## Is there any approximated version of Hilbert 90?

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Suppose $K$ is a local field and $L$ a finite cyclic extension of $K$. By Hilbert 90, we know that if an element $a$ in $L$ such that $N_{L / K}(a) =1$ then $a = b / \sigma(b)$ for some $b$ in $L$ and $\sigma$ a generator of the Galois group.

My question: Suppose $N_{L / K}(a) \simeq 1$, is there some $b$ such that $a \simeq b /\sigma(b)$?

I guess I could have try to make the question more precise, but there seems to be some merits in leaving it a bit vague.

I have accept the the answer by Paul Broussous, which address the situation when the extension is unramified. It is because that is what I need. I am still curious whether something can be done when the extension is totally ramified?

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I think you want $L$ a finite cyclic extension of $K$, not of $L$. – Jamie Weigandt Apr 12 2010 at 18:26
the error is fixed. – Tran Chieu Minh Apr 12 2010 at 19:14

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Doesn't that follow from the standard proof? – Felipe Voloch Apr 12 2010 at 19:35
The problem that I found (if it is really a problem) is that there might be no element in the neighborhood of $a$ which has of norm 1. I try to mimic the original proof for the cyclic case but the independence of automorphism does not work anymore, I mean it still work but I can not bound the $b$ given by this process. – Tran Chieu Minh Apr 13 2010 at 1:53

## 1 Answer

Yes such approximated versions of Hilbert 90 do exist. But you need some technical conditions.

For instance assume that $L/K$ is unramified of degree $d$ and that $a\in {\mathfrak o}_{L}^{\times}$.

Then you condition writes

$N_{L/K}(a)\equiv 1$ modulo $\mathfrak{p}_{K}^{n}$,

for some $n>0$ (I assume that this is what you mean by $\simeq$). This may be rewritten $N_{L/K}(a)=1$ in $U_{L}/U^{n}_L$, where $U$ denotes a unit group. So the map $$\sigma^u\mapsto a\sigma (a) \cdots \sigma^u (a)$$ defines a $1$-cocycle of ${\rm Gal}(L/F)$ in $U_L /U^n_L$ (here $\sigma$ denotes the Frobenius substitution).

So what you want is that this cocycle is split. In fact we have $H^{1}({\rm Gal}(L/K), U_{L}/U^{n}_{L})=1$. This is proved by a standard filtration argument: this is implied by

$$H^{1}({\rm Gal}(L/K), U_L /U^{1}_L ) = H^{1}({\rm Gal}(k_L /k_K ), k_{L}^{\times})=1$$

and

$$H^{1}({\rm Gal}(L/K), U_{L}^{i}/U_{L}^{i+1})=H^{1}({\rm Gal}(k_L /k_K ), k_L )=1$$

here $k$ denotes a residue field. You can find the detail of the proof in, I think, Serre's 'Local fields' or Cassels-Fröhlich's 'Algebréaic Number Theory'.

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Yes, the unramified version is all that I need (but I did not want to ask a very localized question). Homological tools is too much to put in my final year project thesis.  But thanks to your comments I have figured out the correct modification of the original proof.  – Tran Chieu Minh Apr 13 2010 at 2:19
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