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Let $k$ be a local field of characteristic $p$ and $\omega \in k$ a uniformiser. Consider the Artin-Schreier extension $L_n = k[x]/(x^p - x - \omega^n)$ for each $n \in \mathbb{Z}$.

Is there an explicit description for the image of the norm map $N_{L_n/k}: L_n^\times \to k^\times$?

If $n < 0$ it seems that the extension is wildy ramified and I'm not sure how to calculate the image of the norm group in this case.

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  • $\begingroup$ It certainly is wildly ramified if $n<0$ and $p\nmid n$. For the wildly ramified prime cyclic case $L/k$, Serre's Local Fields, Remark at the end of V.§3 seems to say that $(k^\times:NL^\times)=p$. Your question is which subgroup $NL_n^\times$ is precisely? $\endgroup$
    – Arno Fehm
    Dec 2, 2022 at 7:08
  • $\begingroup$ @ArnoFehm; Yes this is exactly what I'm asking for. In fact I think that even the image of the units has index $p$, so the problem is about determining which units are norms. I was hoping given that the situation is so explicit this might be possible! Even if a complete description is not possible, it would be nice to know even some units which are not norms. $\endgroup$ Dec 2, 2022 at 11:05
  • $\begingroup$ Both comments above get right to the point. The action is all in the principal units, clearly. I’ll see whether I can stir my brain up to understand the situation. $\endgroup$
    – Lubin
    Dec 2, 2022 at 13:29
  • $\begingroup$ @Lubin: That would be excellent! Feel free to make some simplifying assumptions, e.g. $n=-1$ or $p=2$. $\endgroup$ Dec 2, 2022 at 14:25
  • $\begingroup$ I’m working computationally, which is getting information; but I think the approach is wrong-headed. Keep tuned — $\endgroup$
    – Lubin
    Dec 2, 2022 at 21:55

2 Answers 2

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Local class field theory has something to say about the norm groups. To set up notation, let's put $k = \kappa((t))$ with uniformiser $\omega = t$. I'll assume $n < 0$ is not divisible by $p$ in the following (without loss).

The element $x \in L_n$ satisfying $x^p - x = t^n$ has valuation $v(x) = n/p$ (where I always normalise the valuation to satisfy $v(t) = 1$). For integers $a, b$ satisfying $ap + bn = 1$, the element $\pi := t^a x^b$ is now a uniformiser of $L_n$. If $\sigma$ is the generator of $G := \operatorname{Gal}(L_n/k)$ sending $x$ to $x+1$, a calculation shows that $v(\sigma(\pi) - \pi) = (1-n)/p$. Hence for the ramification groups in lower numbering we have $G_{-n} =G$, $G_{-n+1} = 1$. For the upper numbering we still have $G^{-n} = G$, $G^{-n+1} = 1$ (the renumbering is trivial up to the relevant point). (See for instance Section II.10 of Neukirch's Algebraische Zahlentheorie for the definition of these groups.)

It follows that the ideal $(t)^{1-n}$ is precisely the conductor of the extension $L_n/k$, and so the subgroup $1 + (t)^{1-n} \leq k^\times$ is contained in $N_{L_n/k}(L_n^\times)$, but $1 + (t)^{-n}$ is not. (See again Neukirch, Sections 1 and 6 of Chapter V.) If $\kappa = \mathbb{F}_p$ is the prime field, then $1 + (t)^{1-n}$ has index $p$ in $1 + (t)^{-n}$, showing that the only elements of $1 + (t)^{-n}$ which are norms from $L_n$ are the ones in $1 + (t)^{1-n}$. Note that this matches Lubin's answer for $n=-1$, and satisfies the desideratum (from the comments to the question) of finding many non-norms in $k$.

I do not know if one can easily identify the norm group more precisely than this. Serre's Corps Locaux has some more material, starting in V.3. The upshot there seems to be that one has a good handle on the norm in the graded components of $\kappa[[t]]^\times$, where $\kappa[[t]]^\times$ is filtered by the subgroups $1 + (t)^m$. This gives little more information than the above, however (it is precisely the above for $\kappa = \mathbb{F}_p$). Piecing things together for an ungraded version is presumably going to be tedious.

Edit: In the comments, Daniel very reasonably asks to characterise the elements of $1 + (t)^{-n}$ which do occur as norms, in the case of a general finite residue field $\kappa$. Here the graded idea does help (see Proposition 5(iii) in V.3 in Serre): Restricting to the graded component $(1 + (t)^{-n})/(1 + (t)^{-n+1})$ of $k^\times$ and identifying it with the additive group $\kappa$ (sending $c \in \kappa$ to the class of $1 + ct^{-n}$), the image of the norm map is the image of an additive polynomial $\alpha X^p + \beta X \in \kappa[X]$. However, we know that this polynomial must in fact be defined over $\mathbb{F}_p$ (since everything we do arises via base change from $\mathbb{F}_p$, see also Serre's discussion in V.4) and not injective on $\mathbb{F}_p$. Hence it is a constant multiple of the Artin-Schreier polynomial $X^p - X$, and can be written as $(\beta X)^p - \beta X$ with $\beta \in \mathbb{F}_p^\times$. This confirms Daniel's suspicion that an element $1 + ct^{-n} + ... \in k^\times$ is a norm if and only if $c$ has the form $d^p - d$ for some $d \in \kappa$.

