Exercise 15.5.7

Question

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When evaluating the multiplication formula for $\varphi(\alpha z)$ at a complex number $z_0$, one needs to worry about poles and vanishing denominators.
\be
\item[(a)] Let $\alpha \in \Z[i]$ be odd, and assume that $z_0$ is a pole of neither $\varphi(z)$ nor $\varphi(\alpha z)$. Prove carefully that $Q_\alpha(\varphi^4(z_0))
e 0$ and that
$$
\varphi(\alpha z_0) = i^\varepsilon \varphi(z_0) \frac{P_\alpha(\varphi^4(z_0))}{Q_\alpha(\varphi^4(z_0))}.
$$
Exercise 9 of Section 5.4 will be useful.
\item[(b)] Let $n$ be odd, and let $p$ be a prime dividing $n$. Then let $\beta$ be a Gaussian prime such that $p=\beta$ if $p \equiv 3 \pmod 4$ and $p = \beta \overline{\beta}$ if $p\equiv 1 \pmod 4$. Use part (a) to prove carefully that
$$
\varphi \left(  \frac{\varpi}{\beta} ight) \in \Q \left( i , \varphi \left(\frac{\varpi}{n} ight) ight).
$$ 
Theorem 15.3.2 will be helpful.

\item[(c)] Let $n$ be odd, and let $p$ be a prime such that $p^2$ divides $n$. Also define $\beta$ as in part (b). Prove that
$$
\varphi \left(  \frac{\varpi}{\beta^2} ight) \in \Q \left( i , \varphi \left(\frac{\varpi}{n} ight) ight).
$$
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\ssi} {si et seulement si }

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\begin{proof} 
\item[(a)] Assume  that $z_0$ is not a pole of $\varphi(z)$. We will prove that $Q_\alpha(\varphi^4(z_0)) = 0$ implies that $z_0$ is a pole of $\varphi(\alpha z)$.

Write $u_0 = \varphi(z_0) \in \C$, and $v_0 = u_0^4 = \varphi^4(z_0)$. Since $Q_\alpha(v_0) = 0$, there is some polynomial $R$ and some $k \in \Z,\ k\geq 1$ such that $Q_n(v) = (v-v_0)^k R(v)$ (we have not proved in the text that $Q_n$ has no multiple root) and $R(v_0) 
e 0$, so that, with $v = u^4$,
$$Q_\alpha(u^4) = (u^4 - u_0^4)^k R(u^4), \qquad R(u_0^4) = R(\varphi^4(z_0)) 
e 0.$$

For all $z \in \C$ such that both members are defined,
\begin{align*}
\varphi(\alpha z) &= i^\varepsilon \varphi(z) \frac{P_\alpha(\varphi^4(z))}{Q_\alpha(\varphi^4(z))}\
&= i^\varepsilon \varphi(z) \frac{P_\alpha(\varphi^4(z))}{(\varphi^4(z) - \varphi^4(z_0))^k R(\varphi^4(z))}\
&= \frac{v(z)}{(\varphi^4(z) - \varphi^4(z_0))^k},
\end{align*}
where $v(z) = i^\varepsilon \varphi(z) \frac{P_\alpha(\varphi^4(z))}{R(\varphi^4(z))}$ is analytic in some neighborhood $U_1$ of $z_0$, since $z_0$ is not a pole of $\varphi$, and since $R(\varphi^4(z_0)) 
e 0$.

Moreover, by Exercise 9 of section 5.4, $u P_\alpha(u)$ and $Q_\alpha(u)$ have no common roots. Since $\varphi^4(u_0)$ is a root of $Q_\alpha(u)$, we have $\varphi^4(z_0) P_\alpha(\varphi^4(z_0)) 
e 0$, so that 
$$v(z_0) 
e 0.$$

\bigskip

If $z_0$ was a zero of $\varphi(z_0)$, then $\varphi(z_0) = 0$ and $Q_\alpha(\varphi^4(z_0)) = Q_\alpha(0) = 1 
e 0$, in contradiction with the hypothesis, thus $\varphi(z_0) 
e 0$.

We write
$$\varphi(z) = \varphi(z_0) + (z-z_0)^l w(z), \qquad w(z_0) 
e 0,$$
where $w(z)$ is analytic in a neighborhood $U_2$ of $z_0$, and $l$ is the order of the zero $z_0$ of $\varphi(z) - \varphi(z_0)$ (thus $l = 1$ or $l = 2$ by the proof of theorem 15.3.3).

Then 
\begin{align*}
\varphi^4(z) - \varphi^4(z_0) &= (\varphi(z)- \varphi(z_0)) (\varphi^3(z) + \varphi(z_0) \varphi^2(z) + \varphi(z_0)^2 \varphi(z) + \varphi^3(z_0))\
&=(\varphi(z)- \varphi(z_0)) s(z)\
&= (z-z_0)^l w(z) s(z)\
&= (z-z_0)^l t(z)
\end{align*}
where $s(z) = \varphi^3(z) + \varphi(z_0) \varphi^2(z) + \varphi(z_0)^2 \varphi(z) + \varphi^3(z_0)$ and $t(z) = s(z) t(z)$. Then $s$ is analytic in a neighborhood $U_3$ of $z_0$, since $z_0$ is not a pole of $\varphi$.  Moreover $s(z_0) = 4 \varphi^3(z_0) 
e 0$. Therefore $t(z) = s(z) w(z)$ is analytic in $U_2 \cap U_3$, and $t(z_0) =s(z_0) w(z_0) 
e 0$.

