cdots + {\rm Tr}(e_ne_n)c_n & = & 0.
\end{eqnarray*}
Since ${\rm Tr}$ is $K$-linear, we can rewrite this system of equations as
\begin{eqnarray*}
{\rm Tr}(e_1(c_1e_1 + \cdots + c_ne_n)) & = & 0 \\
{\rm Tr}(e_2(c_1e_1 + \cdots + c_ne_n)) & = & 0 \\
 & \vdots & \\
{\rm Tr}(e_n(c_1e_1 + \cdots + c_ne_n)) & = & 0.
\end{eqnarray*}
Therefore the number $\alpha = c_1e_1 + \cdots + c_ne_n$ in $L$ has the property that ${\rm Tr}(e_i\alpha) = 0$ for all basis vectors $e_i$. Every element of $L$ is a $K$-linear combination of $e_1, \dots, e_n$, so by $K$-linearity of ${\rm Tr}$ the equations ${\rm Tr}(e_i\alpha) = 0$ for all $i$ imply ${\rm Tr}(\beta\alpha) = 0$ for all $\beta \in L$. The number $\alpha$ is not $0$ because its coefficients $c_1,\dots,c_n$ in the basis we are using are not all $0$. Therefore every element $\gamma$ of $L$ has the form $\beta\alpha$ for some $\beta \in L$ (namely for $\beta := \gamma/\alpha$). Thus ${\rm Tr}(\gamma) = 0$ for all $\gamma$ in $L$, which is saying that the trace mapping from $L$ to $K$ is identically $0$.
A: Assume that $\text{disc}_{\mathbb{Z}/p\mathbb{Z}} \left(\mathcal O_K/\mathcal{P}\right) = \overline 0$. Since $\text{disc}_{\mathbb{Z}/p\mathbb{Z}} \left(\mathcal O_K/\mathcal{P}\right)$ is the determinant of the trace form of $\mathcal O_K / \mathcal P$ over $\mathbb{Z}/p\mathbb{Z}$, this yields that the trace form of $\mathcal O_K / \mathcal P$ over $\mathbb{Z}/p\mathbb{Z}$ is degenerate. In other words, there exists a nonzero $x \in \mathcal O_K / \mathcal P$ such that every $y \in \mathcal O_K / \mathcal P$ satisfies $\operatorname{Tr}_{\left(\mathcal O_K / \mathcal P\right) / \left(\mathbb{Z}/p\mathbb{Z}\right)} \left(xy\right) = 0$. But this $x$ must be invertible (since it is nonzero and since $\mathcal O_K / \mathcal P$ is a field), so that every $z \in \mathcal O_K / \mathcal P$ can be written in the form $xy$ for some $y \in \mathcal O_K / \mathcal P$. Hence, $\operatorname{Tr}_{\left(\mathcal O_K / \mathcal P\right) / \left(\mathbb{Z}/p\mathbb{Z}\right)}\left(z\right) = 0$ for every $z \in \mathcal O_K / \mathcal P$. In other words, $\operatorname{Tr}_{\left(\mathcal O_K / \mathcal P\right) / \left(\mathbb{Z}/p\mathbb{Z}\right)} = 0$.