Edit, later: I somehow missed that Serre actually gives an explicit description of the norms later, in XIV.5. Specifically, he defines a local symbol $[a, b)_v \in \mathbb{Z}/p$ for $a \in k$, $b \in k^\times$, which is essentially equivalent to computing the local Artin symbol in the cyclic Artin-Schreier extension corresponding to $a$. Relevant for us is the fact that $[t^n, b)_v = 0$ iff $b$ is a norm of the extension $L_n/k$. Further, Serre gives the explicit formula (Corollaire to Proposition 15) $[a, b)_v = \operatorname{Tr}_{\kappa/\mathbb{F}_p}(\operatorname{Res}(a \cdot \mathrm{d}b/b))$, where the residue of a differential is defined in the usual way (write the differential as $f \mathrm{d}t$ for some Laurent series $f$, and extract its $t^{-1}$-coefficient.) From this formula we see that an element $b = 1 + ct^n + \dots$ is a norm in $L_n/k$ iff $\operatorname{Tr}_{\kappa/\mathbb{F}_p}(c) = 0$; the latter condition is equivalent to $c$ having the form $d^p - d$ by additive Hilbert 90. The advantage of the formula over what's written above is of course that it also allows to treat a general element $b \in k^\times$.

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  • $\begingroup$ Good, thank you. I thought I could see that the higher ramification information must have something to do with the case, but was stumped. $\endgroup$
    – Lubin
    Dec 7, 2022 at 22:57
  • $\begingroup$ Thanks for the answer, this is really great. I've ran some computations and it seems that an element $1 + at^{-n} + \cdots $ is an norm if and only if $a$ is in the image of the Artin-Schreier map, i.e. $a = b^p - b$ for some $b \in \kappa$. This has index $p$ as expected from local class field theory, so looks very reasonable. Is there some simple way to prove this? $\endgroup$ Dec 8, 2022 at 15:12
  • $\begingroup$ Yes, I've added a proof sketch of that in an edit. $\endgroup$
    – Philip
    Dec 8, 2022 at 17:08
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    $\begingroup$ Sorry, it occurred to me that Serre did have an explicit computation of the Artin symbol somewhere, and so one can indeed characterise the norms nicely - see the second edit. I am afraid that obsoletes much of the previous answer... $\endgroup$
    – Philip
    Dec 8, 2022 at 22:59
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Well, I can give a very satisfying answer to a request you didn’t make, to find something whose norm is trivial, a super-partial answer to the question of identifying at least some non-norms, and a way of looking at the situation that may help you.

First, I’m going to call the chosen uniformizer $t$ rather than $\varpi$ or $\pi$ or $\omega$: too much writing. I’ll be thinking of $k$ as $\kappa((t))$ for a constant ring $\kappa=\Bbb F_q$ for $q=p^s$. At the end, I’ll specialize to the case $\kappa=\Bbb F_p$. We can always let $t$ be the chosen uniformizer in the ring of local integers $\kappa[[t]]$. And I’ll make $n>0$, so that our polynomial is $X^p-X-t^{-n}$. We need $(p,n)=1$ to guarantee irreducibility of this polynomial.

Normalizing the valuation of $t$ to $1$, we get the valuation of a root $1/\lambda$ of the polynomial to be $-n/p$. Then $\lambda$ won’t be a uniformizer except in the case $n=1$, but its minimal polynomial will be $$X^p+t^nX^{p-1}-t^n\,,$$ and the minimal polynomial of $1+\lambda$ will be $$ (X-1)^p+t^n(X-1)^{p-1}-t^n=X^p+t^n(X^{p-1}+X^{p-2}+\cdots+X)-1\,. $$ Thus $\mathbf N^{L_n}_k(1+\lambda)=1$.

Back in the case $n=1$, $\lambda$ is a uniformizer, and you can convince yourself that $\mathbf N^{L_1}_k(1+(\lambda^2))\subset1+(t^2)$, so that no norm starts out $1+t+\cdots$ . Everything else in the group of principal units $1+(t)$ should be a unit, however, in particular $(1+t)^p=1+t^p$. As you see, I’ve left a lot to be verified, but I believe there will be no problems.


But to understand what’s happening, you must have as accurate as possible a grasp of the structure of the multiplicative group $1+(t)$. First, it’s a $\Bbb Z_p$-module, because any series $1+$ (higher terms in $t$)may be raised to a power that’s a $p$-adic integer. Take your $p$-adic integer $z$, and express it in the form $z=\sum_n a_np^n$, where the $a_n$ are natural numbers. Then in $1+(t)$, the powers $(1+t)^{p^n}$ approach $1$, and $\prod_n\bigl((1+t)^{p^n}\bigr)^{a_n}$ is a convergent product.

Not only that, but you see that the binomials $1+t^m$ for $\gcd(p,m)=1$ are independent in the whole module $1+(t)$, and in the very special case that the constant field $\kappa$ is $\Bbb F_p$, that every element $g\in1+(t)$ may be written $$ g=\prod_{\gcd(p,m)=1}(1+t^m)^{z_m}\,, $$ and uniquely so. I won’t go into the general case here, but the dreaded Artin-Hasse Exponential can give some help. It’s all in Hazewinkel’s Formal Groups and Applications, anyway.

Back in the special case $\kappa=\Bbb F_p$, one may get a little insight by writing out the norms of all the $1+t^m$ for $(p,m)=1$ and seeing what’s missing. This is surely not the right approach, however.

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  • $\begingroup$ Thanks for the answer. I need some time to try to digest it. In the meantime I will leave the question open with the hope of obtaining something definitive. $\endgroup$ Dec 6, 2022 at 13:06
  • $\begingroup$ A sensible route to take. Meanwhile, I have a little more to say on the subject, but am busy with other (nonmathematical) matters… $\endgroup$
    – Lubin
    Dec 6, 2022 at 19:52

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