What is $\text{Res}_{\mathbb{F}_p(t)/\mathbb{F}_p(t^p)}(\text{Spec}\,\mathbb{F}_p(t)[x]/(x^p - t))$?

Question

Let $L = \mathbb{F}_p(t)$ and $k = \mathbb{F}_p(t^p)$. Let $X$ be the $L$-scheme $\text{Spec}\,L[x]/(x^p - t)$. What is $\text{Res}_{L/k}\,X$?

Here $\text{Res}$ denotes the Weil restriction of scalars.

Answer 1

The answer is the empty scheme. We use the basis $\{1, t, \ldots, t^{p - 1}\}$ for $L/k$. To find the restriction of scalars, we substitute$$x = y_0 + ty_1 + \ldots + t^{p - 1}y_{p - 1}$$into the equation $x^p - t = 0$ defining $L$, and rewrite the result as$$F_0 + f_1t + \ldots + F_{p - 1} = 0$$with $F_i \in k[y_0, \ldots, y_{p - 1}]$. We get\begin{align*} F_0 & = y_0^p - t^p y_1^p + \ldots + t^{p(p - 1)}y_{p - 1}^p,\\ F_1 & = -1, \\ F_i & = 0 \text{ for }i \ge 2. \end{align*} Then $\text{Res}_{L/k}X$ is $\text{Spec}\,k[y_0, \ldots, y_{p - 1}]/(F_0, \ldots, F_{p - 1})$. This is the empty scheme, since $F_1 = -1$ generates the unit ideal.