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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?

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  • $\begingroup$ Have you considered computing the kernel of the composition of $\phi$ with $\mathbb{Z}_p\rightarrow \mathbb{Z}/p\mathbb{Z}$? $\endgroup$ – WSL Apr 13 '18 at 15:39
  • $\begingroup$ It should be $V(W(\mathbb{F}_p))$, if $V$ denotes the shift map $\endgroup$ – Notone Apr 13 '18 at 15:51
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    $\begingroup$ 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 $\Bbb{Z}_p[\mu_{q-1}]$. $\endgroup$ – Jyrki Lahtonen Apr 13 '18 at 18:46
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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

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

(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 $\Bbb Q_p$.

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  • $\begingroup$ 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 '18 at 7:46

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