Lemma 33.36.10. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a variety over $k$. The following are equivalent

1. $X^{(p)}$ is reduced,

2. $X$ is geometrically reduced,

3. there is a nonempty open $U \subset X$ smooth over $k$.

In this case $X^{(p)}$ is a variety over $k$ and $F_{X/k} : X \to X^{(p)}$ is a finite dominant morphism of degree $p^{\dim (X)}$.

Proof. We have seen the equivalence of (1) and (2) in Lemma 33.36.9. We have seen that (2) implies (3) in Lemma 33.25.7. If (3) holds, then $U$ is geometrically reduced (see for example Lemma 33.12.6) and hence $X$ is geometrically reduced by Lemma 33.6.8. In this way we see that (1), (2), and (3) are equivalent.

Assume (1), (2), and (3) hold. Since $F_{X/k}$ is a homeomorphism (Lemma 33.36.6) we see that $X^{(p)}$ is a variety. Then $F_{X/k}$ is finite by Lemma 33.36.8. It is dominant as it is surjective. To compute the degree (Morphisms, Definition 29.51.8) it suffices to compute the degree of $F_{U/k} : U \to U^{(p)}$ (as $F_{U/k} = F_{X/k}|_ U$ by Lemma 33.36.5). After shrinking $U$ a bit we may assume there exists an étale morphism $h : U \to \mathbf{A}^ n_ k$, see Morphisms, Lemma 29.36.20. Of course $n = \dim (U)$ because $\mathbf{A}^ n_ k \to \mathop{\mathrm{Spec}}(k)$ is smooth of relative dimension $n$, the étale morphism $h$ is smooth of relative dimension $0$, and $U \to \mathop{\mathrm{Spec}}(k)$ is smooth of relative dimension $\dim (U)$ and relative dimensions add up correctly (Morphisms, Lemma 29.29.3). Observe that $h$ is a generically finite dominant morphism of varieties, and hence $\deg (h)$ is defined. By Lemma 33.36.5 we have a commutative diagram

$\xymatrix{ X \ar[rr]_{F_{X/k}} \ar[d]_ h & & X^{(p)} \ar[d]^{h^{(p)}} \\ \mathbf{A}^ n_ k \ar[rr]^{F_{\mathbf{A}^ n_ k/k}} & & (\mathbf{A}^ n_ k)^{(p)} }$

Since $h^{(p)}$ is a base change of $h$ it is étale as well and it follows that $h^{(p)}$ is a generically finite dominant morphism of varieties as well. The degree of $h^{(p)}$ is the degree of the extension $k(X^{(p)})/k((\mathbf{A}^ n_ k)^{(p)})$ which is the same as the degree of the extension $k(X)/k(\mathbf{A}^ n_ k)$ because $h^{(p)}$ is the base change of $h$ (small detail omitted). By multiplicativity of degrees (Morphisms, Lemma 29.51.9) it suffices to show that the degree of $F_{\mathbf{A}^ n_ k/k}$ is $p^ n$. To see this observe that $(\mathbf{A}^ n_ k)^{(p)} = \mathbf{A}^ n_ k$ and that $F_{\mathbf{A}^ n_ k/k}$ is given by the map sending the coordinates to their $p$th powers. $\square$

There are also:

• 4 comment(s) on Section 33.36: Frobenii

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).