Polynomring/Variablenverschiebung/Verschiebung/Aufgabe/Lösung

Es sei ${}\Psi$ die Gesamtabbildung. Dann ist wegen
${}{\begin{aligned}\Psi {\left(\sum _{i=0}^{n}a_{i}X^{i}+\sum _{i=0}^{n}b_{i}X^{i}\right)}&=\Psi {\left(\sum _{i=0}^{n}(a_{i}+b_{i})X^{i}\right)}\\&=\sum _{i=0}^{n}(a_{i}+b_{i})(X+c)^{i}\\&=\sum _{i=0}^{n}a_{i}(X+c)^{i}+\sum _{i=0}^{n}b_{i}(X+c)^{i}\\&=\Psi {\left(\sum _{i=0}^{n}a_{i}X^{i}\right)}+\Psi {\left(\sum _{i=0}^{n}b_{i}X^{i}\right)}\end{aligned}}$

und

${}{\begin{aligned}\Psi {\left(s\sum _{i=0}^{n}a_{i}X^{i}\right)}&=\Psi {\left(\sum _{i=0}^{n}sa_{i}X^{i}\right)}\\&=\sum _{i=0}^{n}sa_{i}(X+c)^{i}\\&=s\sum _{i=0}^{n}a_{i}(X+c)^{i}\\&=s\Psi {\left(\sum _{i=0}^{n}a_{i}X^{i}\right)}\end{aligned}}$

die Abbildung linear.
Dies ist nicht linear, da das Nullpolynom ${}0$ auf ${}c\neq 0$ abgebildet wird.

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