The Stacks project

51.17 Frobenius action

Let $p$ be a prime number. Let $A$ be a ring with $p = 0$ in $A$. The Frobenius endomorphism of $A$ is the map

\[ F : A \longrightarrow A, \quad a \longmapsto a^ p \]

In this section we prove lemmas on modules which have Frobenius actions.

Lemma 51.17.1. Let $p$ be a prime number. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with $p = 0$ in $A$. Let $M$ be a finite $A$-module such that $M \otimes _{A, F} A \cong M$. Then $M$ is finite free.

Proof. Choose a presentation $A^{\oplus m} \to A^{\oplus n} \to M$ which induces an isomorphism $\kappa ^{\oplus n} \to M/\mathfrak m M$. Let $T = (a_{ij})$ be the matrix of the map $A^{\oplus m} \to A^{\oplus n}$. Observe that $a_{ij} \in \mathfrak m$. Applying base change by $F$, using right exactness of base change, we get a presentation $A^{\oplus m} \to A^{\oplus n} \to M$ where the matrix is $T = (a_{ij}^ p)$. Thus we have a presentation with $a_{ij} \in \mathfrak m^ p$. Repeating this construction we find that for each $e \geq 1$ there exists a presentation with $a_{ij} \in \mathfrak m^ e$. This implies the fitting ideals (More on Algebra, Definition 15.8.3) $\text{Fit}_ k(M)$ for $k < n$ are contained in $\bigcap _{e \geq 1} \mathfrak m^ e$. Since this is zero by Krull's intersection theorem (Algebra, Lemma 10.51.4) we conclude that $M$ is free of rank $n$ by More on Algebra, Lemma 15.8.7. $\square$

In this section, we say elements $f_1, \ldots , f_ r$ of a ring $A$ are independent if $\sum a_ if_ i = 0$ implies $a_ i \in (f_1, \ldots , f_ r)$. In other words, with $I = (f_1, \ldots , f_ r)$ we have $I/I^2$ is free over $A/I$ with basis $f_1, \ldots , f_ r$.


Lemma 51.17.2. Let $A$ be a ring. If $f_1, \ldots , f_{r - 1}, f_ rg_ r$ are independent, then $f_1, \ldots , f_ r$ are independent.

Proof. Say $\sum a_ if_ i = 0$. Then $\sum a_ ig_ rf_ i = 0$. Hence $a_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$. Write $a_ r = \sum _{i < r} b_ i f_ i + b f_ rg_ r$. Then $0 = \sum _{i < r} (a_ i + b_ if_ r)f_ i + bf_ r^2g_ r$. Thus $a_ i + b_ i f_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$ which implies $a_ i \in (f_1, \ldots , f_ r)$ as desired. $\square$


Lemma 51.17.3. Let $A$ be a ring. If $f_1, \ldots , f_{r - 1}, f_ rg_ r$ are independent and if the $A$-module $A/(f_1, \ldots , f_{r - 1}, f_ rg_ r)$ has finite length, then

\begin{align*} & \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, f_ rg_ r)) \\ & = \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, f_ r)) + \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, g_ r)) \end{align*}

Proof. We claim there is an exact sequence

\[ 0 \to A/(f_1, \ldots , f_{r - 1}, g_ r) \xrightarrow {f_ r} A/(f_1, \ldots , f_{r - 1}, f_ rg_ r) \to A/(f_1, \ldots , f_{r - 1}, f_ r) \to 0 \]

Namely, if $a f_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$, then $\sum _{i < r} a_ i f_ i + (a + bg_ r)f_ r = 0$ for some $b, a_ i \in A$. Hence $\sum _{i < r} a_ i g_ r f_ i + (a + bg_ r)g_ rf_ r = 0$ which implies $a + bg_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$ which means that $a$ maps to zero in $A/(f_1, \ldots , f_{r - 1}, g_ r)$. This proves the claim. To finish use additivity of lengths (Algebra, Lemma 10.52.3). $\square$


Lemma 51.17.4. Let $(A, \mathfrak m)$ be a local ring. If $\mathfrak m = (x_1, \ldots , x_ r)$ and $x_1^{e_1}, \ldots , x_ r^{e_ r}$ are independent for some $e_ i > 0$, then $\text{length}_ A(A/(x_1^{e_1}, \ldots , x_ r^{e_ r})) = e_1\ldots e_ r$.

Lemma 51.17.5. Let $\varphi : A \to B$ be a flat ring map. If $f_1, \ldots , f_ r \in A$ are independent, then $\varphi (f_1), \ldots , \varphi (f_ r) \in B$ are independent.

Proof. Let $I = (f_1, \ldots , f_ r)$ and $J = \varphi (I)B$. By flatness we have $I/I^2 \otimes _ A B = J/J^2$. Hence freeness of $I/I^2$ over $A/I$ implies freeness of $J/J^2$ over $B/J$. $\square$


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 10.39.18. 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 10.128.1 and 10.106.3.

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 10.60.9. 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 10.52.13) 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. $\square$

Comments (6)

Comment #4209 by Zhiyu Zhang on

Kunz theorem is pretty good, will other properties about Frobenius action (and singularity) be added in the future? For example, Remark 13.6. in Ofer Gabber. Notes on some t-structures. In Geometric aspects of Dwork theory. Vol. I, II, pages 711–734. Walter de Gruyter GmbH & Co. KG, Berlin, 2004. shows that F-finiteness on a noetherian ring with will imply is quotient by a regular ring hence the existence of dualizing complex. The proof is very short and elegant.

Comment #4391 by on

Yes, in the future. If there is a short piece of mathematics you think would be fun to add you could latex it yourself (in the latex file of the chapter) and email me the file...

Comment #6536 by Yijin Wang on

When proving the frobenius map F is flat,I think it has nothing to do with Lemma 10.106.3.

In lemma 51.17.5, J is simply IB,I think.

Comment #6588 by on

@#6536. Why do you say "it has nothing to do with 10.106.3"? In order to apply Lemma 10.128.1 you need to know that the target ring is CM. Then Lemma 10.106.3 (which at the time of writing this comment is 10.106.3) shows a regular ring is CM. Possibly you have another method for showing that the frobenius map of a regular local ring is flat avoiding this intermediate step?

Thanks for pointing out the typo in the proof of the other lemma. See change here.

Comment #7717 by Mabud Ali Sarkar on

What exactly means by a ring in which ? Does it mean is a ring or field of characteristic ?

Comment #7970 by on

The phrase " is a ring with " just means that is a ring and that the image of under the unique ring map is zero. This presupposes that is an integer, which it is in this section, as it is a prime number.

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