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Consider the finite field $\mathbb F_p$ and ring of Witt vectors $W(\mathbb F_p) \cong \mathbb Z_p$, where $\mathbb Z_p$ is the ring of $p$-adic numbers.

Why the Frobenius map on $W(\mathbb F_p)$ fix the prime $p$ ?

Let $a=(a_0,a_1,a_2, \cdots) \in W(\mathbb F_p)$, then the Frobenius $\varphi: W(\mathbb F_p) \to W(\mathbb F_p)$ is defined by $$\varphi(a)=(a_0^p,a_1^p, a_2^p, \cdots),~a_i \in \mathbb F_p.$$ But since $a_i^p=a_i$ in $\mathbb F_p$, we have $\varphi(a)=(a_0,a_1, \cdots)$, i.e., $\varphi$ is identity on $W(\mathbb F_p)$. Thus $\varphi$ fix $p$.

I think we can replace $\mathbb F_p$ by any characteristic $p$ finite field.

The same Frobenius endomorphism extends to the corresponding power series ring $W(\mathbb F_{p^n})[[x]]$ by $\varphi(x)=x^p$ i.e., $$\varphi (f(x))=f(x^p),~~f(x) \in W(\mathbb F_{p^n})[[x]],$$ where $\varphi$ is identity on $W(\mathbb F_p)$ but not identity on $W(\mathbb F_{p^n})$.

I appreciate your comments.

Torsten Schoeneberg
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MAS
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    Indeed, the Frobenius map $W(\phi)$ is the identity on the ring of Witt vectors over $\Bbb{F}_p$. Simply because it is the identity on $\Bbb{F}_p$ and the construction $A\mapsto W(A)$ is a functor from the category of commutative rings of characteristic $p$ to the category of rings. – Jyrki Lahtonen Oct 31 '22 at 13:29
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    Also the ring $\Bbb{Z}_p$ of $p$-adic integers has no non-trivial automorphisms. – Jyrki Lahtonen Oct 31 '22 at 13:31
  • But on the finite extension fields $\Bbb{F}{p^m}$, $m>1$, the Frobenius automorphism is not the identity. Hence it isn' the identity on $W(\Bbb{F}{p^m})$ either. The latter ring can be thought of as the extension ring $\Bbb{Z}_p[\zeta]$ where $\zeta$ is a primitive root of unity of order $p^m-1$. At the level of fields of fractions the Frobenius is a generator of the Galois group $Gal(\Bbb{Q}_p(\zeta)/\Bbb{Q}_p)$, a cyclic group of order $m$. – Jyrki Lahtonen Oct 31 '22 at 13:36
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    I disagree with what you say about the Frobenius on the power series. This may be partly because I don't know whether your ring is $W(\Bbb{F}_p[[x]])$ or $W(\Bbb{F}_p)[[x]]$. It should also affect the coefficients of the power series at least. – Jyrki Lahtonen Oct 31 '22 at 13:40
  • @JyrkiLahtonen, what I know is that if $K$ be a finite extension of the $p$-adic field $\mathbb Q_p$ with ring of integers $O_K$ and uniformizer $\pi$, then for the residue field $k \cong O_K/\pi O_K$, we can define extend the Frobenius endomorphism $\varphi$ on $W(k)[[x]] \to W(k)[[x]]$ by moving $x$ to $x^p$ i.e., $\varphi(x)=x^p$. – MAS Oct 31 '22 at 13:45
  • @JyrkiLahtonen, Next I am saying that the coefficients of $f(x) \in W(k)[[x]]$, which belongs to $W(\mathbb Z_p/p \mathbb Z_p)$, those will be fixed by $\varphi$ while other coefficients belonging to $W(k) \setminus W(\mathbb Z_p/p \mathbb Z_p)$ will be changed accordingly. Am I right ? – MAS Oct 31 '22 at 13:48

1 Answers1

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Maybe first of all it should be noted that if $R$ is any ring which contains the integers $\mathbb Z$, then any ring endomorphism of $R$, by virtue of fixing $1$, fixes all elements of $\mathbb Z$.

In this concrete case, one can also note that if $k$ is any perfect field of characteristic $p$, the element $p$ (i.e. $1+ \dots +1$, $p$ times) in $W(k)$ is actually

$$(0,1,0,0, \dots)$$

which quite obviously is fixed by the Frobenius map

$$(x_0, x_1, \dots ) \mapsto (x_0^p, x_1^p, \dots ) $$

regardless of what other elements might or might not be fixed by the Frobenius. In fact, it follows immediately that $\mathbb Z_p \simeq W(\mathbb F_p) \subset W(k)$ is exactly the fixed point set of the Frobenius map, because $\mathbb F_p \subset k$ is exactly the fixed point set under the classical Frobenius (raising to the $p$-th power).

Torsten Schoeneberg
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