Exercise 7.24

Question

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\def\dcro{\mbox{]\hspace{-.15em}]}}

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Suppose that $f(x) \in \Z/p\Z[x]$ has the property that $f(x+y) = f(x) + f(y) \in \Z/p\Z[x,y]$. Show that $f(x)$ must be of the form $a_0x +a_1x^p+a_2x^{p^2}+\cdots+a_mx^{p^m}.$

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}}

\bigskip

{\bf Lemma} {\it If the prime number $p$ divides all binomial coefficients $\binom{n}{1}, \binom{n}{2},\ldots, \binom{n}{n-1}$, then $n$ is a power of $p$.
\begin{proof}
Let $ u(x) = (x+1)^n -x^n - 1 \in \F_p[x]$. Then $u(x) = \sum\limits_{k=1}^{n-1} \binom{n}{i} x^i = 0$.

Write $n = p^a q$, with $p\wedge q = 1$. We must show that $q=1$. Suppose at the contrary that $q>1$. Then 
$$u(x)= 0 = (x+1)^{p^\alpha q} - x^{p^\alpha q} - 1=(x^{p^\alpha} + 1)^q -x^{p^\alpha q} - 1=\sum\limits_{k=1}^{q-1} \binom{q}{k} x^{kp^a}.$$
Therefore the coefficient $\binom{q}{1} = q$ of $x^{p^a}$ is null in $\F_p$, thus $p \mid q$ : this is absurd. Therefore $q=1$ and $n = p^a$.
\end{proof}

\begin{proof}(Ex. 7.24)

Suppose that $f\in \F_p[x]$ verifies the equality $f(x+y) = f(x)+f(y)$ in $\F_p[x,y]$.

Write $f(x) = \sum\limits_{k=1}^d c_i x^i$.
\begin{align*}
0 = f(x+y) - f(x) -f(y)&= \sum_{n=0}^d c_n[(x+y)^n-x ^n - y^n]\
&=\sum_{n=0}^d \sum_{k=1}^{n-1} c_n\binom{n}{k} x^ky ^{n-k}\
\end{align*}
Therefore,  for all $n$, for all $k$, $1\leq k \leq n-1$, $c_n \binom{n}{k} = 0$ in $\F_p$.

From the lemma, if $n$ is not a power of $p$, there exists a $k$, $1\leq k \leq n-1$ such that $\binom{n}{k} 
ot \equiv 0 \pmod p$, thus $c_n = 0$.
If we write $a_k = c_{p^k}$, then $f(x)$ is of the form
$$f(x) = a_0 x+a_1x^p +a_2 x^{p^2}+\cdots+a_m x^{p^m}.$$
\end{proof}

{ \Large \bf Chapter 8}

Exercise 7.24

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Exercise 7.24

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