Representing the function ${\mathbb{Z}}_{9}\to {\mathbb{Z}}_{9}$, $f(0)=1$, $f(1)=\dots =f(8)=0$ as a polynomial in ${\mathbb{Z}}_{9}[x]$

Josiah Owens
2022-10-21
Answered

Representing the function ${\mathbb{Z}}_{9}\to {\mathbb{Z}}_{9}$, $f(0)=1$, $f(1)=\dots =f(8)=0$ as a polynomial in ${\mathbb{Z}}_{9}[x]$

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Dobricap

Answered 2022-10-22
Author has **14** answers

There is no such polynomial q.

We show this by contradiction. Assume that $q={a}_{k}{x}^{k}+\dots +{a}_{0}{x}^{0}\in {\mathbb{Z}}_{9}[x]$ is a polynomial with $q(0)=1$ and $q(x)=0$ otherwise.

Since ${0}^{2}={3}^{2}={6}^{2}=0$ in ${\mathbb{Z}}_{9}$, we get

$-1=f(3)-f(0)=({a}_{k}{3}^{k}+{a}_{k-1}{3}^{k-1}+\dots +{a}_{1}3+{a}_{0})-({a}_{k}{0}^{k}+{a}_{k-1}{0}^{k-1}+\dots +{a}_{1}0+{a}_{0})=3{a}_{1}.$

But in ${\mathbb{Z}}_{9}$, -1 is not a multiple of 3, so this is a contradiction.

We show this by contradiction. Assume that $q={a}_{k}{x}^{k}+\dots +{a}_{0}{x}^{0}\in {\mathbb{Z}}_{9}[x]$ is a polynomial with $q(0)=1$ and $q(x)=0$ otherwise.

Since ${0}^{2}={3}^{2}={6}^{2}=0$ in ${\mathbb{Z}}_{9}$, we get

$-1=f(3)-f(0)=({a}_{k}{3}^{k}+{a}_{k-1}{3}^{k-1}+\dots +{a}_{1}3+{a}_{0})-({a}_{k}{0}^{k}+{a}_{k-1}{0}^{k-1}+\dots +{a}_{1}0+{a}_{0})=3{a}_{1}.$

But in ${\mathbb{Z}}_{9}$, -1 is not a multiple of 3, so this is a contradiction.

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