This question concerns the explicit law for the Hilbert Symbol given in Sur les lois de réciprocfites explicites I by Henniart. I am trying to deduce the classical value of the Hilbert Symbol in $\mathbb{Q}_2$ with the method explained here but I simply do not get the right answer. I am not a native French speaker so I might be misunderstanding something or making a silly mistake, but I do not see it.
In any case, we want to compute $(u, 2)$. Using the uniformizer $\pi = 2$ we can expand $u$ in a power series and get $u = 1 + a_1*2 + a_2*2^2 + ...$, which in turn produces the series
$$f(X) = 1 + a_1X + a_2X^2 + ...$$
The results of the paper imply that the Hilbert Symbol coincides with a residue at $0$ of a series. The residue being
$$Res\left(\dfrac{\rho l(f)}{X}\right).$$ In here, $l(f)$ is the Artin-Hasse-Shafarevic logarithm and it is defined in the paper as
$$l(f) = \dfrac{1}{2}\log\left(\dfrac{f^2}{\Delta(f)}\right),$$
where $\Delta$ is the action of Frobenius in coefficients and sends $X$ to $X^2$. I have never seen this Artin-Hasse logarithm defined like this before, I see it as $\left(1-\dfrac{\Delta}{2}\right)\log(f)$.
Also $\rho$ is a certain series he constructs, which I will describe below.
To construct $\rho$ we have to follow several steps (the first of which are explained in section 1.8). We begin by finding a series $Z(X)$ such that $$1 + Z(2) = -1$$ ($-1$ being the quadratic root of unity), thus, we have $Z(2) = -2$. Since the $2-$adic expansion of $-1$ is
$$-1 = 1 + 2 + 2^2 + 2^3 + ...,$$ we need $$Z(2) = 2 + 2^2 + 2^3 + ...$$ This means that the series we take is $$Z(X) = X + X^2 + X^3 + ... = \dfrac{X}{1 - X}$$ Furthermore, the integer $e$ he defined in this section is $e = 1$.
Now we define a series $S$ by $$S(X) = (1 + Z)^2 - 1 = \left(1 + \dfrac{X}{X - 1}\right)^2 - 1 = \dfrac{2X - X^2}{(1 - X)^2}.$$
At this point we define $\sigma$ as the inverse of this series, which means $$\sigma = \dfrac{(1 - X)^2}{2X - X^2}$$.
Finally, $\rho$ is defined in section 7, after lemma 7.2, as $$\rho = \sigma + \dfrac{1}{2} + 2^{n-1}\sigma\chi = \sigma + \dfrac{1}{2} + \sigma\chi.$$ For us the $n = 1$ because we are dealing with $p^n = 2^1-$roots of unity.
Here $\chi$ is a certain polynomial satisfying the conditions of lemma 7.2. In that lemma it is explained that the polynomial has the form $$\chi = \displaystyle\sum_{k = 2}^{2^{n-1}e}\chi_kX^{2k-2},$$ for certain coefficients $\chi_k$. However, we see that $2^{n-1}e = 1*1 = 1$, while the sum starts at $k = 2$. I am interpreting this as an empty sum, thus $\chi = 0$. Maybe the interpretation should be different?
In any case, with this interpretation, we get $$\rho = \dfrac{X^2 - 2X + 2}{2(2X - X^2)} = \dfrac{1}{4}(X^2 - 2X + 2)\left(\dfrac{1}{X} + \dfrac{1}{2-X}\right).$$
Using this we have to compute the residue at 0 of $\dfrac{\rho l(f)}{X}$. This is the same as computing the constant term of $\rho l(f)$.
