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?



Is there any approximated version of Hilbert 90?
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'. 