Since the poles of $\varphi(\alpha z)$ are isolated, there is a neighborhood $U_4$ of $z_0$ such that $\varphi(\alpha z)$ is defined on $U_4 \setminus\{z_0\}$.

Then, for all $z$ in the neighborhood $U = U_1 \cap U_2 \cap U_3 \cap U_4$ of $z_0$ such that $z
e z_0$, since both members are defined in $U$, 
$$
\varphi(\alpha z) = \frac{v(z)}{t^k(z)} \frac{1}{(z-z_0)^{kl}}, \qquad z \in U.
$$
Since $v,t$ are analytic, and $t(z_0)
e 0$ , $r(z) = \frac{v(z)}{t^k(z)}$ is analytic in a neighborhood of $z_0$, and
$$\varphi(\alpha z) = \frac{r(z)}{(z-z_0)^{kl}}.$$
Moreover, since $v(z_0) 
e 0$, $r(z_0) =  \frac{v(z_0)}{t^k(z_0)} 
e 0$. This proves that $\varphi(\alpha z)$ has a pole of order $kl \geq 1$.

This proves by contraposition that if $z_0$ is a pole of neither $\varphi(z)$ nor $\varphi(\alpha z)$, then $Q_\alpha(\varphi^4(z_0))
e 0$.

Moreover, since $z_0$ is not a pole of $\varphi(\alpha z)$, $\varphi(\alpha z_0)$ is defined, and since $Q_\alpha(\varphi^4(z_0)) 
e 0$, where $z_0$ is not a pole of $\varphi(z_0)$, the right member is defined, thus
$$
\varphi(\alpha z_0) = i^\varepsilon \varphi(z_0) \frac{P_\alpha(\varphi^4(z_0))}{Q_\alpha(\varphi^4(z_0))}.
$$

\item[(b)] In both cases of the sentence, $\beta$ is a Gaussian prime dividing $n$. Thus there is a Gaussian integer $\alpha$ such that 
$$n = \beta \alpha, \qquad \alpha \in \Z[i].$$
Since $n$ is odd, $\beta$ and $\alpha$ are odd.

To apply part (a), it remains to verify that $z_0 = \frac{\varpi}{n}$ is a pole of neither $\varphi(z)$ nor $\varphi(\alpha z)$.

$\varphi(z)$ has no real poles, thus $ \frac{\varpi}{n} \in \R$ is not a pole of $\varphi(z)$.

If $\frac{\varpi}{n}$ was a pole of $\varpi(\alpha z)$, then $\alpha \frac{\varpi}{n}$ would be a pole of $\varphi(z)$. The Theorem 15.3.2 shows that
$$\alpha \frac{\varpi}{n} = (a+ib) \frac{\varpi}{2}, \qquad a,b 	ext{ odd}.$$
But then $\alpha = \frac{n}{2}(a+ib) 
ot \in \Z[i]$, because $n,a$ and $b$ are odd. This contradicts $\alpha \in \Z[i]$, and this contradiction shows that $\frac{\varpi}{n}$ is not a pole of $\varphi(\alpha z)$.

Therefore, by Theorem 15.4.4 and part (a),
$$\varphi\left(\frac{\varpi}{\beta} ight) = \varphi\left(\alpha \frac {\varpi}{n} ight) = i^\varepsilon \varphi\left(\frac{\varpi}{n}ight) \frac{P_\alpha\left(\varphi^4\left(\frac{\varpi}{n}ight)ight)}{Q_\alpha\left(\varphi^4\left(\frac{\varpi}{n}ight)ight)}.
$$
This shows that $$\varphi\left(\frac{\varpi}{\beta} ight) \in \Q\left(i,\varphi\left( \frac{\varpi}{n}ight)ight).$$

\item[(c)] Now $p^2 \mid n$.  Since $\beta$ is a Gaussian prime dividing $p$, $\beta^2$ divides $n$ in $\Z[i]$, thus there is some $\alpha$ such that
$$n = \beta^2 \alpha,\qquad \alpha \in \Z[i].$$
We know that $n$ and $\beta$ are odd, thus $\alpha$ is odd.

Since $\alpha$ is odd, the same proof as in part (b) shows that  $ \frac{\varpi}{n} \in \R$ is a pole of neither$\varphi(z)$ nor $\varphi(\alpha z)$.

Therefore, by Theorem 15.4.4 and part (a),
$$\varphi\left(\frac{\varpi}{\beta^2} ight) = \varphi\left(\alpha \frac {\varpi}{n} ight) = i^\varepsilon \varphi\left(\frac{\varpi}{n}ight) \frac{P_\alpha\left(\varphi^4\left(\frac{\varpi}{n}ight)ight)}{Q_\alpha\left(\varphi^4\left(\frac{\varpi}{n}ight)ight)}.
$$
This equality shows that
$$\varphi\left(\frac{\varpi}{\beta^2} ight) \in \Q\left(i, \varphi\left(\frac{\varpi}{n}ight)ight).$$
\end{proof}

Exercise 15.5.7

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Exercise 15.5.7

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