Frobenius map in scheme theory Let $\mathbb{K}$ be a field of characteristic $p$ and let $f: \mathbb{A}_{\mathbb{K}}^{1} \mapsto \mathbb{A}_{\mathbb{K}}^{1}$ be the morphism of the form $x \mapsto x^p$. We consider $\mathbb{A}_{\mathbb{K}}^{1}$ to be an affine group scheme (the additive group). Let $Z$ be the kernel of the morphism $f$.
1) Is it an affine group scheme? 
2) If yes, which is an expression of the form $Z=SpecA$?
3) If we consider the multiplicative group $\mathbb{A}_{\mathbb{K}}^{1} \setminus \{0\}$, it's the same of 1) valid?
 A: The kernel of any morphism of affine group $k$-schemes is an affine group $k$-scheme. Indeed, if $f : G \to H$ is a morphism of group $k$-schemes, then the kernel is $G \times_H \{ e \}$, which is affine if $G$ and $H$ are, and it is straightforward (if a little tedious) to see that the group structure on $G$ restricts to a group structure on $G \times_H \{ e \}$.
Recall that $\mathbb{A}^1_k = \operatorname{Spec} k [x]$ and that the Frobenius homomorphism $\mathbb{A}^1_k \to \mathbb{A}^1_k$ comes from the unique homomorphism $k [x] \to k [x]$ sending $x$ to $x^p$. Accordingly, the kernel of the Frobenius morphism must be $\operatorname{Spec} k [x] / (x^p)$. Note that it is irreducible but not reduced.
For the multiplicative group, recall that $\mathbb{A}^1_k \setminus \{ 0 \} = \operatorname{Spec} k [x, x^{-1}]$. The $n$-th power homomorphism $\mathbb{A}^1_k \setminus \{ 0 \} \to \mathbb{A}^1_k \setminus \{ 0 \}$ comes from the unique morphism $k [x, x^{-1}] \to k [x, x^{-1}]$ sending $x$ to $x^n$. The kernel of the $n$-th power map is  $\operatorname{Spec} k [x] / (x^n - 1)$, which (unsurprisingly) is the space of $n$-th roots of unity. Of course, if $n > 0$ and $p \mid n$, then this will not be reduced.
