Thanks Micah, do you think it is possible to give a sharp approximation of $C_N$? So for example, do you think that $C_N=1$ should work?

@Martin, yes you are right but forget for the moment the isomorphism between $L[x]$ and $K[x]$. Assume that you only know that $K$ and $L$ are fields and that you have a group isomorphism between $K^{\times}$ and $L^{\times}$ does this imply that $K$ is isomorphic to $L$ as fields? It could be but I never thought about that.

@Martin, I'm not sure I understand your comment. If $R$ is a domain then you always have $R[x]^{\times}=R^{\times}$ and therefore $R^{\times}\simeq S^{\times}$. But this is just the multiplicative structure. So in a field, for non-zero elements $a,b$ we have $a+b=b(ab^{-1}+1)$ and therefore you still need to know what what it means to add $1$ to a field element... Si if $\phi:K^{\times}\rightarrow L^{\times}$ is a group isomorphism then we get $\phi(a+b)=\phi(a)\phi(1+ba^{-1})$, but this is not clear to me that this is equal to $\phi(a)+\phi(b)$...

Thanks Ralph for the answer, yes indeed, an isomorphism takes algebraic elements over the prime field to algebraic elements over the prime field.

Here is one possible way of constructing such examples: Let $\iota_1:G\rightarrow S_{n}$ and $\iota_2:G\rightarrow S_{m}$ be two embeddings of a finite group $G$ where $S_k$ denotes the symmetric group of degree $k$. Let $K_n$ be the field of rational functions over $\mathbf{Q}$ in $n$ variables then it is easy to see that $K_n^{\iota_1(G)}$ and $K_{m}^{\iota_2(G)}$ are stable isomorphic, but in general I don't see any reason why they should be isomorphic. Of course one needs to choose the group $G$ carefully since for "many" $G$'s $K_n^{\iota_1(G)}$ will always be purely transcendental.

Dear Charlie, so basically my question boils down to compare algebraic Poincare duality with duality of singular cohomology.

Dear Paul, thanks a lot for your effort. There was a confusion in my question. I'm trying to understand the generalization of the so-called Legendre's relation for elliptic curves so I should have assumed in my question that $X$ is projective and I should have taken the algebraic de Rham cohomology rather than the smooth one. Your efforts allowed me to clarify what I wanted.