Lemma 51.17.6 (Kunz). Let $p$ be a prime number. Let $A$ be a Noetherian ring with $p = 0$. The following are equivalent

1. $A$ is regular, and

2. $F : A \to A$, $a \mapsto a^ p$ is flat.

Proof. Observe that $\mathop{\mathrm{Spec}}(F) : \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A)$ is the identity map. Being regular is defined in terms of the local rings and being flat is something about local rings, see Algebra, Lemma 10.38.18. Thus we may and do assume $A$ is a Noetherian local ring with maximal ideal $\mathfrak m$.

Assume $A$ is regular. Let $x_1, \ldots , x_ d$ be a system of parameters for $A$. Applying $F$ we find $F(x_1), \ldots , F(x_ d) = x_1^ p, \ldots , x_ d^ p$, which is a system of parameters for $A$. Hence $F$ is flat, see Algebra, Lemmas 10.127.1 and 10.105.3.

Conversely, assume $F$ is flat. Write $\mathfrak m = (x_1, \ldots , x_ r)$ with $r$ minimal. Then $x_1, \ldots , x_ r$ are independent in the sense defined above. Since $F$ is flat, we see that $x_1^ p, \ldots , x_ r^ p$ are independent, see Lemma 51.17.5. Hence $\text{length}_ A(A/(x_1^ p, \ldots , x_ r^ p)) = p^ r$ by Lemma 51.17.4. Let $\chi (n) = \text{length}_ A(A/\mathfrak m^ n)$ and recall that this is a numerical polynomial of degree $\dim (A)$, see Algebra, Proposition 10.59.8. Choose $n \gg 0$. Observe that

$\mathfrak m^{pn + pr} \subset F(\mathfrak m^ n)A \subset \mathfrak m^{pn}$

as can be seen by looking at monomials in $x_1, \ldots , x_ r$. We have

$A/F(\mathfrak m^ n)A = A/\mathfrak m^ n \otimes _{A, F} A$

By flatness of $F$ this has length $\chi (n) \text{length}_ A(A/F(\mathfrak m)A)$ (Algebra, Lemma 10.51.13) which is equal to $p^ r\chi (n)$ by the above. We conclude

$\chi (pn + pr) \geq p^ r\chi (n) \geq \chi (pn)$

Looking at the leading terms this implies $r = \dim (A)$, i.e., $A$ is regular. $\square$

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