Exercise 10.3

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}}

Suppose that $F$ has $q$ elements. Use the decomposition of $P_n(F)$ into finite points and points at infinity to give another proof of the formula for the number of points in $P_n(F)$.

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}

By exercise 2, the bijection $\psi$ shows that $|\overline{H}| = |P_{n-1}(F)|$. Therefore
$$|P_n(F)| = |P_n(F) \setminus \overline{H} | + |\overline{H} | = |A_n(F)| + |P_{n-1}(F)| = q^n +  |P_{n-1}(F)|.$$
Moreover $|P_0(F)| = 1$. Consequently,
$$|P_n(F)|  = |P_0(F)|  + \sum_{k=1}^n (|P_k(F)| - |P_{k-1}(F)| )= 1 + \sum_{k=1}^n q^k = q^n + q^{n-1} + \cdots + q + 1,$$
This gives another proof of the formula for the number of points in $P_n(F)$.
\end{proof}

Exercise 10.3

Answers

Comments

Exercise 10.3

Answers

Comments

Add answer