Exercise 6.21

Question

\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}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $f(x) = \sum_{n=0}^\infty a_n x^n/n!$ and $g(x) =\sum_{n=0}^\infty b_n x^n/n!$ be power series with $a_n$ and $b_n$integers. If $p$ is a prime such that $p\mid a_i$ for $i = 0, \ldots,
p-1$, show that each coefficient $c_t$ of the product $f(x)g(x) = \sum_{n=0}^\infty c_n x^n$ for $t = 0, \ldots, p-1$ may be written in the form $p(A/B)$, $p
ot | B$.

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}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $k \in \N, 0 \leq k \leq p-1$.
\begin{align*}
c_k &= \sum_{i+j=k} \frac{a_i}{i!}\frac{b_j}{j!}\
&= \sum_{i=0}^k \frac{a_i}{i!}\frac{b_{k-i}}{(k-i)!}\
&=\frac{1}{k!} \sum_{i=0}^k \binom{k}{i} a_i b_{k-i}
\end{align*}
As $k! \wedge p = 1$, and $\sum_{i=0}^k \binom{k}{i} a_i b_{k-i} \equiv 0 \pmod p$ for $k=0,1,\ldots,p-1$,

$c_k = p(A/B) , p \wedge B = 1$.
\end{proof}

Exercise 6.21

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Exercise 6.21

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