I have just started studying Witt vectors and I have questions about the following identity $$W_n(\mathbb{F}_p)\cong \mathbb{Z}/p^n\mathbb{Z}$$

  1. I would like proving this by finding an explicit map $\phi_n: W_n(\mathbb{F}_p)\to \mathbb{Z}/p^n\mathbb{Z}$, but couldn't come up with a reasonable map. Although I have found the isomorphism $$\phi: W(\mathbb{F}_p)\to \mathbb{Z}_p$$ via $(a_0,a_1,...)\mapsto \chi(a_0)+\chi(a_1)p... $, where $\chi$ is the Teichmüller character.

    Can I obtain $\phi_n$ by composing $\phi$ with $pr_n:\mathbb{Z}_p \to \mathbb{Z}/p^n\mathbb{Z}$ and identifying $W_n(\mathbb{F}_p)$ with $(a_0,...,a_{n-1},0,0,...)\in W(\mathbb{F}_p)$? Is there a nice explicit version of this map?

  2. More generally I am interested in the case $A=\mathbb{F}_q$. I know that $W(\mathbb{F}_q)\cong \mathbb{Z}_p[\mu_{q-1}]$ should hold. Is it possible to argue that this is an isomorphism because $\mathbb{Z}_p[\mu_{q-1}]$ and $W(\mathbb{F}_q)$ are both strict $p$-ring with residue field $\mathbb{F}_q$ and as such canonically isomorphic?

  • $\begingroup$ Have you considered computing the kernel of the composition of $\phi$ with $\mathbb{Z}_p\rightarrow \mathbb{Z}/p\mathbb{Z}$? $\endgroup$ Apr 13, 2018 at 15:39
  • $\begingroup$ It should be $V(W(\mathbb{F}_p))$, if $V$ denotes the shift map $\endgroup$
    – Notone
    Apr 13, 2018 at 15:51
  • 3
    $\begingroup$ The multiplicative group of $\Bbb{Z}/p^n\Bbb{Z}$ has a unique cyclic subgroup of order $p-1$. Furthermore, the elements of that subgroup are distinct modulo $p$. So you can use them much the same way that you use the Teichmüller character. In the case $A=\Bbb{F}_q, q=p^r,$ you similar get that $W_n(A)$ is the so called Galois ring $GR(p^n, r)$. This can, indeed, also be constructed by moding out the ideal $\langle p^n\rangle$ of $\Bbb{Z}_p[\mu_{q-1}]$. $\endgroup$ Apr 13, 2018 at 18:46

1 Answer 1


Ad 1) First of all I'm pretty sure that yes, $\phi_n = pr_n\circ\phi$, although usually the index is off by one (so I would have expected $W_{n-1}(\Bbb F_p) \simeq \Bbb Z/p^n$).

To make that map a bit more explicit, I remember a MathOverflow post which might be helpful here. It motivates the Witt polynomials (noted $W(X_0, ..., X_{n-1})$ there), which, if I'm not mistaken, are "more or less" your map $\phi_n$ (annoying index shift again, oh well ...). "More or less" because you have to choose some lifting $$\Bbb F_p \rightarrow \Bbb Z/p^{n}: \quad \bar a \mapsto a.$$ But well, if you just go all the way up to $\Bbb Z$ and work with the old school representatives $\{0, ..., p-1\}$, -- and also remember that in this very special case, the Frobenius $(\cdot)^{\,p}$ is just the identity on $\Bbb F_p$ --, you can explicitly write the map you call $\phi_n$ as

$$(\bar a_0, \bar a_1, ..., \bar a_{n-1}) \mapsto (a_0^{p^{n-1}} +p \cdot a_1^{p^{n-2}} + ... + \,p^{n-1} \cdot a_{n-1}) +p^n\Bbb Z$$

To make it totally explicit, say $p=5, n=3$, and say your element in $W_3(\Bbb F_5)$ is $(\bar 1, \bar 4, \bar2)$, it gets mapped to $$(1^{5^2}+5\cdot 4^5 + 5^2\cdot 2) +5^3\Bbb Z = 5171 + 5^3\Bbb Z = 46 + 5^3\Bbb Z$$

-- and note that crucial and fun fact that the choice of the lifting does not change the result; if e.g. instead you lift $\bar 1$ to $6$, $\bar 4$ to $9$, and $\bar 2$ to $22$, you still get $$(6^{5^2}+5\cdot 9^5 + 5^2\cdot 22) +5^3\Bbb Z = 28430288029929997171 + 5^3\Bbb Z= 46 + 5^3\Bbb Z$$

Added: Let's see the connection to the first part of Jyrki Lahtonen's comment: one can get those Teichmüller-like representatives of $\Bbb F_p$ in $\Bbb Z/p^n$ by raising any set of representatives to the $p^{n-1}$-th power; continuing the example above we get $\chi(1) =1$ and

$$\chi(2) = 2^{5^2} + 5^3\Bbb Z = 33554432 + 5^3\Bbb Z = 57 + 5^3\Bbb Z$$ $$\chi(3) = 3^{5^2} + 5^3\Bbb Z = 847288609443 + 5^3\Bbb Z = 68 + 5^3\Bbb Z$$ $$\chi(4) = 4^{5^2} + 5^3\Bbb Z = 1125899906842624 + 5^3\Bbb Z = 124 + 5^3\Bbb Z = -1 + 5^3\Bbb Z.$$

(Of course we could have noticed $\chi(p-1) = -1$ easier.) Notice how they are multiplicative, indeed they are just the set of standard Teichmüller representatives $\mu_{p-1}(\Bbb Z_p)$ modulo $p^n$, and instead of the earlier computation you can write

$$\phi_n(\bar 1, \bar 4, \bar2) = \chi(1) + 5\cdot \chi(4) + 5^2\cdot\chi(2) +5^3\Bbb Z = 1+5\cdot (-1) + 25\cdot 57 +5^3\Bbb Z = 46 +5^3\Bbb Z.$$

Of course that's all equivalent, but this way you outsource some work into a once-and-for-all-computation of the Teichmüller representatives. Also, beware that as soon as one tries to do the same with $\Bbb F_q$ instead of $\Bbb F_p$, things get more subtle, as one has to put in appropriate $p^i$-th roots at appropriate places.

Ad 2), I think it is totally possible to argue this way, although the proofs of that theorem which I have seen (which would be the one of Bourbaki in Commutative Algebra 9, and of Serre in Local Fields) actually go through the Witt vector machinery and thus might exhibit a bit more structure. I just want to point out that, if $q=p^k$, then another description of $\Bbb Z_p[\mu_{q-1}]$ is: the ring of integers (a.k.a valuation ring) of the unique unramified degree $k$ extension of $\Bbb Q_p$.

  • $\begingroup$ This is exactly what I was looking for, thanks very much! As for 2) I copied the argument infact from Serre's Local fields (c.f. end of chapter 2.6). $\endgroup$
    – Notone
    Apr 15, 2018 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.