Exercise 6.22

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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Show that the relation $\varepsilon \equiv 1 \pmod p$ in Proposition
6.4.4 can also be achieved by replacing $x$ by $1+t$ instead of $e^z$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

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\begin{proof}(solution given by Mikomikon and A.Grounds (agrounds))

We know from the remark after Prop 6.4.3 that
$$g(\chi) = \varepsilon \prod_{k=1}^{(p-1)/2}(\zeta^{2k-1} - \zeta^{-(2k-1)}),$$
where $\varepsilon = \pm 1$.
Let $$f(x) = \sum_{j=1}^{p-1} \chi(j) x^j -\varepsilon \prod_{k=1}^{(p-1)/2}(x^{2k-1} - x^{p-(2k-1)}).$$
Then $f(0) = 0$ and  $f(\zeta) = 0$, therefore $(x^p-1)$ divides $f(x)$. As $f(x) \in \Z[x]$ and $x^p-1 \in \Z[x]$ is monic,  $f(x) = (x^p-1)h(x), h(x) \in \Z[x]$.
If we replace $x$ by $1+t$, we obtain
$$f(1+t) = \sum_{j=1}^{p-1} \chi(j) (1+t)^j -\varepsilon \prod_{k=1}^{(p-1)/2}\left((1+t)^{2k-1} - (1+t)^{p-(2k-1)}ight).$$
We compute the coefficient of $t^{(p-1)/2}$ in the polynomial $f(1+t)$ :
\begin{align*}
\sum_{j=1}^{p-1} \chi(j) (1+t)^j &= \sum_{j=1}^{p-1} \chi(j)  \sum_{i=1}^j \binom{j}{i}t^i\
&= \sum_{i=1}^{p-1} \sum_{j=i}^{p-1} \chi(j) \binom{j}{i}t^i\
\end{align*}
Thus the coefficient of $t^{(p-1)/2}$ in  $\sum\limits_{j=1}^{p-1} \chi(j) (1+t)^j$ is $\sum\limits_{j=(p-1)/2}^{p-1} \chi(j) \binom{j}{(p-1)/2}$.
\begin{align*}
\prod_{k=1}^{(p-1)/2}((1+t)^{2k-1} - (1+t)^{p-(2k-1)})& = \prod_{k=1}^{(p-1)/2} \left((1+ (2k-1) t ) - (1+(p-(2k-1)) t +t^2u(t)ight)\
&=\prod_{k=1}^{(p-1)/2} \left((4k-2-p)t + t^2v(t)ight)\
&=t^{(p-1)/2} \left(\prod_{k=1}^{(p-1)/2} (4k-2-p)ight) + t^{(p+1)/2} w(t),
\end{align*}
where $u(t),v(t),w(t)$ are polynomials. So the coefficient of $t^{(p-1)/2}$ in $f(1+t)$ is
$$c_{(p-1)/2} =  \sum\limits_{j=(p-1)/2}^{p-1} \chi(j) \binom{j}{(p-1)/2} - \varepsilon  \prod_{k=1}^{(p-1)/2} (4k-2-p) .$$
Furthermore, 
\begin{align*}
f(1+t) &= \left((1+t)^p-1ight) h(1+t)\
&=\left[\sum_{i=1}^p \binom{p}{i} t^iight] h(1+t)\
&= \left[\sum_{i=1}^p i!\binom{p}{i} \frac{t^i}{i!}ight] h(1+t)\
&= \left[\sum_{i=0}^p a_i \frac{t^i}{i!}ight] h(1+t),\
\end{align*}
where $a_0 = 0, a_i = i!\binom{p}{i} = \frac{p!}{(p-i)!}$, so $p \mid a_i, i=0,\ldots,p-1$ : the conditions of Ex.21 are verified, so $f(1+t) = \sum_{i=0}^{p-1} c_i t^i$ is such that 
$c_{(p-1)/2} = p (A/B),\ p 
mid B$. Equating these two evaluations of $c_{(p-1)/2}$, we obtain
$$\sum\limits_{j=(p-1)/2}^{p-1} \chi(j) \binom{j}{(p-1)/2} - \varepsilon  \prod_{k=1}^{(p-1)/2} (4k-2-p) = p \frac{A}{B}, \quad p 
mid B.$$
Multiplying by $B (p-1)!/2$, we obtain, as $p
mid B$,
\begin{align*}
\frac{(p-1)!}{2}  \sum_{j=(p-1)/2}^{p-1} \chi(j) \binom{j}{(p-1)/2} &\equiv  \varepsilon  \frac{(p-1)!}{2} \prod_{k=1}^{(p-1)/2} (4k-2) \
&\equiv \varepsilon ( 2\cdot 4\cdot6\cdots(p-1)) \prod_{k=1}^{(p-1)/2} (2k-1) \
&\equiv \varepsilon (p-1)!\
&\equiv -\varepsilon \pmod p
\end{align*}
To prove that $\varepsilon = +1$, it remains to prove
$$S:=\left(\frac{(p-1)}{2}ight)!  \sum_{j=(p-1)/2}^{p-1} \chi(j) \binom{j}{(p-1)/2} \equiv -1 \pmod p$$
The factor of $((p-1)/2)!$ cancels the denominator of $\binom j{(p-1)/2}$, which leaves
\begin{align*}
S &= \sum_{j=(p-1)/2}^{p-1} \chi(j) \cdot j (j-1) \cdots (j-\frac{p-1}{2} + 1)\
&=\sum_{j=1}^{p-1} \chi(j) \cdot j (j-1) \cdots (j-\frac{p-1}{2} + 1).
\end{align*}
The last equality is justified because all terms for $j < \frac{p-1}{2}$ are zero. Collecting powers of $j$, this is
$$S = \sum_{j=1}^{p-1} \sum_{k=0}^{(p-1)/2} \chi(j) \, a_k j^k \equiv \sum_{k=0}^{(p-1)/2} a_k \sum_{j=1}^{p-1} j^{k + \frac{p-1}{2}} \pmod p$$
for some integers $a_k$. It's important to note that $a_{(p-1)/2} = 1$.

Now, we compute $\sum_{j=1}^{p-1} j^n$ mod $p$. Let $T$ denote this sum and let $g$ be a generator of $\Z/p^	imes$. Then, for all positive integer $n$,
$$g^n T = \sum_{j=1}^{p-1} (gj)^n \equiv \sum_{k=1}^{p-1} k^n = T \pmod p.$$
Congruence holds because $gj$ also runs over a complete system of nonzero residues mod $p$. If $g^n 
ot \equiv 1$, that is if $p-1 
mid n$, then $T \equiv 0 \pmod p$. If $g^n \equiv 1$, then $j^n \equiv 1$ for all $j$, hence $T \equiv p-1 \equiv -1 \pmod p$.

Returning to the previous sum, the only nonzero term modulo $p$ is $k = \frac{p-1}{2}$, so
$$S \equiv \sum_{k=0}^{(p-1)/2} a_k \sum_{j=1}^{p-1} j^{k + \frac{p-1}{2}} \equiv a_{(p-1)/2}\cdot (-1) = -1 \pmod p$$
as desired.
\end{proof}

Exercise 6.22

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Exercise 6.22

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