# Differentialoperatoren/Hauptteilring/Ableitungsbeschreibung/Fakt/Beweis

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Beweis

Wir arbeiten mit der Beschreibung

{\displaystyle {}{\begin{aligned}R\otimes _{K}R&=K[X_{1},\ldots ,X_{k}]/\left(F_{1},\ldots ,F_{m}\right)\otimes _{K}K[X_{1},\ldots ,X_{k}]/\left(F_{1},\ldots ,F_{m}\right)\\&=K[X_{1},\ldots ,X_{k},{\tilde {X}}_{1},\ldots ,{\tilde {X}}_{k}]/\left(F_{1},\ldots ,F_{m},{\tilde {F}}_{1},\ldots ,{\tilde {F}}_{m}\right),\end{aligned}}}

wobei ${\displaystyle {}{\tilde {F_{i}}}}$ aus ${\displaystyle {}F_{i}}$ entsteht, indem man ${\displaystyle {}X_{j}}$ durch ${\displaystyle {}{\tilde {X}}_{j}}$ ersetzt. Wir setzen

${\displaystyle {}A_{j}={\tilde {X_{j}}}-X_{j}\,}$

an und schreiben den Ring als

${\displaystyle {}K[X_{1},\ldots ,X_{k},A_{1},\ldots ,A_{k}]/\left(F_{1},\ldots ,F_{m},G_{1},\ldots ,G_{m}\right)=R[A_{1},\ldots ,A_{k}]/\left(G_{1},\ldots ,G_{m}\right)\,,}$

wobei

${\displaystyle {}G_{i}={\tilde {F_{i}}}=F_{i}\left({\tilde {X}}_{1},\ldots ,{\tilde {X}}_{k}\right)=F_{i}\left(X_{1}+A_{1},\ldots ,X_{k}+A_{k}\right)\,}$

ist. Betrachte ein Monom ${\displaystyle {}X_{1}^{\nu _{1}}\cdots X_{k}^{\nu _{k}}}$ aus einem ${\displaystyle {}F}$. In die Gleichung ${\displaystyle {}G}$ geht dies in der Form

${\displaystyle (X_{1}+A_{1})^{\nu _{1}}\cdots (X_{k}+A_{k})^{\nu _{k}}}$

ein. Ausmultiplizieren ergibt

${\displaystyle \sum _{\lambda \leq \nu }{\binom {\nu _{1}}{\lambda _{1}}}\cdots {\binom {\nu _{k}}{\lambda _{k}}}X_{1}^{\nu _{1}-\lambda _{1}}A_{1}^{\lambda _{1}}\cdots X_{k}^{\nu _{k}-\lambda _{k}}A_{k}^{\lambda _{k}}=\sum _{\lambda \leq \nu }{\binom {\nu _{1}}{\lambda _{1}}}\cdots {\binom {\nu _{k}}{\lambda _{k}}}X_{1}^{\nu _{1}-\lambda _{1}}\cdots X_{k}^{\nu _{k}-\lambda _{k}}A_{1}^{\lambda _{1}}\cdots A_{k}^{\lambda _{k}}\,.}$

Auf das Monom ${\displaystyle {}A^{\lambda }}$ in ${\displaystyle {}G}$ bezieht sich also der Term

${\displaystyle {\binom {\nu _{1}}{\lambda _{1}}}\cdots {\binom {\nu _{k}}{\lambda _{k}}}X_{1}^{\nu _{1}-\lambda _{1}}\cdots X_{k}^{\nu _{k}-\lambda _{k}}.}$

Dies stimmt mit

${\displaystyle {\frac {\partial ^{\lambda }}{\lambda !}}\left(X^{\nu }\right)}$

überein.

Zur bewiesenen Aussage