To compute $l(f)$ we follow its definition (I followed the definition of the paper) and we get $$l(f) = \dfrac{1}{2}\left(1 + \dfrac{\sum_{i\neq j}a_ia_jX^{i + j}}{1 + a_1X^2 + a_2X^4 + ...}\right)$$ In here we put $a_0 = 1$ and in the upper sum the indices must be distinct. The coefficients $a_1, a_2,...$ are not raised to the second power because they are $0$ and $1$. Thus we have $$l(f) = \dfrac{1}{2}\displaystyle\sum_{m = 1}^{\infty}\dfrac{(-1)^{m+1}}{m}\left(\dfrac{\sum_{i\neq j}a_ia_jX^{i + j}}{1 + a_1X^2 + a_2X^4 + ...}\right)^m.$$ The series $1 + a_1X^2 + a_2X^4 + ...$ must have an inverse, since its leading coefficient is $1$. Furthermore, because all of its powers are even, the inverse must only have even powers of $X$. Call this inverse $$1 + b_1X^2 + b_2X^4 + ...$$ Using this we can define $c_0, c_1,...$ as $$\left(\sum_{i\neq j}a_ia_jX^{i + j}\right)\left(1 + b_1X^2 + b_2X^2 + ...\right) = c_0 + c_1X + ...$$ Notice that because at least one of $i$ or $j$ is positive, $X^{i+j}$ always has at leats one factor $X$. Thus $c_0 = 0$. Thus we have $$l(f) = \dfrac{1}{2}\displaystyle\sum_{m = 1}^{\infty}\dfrac{(-1)^{m+1}}{m}\left(c_1X + c_2X^2 +...\right)^m$$ We must compute the constant term of $$\dfrac{1}{8}(X^2 - 2X + 2)\left(\dfrac{1}{X} + \dfrac{1}{2-X}\right)\displaystyle\sum_{m = 1}^{\infty}\dfrac{(-1)^{m+1}}{m}\left(c_1X + c_2X^2 +...\right)^m$$ Notice that $$\dfrac{X^2 - 2X + 2}{2-X}\displaystyle\sum_{m = 1}^{\infty}\dfrac{(-1)^{m+1}}{m}\left(c_1X + c_2X^2 +...\right)^m$$ has no pole at $X = 0$ and evaluated there it gives $0$ (since the sum over $m$ starts at $1$). Thus, this term contributes no constant term. The constant term we look for is the one of $$\dfrac{X^2 - 2X + 2}{8X}\displaystyle\sum_{m = 1}^{\infty}\dfrac{(-1)^{m+1}}{m}\left(c_1X + c_2X^2 +...\right)^m.$$ Once more, what comes from the $\dfrac{X^2 - 2X}{8X}$ has no pole at $X = 0$ and so again contributes $0$. Thus we need the constant term of $$\dfrac{2}{8X}\displaystyle\sum_{m = 1}^{\infty}\dfrac{(-1)^{m+1}}{m}\left(c_1X + c_2X^2 +...\right)^m = \dfrac{1}{4}\displaystyle\sum_{m = 1}^{\infty}\dfrac{(-1)^{m+1}}{m}X^{m-1}\left(c_1 + c_2X +...\right)^m.$$ Evaluating at $X = 0$ we get that the constant term is $c_1/4$. Remembering that we defined $$\left(\sum_{i\neq j}a_ia_jX^{i + j}\right)\left(1 + b_1X^2 + b_2X^2 + ...\right) = c_1X + ...$$ we can rewrite this as $$(2a_1X + X^2P(X))(1 + b_1X^2 + b_2X^2 + ...) = c_1X + c_2X^2 + ...,$$ which yields that $c_1 = 2a_1$.
Thus the residue is $$Res\left(\dfrac{\rho l(f)}{X}\right) = \dfrac{a_1}{2}.$$
This is telling that the Hilbert Symbol $(u, 2)_2$ depends only on $a_1$, which translates to it depending on the congruence modulo $4$. This is not true, this symbol depends in modulo $8$.
Something is wrong and I don't know what it is. I tried to repeat this with the usual definition of the logarithm (as I don't know if the two expressions coincide, these are not my usual tools of work) but I obtained the exact same answer and thus the same problem.
If someone can tell me what am I interpreting wrong in all of this or what is my mistake I would be very thankful.
Thank you for your attention.
